I'm trying to write something like this in Haskell:
length . nub . intersect
but it doesn't work.
*Main Data.List> :t intersect
intersect :: Eq a => [a] -> [a] -> [a]
*Main Data.List> :t nub
nub :: Eq a => [a] -> [a]
*Main Data.List> :t length
length :: [a] -> Int
Based on the type, my understanding is that intersect returns a type of [a] and donates to nub , which takes exactly a type of [a] , then also returns a type of [a] to length , then finally the return should be an Int. What's wrong with it?
The problem here is that intersect takes 2 arguments (in a sense)
you can provide one of the arguments explicitly:
> let f a = length . nub . intersect a
> :t f
f :: Eq a => [a] -> [a] -> Int
or you can use a fun little operator like (.:) = (.) . (.):
> let (.:) = (.) . (.)
> :t length .: (nub .: intersect)
length .: (nub .: intersect) :: Eq a => [a] -> [a] -> Int
here is a version where you don't need the parens:
import Data.List
infixr 9 .:
(.:) :: (c -> d) -> (a -> b -> c) -> a -> b -> d
(.:) = (.).(.)
f :: Eq a => [a] -> [a] -> Int
f = length .: nub .: intersect
I guess this is based on the comments in your previous question, where #CarstenKönig mentions (.) . (.).
First of all, length . nub . intersect cannot work. Your types are:
(.) :: (b -> c) -> (a -> b) -> (a -> c)
length :: [a] -> Int
nub :: Eq a => [a] -> [a]
intersect :: Eq a => [a] -> [a] -> [a] ~ [a] -> ([a] -> [a])
As you can see, intersect has the wrong type, in the context of (.), the type parameter b would be replaced by ([a] -> [a]), which isn't the type of nub's first argument.
I'd say this: write your code the "dumb" way at first, and then refactor it to use (.). After some practice the composition operator will then become like second nature.
So you'd first write:
yourFunction xs ys = length (nub (intersect xs ys))
What (.) lets you do is get rid (syntactically) of the last argument of the innermost function, all all the parens. In this case that argument is ys:
yourFunction xs = length . nub . intersect xs
Related
I am having some troubles with function composition and types.
I would like to compose filter (which returns a list) with len, which takes a list as an argument (technically a Foldable but I am simplifying here).
Looking at the types everything is as expected:
> :t length
length :: Foldable t => t a -> Int
> :t filter
filter :: (a -> Bool) -> [a] -> [a]
So now I would expect the type of (len . filter) to be
(length . filter) :: (a -> Bool) -> [a] -> Int
while in reality is
> :t (length . filter)
(length . filter) :: Foldable ((->) [a]) => (a -> Bool) -> Int
So it seems I lost some arguments. Is it included in the the Foldable requirements in some way I am not understanding?
Note that everything works as expected if I do partial application:
> let myFilter = filter odd
> :t myFilter
myFilter :: Integral a => [a] -> [a]
> :t (length . myFilter)
(length . myFilter) :: Integral a => [a] -> Int
> (length . myFilter) [1,2,3]
2
The right composition would be:
(length .) . filter :: (a -> Bool) -> [a] -> Int
which is equivalent to:
\pred xs -> length $ filter pred xs
as in:
\> let count = (length .) . filter
\> :type count
count :: (a -> Bool) -> [a] -> Int
\> count odd [1..3]
2
\> count even [1..3]
1
Definitions:
(.) :: (b -> c) -> (a -> b) -> a -> c
filter :: (m -> Bool) -> [m] -> [m]
length :: Foldable t => t n -> Int
What is u?
length . filter :: u
≡ (.) length filter :: u
Then we must solve a, b, c, t, n:
a -> b ~ (m -> Bool) -> [m] -> [m]
b -> c ~ Foldable t => t n -> Int
It follows:
a ~ m -> Bool
b ~ Foldable t => t n
b ~ [m] -> [m]
c ~ Int
Trivially:
a = m -> Bool
b = [m] -> [m]
c = Int
We have to solve t, n from b ~ Foldable t => t n, i.e. [m] -> [m] ~ Foldable t => t n.
t = ((->) [m])
n = [m]
Therefore, t n = [m] -> [m] which trivially unifies.
Summarising:
(.) :: Foldable ((->) [m]) =>
(([m] -> [m]) -> Int)
-> ((m -> Bool) -> [m] -> [m])
-> (m -> Bool) -> Int
filter :: (m -> Bool) -> [m] -> [m]
length :: Foldable ((->) [m]) => ([m] -> [m]) -> Int
(.) length filter :: Foldable ((->) [m]) => (m -> Bool) -> Int
An easier way to understand why length . filter is not what you want is to look at the definition of (.).
(.) g f x = g(f x)
Therefore:
(.) length filter
≡ \x -> length (filter x)
We know that filter x is not a list.
Pointless versions you may consider:
(length .) . filter
filter >=> return . length
(fmap.fmap) length filter
(id ~> id ~> length) filter -- [1]
filter $* id $$ id *$ length -- [2]
lurryA #N2 (length <$> (filter <$> _1 <*> _2)) -- [3]
Control.Compose
Data.Function.Meld
Data.Function.Tacit
Using "three laws of operator sections", we have
((length .) . filter) x y =
(length .) (filter x) y =
(length . filter x) y =
length (filter x y)
and
((length .) . filter) =
(.) (length .) filter =
(.) ((.) length) filter =
((.) . (.)) length filter
The last bit, ((.).(.)), is sometimes known as "owl operator", also written as .: (length .: filter), or fmap . fmap (for functions, fmap is (.)).
I am trying to compose a function of type (Floating a) => a -> a -> a with a function of type (Floating a) => a -> a to obtain a function of type (Floating a) => a -> a -> a. I have the following code:
test1 :: (Floating a) => a -> a -> a
test1 x y = x
test2 :: (Floating a) => a -> a
test2 x = x
testBoth :: (Floating a) => a -> a -> a
testBoth = test2 . test1
--testBoth x y = test2 (test1 x y)
However, when I compile it in GHCI, I get the following error:
/path/test.hs:8:11:
Could not deduce (Floating (a -> a)) from the context (Floating a)
arising from a use of `test2'
at /path/test.hs:8:11-15
Possible fix:
add (Floating (a -> a)) to the context of
the type signature for `testBoth'
or add an instance declaration for (Floating (a -> a))
In the first argument of `(.)', namely `test2'
In the expression: test2 . test1
In the definition of `testBoth': testBoth = test2 . test1
Failed, modules loaded: none.
Note that the commented-out version of testBoth compiles. The strange thing is that if I remove the (Floating a) constraints from all type signatures or if I change test1 to just take x instead of x and y, testBoth compiles.
I've searched StackOverflow, Haskell wikis, Google, etc. and not found anything about a restriction on function composition relevant to this particular situation. Does anyone know why this is happening?
\x y -> test2 (test1 x y)
== \x y -> test2 ((test1 x) y)
== \x y -> (test2 . (test1 x)) y
== \x -> test2 . (test1 x)
== \x -> (test2 .) (test1 x)
== \x -> ((test2 .) . test1) x
== (test2 .) . test1
These two things are not like each other.
test2 . test1
== \x -> (test2 . test1) x
== \x -> test2 (test1 x)
== \x y -> (test2 (test1 x)) y
== \x y -> test2 (test1 x) y
You're problem doesn't have anything to do with Floating, though the typeclass does make your error harder to understand. Take the below code as an example:
test1 :: Int -> Char -> Int
test1 = undefined
test2 :: Int -> Int
test2 x = undefined
testBoth = test2 . test1
What is the type of testBoth? Well, we take the type of (.) :: (b -> c) -> (a -> b) -> a -> c and turn the crank to get:
b ~ Int (the argument of test2 unified with the first argument of (.))
c ~ Int (the result of test2 unified with the result of the first argument of (.))
a ~ Int (test1 argument 1 unified with argument 2 of (.))
b ~ Char -> Int (result of test1 unified with argument 2 of (.))
but wait! that type variable, 'b' (#4, Char -> Int), has to unify with the argument type of test2 (#1, Int). Oh No!
How should you do this? A correct solution is:
testBoth x = test2 . test1 x
There are other ways, but I consider this the most readable.
Edit: So what was the error trying to tell you? It was saying that unifying Floating a => a -> a with Floating b => b requires an instance Floating (a -> a) ... while that's true, you really didn't want GHC to try and treat a function as a floating point number.
Your problem has nothing to do with Floating, but with the fact that you want to compose a function with two arguments and a function with one argument in a way that doesn't typecheck. I'll give you an example in terms of a composed function reverse . foldr (:) [].
reverse . foldr (:) [] has the type [a] -> [a] and works as expected: it returns a reversed list (foldr (:) [] is essentially id for lists).
However reverse . foldr (:) doesn't type check. Why?
When types match for function composition
Let's review some types:
reverse :: [a] -> [a]
foldr (:) :: [a] -> [a] -> [a]
foldr (:) [] :: [a] -> [a]
(.) :: (b -> c) -> (a -> b) -> a -> c
reverse . foldr (:) [] typechecks, because (.) instantiates to:
(.) :: ([a] -> [a]) -> ([a] -> [a]) -> [a] -> [a]
In other words, in type annotation for (.):
a becomes [a]
b becomes [a]
c becomes [a]
So reverse . foldr (:) [] has the type [a] -> [a].
When types don't match for function composition
reverse . foldr (:) doesn't type check though, because:
foldr (:) :: [a] -> [a] -> [a]
Being the right operant of (.), it would instantiate its type from a -> b to [a] -> ([a] -> [a]). That is, in:
(b -> c) -> (a -> b) -> a -> c
Type variable a would be replaced with [a]
Type variable b would be replaced with [a] -> [a].
If type of foldr (:) was a -> b, the type of (. foldr (:)) would be:
(b -> c) -> a -> c`
(foldr (:) is applied as a right operant to (.)).
But because type of foldr (:) is [a] -> ([a] -> [a]), the type of (. foldr (:)) is:
(([a] -> [a]) -> c) -> [a] -> c
reverse . foldr (:) doesn't type check, because reverse has the type [a] -> [a], not ([a] -> [a]) -> c!
Owl operator
When people first learn function composition in Haskell, they learn that when you have the last argument of function at the right-most of the function body, you can drop it both from arguments and from the body, replacing or parentheses (or dollar-signs) with dots. In other words, the below 4 function definitions are equivalent:
f a x xs = g ( h a ( i x xs))
f a x xs = g $ h a $ i x xs
f a x xs = g . h a . i x $ xs
f a x = g . h a . i x
So people get an intuition that says “I just remove the right-most local variable from the body and from the arguments”, but this intuition is faulty, because once you removed xs,
f a x = g . h a . i x
f a = g . h a . i
are not equivalent! You should understand when function composition typechecks and when it doesn't. If the above 2 were equivalent, then it would mean that the below 2 are also equivalent:
f a x xs = g . h a . i x $ xs
f a x xs = g . h a . i $ x xs
which makes no sense, because x is not a function with xs as a parameter. x is a parameter to function i, and xs is a parameter to function (i x).
There is a trick to make a function with 2 parameters point-free. And that is to use an “owl” operator:
f a x xs = g . h a . i x xs
f a = g . h a .: i
where (.:) = (.).(.)
The above two function definitions are equivalent. Read more on “owl” operator.
References
Haskell programming becomes much easier and straightforward, once you understand functions, types, partial application and currying, function composition and dollar-operator. To nail these concepts, read the following StackOverflow answers:
On types and function composition
On higher-order functions, currying, and function composition
On Haskell type system
On point-free style
On const
On const, flip and types
On curry and uncurry
Read also:
Haskell: difference between . (dot) and $ (dollar sign)
Haskell function composition (.) and function application ($) idioms: correct use
I did it like this – but it is not working:
ma f [] = []
ma f (xs) = foldl (\y ys -> ys++(f y)) [] xs
foldl :: (a -> b -> a) -> a -> [b] -> a
foldr :: (a -> b -> b) -> b -> [a] -> b
Why is there a difference in the function that fold takes. I mean, (a -> b -> a) and (a -> b -> b)?
Is it possible to define map using foldl?
I have another question
I have an expr.
map (:)
I want to know what it will do. I tried to test it but i only get error.
type is map (:) :: [a] -> [[a] -> [a]]
I tried to send in a list of [1,2,3]
Not if you want it to work for infinite as well as finite lists. head $ map id (cycle [1]) must return 1.
foldling over an infinite list diverges (never stops), because foldl is recursive. For example,
foldl g z [a,b,c] = g (g (g z a) b) c
Before g gets a chance to ignore its argument, foldl must reach the last element of the input list, to construct the first call to g. There is no last element in an infinite list.
As for your new question, here's a GHCi transcript that shows that map (:) is a function, and map (:) [1,2,3] is a list of functions, and GHCi just doesn't know how to Show functions:
Prelude> map (:)
<interactive>:1:0:
No instance for (Show ([a] -> [[a] -> [a]]))
Prelude> :t map (:)
map (:) :: [a] -> [[a] -> [a]]
Prelude> map (:) [1,2,3]
<interactive>:1:0:
No instance for (Show ([a] -> [a]))
Prelude> :t map (:) [1,2,3]
map (:) [1,2,3] :: (Num a) => [[a] -> [a]]
Prelude> map ($ [4]) $ map (:) [1,2,3]
[[1,4],[2,4],[3,4]]
Prelude> foldr ($) [4] $ map (:) [1,2,3]
[1,2,3,4]
It becomes more obvious when you swap the type-variable names in one of the functions:
foldl :: (b -> a -> b) -> b -> [a] -> b
foldr :: (a -> b -> b) -> b -> [a] -> b
...because after all, what we need is the result, i.e. [a] -> b. Or, more specially, [a] -> [b], so we might as well substitute that
foldl :: ([b] -> a -> [b]) -> [b] -> [a] -> [b]
foldr :: (a -> [b] -> [b]) -> [b] -> [a] -> [b]
which leaves only one non-list item in each signature, namely the a. That's what we can apply f to, so, in the case of foldl it has to be the 2nd argument of the lambda:
foldl (\ys y -> ys ++ f y)
As Xeo remarks, this isn't done yet, because f y has type b, not [b]. I think you can figure out how to fix that yourself...
ma f [] = []
ma f (xs) = foldl (\ys y -> ys++[(f y)]) [] xs
Works but why does order of arg to lambda matter.
ma f (xs) = foldl (\y ys -> ys++[(f y)]) [] xs gives error
I want to define a function that computes the number of elements in a list that satisfy a given predicate:
number_of_elements :: (a -> Bool) -> [a] -> Int
number_of_elements f xs = length (filter f xs)
For example:
number_of_elements (==2) [2,1,54,1,2]
should return 2.
We can write it shorter:
number_of_elements f = length . filter f
Is it possible to write it without f parameter?
Sure it is:
number_of_elements = (length .) . filter
I don't think you can get more readable than the what you suggested. However, just for the fun of it you can do this:
numberOfElements = (.) (.) (.) length filter
or
(.:) = (.) . (.)
numberOfElements = length .: filter
You might like to read about Semantic Editor Combinators. Take the result combinator from there:
result :: (output -> output') -> (input -> output) -> (input -> output')
result = (.)
The result combinator takes a function and applies it to the result of another function. Now, looking at the functions we have:
filter :: (a -> Bool) -> [a] -> [a]
length :: [a] -> Int
Now, length applies to [a]'s; which, it happens, is the result type of functions of the form foo :: [a] -> [a]. So,
result length :: ([a] -> [a]) -> ([a] -> Int)
But the result of filter is exactly an [a] -> [a] function, so we want to apply result length to the result of filter:
number_of_elements = result (result length) filter
I am trying to compose a function of type (Floating a) => a -> a -> a with a function of type (Floating a) => a -> a to obtain a function of type (Floating a) => a -> a -> a. I have the following code:
test1 :: (Floating a) => a -> a -> a
test1 x y = x
test2 :: (Floating a) => a -> a
test2 x = x
testBoth :: (Floating a) => a -> a -> a
testBoth = test2 . test1
--testBoth x y = test2 (test1 x y)
However, when I compile it in GHCI, I get the following error:
/path/test.hs:8:11:
Could not deduce (Floating (a -> a)) from the context (Floating a)
arising from a use of `test2'
at /path/test.hs:8:11-15
Possible fix:
add (Floating (a -> a)) to the context of
the type signature for `testBoth'
or add an instance declaration for (Floating (a -> a))
In the first argument of `(.)', namely `test2'
In the expression: test2 . test1
In the definition of `testBoth': testBoth = test2 . test1
Failed, modules loaded: none.
Note that the commented-out version of testBoth compiles. The strange thing is that if I remove the (Floating a) constraints from all type signatures or if I change test1 to just take x instead of x and y, testBoth compiles.
I've searched StackOverflow, Haskell wikis, Google, etc. and not found anything about a restriction on function composition relevant to this particular situation. Does anyone know why this is happening?
\x y -> test2 (test1 x y)
== \x y -> test2 ((test1 x) y)
== \x y -> (test2 . (test1 x)) y
== \x -> test2 . (test1 x)
== \x -> (test2 .) (test1 x)
== \x -> ((test2 .) . test1) x
== (test2 .) . test1
These two things are not like each other.
test2 . test1
== \x -> (test2 . test1) x
== \x -> test2 (test1 x)
== \x y -> (test2 (test1 x)) y
== \x y -> test2 (test1 x) y
You're problem doesn't have anything to do with Floating, though the typeclass does make your error harder to understand. Take the below code as an example:
test1 :: Int -> Char -> Int
test1 = undefined
test2 :: Int -> Int
test2 x = undefined
testBoth = test2 . test1
What is the type of testBoth? Well, we take the type of (.) :: (b -> c) -> (a -> b) -> a -> c and turn the crank to get:
b ~ Int (the argument of test2 unified with the first argument of (.))
c ~ Int (the result of test2 unified with the result of the first argument of (.))
a ~ Int (test1 argument 1 unified with argument 2 of (.))
b ~ Char -> Int (result of test1 unified with argument 2 of (.))
but wait! that type variable, 'b' (#4, Char -> Int), has to unify with the argument type of test2 (#1, Int). Oh No!
How should you do this? A correct solution is:
testBoth x = test2 . test1 x
There are other ways, but I consider this the most readable.
Edit: So what was the error trying to tell you? It was saying that unifying Floating a => a -> a with Floating b => b requires an instance Floating (a -> a) ... while that's true, you really didn't want GHC to try and treat a function as a floating point number.
Your problem has nothing to do with Floating, but with the fact that you want to compose a function with two arguments and a function with one argument in a way that doesn't typecheck. I'll give you an example in terms of a composed function reverse . foldr (:) [].
reverse . foldr (:) [] has the type [a] -> [a] and works as expected: it returns a reversed list (foldr (:) [] is essentially id for lists).
However reverse . foldr (:) doesn't type check. Why?
When types match for function composition
Let's review some types:
reverse :: [a] -> [a]
foldr (:) :: [a] -> [a] -> [a]
foldr (:) [] :: [a] -> [a]
(.) :: (b -> c) -> (a -> b) -> a -> c
reverse . foldr (:) [] typechecks, because (.) instantiates to:
(.) :: ([a] -> [a]) -> ([a] -> [a]) -> [a] -> [a]
In other words, in type annotation for (.):
a becomes [a]
b becomes [a]
c becomes [a]
So reverse . foldr (:) [] has the type [a] -> [a].
When types don't match for function composition
reverse . foldr (:) doesn't type check though, because:
foldr (:) :: [a] -> [a] -> [a]
Being the right operant of (.), it would instantiate its type from a -> b to [a] -> ([a] -> [a]). That is, in:
(b -> c) -> (a -> b) -> a -> c
Type variable a would be replaced with [a]
Type variable b would be replaced with [a] -> [a].
If type of foldr (:) was a -> b, the type of (. foldr (:)) would be:
(b -> c) -> a -> c`
(foldr (:) is applied as a right operant to (.)).
But because type of foldr (:) is [a] -> ([a] -> [a]), the type of (. foldr (:)) is:
(([a] -> [a]) -> c) -> [a] -> c
reverse . foldr (:) doesn't type check, because reverse has the type [a] -> [a], not ([a] -> [a]) -> c!
Owl operator
When people first learn function composition in Haskell, they learn that when you have the last argument of function at the right-most of the function body, you can drop it both from arguments and from the body, replacing or parentheses (or dollar-signs) with dots. In other words, the below 4 function definitions are equivalent:
f a x xs = g ( h a ( i x xs))
f a x xs = g $ h a $ i x xs
f a x xs = g . h a . i x $ xs
f a x = g . h a . i x
So people get an intuition that says “I just remove the right-most local variable from the body and from the arguments”, but this intuition is faulty, because once you removed xs,
f a x = g . h a . i x
f a = g . h a . i
are not equivalent! You should understand when function composition typechecks and when it doesn't. If the above 2 were equivalent, then it would mean that the below 2 are also equivalent:
f a x xs = g . h a . i x $ xs
f a x xs = g . h a . i $ x xs
which makes no sense, because x is not a function with xs as a parameter. x is a parameter to function i, and xs is a parameter to function (i x).
There is a trick to make a function with 2 parameters point-free. And that is to use an “owl” operator:
f a x xs = g . h a . i x xs
f a = g . h a .: i
where (.:) = (.).(.)
The above two function definitions are equivalent. Read more on “owl” operator.
References
Haskell programming becomes much easier and straightforward, once you understand functions, types, partial application and currying, function composition and dollar-operator. To nail these concepts, read the following StackOverflow answers:
On types and function composition
On higher-order functions, currying, and function composition
On Haskell type system
On point-free style
On const
On const, flip and types
On curry and uncurry
Read also:
Haskell: difference between . (dot) and $ (dollar sign)
Haskell function composition (.) and function application ($) idioms: correct use