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 idSame 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)To learn more about this library, read the documentation in
the main Control.Foldl module.
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)Contribute a pull request if you have a Fold that you believe other people
would find useful.
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.
Copyright (c) 2016 Gabriel Gonzalez All rights reserved.
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