Map list of functions on list in Haskell

Question

I have a problem with implementing FP (Backus) interpreter in Haskell.

FP functions look like this:

[+,*]:<2,3>

should result with

<+:<2,3>, *:<2,3>> ->
<5,6>

meaning that every function from list on left side should be performed on every every element of a list on a right side.

What I understand is that I need something similar to "map" function, but map applies one function on a list, and I need list of functions on a list of values.

Thank you in advance! :-)

EDIT:

Since I wasn't precise, here is my code that doesn't work:

apply :: [String] -> [Integer] -> [Integer]
apply fs v = [((apply_f f x) | f <- fs | x <- v)]

apply_f :: String -> [Integer] -> [Integer]
apply_f "+" v = [(sum v)]
apply_f "*" v = [(product v)]

I can't figure out why... :-(

EDIT 2:

Sorry, I was too tired from working all day on this. Problem is that I don't need second pipe, just first one:

apply fs v = [ apply_f f v | f <- fs ]

Now, everything works just fine, thank you very much! :-)

What exactly do you mean with "doesn't work"?

Nymer
– Nymer

2012-08-25 18:11:39 +00:00
Commented Aug 25, 2012 at 18:11 — Nymer
– Nymer, Commented Aug 25, 2012 at 18:11

Dervall · Accepted Answer · 2012-08-25 17:46:17Z

18

I believe you are looking for the zipWith function and apply it with the function application operator $.

So, if you have a list of functions funcList and a list of values valueList you would call this with:

zipWith ($) funcList valueList

So, using this would be something like

zipWith ($) [(+ 5),(* 3)] [1,5]

gives you the result [6,15]

To get a sort of cross application you could use

[f a | f <- [(+5), (*3)], a <- [1,5]]

this gives you [6,10,3,15]. Not sure what you mean by the `<, do you want pairs, nested lists or what exactly do you need?

edited Aug 25, 2012 at 17:46

answered Aug 25, 2012 at 17:21

Dervall

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3 Comments

markoub Over a year ago

No, no not exactly what I had in mind. In FP each function on the left side is applied on EACH value of the list on the right side. So, result of that function you mentioned in FP would be: [+5,*3]:<1,5> -> < <6,11>, <3,15> >

Dervall Over a year ago

I see... Let me see what I can do for you :)

markoub Over a year ago

Sorry for "<>" confusion. That's just how FP represents lists. So, just think of it as normal "[]"

Lee · Accepted Answer · 2012-08-25 17:36:56Z

17

It sounds like you want:

import Control.Applicative
apply :: [a -> b] -> [a] -> [b]
apply fs vals = fs <*> vals

Of course this is just the same as the instance of <*> defined for lists.

answered Aug 25, 2012 at 17:36

Lee

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1 Comment

jplatte Over a year ago

Or just apply = (<*>) for the third line (making it obvious it's exactly the same)

huon · Accepted Answer · 2012-08-25 19:53:23Z

6

Do you want something like this?

[f x | f <- [(+3),(*2)], x <- [1..2]]

Output:

[4,5,2,4]

EDIT:

So something like this?

[[f x | f <- [(+3),(*2)]] | x <- [1..2]]
[[4,2],[5,4]]

edited Aug 25, 2012 at 19:53

huon

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answered Aug 25, 2012 at 17:42

Nymer

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4 Comments

markoub Over a year ago

similar, but [ [4,5] , [2,4] ].

Dervall Over a year ago

+1 I think you hit the correct wanted result at the end there :)

markoub Over a year ago

It definitely looks correct, but for some reason it doesn't work... :-( here is my code: apply :: [String] -> [Integer] -> [String] apply fs v = [((apply_f f x) | f <- fs | x <- v)] apply_f :: String -> [Integer] -> [Integer] apply_f "+" v = [(sum v)] apply_f "*" v = [(product v)]

markoub Over a year ago

I posted code in original question, since it is not correctly formatted in comment... :-(

Rotsor · Accepted Answer · 2012-08-25 17:53:44Z

0

I can't figure out anything from your explanation, but here is a possible code, which gives the correct answer for your example:

> map (uncurry foldr1) [((+), [2,3]), ((*), [2,3])]
[5,6]

answered Aug 25, 2012 at 17:53

Rotsor

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Comments

Oleksii Kachaiev · Accepted Answer · 2012-08-25 17:39:10Z

-1

I don't know if it's actually that you need, but

Prelude> [f 2 3 | f <- [(+), (*)]]
[5,6]

If you need lists of <2,3> pairs, you can use something like

Prelude> [zipWith f [2, 20] [3, 30] | f <- [(+), (*)]]
[[5,50],[6,600]]

answered Aug 25, 2012 at 17:39

Oleksii Kachaiev

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