This is an example that shows how foldl and foldr work (from RWH p.93-94): foldl (+) 0 (1:2:3:[]) == foldl (+) (0 + 1) (2:3:[]) == foldl (+) ((0 + 1) + 2) (3:[]) == foldl (+) (((0 + 1) + 2) + 3) [] == (((0 + 1) + 2) + 3) foldr (+) 0 (1:2:3:[]) == 1 + foldr (+) 0 (2:3:[]) == 1 + (2 + foldr (+) 0 (3:[]) == 1 + (2 + (3 + foldr (+) 0 [])) == 1 + (2 + (3 + 0)) The book says on p.94: ----- The difference between foldl and foldr should be clear from looking at where the parentheses and the empty list elements show up. With foldl, the empty list element is on the left, and all the parentheses group to the left. With foldr, the zero value is on the right, and the parentheses group to the right. ---- Huh? With foldl, the only empty list element I see is on the right. Initially, it looked to me ike they did the same thing, and that the only difference was the way they called step. I think "step" is a horrible, non-descriptive name, so I'm going to use "accFunc" instead: foldl calls: accFunc acc x foldr calls: accFunc x acc So it looks like you can define a function using either one and get the same result. Here is a test: --I am going to use odd for pfunc and [1, 2, 3] for xs: myFilter1 pfunc xs = foldl accFunc [] xs where accFunc acc x | pfunc x = acc ++ [x] | otherwise = acc myFilter2 pfunc xs = foldr accFunc [] xs where accFunc x acc | pfunc x = acc ++ [x] | otherwise = acc *Main> myFilter1 odd [1, 2, 3] [1,3] *Main> myFilter2 odd [1, 2, 3] [3,1] Hmmm. So there is a difference. foldr appears to grab elements from the end of the list. Therefore, to get the same result from the function that uses foldr, I did this: myFilter3 pfunc xs = foldr accFunc [] xs where accFunc x acc | pfunc x = x : acc | otherwise = acc *Main> myFilter3 odd [1, 2, 3] [1,3] But then RWH explains that you would never use foldl in practice because it thunks the result, which for large lists can overwhelm the maximum memory alloted for a thunk. But it appears to me the same thunk problem would occur with foldr. So why is foldr used in practice but not foldl?
Am Montag, 9. März 2009 17:46 schrieb 7stud:
This is an example that shows how foldl and foldr work (from RWH p.93-94):
foldl (+) 0 (1:2:3:[]) == foldl (+) (0 + 1) (2:3:[]) == foldl (+) ((0 + 1) + 2) (3:[]) == foldl (+) (((0 + 1) + 2) + 3) [] == (((0 + 1) + 2) + 3)
foldr (+) 0 (1:2:3:[]) == 1 + foldr (+) 0 (2:3:[]) == 1 + (2 + foldr (+) 0 (3:[]) == 1 + (2 + (3 + foldr (+) 0 [])) == 1 + (2 + (3 + 0))
The book says on p.94:
----- The difference between foldl and foldr should be clear from looking at where the parentheses and the empty list elements show up. With foldl, the empty list element is on the left, and all the parentheses group to the left. With foldr, the zero value is on the right, and the parentheses group to the right. ----
Huh? With foldl, the only empty list element I see is on the right.
What they meant was "the value that is the result in case the fold is applied to an empty list", in this case the 0, in the definition fold(l/r) f z xs = ... the 'z'.
Initially, it looked to me ike they did the same thing, and that the only difference was the way they called step. I think "step" is a horrible, non-descriptive name, so I'm going to use "accFunc" instead:
foldl calls: accFunc acc x
foldr calls: accFunc x acc
So it looks like you can define a function using either one and get the same result.
Note that in general the list elements and acc have different types, so only one of accFun acc x and accFun x acc typechecks. If the types are the same, in general accFun x acc /= accFun acc x, so foldr and foldl give different results, too.
Here is a test:
--I am going to use odd for pfunc and [1, 2, 3] for xs:
myFilter1 pfunc xs = foldl accFunc [] xs where accFunc acc x
| pfunc x = acc ++ [x] | otherwise = acc
myFilter2 pfunc xs = foldr accFunc [] xs where accFunc x acc
| pfunc x = acc ++ [x] | otherwise = acc
*Main> myFilter1 odd [1, 2, 3] [1,3] *Main> myFilter2 odd [1, 2, 3] [3,1]
Hmmm. So there is a difference. foldr appears to grab elements from the end of the list. Therefore, to get the same result from the function that uses foldr, I did this:
myFilter3 pfunc xs = foldr accFunc [] xs where accFunc x acc
| pfunc x = x : acc | otherwise = acc
*Main> myFilter3 odd [1, 2, 3] [1,3]
But then RWH explains that you would never use foldl in practice because it thunks the result, which for large lists can overwhelm the maximum memory alloted for a thunk. But it appears to me the same thunk problem would occur with foldr. So why is foldr used in practice but not foldl?
Since with foldr, the parentheses are grouped to the right: x0 `f` (x1 `f` (x2 `f` ... (xn `f` z) ... )), if f can start delivering the result without looking at its second argument, you can start consuming the result before the fold has traversed the whole list. Common examples are things like concat = foldr (++) [], so concat [l1,l2,l3,l4,l5] = l1 ++ (foldr (++) [] [l2,l3,l4,l5]) and the start (l1) can be used before further reducing the fold, and = foldr (&&) True and [True,False,..........] needs only inspect the list until it encounters the first False (if any, otherwise it must of course traverse the whole list) or = foldr (||) False foldr is useful if the combination function is lazy in its second argument. foldl on the other hand can't deliver anything before the whole list is consumed. So since foldl builds thunks (except in some easy cases where the optimiser sees it should be strict), which would have to be evaluated at the end when they've become rather large, foldl isn't as useful and one uses the strict left fold, foldl'.
Daniel Fischer
Am Montag, 9. März 2009 17:46 schrieb 7stud:
This is an example that shows how foldl and foldr work (from RWH p.93-94):
foldl (+) 0 (1:2:3:[]) == foldl (+) (0 + 1) (2:3:[]) == foldl (+) ((0 + 1) + 2) (3:[]) == foldl (+) (((0 + 1) + 2) + 3) [] == (((0 + 1) + 2) + 3)
foldr (+) 0 (1:2:3:[]) == 1 + foldr (+) 0 (2:3:[]) == 1 + (2 + foldr (+) 0 (3:[]) == 1 + (2 + (3 + foldr (+) 0 [])) == 1 + (2 + (3 + 0))
The book says on p.94:
----- The difference between foldl and foldr should be clear from looking at where the parentheses and the empty list elements show up. With foldl, the empty list element is on the left, and all the parentheses group to the left. With foldr, the zero value is on the right, and the parentheses group to the right. ----
Huh? With foldl, the only empty list element I see is on the right.
What they meant was "the value that is the result in case the fold is applied to an empty list", in this case the 0, in the definition
So that's an error right? Or is that correct haskell terminology? The book also says on p. 95: ------------- Like foldl, foldr takes a function and a base case(what to do when the input list is empty) as arguments. ------------- That also does not seem correct. For example: foldrSum xs = foldr accFunc 0 xs where accFunc x acc = acc + x *Main> foldrSum [1, 2, 3] 6 In that example, the first two arguments to foldr are the function accFunc and 0. It does not seem accurate to say that "0 is what to do when the input list is empty". What foldr does when the input list is empty is return the value of the acc parameter variable: foldr _ acc [] = acc In my example, the value of the acc parameter is 6 "when the input list is empty"--not the value 0, which is the argument to foldr.
fold(l/r) f z xs = ...
the 'z'.
Initially, it looked to me ike they did the same thing, and that the only difference was the way they called step. I think "step" is a horrible, non-descriptive name, so I'm going to use "accFunc" instead:
foldl calls: accFunc acc x
foldr calls: accFunc x acc
So it looks like you can define a function using either one and get the same result.
Note that in general the list elements and acc have different types, so only one of accFun acc x and accFun x acc typechecks.
I don't know how that comment is relevant. In my examples, acc and x have different types: *Main> :type [] [] :: [a] *Main> :type 1 1 :: (Num t) => t And both examples work fine.
If the types are the same, in general accFun x acc /= accFun acc x,
Is that correct? Can you give some examples? Here is what I tried: 1) accFunc1 acc x = x + acc accFunc2 x acc = x + acc *Main> let x = 1 *Main> let acc = 3 *Main> accFunc1 acc x 4 *Main> accFunc1 x acc 4 2) accFunc3 acc x = x ++ acc accFunc4 x acc = x ++ acc *Main> let x = [1] *Main> let acc = [2, 3] *Main> accFunc3 acc x [1,2,3] *Main> accFunc4 x acc [1,2,3]
so foldr and foldl give different results, too.
I think the results produced by my example functions myFilter1 and myFilter2 demonstrate that, but the differing results are because of the way foldr and foldl are defined.
Here is a test:
--I am going to use odd for pfunc and [1, 2, 3] for xs:
myFilter1 pfunc xs = foldl accFunc [] xs where accFunc acc x
| pfunc x = acc ++ [x] | otherwise = acc
myFilter2 pfunc xs = foldr accFunc [] xs where accFunc x acc
| pfunc x = acc ++ [x] | otherwise = acc
*Main> myFilter1 odd [1, 2, 3] [1,3] *Main> myFilter2 odd [1, 2, 3] [3,1]
Hmmm. So there is a difference. foldr appears to grab elements from the end of the list. Therefore, to get the same result from the function that uses foldr, I did this:
myFilter3 pfunc xs = foldr accFunc [] xs where accFunc x acc
| pfunc x = x : acc | otherwise = acc
*Main> myFilter3 odd [1, 2, 3] [1,3]
But then RWH explains that you would never use foldl in practice because it thunks the result, which for large lists can overwhelm the maximum memory alloted for a thunk. But it appears to me the same thunk problem would occur with foldr. So why is foldr used in practice but not foldl?
Since with foldr, the parentheses are grouped to the right:
if f can start delivering the result without looking at its second argument, you can start consuming the result before the fold has traversed the whole list.
Ok, that isn't clearly illustrated by the example in the book:
foldl (+) 0 (1:2:3:[]) == foldl (+) (0 + 1) (2:3:[]) == foldl (+) ((0 + 1) + 2) (3:[]) == foldl (+) (((0 + 1) + 2) + 3) [] == (((0 + 1) + 2) + 3)
foldr (+) 0 (1:2:3:[]) == 1 + foldr (+) 0 (2:3:[]) == 1 + (2 + foldr (+) 0 (3:[]) == 1 + (2 + (3 + foldr (+) 0 [])) == 1 + (2 + (3 + 0))
In that example, it doesn't look like anything in foldr can be evaluated until the whole fold has been completed.
Common examples are things like
concat = foldr (++) [], so concat [l1,l2,l3,l4,l5] = l1 ++ (foldr (++) [] [l2,l3,l4,l5]) and the start (l1) can be used before further reducing the fold,
So does haskell store a thunk for everything to the right of l1? You said that when using foldr you can start "consuming" the beginning of the result before the whole result is reduced. I don't quite get that.
and = foldr (&&) True
[to evaluate the expression] and [True,False,..........] [haskell] needs only inspect the list until it encounters the first False (if any), otherwise it must of course traverse the whole list
or = foldr (||) False
foldr is useful if the combination function is lazy in its second argument.
Ok.
foldl on the other hand can't deliver anything before the whole list is consumed. So since foldl builds thunks (except in some easy cases where the optimiser sees it should be strict), which would have to be evaluated at the end when they've become rather large, foldl isn't as useful and one uses the strict left fold, foldl'.
Thanks.
On Tue, Mar 10, 2009 at 3:59 AM, 7stud
------------- Like foldl, foldr takes a function and a base case(what to do when the input list is empty) as arguments. -------------
That also does not seem correct. For example:
foldrSum xs = foldr accFunc 0 xs where accFunc x acc = acc + x
*Main> foldrSum [1, 2, 3] 6
In that example, the first two arguments to foldr are the function accFunc and 0. It does not seem accurate to say that "0 is what to do when the input list is empty". What foldr does when the input list is empty is return the value of the acc parameter variable:
I'm not sure why they explain the base case for fold in that way. At least to me, that is only a trivial result of the 'zero' value's main purpose, which is to be the initial value used in the accumulating fold function. When starting the fold, we have to start somewhere, so the accFunc needs a seed value. That value, and the first element of the list, are fed into accFunc to start things off. Then it just so happens that if the list is empty, the seed value is the result, since no folding can happen.
foldr _ acc [] = acc
In my example, the value of the acc parameter is 6 "when the input list is empty"--not the value 0, which is the argument to foldr.
You're thinking of a slightly different 'empty' here. You're thinking of what happens when you reach the end of the list, after folding it all, and there are no more elements to fold. In this example, you're right that the acc parameter is 6. But what if the list you *first gave to foldr* was empty? Then it would evaluate to 0, the initial seed value. Now you may be thinking, "Why would I ever apply foldr to an empty list? Obviously that would do nothing." Well you may not know whether a list is empty, if it's the result of other calculations. You also may be thinking, "Why do I need to provide a seed value, why can't foldr just start with the first two elements of the list?" That's because the accFunc does not always evaluate to the same type as the elements in the list. For example, you could use a fold to count the number of 'a's in a list of characters. Then the type of accFunc would be "accFunc :: Int -> Char -> Int" for foldl or "accFunc :: Char -> Int -> Int" for foldr. It takes one element from the list, the previous result, and evaluates to a new result. But in the very first fold step, there's no previous result, so you have to provide one. It's called zero as a convention, but it doesn't actually have to *be* zero. It can be any initial value you want. Make your foldSum function only evaluate to results of 10 or larger by doing this: foldSum xs = foldr accFunc 10 xs where accFunc x acc = acc + x [---snip---]
if f can start delivering the result without looking at its second argument, you can start consuming the result before the fold has traversed the whole list.
Ok, that isn't clearly illustrated by the example in the book:
foldl (+) 0 (1:2:3:[]) == foldl (+) (0 + 1) (2:3:[]) == foldl (+) ((0 + 1) + 2) (3:[]) == foldl (+) (((0 + 1) + 2) + 3) [] == (((0 + 1) + 2) + 3)
foldr (+) 0 (1:2:3:[]) == 1 + foldr (+) 0 (2:3:[]) == 1 + (2 + foldr (+) 0 (3:[]) == 1 + (2 + (3 + foldr (+) 0 [])) == 1 + (2 + (3 + 0))
In that example, it doesn't look like anything in foldr can be evaluated until the whole fold has been completed.
You're right, that example doesn't show how you could start using the result without fully evaluating the fold, since addition doesn't give partial results. The concat example is better in that regard.
Common examples are things like
concat = foldr (++) [], so concat [l1,l2,l3,l4,l5] = l1 ++ (foldr (++) [] [l2,l3,l4,l5]) and the start (l1) can be used before further reducing the fold,
So does haskell store a thunk for everything to the right of l1? You said that when using foldr you can start "consuming" the beginning of the result before the whole result is reduced. I don't quite get that.
A thunk is used as a stand-in for most calculations before the result is actually calculated. That way, if you never try to use the result, the calculation never needs to be done, and that means less work. As an example that ties to the concat example above, say your program only wanted to test if the result of the concat fold was an empty list. The function 'null' takes a list and evaluates to True or False, based on whether the list is empty or not. So: someFunc xs = null ( concat xs ) where concat ys = foldr (++) [] ys The 'null' function only needs to test whether the list that is the result of foldConcat has at least one element. Let's say l1 has an element. So it's kind of evaluated like this: someFunc [ l1, l2, l3, l4, l5 ] null (concat [ l1, l2, l3, l4, l5 ] ) null ( l1 ++ ( thunk with rest of fold )) False The rest of the fold doesn't need to be evaluated, since the beginning part is enough for 'null' to tell that the result would have at least one element (because l1 does). That's one way foldr can be used to start consuming the result before the entire fold is done. It depends completely on the accFunc: if it can return partial results, like concat, then you can start consuming the result before a full evaluation. Some accFunc's can't return partial results, like regular addition. In that case, it's probably better to use foldl' (note the apostrophe), which will force the thunks to be evaluated as they are generated and so use less memory. foldl is the same as foldl' except it does generate thunks, and then evaluates them all at the end of the fold, so it uses a bunch of memory to store the thunks in the meantime, which usually isn't useful. Kurt
Kurt Hutchinson
You're thinking of a slightly different 'empty' here. You're thinking of what happens when you reach the end of the list, after folding it all, and there are no more elements to fold. In this example, you're right that the acc parameter is 6. But what if the list you *first gave to foldr* was empty? Then it would evaluate to 0, the initial seed value.
Ok.
if f can start delivering the result without looking at its second argument, you can start consuming the result before the fold has traversed the whole list.
Ok, that isn't clearly illustrated by the example in the book:
foldl (+) 0 (1:2:3:[]) == foldl (+) (0 + 1) (2:3:[]) == foldl (+) ((0 + 1) + 2) (3:[]) == foldl (+) (((0 + 1) + 2) + 3) [] == (((0 + 1) + 2) + 3)
foldr (+) 0 (1:2:3:[]) == 1 + foldr (+) 0 (2:3:[]) == 1 + (2 + foldr (+) 0 (3:[]) == 1 + (2 + (3 + foldr (+) 0 [])) == 1 + (2 + (3 + 0))
In that example, it doesn't look like anything in foldr can be evaluated until the whole fold has been completed.
You're right, that example doesn't show how you could start using the result without fully evaluating the fold, since addition doesn't give partial results. The concat example is better in that regard.
Ok. I guess I did understand something. Therefore, I think the example in the book that uses addition with foldl and foldr is TERRIBLE. I think the sections on folds in RWH are a catastrophe and need to be rewritten. The authors need to get rid of the "bit twiddling" example and provide an example like concat to clearly show the difference between foldl and foldr. I think the example using foldl and foldr with addition should be a secondary example to demonstrate that sometimes it doesn't matter whether you use foldl or foldr.
Common examples are things like
concat = foldr (++) [], so concat [l1,l2,l3,l4,l5] = l1 ++ (foldr (++) [] [l2,l3,l4,l5]) and the start (l1) can be used before further reducing the fold,
So does haskell store a thunk for everything to the right of l1? You said that when using foldr you can start "consuming" the beginning of the before the whole result is reduced. I don't quite get that.
A thunk is used as a stand-in for most calculations before the result is actually calculated. That way, if you never try to use the result, the calculation never needs to be done, and that means less work. As an example that ties to the concat example above, say your program only wanted to test if the result of the concat fold was an empty list. The function 'null' takes a list and evaluates to True or False, based on whether the list is empty or not. So:
someFunc xs = null ( concat xs ) where concat ys = foldr (++) [] ys
The 'null' function only needs to test whether the list that is the result of foldConcat has at least one element. Let's say l1 has an element. So it's kind of evaluated like this: someFunc [ l1, l2, l3, l4, l5 ] null (concat [ l1, l2, l3, l4, l5 ] ) null ( l1 ++ ( thunk with rest of fold )) False
The rest of the fold doesn't need to be evaluated, since the beginning part is enough for 'null' to tell that the result would have at least one element (because l1 does). That's one way foldr can be used to start consuming the result before the entire fold is done.
Ok. The terminology is just a little confusing for me. When you say "start consuming before the fold is done" it sounds to me like somehow you are reading partial results off the front of the result before foldr returns the entire result. But as you explained above what you actually mean is that foldr will return the entire result-- with part of the result being a thunk, and subsequently you can do operations on the result that may never require the thunked part to be evaluated.
It depends completely on the accFunc: if it can return partial results, like concat, then you can start consuming the result before a full evaluation. Some accFunc's can't return partial results, like regular addition. In that case, it's probably better to use foldl' (note the apostrophe), which will force the thunks to be evaluated as they are generated and so use less memory. foldl is the same as foldl' except [foldl] does generate thunks, and then evaluates them all at the end of the fold, so it uses a bunch of memory to store the thunks in the meantime, which usually isn't useful.
Kurt
Thanks to you and Daniel for the great explanations. I'm feeling a lot better about folds now. .... .... ...... .... ... ... .... .... .... .... ... ... ... ....
Am 09.03.2009 um 17:46 schrieb 7stud:
This is an example that shows how foldl and foldr work (from RWH p. 93-94):
foldl (+) 0 (1:2:3:[]) == foldl (+) (0 + 1) (2:3:[]) == foldl (+) ((0 + 1) + 2) (3:[]) == foldl (+) (((0 + 1) + 2) + 3) [] == (((0 + 1) + 2) + 3)
foldr (+) 0 (1:2:3:[]) == 1 + foldr (+) 0 (2:3:[]) == 1 + (2 + foldr (+) 0 (3:[]) == 1 + (2 + (3 + foldr (+) 0 [])) == 1 + (2 + (3 + 0))
The book says on p.94:
----- The difference between foldl and foldr should be clear from looking at where the parentheses and the empty list elements show up. With foldl, the empty list element is on the left, and all the parentheses group to the left. With foldr, the zero value is on the right, and the parentheses group to the right. ----
Huh? With foldl, the only empty list element I see is on the right.
Have a look at foldl.com and foldr.com. With "empty list element" they mean the 0.
But then RWH explains that you would never use foldl in practice because it thunks the result, which for large lists can overwhelm the maximum memory alloted for a thunk. But it appears to me the same thunk problem would occur with foldr. So why is foldr used in practice but not foldl?
The problem is that foldr can lazyly produce a result. Try foldr (:) [] [1..]. It works. Now try foldl (flip (:)) [] [1..]. It breaks. However foldl is tail recursive, so the compiler can optimize the recursion away. In some cases that is beneficial. Notice that there is no difference between foldr g a foldl f a (for appropriate g and f) if g and f are strict in both arguments. Have a look at the "Foldl as foldr" wikipage. Also have a look at this paper
Regards, Adrian
Adrian Neumann wrote:
Notice that there is no difference between
foldr g a foldl f a
(for appropriate g and f) if g and f are strict in both arguments.
Be careful... as apfelmus noted elsewhere in this thread, that's not (in general) true. Prelude> foldr (^) 2 [3,5] 847288609443 Prelude> foldl (^) 2 [3,5] 32768 The reason? Integer exponentiation (^) isn't associative and commutative. So the first is (3 ^ (5^2)) = 3^25, while the second is ((2 ^ 3) ^ 5) = 2^15. Cheers, John
Maybe it helps to visualize it like this. Instead of computing the sum by
using a fold with (+), we just construct data:
data Expr = N Int
| Expr :+: Expr
deriving Show
ns :: [Expr]
ns = map N [1..3]
lf :: Expr
lf = foldl1 (:+:) ns
rf :: Expr
rf = foldr1 (:+:) ns
For simplicity Iused foldl1 and foldr1, which only work on non-empty lists.
(regarding the weird :+: well in Haskell, you can use operators for data
constructors when they start with a colon)
Run this with GHCi, and evaluate lf and rf. You should get
*Main> lf
(N 1 :+: N 2) :+: N 3
*Main> rf
N 1 :+: (N 2 :+: N 3)
So really, foldl "folds on the left", because the parentheses are on the
left side. Similarly for foldr.
Does this help?
On Mon, Mar 9, 2009 at 5:46 PM, 7stud
This is an example that shows how foldl and foldr work (from RWH p.93-94):
foldl (+) 0 (1:2:3:[]) == foldl (+) (0 + 1) (2:3:[]) == foldl (+) ((0 + 1) + 2) (3:[]) == foldl (+) (((0 + 1) + 2) + 3) [] == (((0 + 1) + 2) + 3)
foldr (+) 0 (1:2:3:[]) == 1 + foldr (+) 0 (2:3:[]) == 1 + (2 + foldr (+) 0 (3:[]) == 1 + (2 + (3 + foldr (+) 0 [])) == 1 + (2 + (3 + 0))
The book says on p.94:
----- The difference between foldl and foldr should be clear from looking at where the parentheses and the empty list elements show up. With foldl, the empty list element is on the left, and all the parentheses group to the left. With foldr, the zero value is on the right, and the parentheses group to the right. ----
Huh? With foldl, the only empty list element I see is on the right.
Initially, it looked to me ike they did the same thing, and that the only difference was the way they called step. I think "step" is a horrible, non-descriptive name, so I'm going to use "accFunc" instead:
foldl calls: accFunc acc x
foldr calls: accFunc x acc
So it looks like you can define a function using either one and get the same result. Here is a test:
--I am going to use odd for pfunc and [1, 2, 3] for xs:
myFilter1 pfunc xs = foldl accFunc [] xs where accFunc acc x | pfunc x = acc ++ [x] | otherwise = acc
myFilter2 pfunc xs = foldr accFunc [] xs where accFunc x acc | pfunc x = acc ++ [x] | otherwise = acc
*Main> myFilter1 odd [1, 2, 3] [1,3] *Main> myFilter2 odd [1, 2, 3] [3,1]
Hmmm. So there is a difference. foldr appears to grab elements from the end of the list. Therefore, to get the same result from the function that uses foldr, I did this:
myFilter3 pfunc xs = foldr accFunc [] xs where accFunc x acc | pfunc x = x : acc | otherwise = acc
*Main> myFilter3 odd [1, 2, 3] [1,3]
But then RWH explains that you would never use foldl in practice because it thunks the result, which for large lists can overwhelm the maximum memory alloted for a thunk. But it appears to me the same thunk problem would occur with foldr. So why is foldr used in practice but not foldl?
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7stud wrote:
This is an example that shows how foldl and foldr work (from RWH p.93-94):
foldl (+) 0 (1:2:3:[]) == foldl (+) (0 + 1) (2:3:[]) == foldl (+) ((0 + 1) + 2) (3:[]) == foldl (+) (((0 + 1) + 2) + 3) [] == (((0 + 1) + 2) + 3)
foldr (+) 0 (1:2:3:[]) == 1 + foldr (+) 0 (2:3:[]) == 1 + (2 + foldr (+) 0 (3:[]) == 1 + (2 + (3 + foldr (+) 0 [])) == 1 + (2 + (3 + 0))
The book says on p.94:
----- The difference between foldl and foldr should be clear from looking at where the parentheses and the empty list elements show up. With foldl, the empty list element is on the left, and all the parentheses group to the left. With foldr, the zero value is on the right, and the parentheses group to the right. ----
Huh? With foldl, the only empty list element I see is on the right.
A fold like foldr f z is best understood as a function that replaces each (:) with f and each [] with z . See also the diagrams on http://en.wikipedia.org/wiki/Fold_(higher-order_function)
From this point of view, z "corresponds to the empty list".
Initially, it looked to me like they did the same thing, and that the only difference was the way they called step.
They are only the same when the operation f is associative, i.e. if it satisfies f x (f y z) = f (f x y) z
But then RWH explains that you would never use foldl in practice because it thunks the result, which for large lists can overwhelm the maximum memory alloted for a thunk. But it appears to me the same thunk problem would occur with foldr. So why is foldr used in practice but not foldl?
See also http://en.wikibooks.org/wiki/Haskell/Performance_Introduction#Space Regards, apfelmus -- http://apfelmus.nfshost.com
participants (7)
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7stud -
Adrian Neumann -
Daniel Fischer -
Heinrich Apfelmus -
John Dorsey -
Kurt Hutchinson -
Peter Verswyvelen