使用Haskell在元组中以长度为2 ^ 0，2 ^ 1，…，2 ^ N的列表分割列表

Question

I'm trying to solve Haskell problem, but I don't have any clue where to start. I need to split list in lists with length of 2^0, 2^1, 2^3 .. elements. So if we have list [1,2,3,4,5,6,7,8,9,10,11,12,13] after using our function we should have result [[1],[2,3],[4,5,6,7],[8,9,10,11,12,13]]

Answer 1

You can use the function: splitAt :: Int -> [a] -> ([a],[a]) , and then use recursion:

blocks :: Int -> [a] -> [[a]]
blocks _ [] = []
blocks n ls = la : blocks (2*n) lb
    where ~(la,lb) = splitAt n ls

So in the case we have a blocks 1 [1,2,3,4,5,6] we will obtain [[1],[2,3],[4,5,6]] . In the first case, we look if the list we give to blocks is empty, in which case, there is nothing to split, so we return the empty list. In the recursive case, we splitAt the ls list into la and lb . The la is our first list, and the lb we need to split further. We do the recursion with n*2 as new split length to ensure that the length of the lists will increase like powers of two.

Answer 2

Maybe you can also use zip and groupBy . Seems to be working, but is not so straightforward.

import Data.List
a="Hello World!"
p=[2^n| n<-[0..]]
pa=take (length a) p
b=[elem n pa| n<-[1..length a]]
c=zip a b
d=groupBy (\x y->snd y==False) c
e=map (map (\x->fst x)) d