On Tue, Jul 19, 2016 at 4:08 PM, Niely Boyken <niely.b0y...@gmail.com>
wrote:
> let i = elemIndex toReplace lst in
>
> case i of
> Just i ->
> let z = splitAt i lst
> x = fst z
> y = (snd z)
> in
> init x
> x ++ newNmr
> x ++ y
>
> Nothing -> [5]
>
If I understand what you are trying to do correctly, a more idiomatic (and
syntactically correct code) would be:
case elemIndex toReplace lst of
Just i -> let (xs,_:ys)=splitAt i lst in xs ++ (newNmr:ys)
_ -> [5]
In more general terms, replacing individual elements in standard lists is a
very inefficient operation for large lists. Having you considered using a
Zipper List which allows you to efficiently replace elements at a focal
point (and also to traverse the list forward and backward efficiently)?
An example implementation of a Zipper (just for Lists) is at
https://hackage.haskell.org/package/ListZipper-1.2.0.2/docs/Data-List-Zipper.html,
but implementing your own is an easy and instructive exercise.
Carl Edman
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