Cartesian product of two or more lists: Difference between revisions
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For the n-ary Cartesian product over a list of lists: |
For the n-ary Cartesian product over a list of lists: |
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<lang JavaScript>(() => { |
<lang JavaScript>(() => { |
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// n-ary Cartesian product of a list of lists |
// n-ary Cartesian product of a list of lists. |
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// cartProdN :: [[a]] -> [[a]] |
// cartProdN :: [[a]] -> [[a]] |
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const cartProdN = |
const cartProdN = foldr( |
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xs => as => |
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bind(as)( |
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x => bind(xs)( |
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a => [[a].concat(x)] |
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// |
// TEST ------------------------------------------- |
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[50, 100] |
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show(cartProdN([ |
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[1, 2, 3], |
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[], |
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[50, 100] |
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])) |
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}; |
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// bind :: [a] -> (a -> [b]) -> [b] |
// bind :: [a] -> (a -> [b]) -> [b] |
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const bind = |
const bind = xs => f => xs.flatMap(f); |
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// foldr (a -> b -> b) -> b -> [a] -> b |
// foldr :: (a -> b -> b) -> b -> [a] -> b |
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const foldr = |
const foldr = f => a => xs => { |
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let v = a, |
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i = xs.length; |
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while (i--) v = f(xs[i])(v); |
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return v; |
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}; |
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// intercalate :: String -> [a] -> String |
// intercalate :: String -> [a] -> String |
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const intercalate = |
const intercalate = s => xs => xs.join(s); |
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// map :: (a -> b) -> [a] -> [b] |
// map :: (a -> b) -> [a] -> [b] |
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const map = |
const map = f => xs => xs.map(f); |
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// show :: a -> String |
// show :: a -> String |
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const show = x => JSON.stringify(x); |
const show = x => JSON.stringify(x); |
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return main(); |
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// unlines :: [String] -> String |
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})();</lang> |
})();</lang> |
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{{Out}} |
{{Out}} |
Revision as of 12:33, 18 September 2019
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
Show one or more idiomatic ways of generating the Cartesian product of two arbitrary lists in your language.
Demonstrate that your function/method correctly returns:
- {1, 2} × {3, 4} = {(1, 3), (1, 4), (2, 3), (2, 4)}
and, in contrast:
- {3, 4} × {1, 2} = {(3, 1), (3, 2), (4, 1), (4, 2)}
Also demonstrate, using your function/method, that the product of an empty list with any other list is empty.
- {1, 2} × {} = {}
- {} × {1, 2} = {}
For extra credit, show or write a function returning the n-ary product of an arbitrary number of lists, each of arbitrary length. Your function might, for example, accept a single argument which is itself a list of lists, and return the n-ary product of those lists.
Use your n-ary Cartesian product function to show the following products:
- {1776, 1789} × {7, 12} × {4, 14, 23} × {0, 1}
- {1, 2, 3} × {30} × {500, 100}
- {1, 2, 3} × {} × {500, 100}
11l
<lang 11l>F cart_prod(a, b)
V p = [(0, 0)] * (a.len * b.len) V i = 0 L(aa) a L(bb) b p[i++] = (aa, bb) R p
print(cart_prod([1, 2], [3, 4])) print(cart_prod([3, 4], [1, 2])) [Int] empty_array print(cart_prod([1, 2], empty_array)) print(cart_prod(empty_array, [1, 2]))</lang>
Alternative version
<lang 11l>F cart_prod(a, b)
R multiloop(a, b, (aa, bb) -> (aa, bb))</lang>
- Output:
[(1, 3), (1, 4), (2, 3), (2, 4)] [(3, 1), (3, 2), (4, 1), (4, 2)] [] []
AppleScript
<lang AppleScript>-- CARTESIAN PRODUCTS ---------------------------------------------------------
-- Two lists:
-- cartProd :: [a] -> [b] -> [(a, b)] on cartProd(xs, ys)
script on |λ|(x) script on |λ|(y) x, y end |λ| end script concatMap(result, ys) end |λ| end script concatMap(result, xs)
end cartProd
-- N-ary – a function over a list of lists:
-- cartProdNary :: a -> a on cartProdNary(xss)
script on |λ|(accs, xs) script on |λ|(x) script on |λ|(a) {x & a} end |λ| end script concatMap(result, accs) end |λ| end script concatMap(result, xs) end |λ| end script foldr(result, {{}}, xss)
end cartProdNary
-- TESTS ---------------------------------------------------------------------- on run
set baseExamples to unlines(map(show, ¬ [cartProd({1, 2}, {3, 4}), ¬ cartProd({3, 4}, {1, 2}), ¬ cartProd({1, 2}, {}), ¬ cartProd({}, {1, 2})])) set naryA to unlines(map(show, ¬ cartProdNary([{1776, 1789}, {7, 12}, {4, 14, 23}, {0, 1}]))) set naryB to show(cartProdNary([{1, 2, 3}, {30}, {500, 100}])) set naryC to show(cartProdNary([{1, 2, 3}, {}, {500, 100}])) intercalate(linefeed & linefeed, {baseExamples, naryA, naryB, naryC})
end run
-- GENERIC FUNCTIONS ----------------------------------------------------------
-- concatMap :: (a -> [b]) -> [a] -> [b] on concatMap(f, xs)
set lst to {} set lng to length of xs tell mReturn(f) repeat with i from 1 to lng set lst to (lst & |λ|(item i of xs, i, xs)) end repeat end tell return lst
end concatMap
-- foldr :: (a -> b -> a) -> a -> [b] -> a on foldr(f, startValue, xs)
tell mReturn(f) set v to startValue set lng to length of xs repeat with i from lng to 1 by -1 set v to |λ|(v, item i of xs, i, xs) end repeat return v end tell
end foldr
-- intercalate :: Text -> [Text] -> Text on intercalate(strText, lstText)
set {dlm, my text item delimiters} to {my text item delimiters, strText} set strJoined to lstText as text set my text item delimiters to dlm return strJoined
end intercalate
-- map :: (a -> b) -> [a] -> [b] on map(f, xs)
tell mReturn(f) set lng to length of xs set lst to {} repeat with i from 1 to lng set end of lst to |λ|(item i of xs, i, xs) end repeat return lst end tell
end map
-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Script on mReturn(f)
if class of f is script then f else script property |λ| : f end script end if
end mReturn
-- show :: a -> String on show(e)
set c to class of e if c = list then script serialized on |λ|(v) show(v) end |λ| end script "[" & intercalate(", ", map(serialized, e)) & "]" else if c = record then script showField on |λ|(kv) set {k, ev} to kv "\"" & k & "\":" & show(ev) end |λ| end script "{" & intercalate(", ", ¬ map(showField, zip(allKeys(e), allValues(e)))) & "}" else if c = date then "\"" & iso8601Z(e) & "\"" else if c = text then "\"" & e & "\"" else if (c = integer or c = real) then e as text else if c = class then "null" else try e as text on error ("«" & c as text) & "»" end try end if
end show
-- unlines :: [String] -> String on unlines(xs)
intercalate(linefeed, xs)
end unlines</lang>
- Output:
[[1, 3], [1, 4], [2, 3], [2, 4]] [[3, 1], [3, 2], [4, 1], [4, 2]] [] [] [1776, 7, 4, 0] [1776, 7, 4, 1] [1776, 7, 14, 0] [1776, 7, 14, 1] [1776, 7, 23, 0] [1776, 7, 23, 1] [1776, 12, 4, 0] [1776, 12, 4, 1] [1776, 12, 14, 0] [1776, 12, 14, 1] [1776, 12, 23, 0] [1776, 12, 23, 1] [1789, 7, 4, 0] [1789, 7, 4, 1] [1789, 7, 14, 0] [1789, 7, 14, 1] [1789, 7, 23, 0] [1789, 7, 23, 1] [1789, 12, 4, 0] [1789, 12, 4, 1] [1789, 12, 14, 0] [1789, 12, 14, 1] [1789, 12, 23, 0] [1789, 12, 23, 1] [[1, 30, 500], [1, 30, 100], [2, 30, 500], [2, 30, 100], [3, 30, 500], [3, 30, 100]] []
C
Recursive implementation for computing the Cartesian product of lists. In the pursuit of making it as interactive as possible, the parsing function ended up taking the most space. The product set expression must be supplied enclosed by double quotes. Prints out usage on incorrect invocation. <lang C>
- include<string.h>
- include<stdlib.h>
- include<stdio.h>
void cartesianProduct(int** sets, int* setLengths, int* currentSet, int numSets, int times){ int i,j;
if(times==numSets){ printf("("); for(i=0;i<times;i++){ printf("%d,",currentSet[i]); } printf("\b),"); } else{ for(j=0;j<setLengths[times];j++){ currentSet[times] = sets[times][j]; cartesianProduct(sets,setLengths,currentSet,numSets,times+1); } } }
void printSets(int** sets, int* setLengths, int numSets){ int i,j;
printf("\nNumber of sets : %d",numSets);
for(i=0;i<numSets+1;i++){ printf("\nSet %d : ",i+1); for(j=0;j<setLengths[i];j++){ printf(" %d ",sets[i][j]); } } }
void processInputString(char* str){ int **sets, *currentSet, *setLengths, setLength, numSets = 0, i,j,k,l,start,counter=0; char *token,*holder,*holderToken;
for(i=0;str[i]!=00;i++) if(str[i]=='x') numSets++;
if(numSets==0){ printf("\n%s",str); return; }
currentSet = (int*)calloc(sizeof(int),numSets + 1);
setLengths = (int*)calloc(sizeof(int),numSets + 1);
sets = (int**)malloc((numSets + 1)*sizeof(int*));
token = strtok(str,"x");
while(token!=NULL){ holder = (char*)malloc(strlen(token)*sizeof(char));
j = 0;
for(i=0;token[i]!=00;i++){ if(token[i]>='0' && token[i]<='9') holder[j++] = token[i]; else if(token[i]==',') holder[j++] = ' '; } holder[j] = 00;
setLength = 0;
for(i=0;holder[i]!=00;i++) if(holder[i]==' ') setLength++;
if(setLength==0 && strlen(holder)==0){ printf("\n{}"); return; }
setLengths[counter] = setLength+1;
sets[counter] = (int*)malloc((1+setLength)*sizeof(int));
k = 0;
start = 0;
for(l=0;holder[l]!=00;l++){ if(holder[l+1]==' '||holder[l+1]==00){ holderToken = (char*)malloc((l+1-start)*sizeof(char)); strncpy(holderToken,holder + start,l+1-start); sets[counter][k++] = atoi(holderToken); start = l+2; } }
counter++; token = strtok(NULL,"x"); }
printf("\n{"); cartesianProduct(sets,setLengths,currentSet,numSets + 1,0); printf("\b}");
}
int main(int argC,char* argV[]) { if(argC!=2) printf("Usage : %s <Set product expression enclosed in double quotes>",argV[0]); else processInputString(argV[1]);
return 0; } </lang> Invocation and output :
C:\My Projects\threeJS>cartesianProduct.exe "{1,2} x {3,4}" {(1,3),(1,4),(2,3),(2,4)} C:\My Projects\threeJS>cartesianProduct.exe "{3,4} x {1,2}" {(3,1),(3,2),(4,1),(4,2)} C:\My Projects\threeJS>cartesianProduct.exe "{1,2} x {}" {} C:\My Projects\threeJS>cartesianProduct.exe "{} x {1,2}" {} C:\My Projects\threeJS>cartesianProduct.exe "{1776, 1789} x {7, 12} x {4, 14, 23} x {0, 1}" {(1776,7,4,0),(1776,7,4,1),(1776,7,14,0),(1776,7,14,1),(1776,7,23,0),(1776,7,23,1),(1776,12,4,0),(1776,12,4,1),(1776,12,14,0),(1776,12,14,1),(1776,12,23,0),(1776,12,23,1),(1789,7,4,0),(1789,9,12,14,1),(1789,12,23,0),(1789,12,23,1)} C:\My Projects\threeJS>cartesianProduct.exe "{1, 2, 3} x {30} x {500, 100}" {(1,30,500),(1,30,100),(2,30,500),(2,30,100),(3,30,500),(3,30,100)} C:\My Projects\threeJS>cartesianProduct.exe "{1, 2, 3} x {} x {500, 100}" {}
C++
<lang cpp>
- include <iostream>
- include <vector>
- include <algorithm>
void print(const std::vector<std::vector<int>>& v) {
std::cout << "{ "; for (const auto& p : v) { std::cout << "("; for (const auto& e : p) { std::cout << e << " "; } std::cout << ") "; } std::cout << "}" << std::endl;
}
auto product(const std::vector<std::vector<int>>& lists) {
std::vector<std::vector<int>> result; if (std::find_if(std::begin(lists), std::end(lists), [](auto e) -> bool { return e.size() == 0; }) != std::end(lists)) { return result; } for (auto& e : lists[0]) { result.push_back({ e }); } for (size_t i = 1; i < lists.size(); ++i) { std::vector<std::vector<int>> temp; for (auto& e : result) { for (auto f : lists[i]) { auto e_tmp = e; e_tmp.push_back(f); temp.push_back(e_tmp); } } result = temp; } return result;
}
int main() {
std::vector<std::vector<int>> prods[] = { { { 1, 2 }, { 3, 4 } }, { { 3, 4 }, { 1, 2} }, { { 1, 2 }, { } }, { { }, { 1, 2 } }, { { 1776, 1789 }, { 7, 12 }, { 4, 14, 23 }, { 0, 1 } }, { { 1, 2, 3 }, { 30 }, { 500, 100 } }, { { 1, 2, 3 }, { }, { 500, 100 } } }; for (const auto& p : prods) { print(product(p)); } std::cin.ignore(); std::cin.get(); return 0;
}</lang>
- Output:
{ (1 3) (1 4) (2 3) (2 4) } { (3 1) (3 2) (4 1) (4 2) } { } { } { (1776 7 4 0) (1776 7 4 1) (1776 7 14 0) (1776 7 14 1) (1776 7 23 0) (1776 7 23 1) (1776 12 4 0) (1776 12 4 1) (1776 12 14 0) (1776 12 14 1) (1776 12 23 0) (1776 12 23 1) (1789 7 4 0) (1789 7 4 1) (1789 7 14 0) (1789 7 14 1) (1789 7 23 0) (1789 7 23 1) (1789 12 4 0) (1789 12 4 1) (1789 12 14 0) (1789 12 14 1) (1789 12 23 0) (1789 12 23 1) } { (1 30 500) (1 30 100) (2 30 500) (2 30 100) (3 30 500) (3 30 100) } { }
C#
<lang csharp>using System; public class Program {
public static void Main() { int[] empty = new int[0]; int[] list1 = { 1, 2 }; int[] list2 = { 3, 4 }; int[] list3 = { 1776, 1789 }; int[] list4 = { 7, 12 }; int[] list5 = { 4, 14, 23 }; int[] list6 = { 0, 1 }; int[] list7 = { 1, 2, 3 }; int[] list8 = { 30 }; int[] list9 = { 500, 100 }; foreach (var sequenceList in new [] { new [] { list1, list2 }, new [] { list2, list1 }, new [] { list1, empty }, new [] { empty, list1 }, new [] { list3, list4, list5, list6 }, new [] { list7, list8, list9 }, new [] { list7, empty, list9 } }) { var cart = sequenceList.CartesianProduct() .Select(tuple => $"({string.Join(", ", tuple)})"); Console.WriteLine($"{{{string.Join(", ", cart)}}}"); } }
}
public static class Extensions {
public static IEnumerable<IEnumerable<T>> CartesianProduct<T>(this IEnumerable<IEnumerable<T>> sequences) { IEnumerable<IEnumerable<T>> emptyProduct = new[] { Enumerable.Empty<T>() }; return sequences.Aggregate( emptyProduct, (accumulator, sequence) => from acc in accumulator from item in sequence select acc.Concat(new [] { item })); }
}</lang>
- Output:
{(1, 3), (1, 4), (2, 3), (2, 4)} {(3, 1), (3, 2), (4, 1), (4, 2)} {} {} {(1776, 7, 4, 0), (1776, 7, 4, 1), (1776, 7, 14, 0), (1776, 7, 14, 1), (1776, 7, 23, 0), (1776, 7, 23, 1), (1776, 12, 4, 0), (1776, 12, 4, 1), (1776, 12, 14, 0), (1776, 12, 14, 1), (1776, 12, 23, 0), (1776, 12, 23, 1), (1789, 7, 4, 0), (1789, 7, 4, 1), (1789, 7, 14, 0), (1789, 7, 14, 1), (1789, 7, 23, 0), (1789, 7, 23, 1), (1789, 12, 4, 0), (1789, 12, 4, 1), (1789, 12, 14, 0), (1789, 12, 14, 1), (1789, 12, 23, 0), (1789, 12, 23, 1)} {(1, 30, 500), (1, 30, 100), (2, 30, 500), (2, 30, 100), (3, 30, 500), (3, 30, 100)} {}
If the number of lists is known, LINQ provides an easier solution: <lang csharp>public static void Main() {
///... var cart1 = from a in list1 from b in list2 select (a, b); // C# 7.0 tuple Console.WriteLine($"{{{string.Join(", ", cart1)}}}"); var cart2 = from a in list7 from b in list8 from c in list9 select (a, b, c); Console.WriteLine($"{{{string.Join(", ", cart2)}}}");
}</lang>
- Output:
{(1, 3), (1, 4), (2, 3), (2, 4)} {(1, 30, 500), (1, 30, 100), (2, 30, 500), (2, 30, 100), (3, 30, 500), (3, 30, 100)}
Common Lisp
<lang lisp>(defun cartesian-product (s1 s2)
"Compute the cartesian product of two sets represented as lists" (loop for x in s1
nconc (loop for y in s2 collect (list x y)))) </lang>
Output
<lang lisp> CL-USER> (cartesian-product '(1 2) '(3 4)) ((1 3) (1 4) (2 3) (2 4)) CL-USER> (cartesian-product '(3 4) '(1 2)) ((3 1) (3 2) (4 1) (4 2)) CL-USER> (cartesian-product '(1 2) '()) NIL CL-USER> (cartesian-product '() '(1 2)) NIL </lang>
Extra credit:
<lang lisp>(defun n-cartesian-product (l)
"Compute the n-cartesian product of a list of sets (each of them represented as list). Algorithm: If there are no sets, then produce an empty set of tuples; otherwise, for all the elements x of the first set, concatenate the sets obtained by inserting x at the beginning of each tuple of the n-cartesian product of the remaining sets." (if (null l) (list nil) (loop for x in (car l) nconc (loop for y in (n-cartesian-product (cdr l)) collect (cons x y)))))</lang>
Output:
<lang lisp>CL-USER> (n-cartesian-product '((1776 1789) (7 12) (4 14 23) (0 1))) ((1776 7 4 0) (1776 7 4 1) (1776 7 14 0) (1776 7 14 1) (1776 7 23 0) (1776 7 23 1) (1776 12 4 0) (1776 12 4 1) (1776 12 14 0) (1776 12 14 1) (1776 12 23 0) (1776 12 23 1) (1789 7 4 0) (1789 7 4 1) (1789 7 14 0) (1789 7 14 1) (1789 7 23 0) (1789 7 23 1) (1789 12 4 0) (1789 12 4 1) (1789 12 14 0) (1789 12 14 1) (1789 12 23 0) (1789 12 23 1)) CL-USER> (n-cartesian-product '((1 2 3) (30) (500 100))) ((1 30 500) (1 30 100) (2 30 500) (2 30 100) (3 30 500) (3 30 100)) CL-USER> (n-cartesian-product '((1 2 3) () (500 100))) NIL </lang>
D
<lang D>import std.stdio;
void main() {
auto a = listProduct([1,2], [3,4]); writeln(a);
auto b = listProduct([3,4], [1,2]); writeln(b);
auto c = listProduct([1,2], []); writeln(c);
auto d = listProduct([], [1,2]); writeln(d);
}
auto listProduct(T)(T[] ta, T[] tb) {
struct Result { int i, j;
bool empty() { return i>=ta.length || j>=tb.length; }
T[] front() { return [ta[i], tb[j]]; }
void popFront() { if (++j>=tb.length) { j=0; i++; } } }
return Result();
}</lang>
- Output:
[[1, 3], [1, 4], [2, 3], [2, 4]] [[3, 1], [3, 2], [4, 1], [4, 2]] [] []
F#
The Task
<lang fsharp> //Nigel Galloway February 12th., 2018 let cP2 n g = List.map (fun (n,g)->[n;g]) (List.allPairs n g) </lang>
- Output:
cP2 [1;2] [3;4] -> [[1; 3]; [1; 4]; [2; 3]; [2; 4]] cP2 [3;4] [1;2] -> [[3; 1]; [3; 2]; [4; 1]; [4; 2]] cP2 [1;2] [] -> [] cP2 [] [1;2] -> []
Extra Credit
<lang fsharp> //Nigel Galloway August 14th., 2018 let cP ng=Seq.foldBack(fun n g->[for n' in n do for g' in g do yield n'::g']) ng [[]] </lang>
- Output:
cP [[1;2];[3;4]] -> [[1; 3]; [1; 4]; [2; 3]; [2; 4]] cP [[3;4];[1;2]] -> [[3; 1]; [3; 2]; [4; 1]; [4; 2]] cP [[3;4];[]] ->[] cP [[];[1;2]] ->[] cP [[1776;1789];[7;12];[4;14;23];[0;1]] -> [[1776; 7; 4; 0]; [1776; 7; 4; 1]; [1776; 7; 14; 0]; [1776; 7; 14; 1]; [1776; 7; 23; 0]; [1776; 7; 23; 1]; [1776; 12; 4; 0]; [1776; 12; 4; 1]; [1776; 12; 14; 0]; [1776; 12; 14; 1]; [1776; 12; 23; 0]; [1776; 12; 23; 1]; [1789; 7; 4; 0]; [1789; 7; 4; 1]; [1789; 7; 14; 0]; [1789; 7; 14; 1]; [1789; 7; 23; 0]; [1789; 7; 23; 1]; [1789; 12; 4; 0]; [1789; 12; 4; 1]; [1789; 12; 14; 0]; [1789; 12; 14; 1]; [1789; 12; 23; 0]; [1789; 12; 23; 1]] cP [[1;2;3];[30];[500;100]] -> [[1; 30; 500]; [1; 30; 100]; [2; 30; 500]; [2; 30; 100]; [3; 30; 500]; [3; 30; 100]] cP [[1;2;3];[];[500;100]] -> []
Factor
<lang Factor>IN: scratchpad { 1 2 } { 3 4 } cartesian-product . { { { 1 3 } { 1 4 } } { { 2 3 } { 2 4 } } } IN: scratchpad { 3 4 } { 1 2 } cartesian-product . { { { 3 1 } { 3 2 } } { { 4 1 } { 4 2 } } } IN: scratchpad { 1 2 } { } cartesian-product . { { } { } } IN: scratchpad { } { 1 2 } cartesian-product . { }</lang>
Fōrmulæ
In this page you can see the solution of this task.
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text (more info). Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for transportation effects more than visualization and edition.
The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.
Go
Basic Task <lang go>package main
import "fmt"
type pair [2]int
func cart2(a, b []int) []pair {
p := make([]pair, len(a)*len(b)) i := 0 for _, a := range a { for _, b := range b { p[i] = pair{a, b} i++ } } return p
}
func main() {
fmt.Println(cart2([]int{1, 2}, []int{3, 4})) fmt.Println(cart2([]int{3, 4}, []int{1, 2})) fmt.Println(cart2([]int{1, 2}, nil)) fmt.Println(cart2(nil, []int{1, 2}))
}</lang>
- Output:
[[1 3] [1 4] [2 3] [2 4]] [[3 1] [3 2] [4 1] [4 2]] [] []
Extra credit 1
This solution minimizes allocations and computes and fills the result sequentially. <lang go>package main
import "fmt"
func cartN(a ...[]int) [][]int {
c := 1 for _, a := range a { c *= len(a) } if c == 0 { return nil } p := make([][]int, c) b := make([]int, c*len(a)) n := make([]int, len(a)) s := 0 for i := range p { e := s + len(a) pi := b[s:e] p[i] = pi s = e for j, n := range n { pi[j] = a[j][n] } for j := len(n) - 1; j >= 0; j-- { n[j]++ if n[j] < len(a[j]) { break } n[j] = 0 } } return p
}
func main() {
fmt.Println(cartN([]int{1, 2}, []int{3, 4})) fmt.Println(cartN([]int{3, 4}, []int{1, 2})) fmt.Println(cartN([]int{1, 2}, nil)) fmt.Println(cartN(nil, []int{1, 2}))
fmt.Println() fmt.Println("[") for _, p := range cartN( []int{1776, 1789}, []int{7, 12}, []int{4, 14, 23}, []int{0, 1}, ) { fmt.Println(" ", p) } fmt.Println("]") fmt.Println(cartN([]int{1, 2, 3}, []int{30}, []int{500, 100})) fmt.Println(cartN([]int{1, 2, 3}, []int{}, []int{500, 100}))
fmt.Println() fmt.Println(cartN(nil)) fmt.Println(cartN())
}</lang>
- Output:
[[1 3] [1 4] [2 3] [2 4]] [[3 1] [3 2] [4 1] [4 2]] [] [] [ [1776 7 4 0] [1776 7 4 1] [1776 7 14 0] [1776 7 14 1] [1776 7 23 0] [1776 7 23 1] [1776 12 4 0] [1776 12 4 1] [1776 12 14 0] [1776 12 14 1] [1776 12 23 0] [1776 12 23 1] [1789 7 4 0] [1789 7 4 1] [1789 7 14 0] [1789 7 14 1] [1789 7 23 0] [1789 7 23 1] [1789 12 4 0] [1789 12 4 1] [1789 12 14 0] [1789 12 14 1] [1789 12 23 0] [1789 12 23 1] ] [[1 30 500] [1 30 100] [2 30 500] [2 30 100] [3 30 500] [3 30 100]] [] [] [[]]
Extra credit 2
Code here is more compact, but with the cost of more garbage produced. It produces the same result as cartN above. <lang go>func cartN(a ...[]int) (c [][]int) {
if len(a) == 0 { return [][]int{nil} } r := cartN(a[1:]...) for _, e := range a[0] { for _, p := range r { c = append(c, append([]int{e}, p...)) } } return
}</lang> Extra credit 3
This is a compact recursive version like Extra credit 2 but the result list is ordered differently. This is still a correct result if you consider a cartesian product to be a set, which is an unordered collection. Note that the set elements are still ordered lists. A cartesian product is an unordered collection of ordered collections. It draws attention though to the gloss of using list representations as sets. Any of the functions here will accept duplicate elements in the input lists, and then produce duplicate elements in the result. <lang go>func cartN(a ...[]int) (c [][]int) {
if len(a) == 0 { return [][]int{nil} } last := len(a) - 1 l := cartN(a[:last]...) for _, e := range a[last] { for _, p := range l { c = append(c, append(p, e)) } } return
}</lang>
Groovy
Solution:
The following CartesianCategory class allows for modification of regular Iterable interface behavior, overloading Iterable's multiply (*) operator to perform a Cartesian Product when the second operand is also an Iterable.
<lang groovy>class CartesianCategory {
static Iterable multiply(Iterable a, Iterable b) { assert [a,b].every { it != null } def (m,n) = [a.size(),b.size()] (0..<(m*n)).inject([]) { prod, i -> prod << [a[i.intdiv(n)], b[i%n]].flatten() } }
}</lang>
Test:
The mixin method call is necessary to make the multiply (*) operator work.
<lang groovy>Iterable.metaClass.mixin CartesianCategory
println "\nCore Solution:" println "[1, 2] × [3, 4] = ${[1, 2] * [3, 4]}" println "[3, 4] × [1, 2] = ${[3, 4] * [1, 2]}" println "[1, 2] × [] = ${[1, 2] * []}" println "[] × [1, 2] = ${[] * [1, 2]}"
println "\nExtra Credit:" println "[1776, 1789] × [7, 12] × [4, 14, 23] × [0, 1] = ${[1776, 1789] * [7, 12] * [4, 14, 23] * [0, 1]}" println "[1, 2, 3] × [30] × [500, 100] = ${[1, 2, 3] * [30] * [500, 100]}" println "[1, 2, 3] × [] × [500, 100] = ${[1, 2, 3] * [] * [500, 100]}"
println "\nNon-Numeric Example:" println "[John,Paul,George,Ringo] × [Emerson,Lake,Palmer] × [Simon,Garfunkle] = [" ( ["John","Paul","George","Ringo"] * ["Emerson","Lake","Palmer"] * ["Simon","Garfunkle"] ).each { println "\t${it}," } println "]"</lang> Output:
Core Solution: [1, 2] × [3, 4] = [[1, 3], [1, 4], [2, 3], [2, 4]] [3, 4] × [1, 2] = [[3, 1], [3, 2], [4, 1], [4, 2]] [1, 2] × [] = [] [] × [1, 2] = [] Extra Credit: [1776, 1789] × [7, 12] × [4, 14, 23] × [0, 1] = [[1776, 7, 4, 0], [1776, 7, 4, 1], [1776, 7, 14, 0], [1776, 7, 14, 1], [1776, 7, 23, 0], [1776, 7, 23, 1], [1776, 12, 4, 0], [1776, 12, 4, 1], [1776, 12, 14, 0], [1776, 12, 14, 1], [1776, 12, 23, 0], [1776, 12, 23, 1], [1789, 7, 4, 0], [1789, 7, 4, 1], [1789, 7, 14, 0], [1789, 7, 14, 1], [1789, 7, 23, 0], [1789, 7, 23, 1], [1789, 12, 4, 0], [1789, 12, 4, 1], [1789, 12, 14, 0], [1789, 12, 14, 1], [1789, 12, 23, 0], [1789, 12, 23, 1]] [1, 2, 3] × [30] × [500, 100] = [[1, 30, 500], [1, 30, 100], [2, 30, 500], [2, 30, 100], [3, 30, 500], [3, 30, 100]] [1, 2, 3] × [] × [500, 100] = [] Non-Numeric Example: [John,Paul,George,Ringo] × [Emerson,Lake,Palmer] × [Simon,Garfunkle] = [ [John, Emerson, Simon], [John, Emerson, Garfunkle], [John, Lake, Simon], [John, Lake, Garfunkle], [John, Palmer, Simon], [John, Palmer, Garfunkle], [Paul, Emerson, Simon], [Paul, Emerson, Garfunkle], [Paul, Lake, Simon], [Paul, Lake, Garfunkle], [Paul, Palmer, Simon], [Paul, Palmer, Garfunkle], [George, Emerson, Simon], [George, Emerson, Garfunkle], [George, Lake, Simon], [George, Lake, Garfunkle], [George, Palmer, Simon], [George, Palmer, Garfunkle], [Ringo, Emerson, Simon], [Ringo, Emerson, Garfunkle], [Ringo, Lake, Simon], [Ringo, Lake, Garfunkle], [Ringo, Palmer, Simon], [Ringo, Palmer, Garfunkle], ]
Haskell
Various routes can be taken to Cartesian products in Haskell. For the product of two lists we could write: <lang Haskell>cartProd :: [a] -> [b] -> [(a, b)] cartProd xs ys =
[ (x, y) | x <- xs , y <- ys ]</lang>
more directly: <lang Haskell>cartProd :: [a] -> [b] -> [(a, b)] cartProd xs ys = xs >>= \x -> ys >>= \y -> [(x, y)]</lang>
applicatively: <lang Haskell>cartProd :: [a] -> [b] -> [(a, b)] cartProd xs ys = (,) <$> xs <*> ys</lang>
parsimoniously: <lang Haskell>cartProd :: [a] -> [b] -> [(a, b)] cartProd = (<*>) . fmap (,)</lang>
We might test any of these with: <lang haskell>main :: IO () main =
mapM_ print $ uncurry cartProd <$> [([1, 2], [3, 4]), ([3, 4], [1, 2]), ([1, 2], []), ([], [1, 2])]</lang>
- Output:
[(1,3),(1,4),(2,3),(2,4)] [(3,1),(3,2),(4,1),(4,2)] [] []
For the n-ary Cartesian product of an arbitrary number of lists, we could apply the Prelude's standard sequence function to a list of lists,
<lang haskell>cartProdN :: a -> a
cartProdN = sequence
main :: IO () main = print $ cartProdN [[1, 2], [3, 4], [5, 6]]</lang>
- Output:
[[1,3,5],[1,3,6],[1,4,5],[1,4,6],[2,3,5],[2,3,6],[2,4,5],[2,4,6]]
or we could define ourselves an equivalent function over a list of lists in terms of a fold, for example as: <lang haskell>cartProdN :: a -> a cartProdN = foldr (\xs as -> xs >>= (<$> as) . (:)) [[]]</lang> or, equivalently, as: <lang haskell>cartProdN :: a -> a cartProdN = foldr
(\xs as -> [ x : a | x <- xs , a <- as ]) [[]]</lang>
testing any of these with something like: <lang haskell>main :: IO () main = do
mapM_ print $ cartProdN [[1776, 1789], [7,12], [4, 14, 23], [0,1]] putStrLn "" print $ cartProdN [[1,2,3], [30], [500, 100]] putStrLn "" print $ cartProdN [[1,2,3], [], [500, 100]]</lang>
- Output:
[1776,7,4,0] [1776,7,4,1] [1776,7,14,0] [1776,7,14,1] [1776,7,23,0] [1776,7,23,1] [1776,12,4,0] [1776,12,4,1] [1776,12,14,0] [1776,12,14,1] [1776,12,23,0] [1776,12,23,1] [1789,7,4,0] [1789,7,4,1] [1789,7,14,0] [1789,7,14,1] [1789,7,23,0] [1789,7,23,1] [1789,12,4,0] [1789,12,4,1] [1789,12,14,0] [1789,12,14,1] [1789,12,23,0] [1789,12,23,1] [[1,30,500],[1,30,100],[2,30,500],[2,30,100],[3,30,500],[3,30,100]] []
J
The J primitive catalogue {
forms the Cartesian Product of two or more boxed lists. The result is a multi-dimensional array (which can be reshaped to a simple list of lists if desired).
<lang j> { 1776 1789 ; 7 12 ; 4 14 23 ; 0 1 NB. result is 4 dimensional array with shape 2 2 3 2
┌────────────┬────────────┐
│1776 7 4 0 │1776 7 4 1 │
├────────────┼────────────┤
│1776 7 14 0 │1776 7 14 1 │
├────────────┼────────────┤
│1776 7 23 0 │1776 7 23 1 │
└────────────┴────────────┘
┌────────────┬────────────┐ │1776 12 4 0 │1776 12 4 1 │ ├────────────┼────────────┤ │1776 12 14 0│1776 12 14 1│ ├────────────┼────────────┤ │1776 12 23 0│1776 12 23 1│ └────────────┴────────────┘
┌────────────┬────────────┐
│1789 7 4 0 │1789 7 4 1 │
├────────────┼────────────┤
│1789 7 14 0 │1789 7 14 1 │
├────────────┼────────────┤
│1789 7 23 0 │1789 7 23 1 │
└────────────┴────────────┘
┌────────────┬────────────┐ │1789 12 4 0 │1789 12 4 1 │ ├────────────┼────────────┤ │1789 12 14 0│1789 12 14 1│ ├────────────┼────────────┤ │1789 12 23 0│1789 12 23 1│ └────────────┴────────────┘
{ 1 2 3 ; 30 ; 50 100 NB. result is a 2-dimensional array with shape 2 3
┌───────┬────────┐ │1 30 50│1 30 100│ ├───────┼────────┤ │2 30 50│2 30 100│ ├───────┼────────┤ │3 30 50│3 30 100│ └───────┴────────┘
{ 1 2 3 ; ; 50 100 NB. result is an empty 3-dimensional array with shape 3 0 2
</lang>
Java
<lang Java> import static java.util.Arrays.asList; import static java.util.Collections.emptyList; import static java.util.Optional.of; import static java.util.stream.Collectors.toList;
import java.util.List;
public class CartesianProduct {
public List<?> product(List<?>... a) { if (a.length >= 2) { List<?> product = a[0]; for (int i = 1; i < a.length; i++) { product = product(product, a[i]); } return product; }
return emptyList(); }
private <A, B> List<?> product(List<A> a, List b) { return of(a.stream() .map(e1 -> of(b.stream().map(e2 -> asList(e1, e2)).collect(toList())).orElse(emptyList())) .flatMap(List::stream) .collect(toList())).orElse(emptyList()); }
} </lang>
JavaScript
ES6
Functional
Cartesian products fall quite naturally out of concatMap (Array.flatMap), and its argument-flipped twin bind.
For the Cartesian product of just two lists: <lang JavaScript>(() => {
// CARTESIAN PRODUCT OF TWO LISTS ---------------------
// cartProd :: [a] -> [b] -> a, b const cartProd = xs => ys => xs.flatMap(x => ys.map(y => [x, y]))
// TEST ----------------------------------------------- return [ cartProd([1, 2])([3, 4]), cartProd([3, 4])([1, 2]), cartProd([1, 2])([]), cartProd([])([1, 2]), ].map(JSON.stringify).join('\n');
})();</lang>
- Output:
[[1,3],[1,4],[2,3],[2,4]] [[3,1],[3,2],[4,1],[4,2]] [] []
Abstracting a little more, we can define the cartesian product quite economically in terms of a general applicative operator:
<lang Javascript>(() => {
// CARTESIAN PRODUCT OF TWO LISTS ---------------------
// cartesianProduct :: [a] -> [b] -> [(a, b)] const cartesianProduct = xs => ap(xs.map(Tuple));
// GENERIC FUNCTIONS ----------------------------------
// e.g. [(*2),(/2), sqrt] <*> [1,2,3] // --> ap([dbl, hlf, root], [1, 2, 3]) // --> [2,4,6,0.5,1,1.5,1,1.4142135623730951,1.7320508075688772]
// Each member of a list of functions applied to each // of a list of arguments, deriving a list of new values.
// ap (<*>) :: [(a -> b)] -> [a] -> [b] const ap = fs => xs => // The sequential application of each of a list // of functions to each of a list of values. fs.flatMap( f => xs.map(f) );
// Tuple (,) :: a -> b -> (a, b) const Tuple = a => b => [a, b];
// TEST ----------------------------------------------- return [ cartesianProduct([1, 2])([3, 4]), cartesianProduct([3, 4])([1, 2]), cartesianProduct([1, 2])([]), cartesianProduct([])([1, 2]), ] .map(JSON.stringify) .join('\n');
})();</lang>
- Output:
[[1,3],[1,4],[2,3],[2,4]] [[3,1],[3,2],[4,1],[4,2]] [] []
For the n-ary Cartesian product over a list of lists: <lang JavaScript>(() => {
const main = () => { // n-ary Cartesian product of a list of lists.
// cartProdN :: a -> a const cartProdN = foldr( xs => as => bind(as)( x => bind(xs)( a => [[a].concat(x)] ) ) )([[]]);
// TEST ------------------------------------------- return intercalate('\n\n')([ map(show)( cartProdN([ [1776, 1789], [7, 12], [4, 14, 23], [0, 1] ]) ).join('\n'), show(cartProdN([ [1, 2, 3], [30], [50, 100] ])), show(cartProdN([ [1, 2, 3], [], [50, 100] ])) ]) };
// GENERIC FUNCTIONS ----------------------------------
// bind :: [a] -> (a -> [b]) -> [b] const bind = xs => f => xs.flatMap(f);
// foldr :: (a -> b -> b) -> b -> [a] -> b const foldr = f => a => xs => { let v = a, i = xs.length; while (i--) v = f(xs[i])(v); return v; };
// intercalate :: String -> [a] -> String const intercalate = s => xs => xs.join(s);
// map :: (a -> b) -> [a] -> [b] const map = f => xs => xs.map(f);
// show :: a -> String const show = x => JSON.stringify(x);
return main();
})();</lang>
- Output:
[1776,7,4,0] [1776,7,4,1] [1776,7,14,0] [1776,7,14,1] [1776,7,23,0] [1776,7,23,1] [1776,12,4,0] [1776,12,4,1] [1776,12,14,0] [1776,12,14,1] [1776,12,23,0] [1776,12,23,1] [1789,7,4,0] [1789,7,4,1] [1789,7,14,0] [1789,7,14,1] [1789,7,23,0] [1789,7,23,1] [1789,12,4,0] [1789,12,4,1] [1789,12,14,0] [1789,12,14,1] [1789,12,23,0] [1789,12,23,1] [[1,30,50],[1,30,100],[2,30,50],[2,30,100],[3,30,50],[3,30,100]] []
Imperative
Imperative implementations of Cartesian products are inevitably less compact and direct, but we can certainly write an iterative translation of a fold over nested applications of bind or concatMap:
<lang JavaScript>(() => {
// n-ary Cartesian product of a list of lists // ( Imperative implementation )
// cartProd :: [a] -> [b] -> a, b const cartProd = lists => { let ps = [], acc = [ [] ], i = lists.length; while (i--) { let subList = lists[i], j = subList.length; while (j--) { let x = subList[j], k = acc.length; while (k--) ps.push([x].concat(acc[k])) }; acc = ps; ps = []; }; return acc.reverse(); };
// GENERIC FUNCTIONS ------------------------------------------------------
// intercalate :: String -> [a] -> String const intercalate = (s, xs) => xs.join(s);
// map :: (a -> b) -> [a] -> [b] const map = (f, xs) => xs.map(f);
// show :: a -> String const show = x => JSON.stringify(x);
// unlines :: [String] -> String const unlines = xs => xs.join('\n');
// TEST ------------------------------------------------------------------- return intercalate('\n\n', [show(cartProd([ [1, 2], [3, 4] ])), show(cartProd([ [3, 4], [1, 2] ])), show(cartProd([ [1, 2], [] ])), show(cartProd([ [], [1, 2] ])), unlines(map(show, cartProd([ [1776, 1789], [7, 12], [4, 14, 23], [0, 1] ]))), show(cartProd([ [1, 2, 3], [30], [50, 100] ])), show(cartProd([ [1, 2, 3], [], [50, 100] ])) ]);
})();</lang>
- Output:
[[1,4],[1,3],[2,4],[2,3]] [[3,2],[3,1],[4,2],[4,1]] [] [] [1776,12,4,1] [1776,12,4,0] [1776,12,14,1] [1776,12,14,0] [1776,12,23,1] [1776,12,23,0] [1776,7,4,1] [1776,7,4,0] [1776,7,14,1] [1776,7,14,0] [1776,7,23,1] [1776,7,23,0] [1789,12,4,1] [1789,12,4,0] [1789,12,14,1] [1789,12,14,0] [1789,12,23,1] [1789,12,23,0] [1789,7,4,1] [1789,7,4,0] [1789,7,14,1] [1789,7,14,0] [1789,7,23,1] [1789,7,23,0] [[1,30,50],[1,30,100],[2,30,50],[2,30,100],[3,30,50],[3,30,100]] []
jq
jq is stream-oriented and so we begin by defining a function that will emit a stream of the elements of the Cartesian product of two arrays: <lang jq> def products: .[0][] as $x | .[1][] as $y | [$x,$y]; </lang>
To generate an array of these arrays, one would in practice most likely simply write `[products]`, but to comply with the requirements of this article, we can define `product` as: <lang jq> def product: [products]; </lang>
For the sake of brevity, two illustrations should suffice:
[ [1,2], [3,4] ] | products
produces the stream:
[1,3] [1,4] [2,3] [2,4]
And <lang jq> [[1,2], []] | product </lang> produces:
[]
n-way Cartesian Product
Given an array of two or more arrays as input, `cartesians` as defined here produces a stream of the components of their Cartesian product:
<lang jq> def cartesians:
if length <= 2 then products else .[0][] as $x | (.[1:] | cartesians) as $y | [$x] + $y end;
</lang>
Again for brevity, in the following, we will just show the number of items in the Cartesian products:
[ [1776, 1789], [7, 12], [4, 14, 23], [0, 1]] | [cartesians] | length # 24
[[1, 2, 3], [30], [500, 100] ] | [cartesians] | length # 6
[[1, 2, 3], [], [500, 100] ] | [cartesians] | length # 0
Julia
<lang julia>
- Product {1, 2} × {3, 4}
collect(product([1, 2], [3, 4]))
- Product {3, 4} × {1, 2}
collect(product([3, 4], [1, 2]))
- Product {1, 2} × {}
collect(product([1, 2], []))
- Product {} × {1, 2}
collect(product([], [1, 2]))
- Product {1776, 1789} × {7, 12} × {4, 14, 23} × {0, 1}
collect(product([1776, 1789], [7, 12], [4, 14, 23], [0, 1]))
- Product {1, 2, 3} × {30} × {500, 100}
collect(product([1, 2, 3], [30], [500, 100]))
- Product {1, 2, 3} × {} × {500, 100}
collect(product([1, 2, 3], [], [500, 100])) </lang>
Kotlin
<lang scala>// version 1.1.2
fun flattenList(nestList: List<Any>): List<Any> {
val flatList = mutableListOf<Any>()
fun flatten(list: List<Any>) { for (e in list) { if (e !is List<*>) flatList.add(e) else @Suppress("UNCHECKED_CAST") flatten(e as List<Any>) } }
flatten(nestList) return flatList
}
operator fun List<Any>.times(other: List<Any>): List<List<Any>> {
val prod = mutableListOf<List<Any>>() for (e in this) { for (f in other) { prod.add(listOf(e, f)) } } return prod
}
fun nAryCartesianProduct(lists: List<List<Any>>): List<List<Any>> {
require(lists.size >= 2) return lists.drop(2).fold(lists[0] * lists[1]) { cp, ls -> cp * ls }.map { flattenList(it) }
}
fun printNAryProduct(lists: List<List<Any>>) {
println("${lists.joinToString(" x ")} = ") println("[") println(nAryCartesianProduct(lists).joinToString("\n ", " ")) println("]\n")
}
fun main(args: Array<String>) {
println("[1, 2] x [3, 4] = ${listOf(1, 2) * listOf(3, 4)}") println("[3, 4] x [1, 2] = ${listOf(3, 4) * listOf(1, 2)}") println("[1, 2] x [] = ${listOf(1, 2) * listOf()}") println("[] x [1, 2] = ${listOf<Any>() * listOf(1, 2)}") println("[1, a] x [2, b] = ${listOf(1, 'a') * listOf(2, 'b')}") println() printNAryProduct(listOf(listOf(1776, 1789), listOf(7, 12), listOf(4, 14, 23), listOf(0, 1))) printNAryProduct(listOf(listOf(1, 2, 3), listOf(30), listOf(500, 100))) printNAryProduct(listOf(listOf(1, 2, 3), listOf<Int>(), listOf(500, 100))) printNAryProduct(listOf(listOf(1, 2, 3), listOf(30), listOf('a', 'b')))
}</lang>
- Output:
[1, 2] x [3, 4] = [[1, 3], [1, 4], [2, 3], [2, 4]] [3, 4] x [1, 2] = [[3, 1], [3, 2], [4, 1], [4, 2]] [1, 2] x [] = [] [] x [1, 2] = [] [1, a] x [2, b] = [[1, 2], [1, b], [a, 2], [a, b]] [1776, 1789] x [7, 12] x [4, 14, 23] x [0, 1] = [ [1776, 7, 4, 0] [1776, 7, 4, 1] [1776, 7, 14, 0] [1776, 7, 14, 1] [1776, 7, 23, 0] [1776, 7, 23, 1] [1776, 12, 4, 0] [1776, 12, 4, 1] [1776, 12, 14, 0] [1776, 12, 14, 1] [1776, 12, 23, 0] [1776, 12, 23, 1] [1789, 7, 4, 0] [1789, 7, 4, 1] [1789, 7, 14, 0] [1789, 7, 14, 1] [1789, 7, 23, 0] [1789, 7, 23, 1] [1789, 12, 4, 0] [1789, 12, 4, 1] [1789, 12, 14, 0] [1789, 12, 14, 1] [1789, 12, 23, 0] [1789, 12, 23, 1] ] [1, 2, 3] x [30] x [500, 100] = [ [1, 30, 500] [1, 30, 100] [2, 30, 500] [2, 30, 100] [3, 30, 500] [3, 30, 100] ] [1, 2, 3] x [] x [500, 100] = [ ] [1, 2, 3] x [30] x [a, b] = [ [1, 30, a] [1, 30, b] [2, 30, a] [2, 30, b] [3, 30, a] [3, 30, b] ]
Lua
Functional
An iterator is created to output the product items. <lang lua> local pk,upk = table.pack, table.unpack
local getn = function(t)return #t end local const = function(k)return function(e) return k end end local function attachIdx(f)-- one-time-off function modifier local idx = 0 return function(e)idx=idx+1 ; return f(e,idx)end end local function reduce(t,acc,f) for i=1,t.n or #t do acc=f(acc,t[i])end return acc end local function imap(t,f) local r = {n=t.n or #t, r=reduce, u=upk, m=imap} for i=1,r.n do r[i]=f(t[i])end return r end
local function prod(...) local ts = pk(...) local limit = imap(ts,getn) local idx, cnt = imap(limit,const(1)), 0 local max = reduce(limit,1,function(a,b)return a*b end) local function ret(t,i)return t[idx[i]] end return function() if cnt>=max then return end -- no more output if cnt==0 then -- skip for 1st cnt = cnt + 1 else cnt, idx[#idx] = cnt + 1, idx[#idx] + 1 for i=#idx,2,-1 do -- update index list if idx[i]<=limit[i] then break -- no further update need else -- propagate limit overflow idx[i],idx[i-1] = 1, idx[i-1]+1 end end end return cnt,imap(ts,attachIdx(ret)):u() end end
--- test
for i,a,b in prod({1,2},{3,4}) do print(i,a,b) end print() for i,a,b in prod({3,4},{1,2}) do print(i,a,b) end
</lang>
- Output:
1 1 3 2 1 4 3 2 3 4 2 4 1 3 1 2 3 2 3 4 1 4 4 2
Using coroutines
I have not benchmarked this, but I believe that this should run faster than the functional implementation and also likely the imperative implementation, it has significantly fewer function calls per iteration, and only the stack changes during iteration (no garbage collection during iteration). On the other hand due to avoiding garbage collection, result is reused between returns, so mutating the returned result is unsafe.
It is possible that specialising descend by depth may yield a further improvement in performance, but it would only be able to eliminate the lookup of sets[depth] and the if test, because the reference to result[depth] is required; I doubt the increase in complexity would be worth the (potential) improvement in performance. <lang lua>local function cartesian_product(sets)
local result = {} local set_count = #sets
--[[ I believe that this should make the below go very slightly faster, because it doesn't need to lookup yield in coroutine each time it
yields, though perhaps the compiler optimises the lookup away? ]] local yield = coroutine.yield local function descend(depth) if depth == set_count then for k,v in pairs(sets[depth]) do result[depth] = v yield(result) end else for k,v in pairs(sets[depth]) do result[depth] = v descend(depth + 1) end end end return coroutine.wrap(function() descend(1) end)
end
--- tests local test_cases = {
{{1, 2}, {3, 4}}, {{3, 4}, {1, 2}}, {{1776, 1789}, {7, 12}, {4, 14, 23}, {0,1}}, {{1,2,3}, {30}, {500, 100}}, {{1,2,3}, {}, {500, 100}}
}
local function format_nested_list(list)
if list[1] and type(list[1]) == "table" then local formatted_items = {} for i, item in ipairs(list) do formatted_items[i] = format_nested_list(item) end return format_nested_list(formatted_items) else return "{" .. table.concat(list, ",") .. "}" end
end
for _,test_case in ipairs(test_cases) do
print(format_nested_list(test_case)) for product in cartesian_product(test_case) do print(" " .. format_nested_list(product)) end
end</lang>
Imperative iterator
The functional implementation restated as an imperative iterator, also adjusted to not allocate a new result table on each iteration; this saves time, but makes mutating the returned table unsafe. <lang lua>local function cartesian_product(sets)
local item_counts = {} local indices = {} local results = {} local set_count = #sets local combination_count = 1 for set_index=set_count, 1, -1 do local set = sets[set_index] local item_count = #set item_counts[set_index] = item_count indices[set_index] = 1 results[set_index] = set[1] combination_count = combination_count * item_count end local combination_index = 0 return function() if combination_index >= combination_count then return end -- no more output
if combination_index == 0 then goto skip_update end -- skip first index update indices[set_count] = indices[set_count] + 1 for set_index=set_count, 1, -1 do -- update index list local set = sets[set_index] local index = indices[set_index] if index <= item_counts[set_index] then results[set_index] = set[index] break -- no further update needed else -- propagate item_counts overflow results[set_index] = set[1] indices[set_index] = 1 if set_index > 1 then indices[set_index - 1] = indices[set_index - 1] + 1 end end end ::skip_update:: combination_index = combination_index + 1 return combination_index, results end
end --- tests local test_cases = {
{{1, 2}, {3, 4}}, {{3, 4}, {1, 2}}, {{1776, 1789}, {7, 12}, {4, 14, 23}, {0,1}}, {{1,2,3}, {30}, {500, 100}}, {{1,2,3}, {}, {500, 100}}
}
local function format_nested_list(list)
if list[1] and type(list[1]) == "table" then local formatted_items = {} for i, item in ipairs(list) do formatted_items[i] = format_nested_list(item) end return format_nested_list(formatted_items) else return "{" .. table.concat(list, ",") .. "}" end
end
for _,test_case in ipairs(test_cases) do
print(format_nested_list(test_case)) for i, product in cartesian_product(test_case) do print(i, format_nested_list(product)) end
end</lang>
Maple
<lang Maple> cartmulti := proc ()
local m, v; if [] in {args} then return []; else
m := Iterator:-CartesianProduct(args);
for v in m do printf("%{}a\n", v); end do; end if; end proc;
</lang>
Mathematica
<lang Mathematica>cartesianProduct[args__] := Flatten[Outer[List, args], Length[{args}] - 1]</lang>
Modula-2
<lang modula2>MODULE CartesianProduct; FROM FormatString IMPORT FormatString; FROM Terminal IMPORT WriteString,WriteLn,ReadChar;
PROCEDURE WriteInt(a : INTEGER); VAR buf : ARRAY[0..9] OF CHAR; BEGIN
FormatString("%i", buf, a); WriteString(buf)
END WriteInt;
PROCEDURE Cartesian(a,b : ARRAY OF INTEGER); VAR i,j : CARDINAL; BEGIN
WriteString("["); FOR i:=0 TO HIGH(a) DO FOR j:=0 TO HIGH(b) DO IF (i>0) OR (j>0) THEN WriteString(","); END; WriteString("["); WriteInt(a[i]); WriteString(","); WriteInt(b[j]); WriteString("]") END END; WriteString("]"); WriteLn
END Cartesian;
TYPE
AP = ARRAY[0..1] OF INTEGER; E = ARRAY[0..0] OF INTEGER;
VAR
a,b : AP;
BEGIN
a := AP{1,2}; b := AP{3,4}; Cartesian(a,b);
a := AP{3,4}; b := AP{1,2}; Cartesian(a,b);
(* If there is a way to create an empty array, I do not know of it *)
ReadChar
END CartesianProduct.</lang>
OCaml
The double semicolons are necessary only for the toplevel
Naive but more readable version <lang ocaml> let rec product l1 l2 =
match l1, l2 with | [], _ | _, [] -> [] | h1::t1, h2::t2 -> (h1,h2)::(product [h1] t2)@(product t1 l2)
product [1;2] [3;4];; (*- : (int * int) list = [(1, 3); (1, 4); (2, 3); (2, 4)]*) product [3;4] [1;2];; (*- : (int * int) list = [(3, 1); (3, 2); (4, 1); (4, 2)]*) product [1;2] [];; (*- : (int * 'a) list = []*) product [] [1;2];; (*- : ('a * int) list = []*) </lang>
Implementation with a bit more tail-call optimization, inroducing a helper function. The order of the result is changed but it should not be an issue for most uses. <lang ocaml> let product' l1 l2 =
let rec aux ~acc l1' l2' = match l1', l2' with | [], _ | _, [] -> acc | h1::t1, h2::t2 -> let acc = (h1,h2)::acc in let acc = aux ~acc t1 l2' in aux ~acc [h1] t2 in aux [] l1 l2
product' [1;2] [3;4];; (*- : (int * int) list = [(1, 4); (2, 4); (2, 3); (1, 3)]*) product' [3;4] [1;2];; (*- : (int * int) list = [(3, 2); (4, 2); (4, 1); (3, 1)]*) product' [1;2] [];; (*- : (int * 'a) list = []*) product' [] [1;2];; (*- : ('a * int) list = []*) </lang>
Extra credit function. Since in OCaml a function can return only one type, and because tuples of different arities are different types, this returns a list of lists rather than a list of tuples. <lang ocaml> let rec product l =
(* We need to do the cross product of our current list and all the others * so we define a helper function for that *) let rec aux ~acc l1 l2 = match l1, l2 with | [], _ | _, [] -> acc | h1::t1, h2::t2 -> let acc = (h1::h2)::acc in let acc = (aux ~acc t1 l2) in aux ~acc [h1] t2 (* now we can do the actual computation *) in match l with | [] -> [] | [l1] -> List.map (fun x -> [x]) l1 | l1::tl -> let tail_product = product tl in aux ~acc:[] l1 tail_product
product [[1;2];[3;4]];;
(*- : int list list = [[1; 4]; [2; 4]; [2; 3]; [1; 3]]*)
product [[3;4];[1;2]];;
(*- : int list list = [[3; 2]; [4; 2]; [4; 1]; [3; 1]]*)
product [[1;2];[]];;
(*- : int list list = []*)
product [[];[1;2]];;
(*- : int list list = []*)
product [[1776; 1789];[7; 12];[4; 14; 23];[0; 1]];;
(*
- : int list list =
[[1776; 7; 4; 1]; [1776; 12; 4; 1]; [1776; 12; 14; 1]; [1776; 12; 23; 1];
[1776; 12; 23; 0]; [1776; 12; 14; 0]; [1776; 12; 4; 0]; [1776; 7; 14; 1]; [1776; 7; 23; 1]; [1776; 7; 23; 0]; [1776; 7; 14; 0]; [1789; 7; 4; 1]; [1789; 12; 4; 1]; [1789; 12; 14; 1]; [1789; 12; 23; 1]; [1789; 12; 23; 0]; [1789; 12; 14; 0]; [1789; 12; 4; 0]; [1789; 7; 14; 1]; [1789; 7; 23; 1]; [1789; 7; 23; 0]; [1789; 7; 14; 0]; [1789; 7; 4; 0]; [1776; 7; 4; 0]]
- )
product [[1; 2; 3];[30];[500; 100]];; (* - : int list list =
[[1; 30; 500]; [2; 30; 500]; [3; 30; 500]; [3; 30; 100]; [2; 30; 100];
[1; 30; 100]]
- )
product [[1; 2; 3];[];[500; 100]];; (*- : int list list = []*)
</lang>
Better type
In the latter example, our function has this signature: <lang ocaml> val product : 'a list list -> 'a list list = <fun> </lang> This lacks clarity as those two lists are not equivalent since one replaces a tuple. We can get a better signature by creating a tuple type: <lang ocaml> type 'a tuple = 'a list
let rec product (l:'a list tuple) =
(* We need to do the cross product of our current list and all the others * so we define a helper function for that *) let rec aux ~acc l1 l2 = match l1, l2 with | [], _ | _, [] -> acc | h1::t1, h2::t2 -> let acc = (h1::h2)::acc in let acc = (aux ~acc t1 l2) in aux ~acc [h1] t2 (* now we can do the actual computation *) in match l with | [] -> [] | [l1] -> List.map ~f:(fun x -> ([x]:'a tuple)) l1 | l1::tl -> let tail_product = product tl in aux ~acc:[] l1 tail_product
type 'a tuple = 'a list val product : 'a list tuple -> 'a tuple list = <fun> </lang>
Perl
Iterative
Nested loops, with a short-circuit to quit early if any term is an empty set. <lang perl>sub cartesian {
my $sets = shift @_; for (@$sets) { return [] unless @$_ }
my $products = [[]]; for my $set (reverse @$sets) { my $partial = $products; $products = []; for my $item (@$set) { for my $product (@$partial) { push @$products, [$item, @$product]; } } } $products;
}
sub product {
my($s,$fmt) = @_; my $tuples; for $a ( @{ cartesian( \@$s ) } ) { $tuples .= sprintf "($fmt) ", @$a; } $tuples . "\n";
}
print product([[1, 2], [3, 4] ], '%1d %1d' ). product([[3, 4], [1, 2] ], '%1d %1d' ). product([[1, 2], [] ], '%1d %1d' ). product([[], [1, 2] ], '%1d %1d' ). product([[1,2,3], [30], [500,100] ], '%1d %1d %3d' ). product([[1,2,3], [], [500,100] ], '%1d %1d %3d' ). product([[1776,1789], [7,12], [4,14,23], [0,1]], '%4d %2d %2d %1d')</lang>
- Output:
(1 3) (1 4) (2 3) (2 4) (3 1) (3 2) (4 1) (4 2) (1 30 500) (1 30 100) (2 30 500) (2 30 100) (3 30 500) (3 30 100) (1776 7 4 0) (1776 7 4 1) (1776 7 14 0) (1776 7 14 1) (1776 7 23 0) (1776 7 23 1) (1776 12 4 0) (1776 12 4 1) (1776 12 14 0) (1776 12 14 1) (1776 12 23 0) (1776 12 23 1) (1789 7 4 0) (1789 7 4 1) (1789 7 14 0) (1789 7 14 1) (1789 7 23 0) (1789 7 23 1) (1789 12 4 0) (1789 12 4 1) (1789 12 14 0) (1789 12 14 1) (1789 12 23 0) (1789 12 23 1)
Glob
This being Perl, there's more than one way to do it. A quick demonstration of how glob
, more typically used for filename wildcard expansion, can solve the task.
<lang perl>$tuples = [ map { [split /:/] } glob '{1,2,3}:{30}:{500,100}' ];
for $a (@$tuples) { printf "(%1d %2d %3d) ", @$a; }</lang>
- Output:
(1 30 500) (1 30 100) (2 30 500) (2 30 100) (3 30 500) (3 30 100)
Modules
A variety of modules can do this correctly for an arbitrary number of lists (each of independent length). Arguably using modules is very idiomatic Perl. <lang perl>use ntheory qw/forsetproduct/; forsetproduct { say "@_" } [1,2,3],[qw/a b c/],[qw/@ $ !/];
use Set::Product qw/product/; product { say "@_" } [1,2,3],[qw/a b c/],[qw/@ $ !/];
use Math::Cartesian::Product; cartesian { say "@_" } [1,2,3],[qw/a b c/],[qw/@ $ !/];
use Algorithm::Loops qw/NestedLoops/; NestedLoops([[1,2,3],[qw/a b c/],[qw/@ $ !/]], sub { say "@_"; });</lang>
Perl 6
The cross meta operator X will return the cartesian product of two lists. To apply the cross meta-operator to a variable number of lists, use the reduce cross meta operator [X].
<lang perl6># cartesian product of two lists using the X cross meta-operator say (1, 2) X (3, 4); say (3, 4) X (1, 2); say (1, 2) X ( ); say ( ) X ( 1, 2 );
- cartesian product of variable number of lists using
- the [X] reduce cross meta-operator
say [X] (1776, 1789), (7, 12), (4, 14, 23), (0, 1); say [X] (1, 2, 3), (30), (500, 100); say [X] (1, 2, 3), (), (500, 100);</lang>
- Output:
((1 3) (1 4) (2 3) (2 4)) ((3 1) (3 2) (4 1) (4 2)) () () ((1776 7 4 0) (1776 7 4 1) (1776 7 14 0) (1776 7 14 1) (1776 7 23 0) (1776 7 23 1) (1776 12 4 0) (1776 12 4 1) (1776 12 14 0) (1776 12 14 1) (1776 12 23 0) (1776 12 23 1) (1789 7 4 0) (1789 7 4 1) (1789 7 14 0) (1789 7 14 1) (1789 7 23 0) (1789 7 23 1) (1789 12 4 0) (1789 12 4 1) (1789 12 14 0) (1789 12 14 1) (1789 12 23 0) (1789 12 23 1)) ((1 30 500) (1 30 100) (2 30 500) (2 30 100) (3 30 500) (3 30 100)) ()
Phix
<lang Phix>function cart(sequence s) sequence res = {}
for n=2 to length(s) do for i=1 to length(s[1]) do for j=1 to length(s[2]) do res = append(res,s[1][i]&s[2][j]) end for end for if length(s)=2 then exit end if s[1..2] = {res} res = {} end for return res
end function
?cart({{1,2},{3,4}}) ?cart({{3,4},{1,2}}) ?cart({{1,2},{}}) ?cart({{},{1,2}}) ?cart({{1776, 1789},{7, 12},{4, 14, 23},{0, 1}}) ?cart({{1, 2, 3},{30},{500, 100}}) ?cart({{1, 2, 3},{},{500, 100}})</lang>
- Output:
{{1,3},{1,4},{2,3},{2,4}} {{3,1},{3,2},{4,1},{4,2}} {} {} {{1776,7,4,0},{1776,7,4,1},{1776,7,14,0},{1776,7,14,1},{1776,7,23,0},{1776,7,23,1}, {1776,12,4,0},{1776,12,4,1},{1776,12,14,0},{1776,12,14,1},{1776,12,23,0},{1776,12,23,1}, {1789,7,4,0},{1789,7,4,1},{1789,7,14,0},{1789,7,14,1},{1789,7,23,0},{1789,7,23,1}, {1789,12,4,0},{1789,12,4,1},{1789,12,14,0},{1789,12,14,1},{1789,12,23,0},{1789,12,23,1}} {{1,30,500},{1,30,100},{2,30,500},{2,30,100},{3,30,500},{3,30,100}} {}
PicoLisp
<lang PicoLisp>(de 2lists (L1 L2)
(mapcan '((I) (mapcar '((A) ((if (atom A) list cons) I A)) L2 ) ) L1 ) )
(de reduce (L . @)
(ifn (rest) L (2lists L (apply reduce (rest)))) )
(de cartesian (L . @)
(and L (rest) (pass reduce L)) )
(println
(cartesian (1 2)) )
(println
(cartesian NIL (1 2)) )
(println
(cartesian (1 2) (3 4)) )
(println
(cartesian (3 4) (1 2)) )
(println
(cartesian (1776 1789) (7 12) (4 14 23) (0 1)) )
(println
(cartesian (1 2 3) (30) (500 100)) )
(println
(cartesian (1 2 3) NIL (500 100)) )</lang>
- Output:
NIL NIL ((1 3) (1 4) (2 3) (2 4)) ((3 1) (3 2) (4 1) (4 2)) ((1776 7 4 0) (1776 7 4 1) (1776 7 14 0) (1776 7 14 1) (1776 7 23 0) (1776 7 23 1) (1776 12 4 0) (1776 12 4 1) (1776 12 14 0) (1776 12 14 1) (1776 12 23 0) (1776 12 23 1) (1789 7 4 0) (1789 7 4 1) (1789 7 14 0) (1789 7 14 1) (1789 7 23 0) (1789 7 23 1) (1789 12 4 0) (1789 12 4 1) (1789 12 14 0) (1789 12 14 1) (1789 12 23 0) (1789 12 23 1)) ((1 30 500) (1 30 100) (2 30 500) (2 30 100) (3 30 500) (3 30 100)) NIL
Python
Using itertools
<lang python>import itertools
def cp(lsts):
return list(itertools.product(*lsts))
if __name__ == '__main__':
from pprint import pprint as pp for lists in [[[1,2],[3,4]], [[3,4],[1,2]], [[], [1, 2]], [[1, 2], []], ((1776, 1789), (7, 12), (4, 14, 23), (0, 1)), ((1, 2, 3), (30,), (500, 100)), ((1, 2, 3), (), (500, 100))]: print(lists, '=>') pp(cp(lists), indent=2)
</lang>
- Output:
[[1, 2], [3, 4]] => [(1, 3), (1, 4), (2, 3), (2, 4)] [[3, 4], [1, 2]] => [(3, 1), (3, 2), (4, 1), (4, 2)] [[], [1, 2]] => [] [[1, 2], []] => [] ((1776, 1789), (7, 12), (4, 14, 23), (0, 1)) => [ (1776, 7, 4, 0), (1776, 7, 4, 1), (1776, 7, 14, 0), (1776, 7, 14, 1), (1776, 7, 23, 0), (1776, 7, 23, 1), (1776, 12, 4, 0), (1776, 12, 4, 1), (1776, 12, 14, 0), (1776, 12, 14, 1), (1776, 12, 23, 0), (1776, 12, 23, 1), (1789, 7, 4, 0), (1789, 7, 4, 1), (1789, 7, 14, 0), (1789, 7, 14, 1), (1789, 7, 23, 0), (1789, 7, 23, 1), (1789, 12, 4, 0), (1789, 12, 4, 1), (1789, 12, 14, 0), (1789, 12, 14, 1), (1789, 12, 23, 0), (1789, 12, 23, 1)] ((1, 2, 3), (30,), (500, 100)) => [ (1, 30, 500), (1, 30, 100), (2, 30, 500), (2, 30, 100), (3, 30, 500), (3, 30, 100)] ((1, 2, 3), (), (500, 100)) => []
Using the 'Applicative' abstraction
This task calls for alternative approaches to defining cartesian products, and one particularly compact alternative route to a native cartesian product (in a more mathematically reasoned idiom of programming) is through the Applicative abstraction (see Applicative Functor), which is slightly more general than the possibly better known monad structure. Applicative functions are provided off-the-shelf by languages like Agda, Idris, Haskell and Scala, and can usefully be implemented in any language, including Python, which supports higher-order functions.
If we write ourselves a re-usable Python ap function for the case of lists (applicative functions for other 'data containers' can also be written – this one applies a list of functions to a list of values):
<lang python># ap (<*>) :: [(a -> b)] -> [a] -> [b] def ap(fs):
return lambda xs: foldl( lambda a: lambda f: a + foldl( lambda a: lambda x: a + [f(x)])([])(xs) )([])(fs)</lang>
then one simple use of it will be to define the cartesian product of two lists (of possibly different type) as:
<lang python>ap(map(Tuple, xs))</lang>
where Tuple is a constructor, and xs is bound to the first of two lists. The returned value is a function which can be applied to a second list.
For an nAry product, we can then use a fold (catamorphism) to lift the basic function over two lists cartesianProduct :: [a] -> [b] -> [(a, b)] to a function over a list of lists:
<lang python># nAryCartProd :: [[a], [b], [c] ...] -> [(a, b, c ...)] def nAryCartProd(xxs):
return foldl1(cartesianProduct)( xxs )</lang>
For example:
<lang python># Two lists -> list of tuples
- cartesianProduct :: [a] -> [b] -> [(a, b)]
def cartesianProduct(xs):
return ap(map(Tuple, xs))
- List of lists -> list of tuples
- nAryCartProd :: [[a], [b], [c] ...] -> [(a, b, c ...)]
def nAryCartProd(xxs):
return foldl1(cartesianProduct)( xxs )
- main :: IO ()
def main():
# Product of lists of different types print ( 'Product of two lists of different types:' ) print( cartesianProduct(['a', 'b', 'c'])( [1, 2] ) )
# TESTS OF PRODUCTS OF TWO LISTS
print( '\nSpecified tests of products of two lists:' ) print( cartesianProduct([1, 2])([3, 4]), ' <--> ', cartesianProduct([3, 4])([1, 2]) ) print ( cartesianProduct([1, 2])([]), ' <--> ', cartesianProduct([])([1, 2]) )
# TESTS OF N-ARY CARTESIAN PRODUCTS
print('\nSpecified tests of nAry products:') for xs in nAryCartProd([[1776, 1789], [7, 12], [4, 14, 23], [0, 1]]): print(xs)
for xs in ( map_(nAryCartProd)( [ [[1, 2, 3], [30], [500, 100]], [[1, 2, 3], [], [500, 100]] ] ) ): print( xs )
- GENERIC -------------------------------------------------
- Applicative function for lists
- ap (<*>) :: [(a -> b)] -> [a] -> [b]
def ap(fs):
return lambda xs: foldl( lambda a: lambda f: a + foldl( lambda a: lambda x: a + [f(x)])([])(xs) )([])(fs)
- foldl :: (a -> b -> a) -> a -> [b] -> a
def foldl(f):
def go(v, xs): a = v for x in xs: a = f(a)(x) return a return lambda acc: lambda xs: go(acc, xs)
- foldl1 :: (a -> a -> a) -> [a] -> a
def foldl1(f):
return lambda xs: foldl(f)(xs[0])( xs[1:] ) if xs else None
- map :: (a -> b) -> [a] -> [b]
def map_(f):
return lambda xs: list(map(f, xs))
- Tuple :: a -> b -> (a, b)
def Tuple(x):
return lambda y: ( x + (y,) ) if tuple is type(x) else (x, y)
- TEST ----------------------------------------------------
if __name__ == '__main__':
main()</lang>
- Output:
Product of two lists of different types: [('a', 1), ('a', 2), ('b', 1), ('b', 2), ('c', 1), ('c', 2)] Specified tests of products of two lists: [(1, 3), (1, 4), (2, 3), (2, 4)] <--> [(3, 1), (3, 2), (4, 1), (4, 2)] [] <--> [] Specified tests of nAry products: (1776, 7, 4, 0) (1776, 7, 4, 1) (1776, 7, 14, 0) (1776, 7, 14, 1) (1776, 7, 23, 0) (1776, 7, 23, 1) (1776, 12, 4, 0) (1776, 12, 4, 1) (1776, 12, 14, 0) (1776, 12, 14, 1) (1776, 12, 23, 0) (1776, 12, 23, 1) (1789, 7, 4, 0) (1789, 7, 4, 1) (1789, 7, 14, 0) (1789, 7, 14, 1) (1789, 7, 23, 0) (1789, 7, 23, 1) (1789, 12, 4, 0) (1789, 12, 4, 1) (1789, 12, 14, 0) (1789, 12, 14, 1) (1789, 12, 23, 0) (1789, 12, 23, 1) [(1, 30, 500), (1, 30, 100), (2, 30, 500), (2, 30, 100), (3, 30, 500), (3, 30, 100)] []
R
<lang R> one_w_many <- function(one, many) lapply(many, function(x) c(one,x))
- Let's define an infix operator to perform a cartesian product.
"%p%" <- function( a, b ) {
p = c( sapply(a, function (x) one_w_many(x, b) ) ) if (is.null(unlist(p))) list() else p}
display_prod <-
function (xs) { for (x in xs) cat( paste(x, collapse=", "), "\n" ) }
fmt_vec <- function(v) sprintf("(%s)", paste(v, collapse=', '))
go <- function (...) {
cat("\n", paste( sapply(list(...),fmt_vec), collapse=" * "), "\n") prod = Reduce( '%p%', list(...) ) display_prod( prod ) }
</lang>
Simple tests:
<lang R> > display_prod( c(1, 2) %p% c(3, 4) ) 1, 3 1, 4 2, 3 2, 4 > display_prod( c(3, 4) %p% c(1, 2) ) 3, 1 3, 2 4, 1 4, 2 > display_prod( c(3, 4) %p% c() ) > </lang>
Tougher tests:
<lang R> go( c(1776, 1789), c(7, 12), c(4, 14, 23), c(0, 1) ) go( c(1, 2, 3), c(30), c(500, 100) ) go( c(1, 2, 3), c(), c(500, 100) ) </lang>
- Output:
(1776, 1789) * (7, 12) * (4, 14, 23) * (0, 1) 1776, 7, 4, 0 1776, 7, 4, 1 1776, 7, 14, 0 1776, 7, 14, 1 1776, 7, 23, 0 1776, 7, 23, 1 1776, 12, 4, 0 1776, 12, 4, 1 1776, 12, 14, 0 1776, 12, 14, 1 1776, 12, 23, 0 1776, 12, 23, 1 1789, 7, 4, 0 1789, 7, 4, 1 1789, 7, 14, 0 1789, 7, 14, 1 1789, 7, 23, 0 1789, 7, 23, 1 1789, 12, 4, 0 1789, 12, 4, 1 1789, 12, 14, 0 1789, 12, 14, 1 1789, 12, 23, 0 1789, 12, 23, 1 (1, 2, 3) * (30) * (500, 100) 1, 30, 500 1, 30, 100 2, 30, 500 2, 30, 100 3, 30, 500 3, 30, 100 (1, 2, 3) * () * (500, 100)
Racket
Racket has a built-in "cartesian-product" function:
<lang>#lang racket/base (require rackunit
;; usually, included in "racket", but we're using racket/base so we ;; show where this comes from (only-in racket/list cartesian-product))
- these tests will pass silently
(check-equal? (cartesian-product '(1 2) '(3 4))
'((1 3) (1 4) (2 3) (2 4)))
(check-equal? (cartesian-product '(3 4) '(1 2))
'((3 1) (3 2) (4 1) (4 2)))
(check-equal? (cartesian-product '(1 2) '()) '()) (check-equal? (cartesian-product '() '(1 2)) '())
- these will print
(cartesian-product '(1776 1789) '(7 12) '(4 14 23) '(0 1)) (cartesian-product '(1 2 3) '(30) '(500 100)) (cartesian-product '(1 2 3) '() '(500 100))</lang>
- Output:
'((1776 7 4 0) (1776 7 4 1) (1776 7 14 0) (1776 7 14 1) (1776 7 23 0) (1776 7 23 1) (1776 12 4 0) (1776 12 4 1) (1776 12 14 0) (1776 12 14 1) (1776 12 23 0) (1776 12 23 1) (1789 7 4 0) (1789 7 4 1) (1789 7 14 0) (1789 7 14 1) (1789 7 23 0) (1789 7 23 1) (1789 12 4 0) (1789 12 4 1) (1789 12 14 0) (1789 12 14 1) (1789 12 23 0) (1789 12 23 1)) '((1 30 500) (1 30 100) (2 30 500) (2 30 100) (3 30 500) (3 30 100)) '()
REXX
version 1
This REXX version isn't limited by the number of lists or the number of sets within a list. <lang rexx>/*REXX program calculates the Cartesian product of two arbitrary-sized lists. */ @.= /*assign the default value to @. array*/ parse arg @.1 /*obtain the optional value of @.1 */ if @.1= then do; @.1= "{1,2} {3,4}" /*Not specified? Then use the defaults*/
@.2= "{3,4} {1,2}" /* " " " " " " */ @.3= "{1,2} {}" /* " " " " " " */ @.4= "{} {3,4}" /* " " " " " " */ @.5= "{1,2} {3,4,5}" /* " " " " " " */ end /* [↓] process each of the @.n values*/ do n=1 while @.n \= /*keep processing while there's a value*/ z= translate( space( @.n, 0), , ',') /*translate the commas to blanks. */ do #=1 until z== /*process each elements in first list. */ parse var z '{' x.# '}' z /*parse the list (contains elements). */ end /*#*/ $= do i=1 for #-1 /*process the subsequent lists. */ do a=1 for words(x.i) /*obtain the elements of the first list*/ do j=i+1 for #-1 /* " " subsequent lists. */ do b=1 for words(x.j) /* " " elements of subsequent list*/ $=$',('word(x.i, a)","word(x.j, b)')' /*append partial Cartesian product ──►$*/ end /*b*/ end /*j*/ end /*a*/ end /*i*/ say 'Cartesian product of ' space(@.n) " is ───► {"substr($, 2)'}' end /*n*/ /*stick a fork in it, we're all done. */</lang>
- output when using the default lists:
Cartesian product of {1,2} {3,4} is ───► {(1,3),(1,4),(2,3),(2,4)} Cartesian product of {3,4} {1,2} is ───► {(3,1),(3,2),(4,1),(4,2)} Cartesian product of {1,2} {} is ───► {} Cartesian product of {} {3,4} is ───► {} Cartesian product of {1,2} {3,4,5} is ───► {(1,3),(1,4),(1,5),(2,3),(2,4),(2,5)}
version 2
<lang rexx>/* REXX computes the Cartesian Product of up to 4 sets */ Call cart '{1, 2} x {3, 4}' Call cart '{3, 4} x {1, 2}' Call cart '{1, 2} x {}' Call cart '{} x {1, 2}' Call cart '{1776, 1789} x {7, 12} x {4, 14, 23} x {0, 1}' Call cart '{1, 2, 3} x {30} x {500, 100}' Call cart '{1, 2, 3} x {} x {500, 100}' Exit
cart:
Parse Arg sl Say sl Do i=1 By 1 while pos('{',sl)>0 Parse Var sl '{' list '}' sl Do j=1 By 1 While list<> Parse Var list e.i.j . ',' list End n.i=j-1 If n.i=0 Then Do /* an empty set */ Say '{}' Say Return End End n=i-1 ct2.=0 Do i=1 To n.1 Do j=1 To n.2 z=ct2.0+1 ct2.z=e.1.i e.2.j ct2.0=z End End If n<3 Then Return output(2) ct3.=0 Do i=1 To ct2.0 Do k=1 To n.3 z=ct3.0+1 ct3.z=ct2.i e.3.k ct3.0=z End End If n<4 Then Return output(3) ct4.=0 Do i=1 To ct3.0 Do l=1 To n.4 z=ct4.0+1 ct4.z=ct3.i e.4.l ct4.0=z End End Return output(4)
output:
Parse Arg u Do v=1 To value('ct'u'.0') res='{'translate(value('ct'u'.'v),',',' ')'}' Say res End Say ' ' Return 0</lang>
- Output:
{1, 2} x {3, 4} {1,3} {1,4} {2,3} {2,4} {3, 4} x {1, 2} {3,1} {3,2} {4,1} {4,2} {1, 2} x {} {} {} x {1, 2} {} {1776, 1789} x {7, 12} x {4, 14, 23} x {0, 1} {1776,7,4,0} {1776,7,4,1} {1776,7,14,0} {1776,7,14,1} {1776,7,23,0} {1776,7,23,1} {1776,12,4,0} {1776,12,4,1} {1776,12,14,0} {1776,12,14,1} {1776,12,23,0} {1776,12,23,1} {1789,7,4,0} {1789,7,4,1} {1789,7,14,0} {1789,7,14,1} {1789,7,23,0} {1789,7,23,1} {1789,12,4,0} {1789,12,4,1} {1789,12,14,0} {1789,12,14,1} {1789,12,23,0} {1789,12,23,1} {1, 2, 3} x {30} x {500, 100} {1,30,500} {1,30,100} {2,30,500} {2,30,100} {3,30,500} {3,30,100} {1, 2, 3} x {} x {500, 100} {}
Ring
<lang ring>
- Project : Cartesian product of two or more lists
list1 = [[1,2],[3,4]] list2 = [[3,4],[1,2]] cartesian(list1) cartesian(list2)
func cartesian(list1)
for n = 1 to len(list1[1]) for m = 1 to len(list1[2]) see "(" + list1[1][n] + ", " + list1[2][m] + ")" + nl next next see nl
</lang> Output:
(1, 3) (1, 4) (2, 3) (2, 4) (3, 1) (3, 2) (4, 1) (4, 2)
Ruby
"product" is a method of arrays. It takes one or more arrays as argument and results in the Cartesian product: <lang ruby>p [1, 2].product([3, 4]) p [3, 4].product([1, 2]) p [1, 2].product([]) p [].product([1, 2]) p [1776, 1789].product([7, 12], [4, 14, 23], [0, 1]) p [1, 2, 3].product([30], [500, 100]) p [1, 2, 3].product([], [500, 100]) </lang>
- Output:
[[1, 3], [1, 4], [2, 3], [2, 4]][[3, 1], [3, 2], [4, 1], [4, 2]] [] [] [[1776, 7, 4, 0], [1776, 7, 4, 1], [1776, 7, 14, 0], [1776, 7, 14, 1], [1776, 7, 23, 0], [1776, 7, 23, 1], [1776, 12, 4, 0], [1776, 12, 4, 1], [1776, 12, 14, 0], [1776, 12, 14, 1], [1776, 12, 23, 0], [1776, 12, 23, 1], [1789, 7, 4, 0], [1789, 7, 4, 1], [1789, 7, 14, 0], [1789, 7, 14, 1], [1789, 7, 23, 0], [1789, 7, 23, 1], [1789, 12, 4, 0], [1789, 12, 4, 1], [1789, 12, 14, 0], [1789, 12, 14, 1], [1789, 12, 23, 0], [1789, 12, 23, 1]] [[1, 30, 500], [1, 30, 100], [2, 30, 500], [2, 30, 100], [3, 30, 500], [3, 30, 100]] []
Rust
<lang rust>type List = Vec<Vec<u32>>;
fn cartesian_product(lists: &List) -> List {
let mut res: List = vec![]; if lists.len() < 2 || lists.iter().any(|x| x.len() == 0) { return res }
let mut list_iter = lists.iter(); if let Some(first_list) = list_iter.next() { for &i in first_list { res.push(vec![i]); } } for l in list_iter { let mut tmp: List = List::new(); for r in res { for &el in l { let mut tmp_el = r.clone(); tmp_el.push(el); tmp.push(tmp_el); } } res = tmp; } res
}
fn print_list(list: &List) {
print!("{{ "); for inner_list in list { print!("( "); for el in inner_list { print!("{} ", el); } print!(") "); } println!("}}");
}
fn main() {
println!("Template:1, 2 x Template:3, 4"); print_list(&cartesian_product(&vec![vec![1, 2], vec![3, 4]])); println!("\nTemplate:3, 4 x Template:1, 2"); print_list(&cartesian_product(&vec![vec![3, 4], vec![1, 2]])); println!("\nTemplate:1, 2 x {{}}"); print_list(&cartesian_product(&vec![vec![1, 2], vec![]])); println!("\n{{}} x Template:1, 2"); print_list(&cartesian_product(&vec![vec![], vec![1, 2]])); println!("\nTemplate:1776, 1789 × Template:7, 12 × Template:4, 14, 23 × Template:0, 1"); print_list(&cartesian_product(&vec![vec![1776, 1789], vec![7, 12], vec![4, 14, 23], vec![0, 1]])); println!("\nTemplate:1, 2, 3 × Template:30 × Template:500, 100"); print_list(&cartesian_product(&vec![vec![1, 2, 3], vec![30], vec![500, 100]])); println!("\nTemplate:1, 2, 3 × {{}} × Template:500, 100"); print_list(&cartesian_product(&vec![vec![1, 2, 3], vec![], vec![500, 100]]));
} </lang>
- Output:
{1, 2} x {3, 4}{ ( 1 3 ) ( 1 4 ) ( 2 3 ) ( 2 4 ) }
{3, 4} x {1, 2} { ( 3 1 ) ( 3 2 ) ( 4 1 ) ( 4 2 ) }
{1, 2} x {} { }
{} x {1, 2} { }
{1776, 1789} × {7, 12} × {4, 14, 23} × {0, 1} { ( 1776 7 4 0 ) ( 1776 7 4 1 ) ( 1776 7 14 0 ) ( 1776 7 14 1 ) ( 1776 7 23 0 ) ( 1776 7 23 1 ) ( 1776 12 4 0 ) ( 1776 12 4 1 ) ( 1776 12 14 0 ) ( 1776 12 14 1 ) ( 1776 12 23 0 ) ( 1776 12 23 1 ) ( 1789 7 4 0 ) ( 1789 7 4 1 ) ( 1789 7 14 0 ) ( 1789 7 14 1 ) ( 1789 7 23 0 ) ( 1789 7 23 1 ) ( 1789 12 4 0 ) ( 1789 12 4 1 ) ( 1789 12 14 0 ) ( 1789 12 14 1 ) ( 1789 12 23 0 ) ( 1789 12 23 1 ) }
{1, 2, 3} × {30} × {500, 100} { ( 1 30 500 ) ( 1 30 100 ) ( 2 30 500 ) ( 2 30 100 ) ( 3 30 500 ) ( 3 30 100 ) }
{1, 2, 3} × {} × {500, 100} { }
Scala
Function returning the n-ary product of an arbitrary number of lists, each of arbitrary length:
<lang scala>def cartesianProduct[T](lst: List[T]*): List[List[T]] = {
/** * Prepend single element to all lists of list * @param e single elemetn * @param ll list of list * @param a accumulator for tail recursive implementation * @return list of lists with prepended element e */ def pel(e: T, ll: List[List[T]], a: List[List[T]] = Nil): List[List[T]] = ll match { case Nil => a.reverse case x :: xs => pel(e, xs, (e :: x) :: a ) }
lst.toList match { case Nil => Nil case x :: Nil => List(x) case x :: _ => x match { case Nil => Nil case _ => lst.par.foldRight(List(x))( (l, a) => l.flatMap(pel(_, a)) ).map(_.dropRight(x.size)) } }
}</lang> and usage: <lang scala>cartesianProduct(List(1, 2), List(3, 4))
.map(_.mkString("(", ", ", ")")).mkString("{",", ","}")</lang>
- Output:
{(1, 3), (1, 4), (2, 3), (2, 4)}
<lang scala>cartesianProduct(List(3, 4), List(1, 2))
.map(_.mkString("(", ", ", ")")).mkString("{",", ","}")</lang>
- Output:
{(3, 1), (3, 2), (4, 1), (4, 2)}
<lang scala>cartesianProduct(List(1, 2), List.empty)
.map(_.mkString("(", ", ", ")")).mkString("{",", ","}")</lang>
- Output:
{}
<lang scala>cartesianProduct(List.empty, List(1, 2))
.map(_.mkString("(", ", ", ")")).mkString("{",", ","}")</lang>
- Output:
{}
<lang scala>cartesianProduct(List(1776, 1789), List(7, 12), List(4, 14, 23), List(0, 1))
.map(_.mkString("(", ", ", ")")).mkString("{",", ","}")</lang>
- Output:
{(1776, 7, 4, 0), (1776, 7, 4, 1), (1776, 7, 14, 0), (1776, 7, 14, 1), (1776, 7, 23, 0), (1776, 7, 23, 1), (1776, 12, 4, 0), (1776, 12, 4, 1), (1776, 12, 14, 0), (1776, 12, 14, 1), (1776, 12, 23, 0), (1776, 12, 23, 1), (1789, 7, 4, 0), (1789, 7, 4, 1), (1789, 7, 14, 0), (1789, 7, 14, 1), (1789, 7, 23, 0), (1789, 7, 23, 1), (1789, 12, 4, 0), (1789, 12, 4, 1), (1789, 12, 14, 0), (1789, 12, 14, 1), (1789, 12, 23, 0), (1789, 12, 23, 1)}
<lang scala>cartesianProduct(List(1, 2, 3), List(30), List(500, 100))
.map(_.mkString("(", ", ", ")")).mkString("{",", ","}")</lang>
- Output:
{(1, 30, 500), (1, 30, 100), (2, 30, 500), (2, 30, 100), (3, 30, 500), (3, 30, 100)}
<lang scala>cartesianProduct(List(1, 2, 3), List.empty, List(500, 100))
.map(_.mkString("[", ", ", "]")).mkString("\n")</lang>
- Output:
{}
Sidef
In Sidef, the Cartesian product of an arbitrary number of arrays is built-in as Array.cartesian(): <lang ruby>cartesian([[1,2], [3,4], [5,6]]).say cartesian([[1,2], [3,4], [5,6]], {|*arr| say arr })</lang>
Alternatively, a simple recursive implementation: <lang ruby>func cartesian_product(*arr) {
var c = [] var r = []
func { if (c.len < arr.len) { for item in (arr[c.len]) { c.push(item) __FUNC__() c.pop } } else { r.push([c...]) } }()
return r
}</lang>
Completing the task: <lang ruby>say cartesian_product([1,2], [3,4]) say cartesian_product([3,4], [1,2])</lang>
- Output:
[[1, 3], [1, 4], [2, 3], [2, 4]] [[3, 1], [3, 2], [4, 1], [4, 2]]
The product of an empty list with any other list is empty: <lang ruby>say cartesian_product([1,2], []) say cartesian_product([], [1,2])</lang>
- Output:
[] []
Extra credit: <lang ruby>cartesian_product([1776, 1789], [7, 12], [4, 14, 23], [0, 1]).each{ .say }</lang>
- Output:
[1776, 7, 4, 0] [1776, 7, 4, 1] [1776, 7, 14, 0] [1776, 7, 14, 1] [1776, 7, 23, 0] [1776, 7, 23, 1] [1776, 12, 4, 0] [1776, 12, 4, 1] [1776, 12, 14, 0] [1776, 12, 14, 1] [1776, 12, 23, 0] [1776, 12, 23, 1] [1789, 7, 4, 0] [1789, 7, 4, 1] [1789, 7, 14, 0] [1789, 7, 14, 1] [1789, 7, 23, 0] [1789, 7, 23, 1] [1789, 12, 4, 0] [1789, 12, 4, 1] [1789, 12, 14, 0] [1789, 12, 14, 1] [1789, 12, 23, 0] [1789, 12, 23, 1]
<lang ruby>say cartesian_product([1, 2, 3], [30], [500, 100]) say cartesian_product([1, 2, 3], [], [500, 100])</lang>
- Output:
[[1, 30, 500], [1, 30, 100], [2, 30, 500], [2, 30, 100], [3, 30, 500], [3, 30, 100]] []
SQL
If we create lists as tables with one column, cartesian product is easy. <lang sql>-- set up list 1 create table L1 (value integer); insert into L1 values (1); insert into L1 values (2); -- set up list 2 create table L2 (value integer); insert into L2 values (3); insert into L2 values (4); -- get the product select * from L1, L2;</lang>
- Output:
VALUE VALUE ---------- ---------- 1 3 1 4 2 3 2 4
You should be able to be more explicit should get the same result:<lang sql>select * from L1 cross join L2;</lang>
Product with an empty list works as expected (using the tables created above): <lang sql>delete from L2; select * from L1, L2;</lang>
- Output:
no rows selected
I don't think "extra credit" is meaningful here because cartesian product is so hard-baked into SQL, so here's just one of the extra credit examples (again using the tables created above):<lang sql>insert into L1 values (3); insert into L2 values (30); create table L3 (value integer); insert into L3 values (500); insert into L3 values (100); -- product works the same for as many "lists" as you'd like select * from L1, L2, L3;</lang>
- Output:
VALUE VALUE VALUE ---------- ---------- ---------- 1 30 500 2 30 500 3 30 500 1 30 100 2 30 100 3 30 100
Standard ML
<lang sml>fun prodList (nil, _) = nil
| prodList ((x::xs), ys) = map (fn y => (x,y)) ys @ prodList (xs, ys)
fun naryProdList zs = foldl (fn (xs, ys) => map op:: (prodList (xs, ys))) [[]] (rev zs)</lang>
- Output:
- prodList ([1, 2], [3, 4]); val it = [(1,3),(1,4),(2,3),(2,4)] : (int * int) list - prodList ([3, 4], [1, 2]); val it = [(3,1),(3,2),(4,1),(4,2)] : (int * int) list - prodList ([1, 2], []); stdIn:8.1-8.22 Warning: type vars not generalized because of value restriction are instantiated to dummy types (X1,X2,...) val it = [] : (int * ?.X1) list - naryProdList [[1776, 1789], [7, 12], [4, 14, 23], [0, 1]]; val it = [[1776,7,4,0],[1776,7,4,1],[1776,7,14,0],[1776,7,14,1],[1776,7,23,0], [1776,7,23,1],[1776,12,4,0],[1776,12,4,1],[1776,12,14,0],[1776,12,14,1], [1776,12,23,0],[1776,12,23,1],[1789,7,4,0],[1789,7,4,1],[1789,7,14,0], [1789,7,14,1],[1789,7,23,0],[1789,7,23,1],[1789,12,4,0],[1789,12,4,1], [1789,12,14,0],[1789,12,14,1],[1789,12,23,0],[1789,12,23,1]] : int list list - naryProdList [[1, 2, 3], [30], [500, 100]]; val it = [[1,30,500],[1,30,100],[2,30,500],[2,30,100],[3,30,500],[3,30,100]] : int list list - naryProdList [[1, 2, 3], [], [500, 100]]; val it = [] : int list list
Stata
In Stata, the command fillin may be used to expand a dataset with all combinations of a number of variables. Thus it's easy to compute a cartesian product.
<lang stata>. list
+-------+ | a b | |-------| 1. | 1 3 | 2. | 2 4 | +-------+
. fillin a b . list
+-----------------+ | a b _fillin | |-----------------| 1. | 1 3 0 | 2. | 1 4 1 | 3. | 2 3 1 | 4. | 2 4 0 | +-----------------+</lang>
The other way around:
<lang stata>. list
+-------+ | a b | |-------| 1. | 3 1 | 2. | 4 2 | +-------+
. fillin a b . list
+-----------------+ | a b _fillin | |-----------------| 1. | 3 1 0 | 2. | 3 2 1 | 3. | 4 1 1 | 4. | 4 2 0 | +-----------------+</lang>
Note, however, that this is not equivalent to a cartesian product when one of the variables is "empty" (that is, only contains missing values).
<lang stata>. list
+-------+ | a b | |-------| 1. | 1 . | 2. | 2 . | +-------+
. fillin a b . list
+-----------------+ | a b _fillin | |-----------------| 1. | 1 . 0 | 2. | 2 . 0 | +-----------------+</lang>
This command works also if the varaibles have different numbers of nonmissing elements. However, this requires additional code to remove the observations with missing values.
<lang stata>. list
+-----------+ | a b c | |-----------| 1. | 1 4 6 | 2. | 2 5 . | 3. | 3 . . | +-----------+
. fillin a b c . list
+---------------------+ | a b c _fillin | |---------------------| 1. | 1 4 6 0 | 2. | 1 4 . 1 | 3. | 1 5 6 1 | 4. | 1 5 . 1 | 5. | 1 . 6 1 | |---------------------| 6. | 1 . . 1 | 7. | 2 4 6 1 | 8. | 2 4 . 1 | 9. | 2 5 6 1 | 10. | 2 5 . 0 | |---------------------| 11. | 2 . 6 1 | 12. | 2 . . 1 | 13. | 3 4 6 1 | 14. | 3 4 . 1 | 15. | 3 5 6 1 | |---------------------| 16. | 3 5 . 1 | 17. | 3 . 6 1 | 18. | 3 . . 0 | +---------------------+
. foreach var of varlist _all {
quietly drop if missing(`var') }
. list
+---------------------+ | a b c _fillin | |---------------------| 1. | 1 4 6 0 | 2. | 1 5 6 1 | 3. | 2 4 6 1 | 4. | 2 5 6 1 | 5. | 3 4 6 1 | |---------------------| 6. | 3 5 6 1 | +---------------------+</lang>
Swift
<lang swift>func + <T>(el: T, arr: [T]) -> [T] {
var ret = arr
ret.insert(el, at: 0)
return ret
}
func cartesianProduct<T>(_ arrays: [T]...) -> T {
guard let head = arrays.first else { return [] }
let first = Array(head)
func pel( _ el: T, _ ll: T, _ a: T = [] ) -> T { switch ll.count { case 0: return a.reversed() case _: let tail = Array(ll.dropFirst()) let head = ll.first!
return pel(el, tail, el + head + a) } }
return arrays.reversed() .reduce([first], {res, el in el.flatMap({ pel($0, res) }) }) .map({ $0.dropLast(first.count) })
}
print(cartesianProduct([1, 2], [3, 4]))
print(cartesianProduct([3, 4], [1, 2]))
print(cartesianProduct([1, 2], []))
print(cartesianProduct([1776, 1789], [7, 12], [4, 14, 23], [0, 1]))
print(cartesianProduct([1, 2, 3], [30], [500, 100]))
print(cartesianProduct([1, 2, 3], [], [500, 100])</lang>
- Output:
[[1, 3], [1, 4], [2, 3], [2, 4]] [[3, 1], [3, 2], [4, 1], [4, 2]] [] [[1776, 7, 4, 0], [1776, 7, 4, 1], [1776, 7, 14, 0], [1776, 7, 14, 1], [1776, 7, 23, 0], [1776, 7, 23, 1], [1776, 12, 4, 0], [1776, 12, 4, 1], [1776, 12, 14, 0], [1776, 12, 14, 1], [1776, 12, 23, 0], [1776, 12, 23, 1], [1789, 7, 4, 0], [1789, 7, 4, 1], [1789, 7, 14, 0], [1789, 7, 14, 1], [1789, 7, 23, 0], [1789, 7, 23, 1], [1789, 12, 4, 0], [1789, 12, 4, 1], [1789, 12, 14, 0], [1789, 12, 14, 1], [1789, 12, 23, 0], [1789, 12, 23, 1]] [[1, 30, 500], [1, 30, 100], [2, 30, 500], [2, 30, 100], [3, 30, 500], [3, 30, 100]] []
Tcl
<lang tcl> proc cartesianProduct {l1 l2} {
set result {} foreach el1 $l1 { foreach el2 $l2 { lappend result [list $el1 $el2] } } return $result
}
puts "simple" puts "result: [cartesianProduct {1 2} {3 4}]" puts "result: [cartesianProduct {3 4} {1 2}]" puts "result: [cartesianProduct {1 2} {}]" puts "result: [cartesianProduct {} {3 4}]"
proc cartesianNaryProduct {lists} {
set result {{}} foreach l $lists { set res {} foreach comb $result { foreach el $l { lappend res [linsert $comb end $el] } } set result $res } return $result
}
puts "n-ary" puts "result: [cartesianNaryProduct {{1776 1789} {7 12} {4 14 23} {0 1}}]" puts "result: [cartesianNaryProduct {{1 2 3} {30} {500 100}}]" puts "result: [cartesianNaryProduct {{1 2 3} {} {500 100}}]"
</lang>
- Output:
simple result: {1 3} {1 4} {2 3} {2 4} result: {3 1} {3 2} {4 1} {4 2} result: result: n-ary result: {1776 7 4 0} {1776 7 4 1} {1776 7 14 0} {1776 7 14 1} {1776 7 23 0} {1776 7 23 1} {1776 12 4 0} {1776 12 4 1} {1776 12 14 0} {1776 12 14 1} {1776 12 23 0} {1776 12 23 1} {1789 7 4 0} {1789 7 4 1} {1789 7 14 0} {1789 7 14 1} {1789 7 23 0} {1789 7 23 1} {1789 12 4 0} {1789 12 4 1} {1789 12 14 0} {1789 12 14 1} {1789 12 23 0} {1789 12 23 1} result: {1 30 500} {1 30 100} {2 30 500} {2 30 100} {3 30 500} {3 30 100} result:
Visual Basic .NET
<lang vbnet>Imports System.Runtime.CompilerServices
Module Module1
<Extension()> Function CartesianProduct(Of T)(sequences As IEnumerable(Of IEnumerable(Of T))) As IEnumerable(Of IEnumerable(Of T)) Dim emptyProduct As IEnumerable(Of IEnumerable(Of T)) = {Enumerable.Empty(Of T)} Return sequences.Aggregate(emptyProduct, Function(accumulator, sequence) From acc In accumulator From item In sequence Select acc.Concat({item})) End Function
Sub Main() Dim empty(-1) As Integer Dim list1 = {1, 2} Dim list2 = {3, 4} Dim list3 = {1776, 1789} Dim list4 = {7, 12} Dim list5 = {4, 14, 23} Dim list6 = {0, 1} Dim list7 = {1, 2, 3} Dim list8 = {30} Dim list9 = {500, 100}
For Each sequnceList As Integer()() In { ({list1, list2}), ({list2, list1}), ({list1, empty}), ({empty, list1}), ({list3, list4, list5, list6}), ({list7, list8, list9}), ({list7, empty, list9}) } Dim cart = sequnceList.CartesianProduct().Select(Function(tuple) $"({String.Join(", ", tuple)})") Console.WriteLine($"{{{String.Join(", ", cart)}}}") Next End Sub
End Module</lang>
- Output:
{(1, 3), (1, 4), (2, 3), (2, 4)} {(3, 1), (3, 2), (4, 1), (4, 2)} {} {} {(1776, 7, 4, 0), (1776, 7, 4, 1), (1776, 7, 14, 0), (1776, 7, 14, 1), (1776, 7, 23, 0), (1776, 7, 23, 1), (1776, 12, 4, 0), (1776, 12, 4, 1), (1776, 12, 14, 0), (1776, 12, 14, 1), (1776, 12, 23, 0), (1776, 12, 23, 1), (1789, 7, 4, 0), (1789, 7, 4, 1), (1789, 7, 14, 0), (1789, 7, 14, 1), (1789, 7, 23, 0), (1789, 7, 23, 1), (1789, 12, 4, 0), (1789, 12, 4, 1), (1789, 12, 14, 0), (1789, 12, 14, 1), (1789, 12, 23, 0), (1789, 12, 23, 1)} {(1, 30, 500), (1, 30, 100), (2, 30, 500), (2, 30, 100), (3, 30, 500), (3, 30, 100)} {}
zkl
Cartesian product is build into iterators or can be done with nested loops. <lang zkl>zkl: Walker.cproduct(List(1,2),List(3,4)).walk().println(); L(L(1,3),L(1,4),L(2,3),L(2,4)) zkl: foreach a,b in (List(1,2),List(3,4)){ print("(%d,%d) ".fmt(a,b)) } (1,3) (1,4) (2,3) (2,4)
zkl: Walker.cproduct(List(3,4),List(1,2)).walk().println(); L(L(3,1),L(3,2),L(4,1),L(4,2))</lang>
The walk method will throw an error if used on an empty iterator but the pump method doesn't. <lang zkl>zkl: Walker.cproduct(List(3,4),List).walk().println(); Exception thrown: TheEnd(Ain't no more)
zkl: Walker.cproduct(List(3,4),List).pump(List).println(); L() zkl: Walker.cproduct(List,List(3,4)).pump(List).println(); L()</lang> <lang zkl>zkl: Walker.cproduct(L(1776,1789),L(7,12),L(4,14,23),L(0,1)).walk().println(); L(L(1776,7,4,0),L(1776,7,4,1),L(1776,7,14,0),L(1776,7,14,1),L(1776,7,23,0),L(1776,7,23,1),L(1776,12,4,0),L(1776,12,4,1),L(1776,12,14,0),L(1776,12,14,1),L(1776,12,23,0),L(1776,12,23,1),L(1789,7,4,0),L(1789,7,4,1),L(1789,7,14,0),L(1789,7,14,1),L(1789,7,23,0),L(1789,7,23,1),L(1789,12,4,0),L(1789,12,4,1),...)
zkl: Walker.cproduct(L(1,2,3),L(30),L(500,100)).walk().println(); L(L(1,30,500),L(1,30,100),L(2,30,500),L(2,30,100),L(3,30,500),L(3,30,100))
zkl: Walker.cproduct(L(1,2,3),List,L(500,100)).pump(List).println(); L()</lang>