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
Related
So I'm learning about Haskell at the moment, and I came across this question:
The type of a function f is supposed to be a->[a]->a. The
following definitions of f are incorrect because their types are all
different from a->[a]->a:
i. f x xs = xs
ii. f x xs = x+1
iii. f x xs = x ++ xs
iv. f x (y:ys) = y
My answers as I see it are:
i) f :: a -> a -> a
This is because x or xs can be of any value and is not a list as it does not contain the ':' operator.
ii) f :: Int -> a -> Int
This is because the + operator is used on x, meaning x is of type Int.
iii) f :: Eq a => a -> a -> a
The ++ operators are used, therefore in order to concatenate they must be of the same type..?
iv) f :: a -> [a] -> a
f returns an element from the list.
The last one is definitely wrong, because it can't be of type a -> [a] -> a. Are there any others I did wrong, and why? I'm hoping I can fully understand types and how to find out the types of functions.
i) f :: a -> a -> a
f x xs = xs
This is because x or xs can be of any value and is not a list as it does not contain the ':' operator.
True, but it also does not have to be of the same type!
So, it's actually f :: a -> b -> b.
ii) f :: Int -> a -> Int
f x xs = x+1
This is because the + operator is used on x, meaning x is of type Int.
Correct. (Actually, in Haskell we get Num b => b -> a -> b which generalized the Int to any numeric type, but it's not that important.)
iii) f :: Eq a => a -> a -> a
f x xs = x ++ xs
The ++ operators are used, therefore in order to concatenate they must be of the same type..?
True, but they must be lists. Also, Eq is only needed if you use == or something which compares values.
Here, f :: [a] -> [a] -> [a].
iv) f :: a -> [a] -> a
f x (y:ys) = y
f returns an element from the list.
The type of x does not have to be the same. We get f :: b -> [a] -> a.
i. f x xs = xs
(...)
i) f :: a -> a -> a
Although this can be a type signature, you make it too restrictive. The function takes two parameters x and xs. Initially we can reason that x and xs can have different types, so we say that x :: a, and xs :: b. Since the function returns xs, the return type is b as well, so the type is:
f :: a -> b -> b
f x xs = xs
ii. f x xs = x+1
(...)
ii) f :: Int -> a -> Int
Again you make the function too restrictive. Let us again assume that x :: a and xs :: b have different types. We see that we return x + 1 (or in more canonical form (+) x 1. Since (+) has signature (+) :: Num c => c -> c -> c (we here use c since a is already used), and 1 has signature 1 :: Num d => d, we thus see that we call (+) with x and 1, as a result we know that a ~ c (a and c are the same type), and c ~ d, so as a result we obtain the signature:
f :: Num c => c -> b -> c
f x xs = x+1
iii. f x xs = x ++ xs
(...)
iii) f :: Eq a => a -> a -> a
This is wrong: we here see that f has two parameters, x :: a and xs :: b. We see that we return (++) x xs. Since (++) has signature (++) :: [c] -> [c] -> [c], we thus know that a ~ [c] and b ~ [c], so the type is:
f :: [c] -> [c] -> [c]
f x xs = x ++ xs
iv. f x (y:ys) = y
(...)
iv) f :: a -> [a] -> a
This is again too restrictive. Here we see again two parameters: x and (y:ys). We first generate a type a for x :: a, and (y:ys) :: b, since the pattern of the second parameter is (y:ys), this is a list constructor with as parameters (:) :: c -> [c] -> [c]. As a result we can derive that y :: c, and ys :: [c], furthermore the pattern (y:ys) has type [c]. Since the function returns y, we know that the return type is c, so:
f :: a -> [c] -> c
f x (y:ys) = y
Note: you can let Haskell derive the type of the function itself. In GHCi you can use the :t command to query the type of an expression. For example:
Prelude> f x (y:ys) = y
Prelude> :t f
f :: t1 -> [t] -> t
I am writing a function my_map which takes a unary function and a list and returns the list resulting from mapping the function over all elements of the input list.
Main> my_map (^3) [1..5]
[1,8,27,64,125]
I tried it like this:
my_map :: (a -> b) -> [a] -> [b]
my_map f [] = []
my_map f (x:xs) = foldr (\x xs -> (f x):xs) [] xs
But after running above, I get only [8,27,64,125]. the first number 1 is not displaying in output.
Can anybody help me?
You are using the (x:xs) pattern in your arguments, but when you apply the fold, you only apply it to the xs part, which means your first element i.e. the one that x represents never gets processed. You need to change it to this:
my_map :: (a -> b) -> [a] -> [b]
my_map f xs = foldr (\y ys -> (f y):ys) [] xs
Since you are using foldr, you do not need to explicitly handle the empty list case. Moreoever, you do not need to specify the list in (x:xs) format.
Finally, my own preference is to avoid using the same name for function inputs and any helper functions or expressions in the function definition.That is why, I have used xs for the input list and y and ys for the parameters passed to the lambda.
"shree.pat18" is perfectly right, and also the comments are valuable. I learned a lot from that. Just make it better visible, and to explain the alternatives...
Answer
-- The problem is here ....................... vv
my_map f (x:xs) = foldr (\x xs -> (f x):xs) [] xs
-- --
The remaining part xs is aplied to foldr.
To fix just this, apply the whole list. This can be done by placing xx# before (x:xs). By that, the whole list is bound to xx.
-- vvv ........... see here ............... vv
my_map f xx#(x:xs) = foldr (\x xs -> (f x):xs) [] xx
-- --- --
Recommended impovement
Note: foldr can already deal with [] as input. Hence, my_map f [] = [] is not needed. But foldr would not be called when you apply [] to my_map. To get rid of my_map f [] = [], you need to remove the pattern matching, because (x:xs) matches only to lists with at least one element.
main :: IO ()
main = print $ my_map (^(3 :: Int)) ([1..5] :: [Integer])
my_map :: (a -> b) -> [a] -> [b]
my_map f xx = foldr (\x xs -> (f x):xs) [] xx
The answer is complete here. The rest below is for pleasure.
Further reductions
Simple expression instead of lambda expression
If you want to reduce the lambda expression (\x xs -> (f x):xs), as suggested by "Aadit M Shah"...
(:) is equal to (\x xs -> x:xs), because : is an operator and its function is (:)
. can be used to combine the function f with (:), hence (\x xs -> (f x):xs) is equal to ((:) . f)
main :: IO ()
main = print $ my_map (^(3 :: Int)) ([] :: [Integer])
my_map :: (a -> b) -> [a] -> [b]
my_map f xx = foldr ((:) . f) [] xx
Currying
A function of the form
-- v v
f a b c = .... c
can be reduced to
-- v v
f a b = ....
and a function of the form
-- v v v v
f a b c = .... b c
can be reduced to
-- v v v v
f a = ....
and so on, by currying.
Hence, my_map f xx = foldr ((:) . f) [] xx equals my_map f = foldr ((:) . f) [].
Combination and flip
flip flips the first two parameters.
Example, the following functions are equal:
f' a b c = (\c' b' a' -> ((a' - b') / c')) b a c
f'' a b c = flip (\c' b' a' -> ((a' - b') / c')) a b c
f''' = flip (\c' b' a' -> ((a' - b') / c'))
Hence, the following code works as well.
main :: IO ()
main = print $ my_map (^(3 :: Int)) ([1..5] :: [Integer])
my_map :: (a -> b) -> [a] -> [b]
my_map f = flip foldr [] ((:) . f)
But we can not get rid of f as above, because of the form in the expression flip foldr [] ((:) . f).
If we remove f ...
`((:) . f)` has type `a -> [a] -> [a]
-- v
`((:) . )` has type `(a -> a) -> a -> [a] -> [a]`
and
`flip foldr []` has type `Foldable t => (a1 -> [a2] -> [a2]) -> t a1 -> [a2]`
hence
f :: a -> a
is passed to
((:) . )
becomming
a -> [a] -> [a]
is passed to
flip foldr []
becomming
t a1 -> [a2]
Hence,
main :: IO ()
main = print $ my_map (^(3 :: Int)) ([1..5] :: [Integer])
my_map :: (a -> b) -> [a] -> [b]
my_map = flip foldr [] . ((:) . )
works nicely.
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'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
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