How to create a new list of lists from a list of lists in Haskell?

Question

I've a field made of list of lists like this:

let list1 = ["a", "abcdefgh", "abc"]

I want to get a new field 5x5 such that there could be more then 5 lines or columns, but no less. It must be produced from the previous one by addition spaces ' ' or keeping them as they are, if there are more then 5 symbols. For additional lines no spaces needed. For example:

list2 = ["a    ", "abcdefgh", "abc  ", "     ", "     "]

or:

let list1 = ["a", "abcdefgh", "c", "d", "e", "f"]
list2 = ["a    ", "abcdefgh", "c    ", "d    ", "e    ", "f"]

I thought about map, replicate, and length, but I'm a little bit confused. If it's possible, don't use libs (or use Data.Map only). Thank you

I made sth. like this: (map (++ (replicate 5 ' ')) linesOfFiles) ++ (replicate (5- length linesOfFiles) (replicate 5' ')) But that's not what I want — TheWorstHaskellProger
– TheWorstHaskellProger, Commented Apr 3, 2021 at 9:36
Your map will each time add five spaces to each line. You should determine the length of the string, and append accordingly... — willeM_ Van Onsem
– willeM_ Van Onsem, Commented Apr 3, 2021 at 9:37
@WillemVanOnsem Yes, I know, but I don't know how I can determine the length of the string inside the map function — TheWorstHaskellProger
– TheWorstHaskellProger, Commented Apr 3, 2021 at 9:39

willeM_ Van Onsem · Accepted Answer · 2021-04-03 09:51:44Z

1

You need to determine the length of the strings. You can do this for example with a lambda expression, so then the mapping looks like:

func :: [String] -> [String]
func mylines = map (\x -> … ) linesOfFiles ++ replicate (5-length mylines) (replicate 5' ')

where x is thus string that will be mapped, and … is an expression to what that value is mapped. I leave filling in … as an exercise.

Usually it is better not to work with length: length takes linear time, and for infinite lists, this will even get the program into an infinite loop.

You can work with recursion to perform padding, something like:

padding :: Int -> a -> [a] -> [a]
padding 0 _ xs = …
padding n x [] = …
padding n x (x:xs) = …

edited Apr 3, 2021 at 9:51

answered Apr 3, 2021 at 9:46

willeM_ Van Onsem

482k33 gold badges483 silver badges624 bronze badges

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

4 Comments

TheWorstHaskellProger Over a year ago

Thank you, I already understood this after your first comment. I'm just trying to do it now

TheWorstHaskellProger Over a year ago

Thanks again, I've not so much pracrice in lambda usage.I know about the linear time, but I get the list form the input file, so it doesn't take so much time. I made: (map (\x -> (x ++ (replicate (5 - (length x)) ' '))) linesOfFiles) ++ (replicate (5- length linesOfFiles) (replicate 5' '))

Jon Purdy Over a year ago

@TheWorstHaskellProger: Instead of computing the number of spaces, one way to avoid length is to think of the problem as: 1. pad the input with spaces in both directions, then 2. trim it to the desired size with take. Look at the result of take 5 ("abc" ++ repeat ' '), and consider how you could use this pattern in your code. Also, notice how take 5 (x ++ repeat ' ') is different from x ++ replicate (5 - length x) ' ' when length x is greater than 5, as in "abcdefg". Your question and your sample input/output are a little unclear about which of these results you want.

TheWorstHaskellProger Over a year ago

@JonPurdy Sorry, I made a mistake when I wrote "there could be more then 5 lines or columns, but no more" I should have written "no less". So you see, replicate (length (-10)) ' ' will not change my line, but take const will cut my lines. I'll correct the error right now. Thank you for the answer

Collectives™ on Stack Overflow

How to create a new list of lists from a list of lists in Haskell?

1 Answer 1

4 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

4 Comments

Your Answer

Sign up or log in

Post as a guest

Related