Range consolidation: Difference between revisions
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Defining consolidation as a fold over a list of tuples: |
Defining consolidation as a fold over a list of tuples: |
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{{Works with|Python|3.7}} |
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<lang python>'''Consolidated ranges''' |
<lang python>'''Consolidated ranges''' |
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Revision as of 23:09, 28 March 2019
You are encouraged to solve this task according to the task description, using any language you may know.
Define a range of numbers R, with bounds b0 and b1 covering all numbers between and including both bounds. That range can be shown as:
[b0, b1]
or equally as:
[b1, b0].
Given two ranges, the act of consolidation between them compares the two ranges:
- If one range covers all of the other then the result is that encompassing range.
- If the ranges touch or intersect then the result is one new single range covering the overlapping ranges.
- Otherwise the act of consolidation is to return the two non-touching ranges.
Given N ranges where N>2 then the result is the same as repeatedly replacing all combinations of two ranges by their consolidation until no further consolidation between range pairs is possible. If N<2 then range consolidation has no strict meaning and the input can be returned.
- Example 1:
- Given the two ranges [1, 2.5] and [3, 4.2] then there is no
- common area between the ranges and the result is the same as the input.
- Example 2:
- Given the two ranges [1, 2.5] and [1.8, 4.7] then there is
- an overlap [2.5, 1.8] between the ranges and the result is the single
- range [1, 4.7]. Note that order of bounds in a range is not, (yet), stated.
- Example 3:
- Given the two ranges [6.1, 7.2] and [7.2, 8.3] then they
- touch at 7.2 and the result is the single range [6.1, 8.3].
- Example 4:
- Given the three ranges [1, 2] and [4, 8] and [2, 5]
- then there is no intersection of the ranges [1, 2] and [4, 8]
- but the ranges [1, 2] and [2, 5] overlap and consolidate to
- produce the range [1, 5]. This range, in turn, overlaps the other range
- [4, 8], and so consolidates to the final output of the single range
- [1, 8]
- Task:
Let a normalized range display show the smaller bound to the left; and show the range with the smaller lower bound to the left of other ranges when showing multiple ranges.
Output the normalised result of applying consolidation to these five sets of ranges:
[1.1, 2.2]
[6.1, 7.2], [7.2, 8.3]
[4, 3], [2, 1]
[4, 3], [2, 1], [-1, -2], [3.9, 10]
[1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6]
Show output here.
See also
C#
<lang csharp>using static System.Math; using System.Linq; using System;
public static class RangeConsolidation {
public static void Main() {
foreach (var list in new [] {
new[] { (1.1, 2.2) }.ToList(),
new[] { (6.1, 7.2), (7.2, 8.3) }.ToList(),
new[] { (4d, 3d), (2, 1) }.ToList(),
new[] { (4d, 3d), (2, 1), (-1, 2), (3.9, 10) }.ToList(),
new[] { (1d, 3d), (-6, -1), (-4, -5), (8, 2), (-6, -6) }.ToList()
})
{
for (int z = list.Count-1; z >= 1; z--) {
for (int y = z - 1; y >= 0; y--) {
if (Overlap(list[z], list[y])) {
list[y] = Consolidate(list[z], list[y]);
list.RemoveAt(z);
break;
}
}
}
Console.WriteLine(string.Join(", ", list.Select(Normalize).OrderBy(range => range.s)));
}
}
private static bool Overlap((double s, double e) left, (double s, double e) right) =>
Max(left.s, left.e) > Max(right.s, right.e)
? Max(right.s, right.e) >= Min(left.s, left.e)
: Max(left.s, left.e) >= Min(right.s, right.e);
private static (double s, double e) Consolidate((double s, double e) left, (double s, double e) right) =>
(Min(Min(left.s, left.e), Min(right.s, right.e)), Max(Max(left.s, left.e), Max(right.s, right.e)));
private static (double s, double e) Normalize((double s, double e) range) =>
(Min(range.s, range.e), Max(range.s, range.e));
}</lang>
- Output:
(1.1, 2.2) (6.1, 8.3) (1, 2), (3, 4) (-1, 2), (3, 10) (-6, -1), (1, 8)
Go
<lang go>package main
import (
"fmt" "math" "sort"
)
type Range struct{ Lower, Upper float64 }
func (r Range) Norm() Range {
if r.Lower > r.Upper {
return Range{r.Upper, r.Lower}
}
return r
}
func (r Range) String() string {
return fmt.Sprintf("[%g, %g]", r.Lower, r.Upper)
}
func (r1 Range) Union(r2 Range) []Range {
if r1.Upper < r2.Lower {
return []Range{r1, r2}
}
r := Range{r1.Lower, math.Max(r1.Upper, r2.Upper)}
return []Range{r}
}
func consolidate(rs []Range) []Range {
for i := range rs {
rs[i] = rs[i].Norm()
}
le := len(rs)
if le < 2 {
return rs
}
sort.Slice(rs, func(i, j int) bool {
return rs[i].Lower < rs[j].Lower
})
if le == 2 {
return rs[0].Union(rs[1])
}
for i := 0; i < le-1; i++ {
for j := i + 1; j < le; j++ {
ru := rs[i].Union(rs[j])
if len(ru) == 1 {
rs[i] = ru[0]
copy(rs[j:], rs[j+1:])
rs = rs[:le-1]
le--
i--
break
}
}
}
return rs
}
func main() {
rss := [][]Range{
Template:1.1, 2.2,
{{6.1, 7.2}, {7.2, 8.3}},
{{4, 3}, {2, 1}},
{{4, 3}, {2, 1}, {-1, -2}, {3.9, 10}},
{{1, 3}, {-6, -1}, {-4, -5}, {8, 2}, {-6, -6}},
}
for _, rs := range rss {
s := fmt.Sprintf("%v", rs)
fmt.Printf("%40s => ", s[1:len(s)-1])
rs2 := consolidate(rs)
s = fmt.Sprintf("%v", rs2)
fmt.Println(s[1 : len(s)-1])
}
}</lang>
- Output:
[1.1, 2.2] => [1.1, 2.2]
[6.1, 7.2] [7.2, 8.3] => [6.1, 8.3]
[4, 3] [2, 1] => [1, 2] [3, 4]
[4, 3] [2, 1] [-1, -2] [3.9, 10] => [-2, -1] [1, 2] [3, 10]
[1, 3] [-6, -1] [-4, -5] [8, 2] [-6, -6] => [-6, -1] [1, 8]
J
Solution: <lang j>ensure2D=: ,:^:(1 = #@$) NB. if list make 1 row table normalise=: ([: /:~ /:~"1)@ensure2D NB. normalises list of ranges merge=: ,:`(<.&{. , >.&{:)@.(>:/&{: |.) NB. merge ranges x and y consolidate=: (}.@] ,~ (merge {.)) ensure2D</lang> Required Examples: <lang j> tests=: <@".;._2 noun define 1.1 2.2 6.1 7.2 ,: 7.2 8.3 4 3 ,: 2 1 4 3 , 2 1 , _1 _2 ,: 3.9 10 1 3 , _6 _1 , _4 _5 , 8 2 ,: _6 _6 )
consolidate/@normalise&.> tests
+-------+-------+---+-----+-----+ |1.1 2.2|6.1 8.3|1 2|_2 _1|_6 _1| | | |3 4| 1 2| 1 8| | | | | 3 10| | +-------+-------+---+-----+-----+</lang>
Julia
In Julia, a Range is a type of iterator, generally one over a specified interval. The task as specified is orthogonal to the iteration purpose of a Julia Range, since the task is about merging sets of numbers, not iterations. Therefore, a translation of the Python code is done, rather than using a native Julia Range.
<lang julia>normalize(s) = sort([sort(bounds) for bounds in s])
function consolidate(ranges)
norm = normalize(ranges)
for (i, r1) in enumerate(norm)
if !isempty(r1)
for r2 in norm[i+1:end]
if !isempty(r2) && r1[end] >= r2[1] # intersect?
r1 .= [r1[1], max(r1[end], r2[end])]
empty!(r2)
end
end
end
end
[r for r in norm if !isempty(r)]
end
function testranges()
for s in [[[1.1, 2.2]], [[6.1, 7.2], [7.2, 8.3]], [[4, 3], [2, 1]],
[[4, 3], [2, 1], [-1, -2], [3.9, 10]],
[[1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6]]]
println("$s => $(consolidate(s))")
end
end
testranges()
</lang>
- Output:
Array{Float64,1}[[1.1, 2.2]] => Array{Float64,1}[[1.1, 2.2]]
Array{Float64,1}[[6.1, 7.2], [7.2, 8.3]] => Array{Float64,1}[[6.1, 8.3]]
Array{Float64,1}[[4.0, 3.0], [2.0, 1.0]] => Array{Float64,1}[[1.0, 2.0], [3.0, 4.0]]
Array{Float64,1}[[4.0, 3.0], [2.0, 1.0], [-1.0, -2.0], [3.9, 10.0]] => Array{Float64,1}[[-2.0, -1.0], [1.0, 2.0], [3.0, 10.0]]
Array{Float64,1}[[1.0, 3.0], [-6.0, -1.0], [-4.0, -5.0], [8.0, 2.0], [-6.0, -6.0]] => Array{Float64,1}[[-6.0, -1.0], [1.0, 8.0]]
Perl
<lang perl>use strict; use warnings;
use List::Util qw(min max);
sub consolidate {
our @arr; local *arr = shift;
my @sorted = sort { @$a[0] <=> @$b[0] } map { [sort { $a <=> $b } @$_] } @arr;
my @merge = shift @sorted;
for my $i (@sorted) {
if ($merge[-1][1] >= @$i[0]) {
$merge[-1][0] = min($merge[-1][0], @$i[0]);
$merge[-1][1] = max($merge[-1][1], @$i[1]);
} else {
push @merge, $i;
}
}
return @merge;
}
for my $intervals (
[[1.1, 2.2],],
[[6.1, 7.2], [7.2, 8.3]],
[[4, 3], [2, 1]],
[[4, 3], [2, 1], [-1, -2], [3.9, 10]],
[[1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6]]) {
my($in,$out);
$in = join ', ', map { '[' . join(', ', @$_) . ']' } @$intervals;
$out .= join('..', @$_). ' ' for consolidate($intervals);
printf "%44s => %s\n", $in, $out;
}</lang>
- Output:
[1.1, 2.2] => 1.1..2.2
[6.1, 7.2], [7.2, 8.3] => 6.1..8.3
[4, 3], [2, 1] => 1..2 3..4
[4, 3], [2, 1], [-1, -2], [3.9, 10] => -2..-1 1..2 3..10
[1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6] => -6..-1 1..8
Perl 6
In Perl 6, a Range is a first class object with its own specialized notation. Perl 6 Ranges allow for exclusion of the boundary numbers. This example doesn't since it isn't a requirement in this task. Much of the logic is lifted from the Set_of_real_numbers task with simplified logic for the much simpler requirements.
Note: the output is in standard Perl 6 notation for Ranges.
<lang perl6># Union sub infix:<∪> (Range $a, Range $b) { Range.new($a.min,max($a.max,$b.max)) }
- Intersection
sub infix:<∩> (Range $a, Range $b) { so $a.max >= $b.min }
multi consolidate() { () }
multi consolidate($this is copy, **@those) {
gather {
for consolidate |@those -> $that {
if $this ∩ $that { $this ∪= $that }
else { take $that }
}
take $this;
}
}
for [[1.1, 2.2],],
[[6.1, 7.2], [7.2, 8.3]], [[4, 3], [2, 1]], [[4, 3], [2, 1], [-1, -2], [3.9, 10]], [[1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6]], [[], [2], [4,6], [5,9,8]]
-> @intervals {
printf "%46s => ", @intervals.perl;
say reverse consolidate |@intervals.grep(*.elems)».sort.sort({ [.[0], .[*-1]] }).map: { Range.new(.[0], .[*-1]) }
}</lang>
- Output:
[[1.1, 2.2],] => (1.1..2.2)
[[6.1, 7.2], [7.2, 8.3]] => (6.1..8.3)
[[4, 3], [2, 1]] => (1..2 3..4)
[[4, 3], [2, 1], [-1, -2], [3.9, 10]] => (-2..-1 1..2 3..10)
[[1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6]] => (-6..-1 1..8)
[[], [2], [4, 6], [5, 9, 8]] => (2..2 4..9)
Python
<lang python>def normalize(s):
return sorted(sorted(bounds) for bounds in s if bounds)
def consolidate(ranges):
norm = normalize(ranges)
for i, r1 in enumerate(norm):
if r1:
for r2 in norm[i+1:]:
if r2 and r1[-1] >= r2[0]: # intersect?
r1[:] = [r1[0], max(r1[-1], r2[-1])]
r2.clear()
return [rnge for rnge in norm if rnge]
if __name__ == '__main__':
for s in [
1.1, 2.2,
[[6.1, 7.2], [7.2, 8.3]],
[[4, 3], [2, 1]],
[[4, 3], [2, 1], [-1, -2], [3.9, 10]],
[[1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6]],
]:
print(f"{str(s)[1:-1]} => {str(consolidate(s))[1:-1]}")
</lang>
- Output:
[1.1, 2.2] => [1.1, 2.2] [6.1, 7.2], [7.2, 8.3] => [6.1, 8.3] [4, 3], [2, 1] => [1, 2], [3, 4] [4, 3], [2, 1], [-1, -2], [3.9, 10] => [-2, -1], [1, 2], [3, 10] [1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6] => [-6, -1], [1, 8]
Functional
Defining consolidation as a fold over a list of tuples:
<lang python>Consolidated ranges
from functools import (reduce)
- consolidated :: [(Float, Float)] -> [(Float, Float)]
def consolidated(xs):
A consolidated list of
floating point ranges.
def fused(a, b):
return a[0], max(a[1], b[1])
ys = sorted(map(tupleSort, xs), reverse=True)
return reduce(
lambda a, x: [fused(x, a[0])] + a[1:] if (
x[1] >= a[0][0]
) else [x] + a,
ys[1:],
[ys[0]]
)
- tupleSort :: (a, a) -> (a, a)
def tupleSort(ab):
A pair of values sorted in
ascending order.
a, b = ab
return ab if a <= b else (b, a)
- GENERIC FUNCTIONS FOR TESTS -----------------------------
- compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
def compose(g):
Right to left function composition. return lambda f: lambda x: g(f(x))
- tabulated :: String -> (a -> String) ->
- (b -> String) ->
- (a -> b) -> [a] -> String
def tabulated(s):
Heading -> x display function -> fx display function ->
f -> value list -> tabular string.
def go(xShow, fxShow, f, xs):
w = max(map(compose(len)(xShow), xs))
return s + '\n' + '\n'.join([
xShow(x).rjust(w, ' ') + ' -> ' + fxShow(f(x)) for x in xs
])
return lambda xShow: lambda fxShow: (
lambda f: lambda xs: go(
xShow, fxShow, f, xs
)
)
- TESTS ---------------------------------------------------
- main :: ()
def main():
Tests
print(
tabulated('Consolidation of numeric ranges:')(str)(str)(
consolidated
)([
[(1.1, 2.2)],
[(6.1, 7.2), (7.2, 8.3)],
[(4, 3), (2, 1)],
[(4, 3), (2, 1), (-1, -2), (3.9, 10)],
[(1, 3), (-6, -1), (-4, -5), (8, 2), (-6, -6)]
])
)
- MAIN ---
if __name__ == '__main__':
main()</lang>
- Output:
Consolidation of numeric ranges:
[(1.1, 2.2)] -> [(1.1, 2.2)]
[(6.1, 7.2), (7.2, 8.3)] -> [(6.1, 8.3)]
[(4, 3), (2, 1)] -> [(1, 2), (3, 4)]
[(4, 3), (2, 1), (-1, -2), (3.9, 10)] -> [(-2, -1), (1, 2), (3, 10)]
[(1, 3), (-6, -1), (-4, -5), (8, 2), (-6, -6)] -> [(-6, -1), (1, 8)]
REXX
Most of the REXX code was testing (and rebuilding) the syntax (insuring blanks after commas), and handling of a null set.
The actual logic for the range consolidation is marked with the comments: /*■■■■►*/ <lang rexx>/*REXX program performs range consolidation (they can be [equal] ascending/descending). */
- .= /*define the default for range sets. */
parse arg #.1 /*obtain optional arguments from the CL*/ if #.1= then do /*Not specified? Then use the defaults*/
#.1= '[1.1, 2.2]'
#.2= '[6.1, 7.2], [7.2, 8.3]'
#.3= '[4, 3], [2, 1]'
#.4= '[4, 3], [2, 1], [-1, -2], [3.9, 10]'
#.5= '[1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6]'
#.6= '[]'
end
do j=1 while #.j\==; $= #.j /*process each of the range sets. */
say copies('═', 75) /*display a fence between range sets. */
say ' original ranges:' $ /*display the original range set. */
$= order($) /*order low and high ranges; normalize.*/
call xSort words($) /*sort the ranges using a simple sort. */
$= merge($) /*consolidate the ranges. */
say ' consolidated ranges:' $ /*display the consolidated range set. */
end /*j*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ merge: procedure expose @.; parse arg y
if words(y)<2 then signal build /*Null or only 1 range? Skip merging. */
do j=1 to @.0-1; if @.j== then iterate /*skip deleted ranges.*/
do k=j+1 to @.0; if @.k== then iterate /* " " " */
parse var @.j a b; parse var @.k aa bb /*extract low and high*/
/*■■■■►*/ if a<=aa & b>=bb then do; @.k=; iterate; end /*within a range*/ /*■■■■►*/ if a<=aa & b>=aa then do; @.j= a bb; @.k=; iterate; end /*abutted ranges*/
end /*k*/
end /*j*/
build: z=
do r=1 for @.0; z= z translate(@.r, ',', " "); end /*r*/ /*add comma*/
f=; do s=1 for words(z); f= f '['word(z, s)"], "; end /*s*/ /*add [ ], */
if f== then return '[]' /*null set.*/
return space( changestr(',', strip( space(f), 'T', ","), ", ") ) /*add blank*/
/*──────────────────────────────────────────────────────────────────────────────────────*/ order: procedure expose @.; parse arg y,,z; @.= /*obtain arguments from the invocation.*/
y= space(y, 0) /*elide superfluous blanks in the sets.*/
do k=1 while y\== & y\=='[]' /*process ranges while range not blank.*/
if left(y,1)==',' then y= substr(y,2) /*elide commas between sets of ranges. */
parse var y '[' L "," H ']' y /*extract the "low" and "high" values.*/
if H<L then parse value L H with H L /*order " " " " " */
L= L / 1; H= H / 1 /*normalize the L and the H values.*/
@.k= L H; z= z L','H /*re─build the set w/o and with commas.*/
end /*k*/ /* [↓] at this point, K is one to big.*/
@.0= k - 1 /*keep track of the number of ranges. */
return strip(z) /*elide the extra leading blank in set.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/ xSort: procedure expose @.; parse arg n /*a simple sort for small set of ranges*/
do j=1 to n-1; _= @.j
do k=j+1 to n; if word(@.k,1)>=word(_,1) then iterate; @.j=@.k; @.k=_; _=@.j
end /*k*/</lang>
- output when using the default inputs:
═══════════════════════════════════════════════════════════════════════════
original ranges: [1.1, 2.2]
consolidated ranges: [1.1, 2.2]
═══════════════════════════════════════════════════════════════════════════
original ranges: [6.1, 7.2], [7.2, 8.3]
consolidated ranges: [6.1, 8.3]
═══════════════════════════════════════════════════════════════════════════
original ranges: [4, 3], [2, 1]
consolidated ranges: [1, 2], [3, 4]
═══════════════════════════════════════════════════════════════════════════
original ranges: [4, 3], [2, 1], [-1, -2], [3.9, 10]
consolidated ranges: [-2, -1], [1, 2], [3, 10]
═══════════════════════════════════════════════════════════════════════════
original ranges: [1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6]
consolidated ranges: [-6, -1], [1, 8]
═══════════════════════════════════════════════════════════════════════════
original ranges: []
consolidated ranges: []
zkl
<lang zkl>fcn consolidate(rs){
(s:=List()).append(
normalize(rs).reduce('wrap(ab,cd){
if(ab[1]>=cd[0]) L(ab[0],ab[1].max(cd[1])); // consolidate else{ s.append(ab); cd } // no overlap
}) )
} fcn normalize(s){ s.apply("sort").sort(fcn(a,b){ a[0]<b[0] }) }</lang> <lang zkl>foreach rs in (L(
L(L(1.1, 2.2)), L(L(6.1, 7.2), L(7.2, 8.3)), L(L(4, 3), L(2, 1)),
L(L(4.0, 3.0), L(2.0, 1.0), L(-1.0, -2.0), L(3.9, 10.0)),
L(L(1, 3), L(-6, -1), L(-4, -5), L(8, 2), L(-6, -6)),
)){ println(ppp(rs),"--> ",ppp(consolidate(rs))) }
fcn ppp(ll){ ll.pump(String,fcn(list){ list.concat(", ", "[", "] ") }) }</lang>
- Output:
[1.1, 2.2] --> [1.1, 2.2] [6.1, 7.2] [7.2, 8.3] --> [6.1, 8.3] [4, 3] [2, 1] --> [1, 2] [3, 4] [4, 3] [2, 1] [-1, -2] [3.9, 10] --> [-2, -1] [1, 2] [3, 10] [1, 3] [-6, -1] [-4, -5] [8, 2] [-6, -6] --> [-6, -1] [1, 8]