On 01/12/2006, at 12:47 PM, Huazhi (Hank) Gong wrote:
Like given a string list s="This is the string I want to test", I want to get the substring. In ruby or other language, it's simple like s [2..10], but how to do it in Haskell?
If your indices are in ascending order, and unique, then something like this might do the trick: els1 indexes list = els' (zip [0..] list) indexes where els' [] _ = [] els' _ [] = [] els' ((j,x):xs) indexes@(i:is) | i == j = x : els' xs is | otherwise = els' xs indexes Of course this is a right fold, so you ought to be able to use foldr. Here's an attempt: els2 indexes list = foldr comb undefined [0..] list indexes where comb _ _ [] _ = [] comb _ _ _ [] = [] comb j rec (x:xs) indexes@(i:is) | j == i = x : rec xs is | otherwise = rec xs indexes Bonus marks for figuring out why I used "undefined". Warning: this is largely untested code. Cheers, Bernie.