Mapping which holds and passes previous result

When solving system of linear equations by Tridiagonal matrix algorithm in Haskell I met following problem.

We have three vectors: a, b and c, and we want to make a third vector c' which is a combination of them:

c'[i] = c[i] / b[i], i = 0
c'[i] = c[i] / (b[i] - a[i] * c'[i-1]), 0 < i < n - 1
c'[i] = undefined, i = n - 1

Naive implementation of the formula above in Haskell is a开发者_运维问答s follows:

calcC' a b c = Data.Vector.generate n f
  where
    n = Data.Vector.length a
    f i = 
      | i == 0 = c!0 / b!0 
      | i == n - 1 = 0
      | otherwise = c!i / (b!i - a!i * f (i - 1))

It looks like this function calcC' has complexity O(n²) due to recurrence. But all we actualy need is to pass to inner function f one more parameter with previously generated value.

I wrote my own version of generate with complexity O(n) and helper function mapP:

mapP f xs = mapP' xs Nothing
  where
    mapP' [] _ = []
    mapP' (x:xs) xp = xn : mapP' xs (Just xn)
      where
        xn = f x xp

generateP n f = Data.Vector.fromList $ mapP f [0 .. n-1]

As one can see, mapP acts like a standard map, but also passes to mapping function previously generated value or Nothing for first call.

My question: is there any pretty standard ways to do this in Haskell? Don't I reinvent the weel?

Thanks.

---

There are two standard function called mapAccumL and mapAccumR that do precisely what you want.

mapAccumL :: (acc -> x -> (acc, y)) -> acc -> [x] -> (acc, [y])
mapAccumR :: (acc -> x -> (acc, y)) -> acc -> [x] -> (acc, [y])

Basically, they behave like a combination of fold and map.

map   f   = snd . mapAccumL (\_ x -> (()   , f x) ()
foldl f b = fst . mapAccumL (\b x -> (f b x, () ) b

---

If you use Data.Array, which is lazy, you can express the recurrence directly by referring to c' while defining c'.

---

Following code seems to be the simplest implementation of formula above in my case:

import qualified Data.Vector.Generic as V

calcC' a b c = V.postscanl' f 0.0 $ V.zip3 a b c
  where
    f c' (a, b, c) = c / (b - a * c')

Thanks to the authors of Vector who added helpfull postscanl' method.