Difference between revisions of "Foldl as foldr alternative"
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This page explains how <hask>foldl</hask> can be written using <hask>foldr</hask>. Yes, there is already [[Foldl as foldr|such a page]]! This one explains it differently. |
This page explains how <hask>foldl</hask> can be written using <hask>foldr</hask>. Yes, there is already [[Foldl as foldr|such a page]]! This one explains it differently. |
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The usual definition of <hask>foldl</hask> looks like this: |
The usual definition of <hask>foldl</hask> looks like this: |
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− | Now the <hask>f</hask> never changes in the recursion |
+ | Now the <hask>f</hask> never changes in the recursion. It turns out things will be simpler later if we pull it out: |
<haskell> |
<haskell> |
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− | + | foldl :: (a -> x -> r) -> a -> [x] -> r |
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− | f a |
+ | foldl f a list = go a list |
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</haskell> |
</haskell> |
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− | While we're at it, let's give a name to <hask>foldl f</hask>: <hask>stuff</hask>. So |
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− | <haskell> |
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− | stuff :: Ord x => Set x -> [x] -> Set x |
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− | </haskell> |
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− | takes all the elements of the list it's given and stuffs them into the <hask>Set</hask> it's given. |
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− | For some reason (maybe we're crazy; maybe we want to do weird things with fusion; who knows?) we want to write this using <hask>foldr</hask>. Haskell programmers like curry, so it's natural to see <hask> |
+ | For some reason (maybe we're crazy; maybe we want to do weird things with fusion; who knows?) we want to write this using <hask>foldr</hask>. Haskell programmers like curry, so it's natural to see <hask>go a xs</hask> as <hask>(go a) xs</hask>—that is, to see <hask>go a</hask> as a function that takes a list and returns the result of folding <hask>f</hask> into the list starting with an accumulator value of <hask>a</hask>. This perspective, however, is the ''wrong one'' for what we're trying to do here. So let's change the order of the arguments of the helper: |
<haskell> |
<haskell> |
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− | + | foldl :: (a -> x -> r) -> a -> [x] -> r |
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− | + | foldl f a list = go2 list a |
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+ | go2 [] a = a |
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</haskell> |
</haskell> |
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− | So now we see that <hask> |
+ | So now we see that <hask>go2 xs</hask> is a function that takes an accumulator and uses it as the initial value to fold <hask>f</hask> into <hask>xs</hask>. With this shift of perspective, we can rewrite <hask>go2</hask> just a little: |
<haskell> |
<haskell> |
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− | + | foldl :: (a -> x -> r) -> a -> [x] -> r |
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− | + | foldl f a list = go2 list a |
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+ | where |
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</haskell> |
</haskell> |
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<haskell> |
<haskell> |
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− | stuffy :: Ord x => [x] -> Set x -> Set x |
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+ | where |
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− | + | go2 [] = (\a -> a) |
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</haskell> |
</haskell> |
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− | This isn't an academic paper, so we won't mention Graham Hutton's "Tuturial on the Universality and Expressiveness of Fold", but <hask> |
+ | This isn't an academic paper, so we won't mention Graham Hutton's "Tuturial on the Universality and Expressiveness of Fold", but <hask>go2</hask> fits the <hask>foldr</hask> pattern: |
<haskell> |
<haskell> |
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− | + | go2 ys = foldr whatsit (\a -> a) ys |
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− | stuffy ys = foldr whatsit (\a -> a) ys |
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where |
where |
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whatsit x r = \a -> r (f a x) |
whatsit x r = \a -> r (f a x) |
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<haskell> |
<haskell> |
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− | + | foldl f a list = (foldr whatsit (\a -> a) list) a |
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− | stuffy list a = (foldr whatsit (\a -> a) list) a |
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where |
where |
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whatsit x r = \a -> r (f a x) |
whatsit x r = \a -> r (f a x) |
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⚫ | And that's all she wrote! One way to look at this final expression is that <hask>whatsit</hask> takes an element of the list, a function produced by folding over the rest of the list, and the value of an accumulator. It applies <hask>f</hask> to the accumulator it's given and the list element, and passes the result forward to the function it got. |
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− | And that's just about it! We wanted <hask>stuff</hask>, however, not <hask>stuffy</hask>, so let's swap the argument order again: |
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− | <haskell> |
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− | stuff :: Ord x => Set a -> [x] -> Set x |
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− | stuff a list = (foldr whatsit (\a -> a) list) a |
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− | whatsit x r = \a -> r (f a x) |
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− | </haskell> |
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− | Now since we do want to be able to use general <hask>foldl</hask> forms, we should generalize it again: |
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− | <haskell> |
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− | foldl :: (a -> x -> r) -> a -> [x] -> r |
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− | </haskell> |
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⚫ |
Revision as of 05:42, 4 September 2014
This page explains how foldl
can be written using foldr
. Yes, there is already such a page! This one explains it differently.
The usual definition of foldl
looks like this:
foldl :: (a -> x -> r) -> a -> [x] -> r
foldl f a [] = a
foldl f a (x : xs) = foldl f (f a x) xs
Now the f
never changes in the recursion. It turns out things will be simpler later if we pull it out:
foldl :: (a -> x -> r) -> a -> [x] -> r
foldl f a list = go a list
where
go a [] = a
go a (x : xs) = go (f a x) xs
For some reason (maybe we're crazy; maybe we want to do weird things with fusion; who knows?) we want to write this using foldr
. Haskell programmers like curry, so it's natural to see go a xs
as (go a) xs
—that is, to see go a
as a function that takes a list and returns the result of folding f
into the list starting with an accumulator value of a
. This perspective, however, is the wrong one for what we're trying to do here. So let's change the order of the arguments of the helper:
foldl :: (a -> x -> r) -> a -> [x] -> r
foldl f a list = go2 list a
where
go2 [] a = a
go2 (x : xs) a = go2 xs (f a x)
So now we see that go2 xs
is a function that takes an accumulator and uses it as the initial value to fold f
into xs
. With this shift of perspective, we can rewrite go2
just a little:
foldl :: (a -> x -> r) -> a -> [x] -> r
foldl f a list = go2 list a
where
go2 [] = \a -> a
go2 (x : xs) = \a -> go2 xs (f a x)
Believe it or not, we're almost done! How is that? Let's parenthesize a bit for emphasis:
foldl f a list = go2 list a
where
go2 [] = (\a -> a)
go2 (x : xs) = \a -> (go2 xs) (f a x)
This isn't an academic paper, so we won't mention Graham Hutton's "Tuturial on the Universality and Expressiveness of Fold", but go2
fits the foldr
pattern:
go2 ys = foldr whatsit (\a -> a) ys
where
whatsit x r = \a -> r (f a x)
Substituting this in,
foldl f a list = (foldr whatsit (\a -> a) list) a
where
whatsit x r = \a -> r (f a x)
And that's all she wrote! One way to look at this final expression is that whatsit
takes an element of the list, a function produced by folding over the rest of the list, and the value of an accumulator. It applies f
to the accumulator it's given and the list element, and passes the result forward to the function it got.