Replacing empty lists with a value

Question

I have a function that returns a list like this:

[ [1, 2, 3], [], [5], [5,6], []]

But I want to replace the empty lists with 0, so that it looks like this

[ [1, 2, 3], [0], [5], [5,6], [0]]

So far I have tried filter and map with little success. Can someone please point me in the right direction?

Here is the code in question:

knightPlace:: [Int] -> [[Int]]
knightPlace n = makeboard n

    where
    makeboard n =  [x | i<-[0..(length n -1 )], x <- [checkPos i]]
    -- checkPos add zero
    checkPos i = [j+1 | j<-[0..(length n -1 )], queenFilter n i j]
    -- filters all rows, cols and diags, also knights
    queenFilter n i j = and [n!!(i) == 0 && n!!(k) /=(j+1) && (n!!(k)==0 || (abs(n!!(k)-(j+1))/=abs(i-k))) && (n!!(k)==0 ||not( ((abs(i-k)==2)&& (abs(n!!(k)-(j+1))==1)) ||((abs(i-k)==1)&& (abs(n!!(k)-(j+1))==2))))       | k<-[0..(length n - 1)] ]

Called like

 knightPlace [0, 0, 6, 0, 0, 4, 0, 0]

Sarah · Accepted Answer · 2015-04-19 23:11:25Z

5

Consider that if you have a function foo that can transform [] to [0] and return every other list untouched, map foo is the final function that you want.

answered Apr 19, 2015 at 23:11

Sarah

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Comments

elm · Accepted Answer · 2015-04-20 06:59:35Z

2

Replace xs with ys in l,

replaceWith :: [a] -> [a] -> [[a]]
replaceWith xs ys l = map (\x -> if x == xs then ys else x) l

In the case here,

replaceWith [] [0] [[1, 2, 3],[],[5],[5,6],[]]

edited Apr 20, 2015 at 6:59

answered Apr 20, 2015 at 4:29

elm

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Comments

mariop · Accepted Answer · 2015-04-19 23:17:57Z

2

Change your function with

knightPlace n = map (\x -> if null x then [0] else x) $ makeboard n

Also consider using pointfree style

knightPlace = map (\x -> if null x then [0] else x) . makeboard

answered Apr 19, 2015 at 23:17

mariop

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2 Comments

kqr Over a year ago

Did you try to compile that last function in the comments? You should get a scope error. ;)

mariop Over a year ago

Oh right, I write it while on the train and didn't check it.

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