Dynamically build list comprehension in Haskell
I am curious if it is possible to dynamically build a list comprehension in Haskell.
As an example, if I have the following:
all_pows (a,a') (b,b') = [ a^y * b^z | y <- take a' [0..], z <- take b' [0..] ]
I get what I am after
*Main> List.sort $ all_pows (2,3) (5,3)
[1,2,4,5,10,20,25,50,100]
However, what I'd reall开发者_如何转开发y like is to have something like
all_pows [(Int,Int)] -> [Integer]
So that I can support N
pairs of arguments without building N
versions of all_pows
. I'm still pretty new to Haskell so I may have overlooked something obvious. Is this even possible?
The magic of the list monad:
ghci> let powers (a, b) = [a ^ n | n <- [0 .. b-1]] ghci> powers (2, 3) [1,2,4] ghci> map powers [(2, 3), (5, 3)] [[1,2,4],[1,5,25]] ghci> sequence it [[1,1],[1,5],[1,25],[2,1],[2,5],[2,25],[4,1],[4,5],[4,25]] ghci> mapM powers [(2, 3), (5, 3)] [[1,1],[1,5],[1,25],[2,1],[2,5],[2,25],[4,1],[4,5],[4,25]] ghci> map product it [1,5,25,2,10,50,4,20,100] ghci> let allPowers list = map product $ mapM powers list ghci> allPowers [(2, 3), (5, 3)] [1,5,25,2,10,50,4,20,100]
This probably deserves a bit more explanation.
You could have written your own
cartesianProduct :: [[a]] -> [[a]]
cartesianProduct [] = [[]]
cartesianProduct (list:lists)
= [ (x:xs) | x <- list, xs <- cartesianProduct lists ]
such that cartesianProduct [[1],[2,3],[4,5,6]]
⇒ [[1,2,4],[1,2,5],[1,2,6],[1,3,4],[1,3,5],[1,3,6]]
.
However, comprehensions and monads are intentionally similar. The standard Prelude has sequence :: Monad m => [m a] -> m [a]
, and when m
is the list monad []
, it actually does exactly what we wrote above.
As another shortcut, mapM :: Monad m => (a -> m b) -> [a] -> m [b]
is simply a composition of sequence
and map
.
For each inner list of varying powers of each base, you want to multiply them to a single number. You could write this recursively
product list = product' 1 list
where product' accum [] = accum
product' accum (x:xs)
= let accum' = accum * x
in accum' `seq` product' accum' xs
or using a fold
import Data.List
product list = foldl' (*) 1 list
but actually, product :: Num a => [a] -> a
is already defined! I love this language ☺☺☺
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