[Haskell-beginners] short function to remove and replace an item

[Zachary Turner]

On Wed, Apr 1, 2009 at 9:50 PM, Zachary Turner <divisortheory at gmail.com>wrote:

>
>
> On Wed, Apr 1, 2009 at 7:45 PM, Michael P Mossey <mpm at alumni.caltech.edu>wrote:
>
>> What if I have a list xs, and I want to remove one item and replace it
>> with another? The item to remove can be detected by a predicate function.
>> The order of the items doesn't matter. How about this:
>>
>> replaceItem :: (a -> Bool) -> a -> [a] -> [a]
>> replaceItem p new xs = new : snd (partition p xs)
>>
>> This will actually replace all items that match the predicate with one
>> copy of 'new'. It will also prepend 'new' even if no items match the
>> predicate.
>>
>> For another challenge, can someone explain to me how to write this in
>> point-free style?
>>
>
> I used an intermediate helper function, but the replaceItem function is
> point free
>
> choose :: (a -> Bool) -> (a -> b) -> (a -> b) -> a -> b
> choose pred yes _ val | (pred val) = yes val
> choose _ _ no val = no val
>
> replaceItem :: (a -> Bool) -> a -> [a] -> [a]
> replaceItem pred rep = map (choose pred id (const rep))
>

Ok I looked at the OP again and apparently I didn't understand the wording
of the question :P  I just thought it said to replace each element in the
list that matched the predicate with the new item.

How about this?

replaceItem :: [a] -> (a -> Bool) -> a -> [a]
let replaceItem xs pred = (: filter (not.pred) xs)

Note that I changed the order of the arguments, the value now comes last.
But I think this still should be ok?
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