问题

So i'm not too sure how to phrase this properly, but say I wanted to get the sum of all odd numbers in a list, do I have two functions (sumList and getOddNumbers) and combine them into sumOddList or is there a way to put these two together in a single function? If there isnt a better function, how exactly would I combine them into sumOddList?

getOddNumbers :: [Integer] -> [Integer]
getOddNumbers [] = []
getOddNumbers (x:xs)
    |odd x = x:getOddNumbers xs
    |otherwise = getOddNumbers xs

sumList :: [Integer] -> Integer
sumList list = case list of
   [] -> 0
   (x:xs) -> x + (sumList xs)

I also ask mainly because putting two diff functions together is something I struggled with before, when putting a colour and a shape using CodeWorld to output a shape of that colour.

Thank you

(Note: I've been using Haskell for just over 5 weeks now and I'm a total noob clearly)

回答1:

Passing output as input to (another) function

Well what you basically want to do is use the output of the getOddNumbers as input for the sumList function, so we can define a sumOddList function as:

sumOddList :: [Integer] -> Integer
sumOddList l = sumList (getOddNumbers l)

Here l is the list we want to process, and the result is thus a function application on the result of getOddNumbers l (with sumList the function).

Chaining functions: the `(.)` function

The above pattern is quite common: frequently we want to pass data first through a function g, and the result through a function f. Haskell has the (.) :: (b -> c) -> (a -> b) -> a -> c function to "chain" functions. We can thus chain sumList and getOddNumbers together like:

sumOddList :: [Integer] -> Integer
sumOddList = (.) sumList getOddNumbers

Notice that we no longer use an l parameter here. sumOddList is here defined as a "pipeline" where data is first passed to the getOddNumbers, and then is "post-processed" by the sumList function.

The (.) function can also be used as an infix operator:

sumOddList :: [Integer] -> Integer
sumOddList = sumList . getOddNumbers

回答2:

In the following are three equivalent ways to write the function oddSum :: [Integer] -> Integer:

oddSum xs = sumList (getOddNumbers xs)

oddSum xs = sumList $ getOddNumbers xs

oddSum = sumList . getOddNumbers

Btw, have a look at the filter and sum functions in the Prelude with which you could replace getOddNumbers and sumList respectively.

回答3:

or is there a way to put these two together in a single function ... sumOddList?

Yes there is.

Chaining functions by using one's output as the other's input works, under lazy evaluation especially, but leaves us reliant on the fusion to be performed by a compiler. Which after all is not guaranteed to happen (and often doesn't).

Instead, what you said&hairsp;:

mapping f cons x xs = cons (f x) xs

filtering p cons x xs = if (p x) then (cons x xs) else xs

transduce xf cons z xs = foldr (xf cons) z xs

sumOddList xs = transduce (filtering odd) (+) 0 xs

Thus,

> sumOddList [1..10]
25
> sum [1,3..10]
25
> transduce (mapping (+1) . filtering odd) (+) 0 [1..10]
35
> sum . filter odd . map (+1) $ [1..10]
35
> sum . map (+1) . filter odd $ [1..10]
30
> transduce (filtering odd . mapping (+1)) (+) 0 [1..10]
30

This works because folds fuse by composing their reducer functions' transformers (like the mapping and the filtering above which are transforming their reducer argument cons):

foldr (+) 0
   . foldr (\x r -> x+1 : r) []
       . foldr (\x r -> if odd x then x : r else r) [] 
         $ [1..10]
=
foldr (+) 0
   . foldr ((\cons x r -> cons (x+1) r) (:)) []
       . foldr ((\cons x r -> if odd x then cons x r else r) (:)) [] 
         $ [1..10]
=
     foldr ((\cons x r -> cons (x+1) r) (+)) 0
       . foldr ((\cons x r -> if odd x then cons x r else r) (:)) [] 
         $ [1..10]
=
         foldr ((\cons x r -> if odd x then cons x r else r) 
                  ((\cons x r -> cons (x+1) r)   (+))) 0 
         $ [1..10]
=
         foldr ( ( (\cons x r -> if odd x then cons x r else r) 
                 . (\cons x r -> cons (x+1) r) ) (+))  0 
         $ [1..10]
=
         foldr ( (filtering odd . mapping (+1))  (+))  0 
         $ [1..10]
=
         foldr (  filtering odd ( mapping (+1)   (+))) 0 
         $ [1..10]
=
         30

One foldr doing the work of the three. Fusion explicitly achieved, by composing the reducer functions after the cons operation has been abstracted from them, each such changed function becoming a cons transformer, thus, liable to be composed with other such cons-transforming functions.

来源：https://stackoverflow.com/questions/52034032/using-a-filtered-list-in-a-new-function-in-haskell

标签

list