List comprehensions: Difference between revisions
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<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]</python> |
<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]</python> |
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Generator comprehension: |
Generator comprehension (note the outer round brackets): |
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<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)</python> |
<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)</python> |
Revision as of 10:18, 3 August 2008
![Task](http://static.miraheze.org/rosettacodewiki/thumb/b/ba/Rcode-button-task-crushed.png/64px-Rcode-button-task-crushed.png)
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.
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:
<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]</python>
Generator comprehension (note the outer round brackets):
<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)</python>
Generator function:
<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)</python>