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Foldl as foldr

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(foldl using Update monoid)
(foldlMaybeMonoid)
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By the way:
By the way:
 +
<hask>Update a</hask> is just <hask>Dual (Endo a)</hask>.
If you use a <hask>State</hask> monad instead of a monoid,
If you use a <hask>State</hask> monad instead of a monoid,
you obtain an alternative implementation of <hask>mapAccumL</hask>.
you obtain an alternative implementation of <hask>mapAccumL</hask>.
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The answer to the second question is:
The answer to the second question is:
-
We can write a <hask>foldl</hask> that may stop before reaching the end of the input list
+
Using the <hask>foldr</hask> expression we can write variants of <hask>foldl</hask>
 +
that behave slightly different from the original one.
 +
E.g. we can write a <hask>foldl</hask> that can stop before reaching the end of the input list
and thus may also terminate on infinite input.
and thus may also terminate on infinite input.
The function <hask>foldlMaybe</hask> terminates with <hask>Nothing</hask> as result
The function <hask>foldlMaybe</hask> terminates with <hask>Nothing</hask> as result
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</haskell>
</haskell>
 +
Maybe the monoidic version is easier to understand.
 +
The implementation of the fold is actually the same, we do only use a different monoid.
 +
<haskell>
 +
import Control.Monad ((>=>), )
 +
 +
newtype UpdateMaybe a = UpdateMaybe {evalUpdateMaybe :: a -> Maybe a}
 +
 +
instance Monoid (UpdateMaybe a) where
 +
mempty = UpdateMaybe Just
 +
mappend (UpdateMaybe x) (UpdateMaybe y) = UpdateMaybe (x>=>y)
 +
 +
foldlMaybeMonoid :: (a -> b -> Maybe a) -> a -> [b] -> Maybe a
 +
foldlMaybeMonoid f a bs =
 +
flip evalUpdateMaybe a $
 +
mconcat $
 +
map (UpdateMaybe . flip f) bs
 +
</haskell>
[[Category:Idioms]]
[[Category:Idioms]]

Revision as of 23:12, 16 February 2009

When you wonder whether to choose foldl or foldr you may remember,

that both
foldl
and
foldl'
can be expressed as
foldr
. (
foldr
may lean so far right it came back left again.) The converse is not true, since
foldr
may work on infinite lists, which
foldl
variants never can do.

It holds 

foldl :: (a -> b -> a) -> a -> [b] -> a
foldl f a bs =
   foldr (\b g x -> g (f x b)) id bs a

Now the question are:

  • How can someone find a convolved expression like this?
  • How can we benefit from this rewrite?


1 Folding by concatenating updates

Instead of thinking in terms of
foldr
and a function
g
as argument to the accumulator function,

I find it easier to imagine a fold as a sequence of updates. An update is a function mapping from an old value to an updated new value. 

newtype Update a = Update {evalUpdate :: a -> a}

We need a way to assemble several updates.

To this end we define a
Monoid
instance. 
instance Monoid (Update a) where
   mempty = Update id
   mappend (Update x) (Update y) = Update (y.x)

Now left-folding is straight-forward. 

foldlMonoid :: (a -> b -> a) -> a -> [b] -> a
foldlMonoid f a bs =
   flip evalUpdate a $
   mconcat $
   map (Update . flip f) bs
Now, where is the
foldr
? It is hidden in
mconcat
. 
mconcat :: Monoid a => [a] -> a
mconcat = foldr mappend mempty
Since
mappend
must be associative (and is actually associative for our
Update
monoid),
mconcat
could also be written as
foldl
, but this is avoided, precisely
foldl
fails on infinite lists.

By the way:

Update a
is just
Dual (Endo a)
. If you use a
State
monad instead of a monoid, you obtain an alternative implementation of
mapAccumL
.


2 foldl which may terminate early

The answer to the second question is:

Using the
foldr
expression we can write variants of
foldl

that behave slightly different from the original one.

E.g. we can write a
foldl
that can stop before reaching the end of the input list

and thus may also terminate on infinite input.

The function
foldlMaybe
terminates with
Nothing
as result when it encounters a
Nothing
as interim accumulator result. 
foldlMaybe :: (a -> b -> Maybe a) -> a -> [b] -> Maybe a
foldlMaybe f a bs =
   foldr (\b g x -> f x b >>= g) Just bs a

Maybe the monoidic version is easier to understand. The implementation of the fold is actually the same, we do only use a different monoid. 

import Control.Monad ((>=>), )
 
newtype UpdateMaybe a = UpdateMaybe {evalUpdateMaybe :: a -> Maybe a}
 
instance Monoid (UpdateMaybe a) where
   mempty = UpdateMaybe Just
   mappend (UpdateMaybe x) (UpdateMaybe y) = UpdateMaybe (x>=>y)
 
foldlMaybeMonoid :: (a -> b -> Maybe a) -> a -> [b] -> Maybe a
foldlMaybeMonoid f a bs =
   flip evalUpdateMaybe a $
   mconcat $
   map (UpdateMaybe . flip f) bs
Retrieved from "http://www.haskell.org/haskellwiki/Foldl_as_foldr"

Category: Idioms