List comprehensions: Difference between revisions
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Write a list comprehension that builds the list of all pythagorean triples with elements between 1 and n. If the language has multiple ways for expressing such a construct (for example, direct list comprehensions and generators), write one example for each.
=={{header|ALGOL 68}}==
ALGOL 68 does not have list comprehension, however it is ''sometimes'' reasonably
generous about where a '''flex''' array is declared. And with the addition
of an append '''operator''' "+:=" for lists they can be similarly manipulated.
{{trans|Python}}
{{works with|ALGOL 68|Standard - no extensions to language used}}
{{works with|ALGOL 68G|Any - tested with release mk15-0.8b.fc9.i386}}
{{works with|ELLA ALGOL 68|Any (with appropriate job cards) - tested with release 1.8.8d.fc9.i386}}
<lang algol>MODE XYZ = STRUCT(INT x,y,z);
OP +:= = (REF FLEX[]XYZ lhs, XYZ rhs)FLEX[]XYZ: (
[UPB lhs+1]XYZ out;
out[:UPB lhs] := lhs;
out[UPB out] := rhs;
lhs := out
);
INT n = 20;
print (([]XYZ(
FLEX[0]XYZ xyz;
FOR x TO n DO FOR y FROM x+1 TO n DO FOR z FROM y+1 TO n DO IF x*x + y*y = z*z THEN xyz +:= XYZ(x,y,z) FI OD OD OD;
xyz), new line
))</lang>
Output:
<pre>
+3 +4 +5 +5 +12 +13 +6 +8 +10 +8 +15 +17 +9 +12 +15 +12 +16 +20
</pre>
=={{header|Clojure}}==
(for [x (range 1 21) y (range x 21) z (range y 21) :when (= (+ (* x x) (* y y)) (* z z))] [x y z])
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Revision as of 12:16, 5 March 2009
You are encouraged to solve this task according to the task description, using any language you may know.
A list comprehension is a special syntax in some programming languages to describe lists. It is similar to the way mathematicians describe sets, with a set comprehension, hence the name.
Write a list comprehension that builds the list of all pythagorean triples with elements between 1 and n. If the language has multiple ways for expressing such a construct (for example, direct list comprehensions and generators), write one example for each.
ALGOL 68
ALGOL 68 does not have list comprehension, however it is sometimes reasonably generous about where a flex array is declared. And with the addition of an append operator "+:=" for lists they can be similarly manipulated.
<lang algol>MODE XYZ = STRUCT(INT x,y,z);
OP +:= = (REF FLEX[]XYZ lhs, XYZ rhs)FLEX[]XYZ: (
[UPB lhs+1]XYZ out; out[:UPB lhs] := lhs; out[UPB out] := rhs; lhs := out
);
INT n = 20; print (([]XYZ(
FLEX[0]XYZ xyz; FOR x TO n DO FOR y FROM x+1 TO n DO FOR z FROM y+1 TO n DO IF x*x + y*y = z*z THEN xyz +:= XYZ(x,y,z) FI OD OD OD; xyz), new line
))</lang> Output:
+3 +4 +5 +5 +12 +13 +6 +8 +10 +8 +15 +17 +9 +12 +15 +12 +16 +20
Clojure
(for [x (range 1 21) y (range x 21) z (range y 21) :when (= (+ (* x x) (* y y)) (* z z))] [x y z])
E
pragma.enable("accumulator") # considered experimental accum [] for x in 1..n { for y in x..n { for z in y..n { if (x**2 + y**2 <=> z**2) { _.with([x,y,z]) } } } }
Erlang
pythag(N) -> [ {A,B,C} || A <- lists:seq(1,N), B <- lists:seq(1,N), C <- lists:seq(1,N), A+B+C =< N, A*A+B*B == C*C ].
Haskell
pyth n = [(x,y,z) | x <- [1..n], y <- [x..n], z <- [y..n], x^2 + y^2 == z^2]
Since lists are Monads, one can alternatively also use the do-notation (which is practical if the comprehension is large):
import Control.Monad pyth n = do x <- [1..n] y <- [x..n] z <- [y..n] guard $ x^2 + y^2 == z^2 return (x,y,z)
Mathematica
Select[Tuples[Range[n], 3], #1[[1]]^2 + #1[[2]]^2 == #1[[3]]^2 &]
Pop11
lvars n = 10, i, j, k; [% for i from 1 to n do for j from 1 to n do for k from 1 to n do if i*i + j*j = k*k then [^i ^j ^k]; endif; endfor; endfor; endfor %] =>
Python
List comprehension:
<lang python>[(x,y,z) for x in xrange(1,n+1) for y in xrange(x,n+1) for z in xrange(y,n+1) if x**2 + y**2 == z**2]</lang>
Generator comprehension (note the outer round brackets):
<lang python>((x,y,z) for x in xrange(1,n+1) for y in xrange(x,n+1) for z in xrange(y,n+1) if x**2 + y**2 == z**2)</lang>
Generator function:
<lang python>def gentriples(n): for x in xrange(1,n+1): for y in xrange(x,n+1): for z in xrange(y,n+1): if x**2 + y**2 == z**2: yield (x,y,z)</lang>