How does foldr work?

问题

Can anybody explain how does foldr work?

Take these examples:

Prelude> foldr (-) 54 [10, 11]
53
Prelude> foldr (\x y -> (x+y)/2) 54 [12, 4, 10, 6]
12.0

I am confused about these executions. Any suggestions?

回答1:

foldr begins at the right-hand end of the list and combines each list entry with the accumulator value using the function you give it. The result is the final value of the accumulator after "folding" in all the list elements. Its type is:

foldr :: (a -> b -> b) -> b -> [a] -> b

and from this you can see that the list element (of type a) is the first argument to the given function, and the accumulator (of type b) is the second.

For your first example:

Starting accumulator = 54
11 -   54  = -43
10 - (-43) =  53

        ^  Result from the previous line

 ^ Next list item

So the answer you got was 53.

The second example:

Starting accumulator = 54
(6  + 54) / 2 = 30
(10 + 30) / 2 = 20
(4  + 20) / 2 = 12
(12 + 12) / 2 = 12

So the result is 12.

Edit: I meant to add, that's for finite lists. foldr can also work on infinite lists but it's best to get your head around the finite case first, I think.

回答2:

The easiest way to understand foldr is to rewrite the list you're folding over without the sugar.

[1,2,3,4,5] => 1:(2:(3:(4:(5:[]))))

now what foldr f x does is that it replaces each : with f in infix form and [] with x and evaluates the result.

For example:

sum [1,2,3] = foldr (+) 0 [1,2,3]

[1,2,3] === 1:(2:(3:[]))

so

sum [1,2,3] === 1+(2+(3+0)) = 6

回答3:

It helps to understand the distinction between foldr and foldl. Why is foldr called "fold right"?

Initially I thought it was because it consumed elements from right to left. Yet both foldr and foldl consume the list from left to right.

• foldl evaluates from left to right (left-associative)
• foldr evaluates from right to left (right-associative)

We can make this distinction clear with an example that uses an operator for which associativity matters. We could use a human example, such as the operator, "eats":

foodChain = (human : (shark : (fish : (algae : []))))

foldl step [] foodChain
  where step eater food = eater `eats` food  -- note that "eater" is the accumulator and "food" is the element

foldl `eats` [] (human : (shark : (fish : (algae : []))))
  == foldl eats (human `eats` shark)                              (fish : (algae : []))
  == foldl eats ((human `eats` shark) `eats` fish)                (algae : [])
  == foldl eats (((human `eats` shark) `eats` fish) `eats` algae) []
  ==            (((human `eats` shark) `eats` fish) `eats` algae)

The semantics of this foldl is: A human eats some shark, and then the same human who has eaten shark then eats some fish, etc. The eater is the accumulator.

Contrast this with:

foldr step [] foodChain
    where step food eater = eater `eats` food.   -- note that "eater" is the element and "food" is the accumulator

foldr `eats` [] (human : (shark : (fish : (algae : []))))
  == foldr eats (human `eats` shark)                              (fish : (algae : []))))
  == foldr eats (human `eats` (shark `eats` (fish))               (algae : [])
  == foldr eats (human `eats` (shark `eats` (fish `eats` algae))) []
  ==            (human `eats` (shark `eats` (fish `eats` algae) 

The semantics of this foldr is: A human eats a shark which has already eaten a fish, which has already eaten some algae. The food is the accumulator.

Both foldl and foldr "peel off" eaters from left to right, so that's not the reason we refer to foldl as "left fold". Instead, the order of evaluation matters.

回答4:

Think about foldr's very definition:

 -- if the list is empty, the result is the initial value z
 foldr f z []     = z                  
 -- if not, apply f to the first element and the result of folding the rest 
 foldr f z (x:xs) = f x (foldr f z xs)

So for example foldr (-) 54 [10,11] must equal (-) 10 (foldr (-) 54 [11]), i.e. expanding again, equal (-) 10 ((-) 11 54). So the inner operation is 11 - 54, that is, -43; and the outer operation is 10 - (-43), that is, 10 + 43, therefore 53 as you observe. Go through similar steps for your second case, and again you'll see how the result forms!

回答5:

foldr means fold from the right, so foldr (-) 0 [1, 2, 3] produces (1 - (2 - (3 - 0))). In comparison foldl produces (((0 - 1) - 2) - 3).

When the operators are not commutative foldl and foldr will get different results.

In your case, the first example expands to (10 - (11 - 54)) which gives 53.

回答6:

An easy way to understand foldr is this: It replaces every list constructor with an application of the function provided. Your first example would translate to:

10 - (11 - 54)

from:

10 : (11 : [])

A good piece of advice that I got from the Haskell Wikibook might be of some use here:

As a rule you should use foldr on lists that might be infinite or where the fold is building up a data structure, and foldl' if the list is known to be finite and comes down to a single value. foldl (without the tick) should rarely be used at all.

回答7:

I've always thought http://foldr.com to be a fun illustration. See the Lambda the Ultimate post.

回答8:

I think that implementing map, foldl and foldr in a simple fashion helps explain how they work. Worked examples also aid in our understanding.

  myMap f [] = []
  myMap f (x:xs) = f x : myMap f xs

  myFoldL f i [] = i
  myFoldL f i (x:xs) = myFoldL f (f i x) xs

  > tail [1,2,3,4] ==> [2,3,4]
  > last [1,2,3,4] ==> 4
  > head [1,2,3,4] ==> 1
  > init [1,2,3,4] ==> [1,2,3]

  -- where f is a function,
  --  acc is an accumulator which is given initially
  --  l is a list.
  --
  myFoldR' f acc [] = acc
  myFoldR' f acc l = myFoldR' f (f acc (last l)) (init l)

  myFoldR f z []     = z
  myFoldR f z (x:xs) = f x (myFoldR f z xs)

  > map (\x -> x/2) [12,4,10,6] ==> [6.0,2.0,5.0,3.0]
  > myMap (\x -> x/2) [12,4,10,6] ==> [6.0,2.0,5.0,3.0]

  > foldl (\x y -> (x+y)/2) 54 [12, 4, 10, 6] ==> 10.125
  > myFoldL (\x y -> (x+y)/2) 54 [12, 4, 10, 6] ==> 10.125

    foldl from above: Starting accumulator = 54
      (12  + 54) / 2 = 33
      (4 + 33) / 2 = 18.5
      (10  + 18.5) / 2 = 14.25
      (6 + 14.25) / 2 = 10.125`

 > foldr (++) "5" ["1", "2", "3", "4"] ==> "12345"

 > foldl (++) "5" ["1", "2", "3", "4"] ==> “51234"

 > foldr (\x y -> (x+y)/2) 54 [12,4,10,6] ==> 12
 > myFoldR' (\x y -> (x+y)/2) 54 [12,4,10,6] ==> 12
 > myFoldR (\x y -> (x+y)/2) 54 [12,4,10,6] ==> 12

    foldr from above: Starting accumulator = 54
        (6  + 54) / 2 = 30
        (10 + 30) / 2 = 20
        (4  + 20) / 2 = 12
        (12 + 12) / 2 = 12

回答9:

Ok, lets look at the arguments:

• a function (that takes a list element and a value (a possible partial result) of the same kind of the value it returns);
• a specification of the initial result for the empty list special case
• a list;

return value:

• some final result

It first applies the function to the last element in the list and the empty list result. It then reapplies the function with this result and the previous element, and so forth until it takes some current result and the first element of the list to return the final result.

Fold "folds" a list around an initial result using a function that takes an element and some previous folding result. It repeats this for each element. So, foldr does this starting at the end off the list, or the right side of it.

folr f emptyresult [1,2,3,4] turns into f(1, f(2, f(3, f(4, emptyresult) ) ) ) . Now just follow parenthesis in evaluation and that's it.

One important thing to notice is that the supplied function f must handle its own return value as its second argument which implies both must have the same type.

Source: my post where I look at it from an imperative uncurried javascript perspective if you think it might help.

来源：https://stackoverflow.com/questions/1757740/how-does-foldr-work