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Tags: Control     Foldl

foldl alternatives and similar packages

Based on the "foldl" category.
Alternatively, view foldl alternatives based on common mentions on social networks and blogs.

• foldl-transduce

Transducers for foldl folds.
• foldl-exceptions

Exception handling with FoldM
• Deliver Cleaner and Safer Code - Right in Your IDE of Choice!

SonarLint is a free and open source IDE extension that identifies and catches bugs and vulnerabilities as you code, directly in the IDE. Install from your favorite IDE marketplace today.
• foldl-transduce-attoparsec

Attoparsec and foldl-transduce integration.
• foldl-incremental

incremental folds

Do you think we are missing an alternative of foldl or a related project?

foldl

Use this foldl library when you want to compute multiple folds over a collection in one pass over the data without space leaks.

For example, suppose that you want to simultaneously compute the sum of the list and the length of the list. Many Haskell beginners might write something like this:

sumAndLength :: Num a => [a] -> (a, Int)
sumAndLength xs = (sum xs, length xs)

However, this solution will leak space because it goes over the list in two passes. If you demand the result of sum the Haskell runtime will materialize the entire list. However, the runtime cannot garbage collect the list because the list is still required for the call to length.

Usually people work around this by hand-writing a strict left fold that looks something like this:

{-# LANGUAGE BangPatterns #-}

import Data.List (foldl')

sumAndLength :: Num a => [a] -> (a, Int)
sumAndLength xs = foldl' step (0, 0) xs
where
step (x, y) n = (x + n, y + 1)

That now goes over the list in one pass, but will still leak space because the tuple is not strict in both fields! You have to define a strict Pair type to fix this:

{-# LANGUAGE BangPatterns #-}

import Data.List (foldl')

data Pair a b = Pair !a !b

sumAndLength :: Num a => [a] -> (a, Int)
sumAndLength xs = done (foldl' step (Pair 0 0) xs)
where
step (Pair x y) n = Pair (x + n) (y + 1)

done (Pair x y) = (x, y)

However, this is not satisfactory because you have to reimplement the guts of every fold that you care about and also define a custom strict data type for your fold. Hand-writing the step function, accumulator, and strict data type for every fold that you want to use gets tedious fast. For example, implementing something like reservoir sampling over and over is very error prone.

What if you just stored the step function and accumulator for each individual fold and let some high-level library do the combining for you? That's exactly what this library does! Using this library you can instead write:

import qualified Control.Foldl as Fold

sumAndLength :: Num a => [a] -> (a, Int)
sumAndLength xs = Fold.fold ((,) <\$> Fold.sum <*> Fold.length) xs

-- or, more concisely:
sumAndLength = Fold.fold ((,) <\$> Fold.sum <*> Fold.length)

To see how this works, the Fold.sum value is just a datatype storing the step function and the starting state (and a final extraction function):

sum :: Num a => Fold a a
sum = Fold (+) 0 id

Same thing for the Fold.length value:

length :: Fold a Int
length = Fold (\n _ -> n + 1) 0 id

... and the Applicative operators combine them into a new datatype storing the composite step function and starting state:

(,) <\$> Fold.sum <*> Fold.length = Fold step (Pair 0 0) done
where
step (Pair x y) = Pair (x + n) (y + 1)

done (Pair x y) = (x, y)

... and then fold just transforms that to a strict left fold:

fold (Fold step begin done) = done (foldl' step begin)

Since we preserve the step function and accumulator, we can use the Fold type to fold things other than pure collections. For example, we can fold a Producer from pipes using the same Fold:

Fold.purely Pipes.Prelude.fold ((,) <\$> sum <*> length)
:: (Monad m, Num a) => Producer a m () -> m (a, Int)

Quick start

Install the stack tool and then run:

\$ stack setup
\$ stack ghci foldl
Prelude> import qualified Control.Foldl as Fold
Prelude Fold> Fold.fold ((,) <\$> Fold.sum <*> Fold.length) [1..1000000]
(500000500000,1000000)

How to contribute

Contribute a pull request if you have a Fold that you believe other people would find useful.

Development Status The foldl library is pretty stable at this point. I don't expect there to be breaking changes to the API from this point forward unless people discover new bugs.