CODEWITHSUNDEEP
listfunctionhaskell
user2999349
user2999349

Reputation: 879

Haskell : A function that updates a state by calculating with each value of a list

I need a function that updates a state multiple times depending on a list of values: for each value of the list it may possibly update the state or leave it unchanged. So I figured I'd need a function of a type like:

[a] -> b -> (a->b->b) -> b

Where [a] is a list of values for which goes that for each of them the state b may be updated (depending of the value of a). Then the resulting b is a new state that has all necessary updates applied.

However I could not find any function that on Hoogle that does this, so I figure I'd have to make one myself however I have no idea how I could do this. Are there any functions existing that I can use to accomplish such function?

If anyone could help me out on this it'd be much appreciated!

Best regards, Skyfe.

EDIT: A (simplified) example of what I could have & need:

[a] = [1, 5, 3, 6]
b   = State{x, y, z}
f :: (a->b->b)
f a b = if someAlgorithm a then b{x=someFunc x, y=y+1} else b

=> Then the function I need should execute f on all a's and for each time it does so, it should return the new (updated or unchanged) b as argument for the new call to f along with the next element from the list [a], and so on untill it has done this for all elements and results into a final b with all updates applied.

Upvotes: 1

Views: 320

Answers (1)

Cubic
Cubic

Reputation: 15693

What you want is foldr f b a after applying filter someAlgorithm on your list. If you don't know these functions, look them up on hoogle.

Upvotes: 2

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