Multiple functions on one list?

Question

i have a list

[1,1,1,1,1]

and i am trying to write function which will return list

[2,3,4,5,6]

i want to use function map like this

map (+1) [1,1,1,1,1]

which will return

[2,2,2,2,2]

after that i want to call map function on last four elements of returned list so after i get [2,2,2,2,2] i want to use map on last four [2,2,2,2] that will return [3,3,3,3] and replace last four elements from first map call so i get [2,3,3,3,3] etc..

map (+1)[1,1,1,1,1] 
map (+1)  [2,2,2,2] 
map (+1)    [3,3,3] 
map (+1)      [4,4] 
map (+1)        [5]

returned:

[2,2,2,2,2]
[2,3,3,3,3] 
[2,3,4,4,4]
[2,3,4,5,5]
[2,3,4,5,6]

i need to return only last list... btw this is only Simplified version, originaly i have list of lists ... i just cant figure how to call function how i described.. thanks.

Should all immediate steps be returned? Or only the last list, [2,3,4,5,6] — kaan
– kaan, Commented Mar 18, 2013 at 16:47

Ingo · Accepted Answer · 2013-03-18 16:55:49Z

8

I think you want something like

mapTails f [] = []
mapTails f (x:xs) = f x : mapTails f (map f xs)

answered Mar 18, 2013 at 16:55

Ingo

36.4k6 gold badges57 silver badges102 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

leftaroundabout · Accepted Answer · 2013-03-18 17:26:10Z

4

IMO the most elegant way would be

zipWith($) $ iterate((+1).) id

answered Mar 18, 2013 at 17:26

leftaroundabout

121k6 gold badges185 silver badges337 bronze badges

2 Comments

nymk Over a year ago

Fantastic! I think it should be zipWith($) $ tail $ iterate((+1).) id.

ertes Over a year ago

I like this solution most. The others are either not generic enough or not very elegant. +1. Although I would use more haskelly syntax and no ($): zipWith ($) (iterate (succ .) succ) (replicate 5 1).

Daniel Lyons · Accepted Answer · 2013-03-18 16:54:27Z

3

scanl almost does what you want:

Prelude> scanl (+) 1 [1,1,1,1,1]
[1,2,3,4,5,6]

You could drop the first item, which is just the initial state value we're passing in:

Prelude> tail $ scanl (+) 1 [1,1,1,1,1]
[2,3,4,5,6]

answered Mar 18, 2013 at 16:54

Daniel Lyons

22.9k3 gold badges54 silver badges82 bronze badges

1 Comment

nymk Over a year ago

I don't think OP is seeking this function, though it yields the right answer for the case (+1). For other input function, e.g. (+2), it won't work.

Xophmeister · Accepted Answer · 2013-03-18 16:56:42Z

1

You can accomplish what you're looking for using a recursive function, instead:

myFn :: Num a => [a] -> [a]
myFn []     = []
myFn (x:xs) = x + 1 : (myFn $ map (+1) xs)

main = print $ myFn [1,1,1,1,1]  -- Prints [2,3,4,5,6]

See http://codepad.org/wBwynlGt

answered Mar 18, 2013 at 16:56

Xophmeister

9,2495 gold badges52 silver badges99 bronze badges

Comments

Ramon Snir · Accepted Answer · 2013-03-18 16:59:02Z

0

Would something like this do what you want?

startList = [1,1,1,1] -- orwhatever you want it to be 
map (\(x,i) -> x + i) $ zip startList [1..]

The zip basically pairs each element in the list with what you want to add to it, and the map function then adds each element in the list to that value to get the result you want.

edited Mar 18, 2013 at 16:59

Ramon Snir

7,5603 gold badges45 silver badges62 bronze badges

answered Mar 18, 2013 at 16:50

tredontho

1,28110 silver badges14 bronze badges

Comments

kyticka · Accepted Answer · 2013-03-18 17:01:36Z

0

Your algorithm version O(n²) time:

plusSlow :: [Int] -> [Int]
plusSlow [] = []
plusSlow (x:xs) = (head mapped):(plusSlow $ tail mapped)
                 where mapped = map (+1) (x:xs)

Faster version O(n) time:

plusFast :: [Int] -> [Int]
plusFast x = pf x 1

pf :: [Int] -> Int -> [Int]
pf [] _ = []
pf (x:xs) n = (x+n):(pf xs (n+1))

edited Mar 18, 2013 at 17:01

answered Mar 18, 2013 at 16:55

kyticka

6149 silver badges19 bronze badges

Collectives™ on Stack Overflow

Multiple functions on one list?

6 Answers 6

Comments

2 Comments

1 Comment

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

6 Answers 6

Comments

2 Comments

1 Comment

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related