Dependent and independent variables in foldl and foldr
I have this myLength1 = foldl (\n _ -> n + 1) 0 and this myLength2 = foldr (\_ n -> n + 1) 0 I am guessing that foldl knows to assign the accumulator-seed argument to the dependent variable and the list argument's elements recursively to the independent variable; and with foldr to do the opposite. Is this a fair assumption? BTW, where can I get a look at the code for fold functions; or does the type definition answer my original question? Not really able to decipher it so well :t foldl foldl :: Foldable t => (b -> a -> b) -> b -> t a -> b LB
Il 16 gennaio 2021 alle 16:10 Lawrence Bottorff ha scritto:
I have this
myLength1 = foldl (\n _ -> n + 1) 0
and this
myLength2 = foldr (\_ n -> n + 1) 0
I am guessing that foldl knows to assign the accumulator-seed argument to the dependent variable and the list argument's elements recursively to the independent variable; and with foldr to do the opposite. Is this a fair assumption? BTW, where can I get a look at the code for fold functions; or does the type definition answer my original question? Not really able to decipher it so well
:t foldl foldl :: Foldable t => (b -> a -> b) -> b -> t a -> b
foldl and foldr have slightly different signatures, λ> :t +d foldl foldl :: (b -> a -> b) -> b -> [a] -> b λ> :t +d foldr foldr :: (a -> b -> b) -> b -> [a] -> b (Notice `b -> a -> b` vs. `a -> b -> b`), hence the lambdas have a different non-matched parameter. Does this answer your question? —F
This is the definition of list foldr
foldr :: (a -> b -> b) -> b -> [a] -> b
foldr _ z [] = z
foldr f z (x:xs) = f x (foldr f z xs)
In both foldl and foldr in the OP the n variable in lambda functions would
seem to be for the accumulator, hence, I assume the n is considered the
free variable? And then the wildcard in each lambda function refers to the
bound variable, i.e., the list argument's elements to be folded. So I can
recreate
foldr (+) 5 [1,2,3,4]
with
foldr (\x n -> x + n) 5 [1,2,3,4]
They both return 15. The first one results in
(+) 1 ((+) 2 ((+) 3 ((+) 4 5)))
but the second example I'm not sure how the (\x n -> x + n) is being
applied in the form . . . f x (foldr f z xs) It obviously must be doing the
same (+) 1 ((+) 2 ((+) 3 ((+) 4 5))) but how the function is being applied
I don't understand. Beta reduction doesn't get me very far
\x n -> x + n (5)([1,2,3,4])
\x 5 -> x + 5 ([1,2,3,4])
but obviously the enclosing lambda calc for foldr is doing something to
create the (+) 1 ((+) 2 ((+) 3 ((+) 4 5))) form.
BTW, is the t a format in
:type foldr
foldr :: Foldable t => (a -> b -> b) -> b -> t a -> b
something from category theory, i.e., for the list instance, t a is [a] What
is the algebraic syntax where t a becomes [a] in the case of lists? It
would be nice to understand some day exactly what :i Foldable is saying
On Sat, Jan 16, 2021 at 4:36 PM Francesco Ariis
I have this
myLength1 = foldl (\n _ -> n + 1) 0
and this
myLength2 = foldr (\_ n -> n + 1) 0
I am guessing that foldl knows to assign the accumulator-seed argument to the dependent variable and the list argument's elements recursively to
Il 16 gennaio 2021 alle 16:10 Lawrence Bottorff ha scritto: the
independent variable; and with foldr to do the opposite. Is this a fair assumption? BTW, where can I get a look at the code for fold functions; or does the type definition answer my original question? Not really able to decipher it so well
:t foldl foldl :: Foldable t => (b -> a -> b) -> b -> t a -> b
foldl and foldr have slightly different signatures,
λ> :t +d foldl foldl :: (b -> a -> b) -> b -> [a] -> b λ> :t +d foldr foldr :: (a -> b -> b) -> b -> [a] -> b
(Notice `b -> a -> b` vs. `a -> b -> b`), hence the lambdas have a different non-matched parameter. Does this answer your question? —F _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
Il 16 gennaio 2021 alle 23:32 Lawrence Bottorff ha scritto:
This is the definition of list foldr
foldr :: (a -> b -> b) -> b -> [a] -> b foldr _ z [] = z foldr f z (x:xs) = f x (foldr f z xs)
In both foldl and foldr in the OP the n variable in lambda functions would seem to be for the accumulator, hence, I assume the n is considered the free variable? And then the wildcard in each lambda function refers to the bound variable, i.e., the list argument's elements to be folded.
The ‘_’ means «do not bind this parameter». Hence λ> (\x _ -> x + 1) 10 20 11
but the second example I'm not sure how the (\x n -> x + n) is being applied in the form . . . f x (foldr f z xs) It obviously must be doing the same (+) 1 ((+) 2 ((+) 3 ((+) 4 5))) but how the function is being applied I don't understand.
foldr (+) 0 ([1,2,3]) (+) 1 (foldr (+) 0 [2,3]) (+) 1 ((+) 2 (foldr (+) 0 [3])) (+) 1 ((+) 2 ((+) 3 (foldr (+) 0 []))) (+) 1 ((+) 2 ((+) 3 0)) (+) 1 ((+) 2 3) (+) 1 5 6
BTW, is the t a format in
:type foldr foldr :: Foldable t => (a -> b -> b) -> b -> t a -> b
something from category theory, i.e., for the list instance, t a is [a] What is the algebraic syntax where t a becomes [a] in the case of lists? It would be nice to understand some day exactly what :i Foldable is saying
What book are you studying on that does not talk about Typeclasses? I feel you are making your life harder by not following a good study program.
participants (2)
-
Francesco Ariis -
Lawrence Bottorff