Difference between revisions of "99 questions/Solutions/16"
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< 99 questions  Solutions
(a slight variation using guards) 

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</haskell> 
</haskell> 

−  +  An alternative iterative solution: 

+  
<haskell> 
<haskell> 

dropEvery :: [a] > Int > [a] 
dropEvery :: [a] > Int > [a] 

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</haskell> 
</haskell> 

−  +  Yet another iterative solution which divides lists using Prelude: 

+  
<haskell> 
<haskell> 

dropEvery :: [a] > Int > [a] 
dropEvery :: [a] > Int > [a] 

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</haskell> 
</haskell> 

−  or using zip: 

+  A similar approach using guards: 

+  
+  <haskell> 

+  dropEvery :: [a] > Int > [a] 

+  dropEvery xs n 

+   length xs < n = xs 

+   otherwise = take (n1) xs ++ dropEvery (drop n xs) n 

+  </haskell> 

+  
+  Using zip: 

+  
<haskell> 
<haskell> 

dropEvery n = map snd . filter ((n/=) . fst) . zip (cycle [1..n]) 
dropEvery n = map snd . filter ((n/=) . fst) . zip (cycle [1..n]) 
Revision as of 20:01, 15 July 2010
(**) Drop every N'th element from a list.
dropEvery :: [a] > Int > [a]
dropEvery [] _ = []
dropEvery (x:xs) n = dropEvery' (x:xs) n 1 where
dropEvery' (x:xs) n i = (if (n `divides` i) then
[] else
[x])
++ (dropEvery' xs n (i+1))
dropEvery' [] _ _ = []
divides x y = y `mod` x == 0
An alternative iterative solution:
dropEvery :: [a] > Int > [a]
dropEvery list count = helper list count count
where helper [] _ _ = []
helper (x:xs) count 1 = helper xs count count
helper (x:xs) count n = x : (helper xs count (n  1))
Yet another iterative solution which divides lists using Prelude:
dropEvery :: [a] > Int > [a]
dropEvery [] _ = []
dropEvery list count = (take (count1) list) ++ dropEvery (drop count list) count
A similar approach using guards:
dropEvery :: [a] > Int > [a]
dropEvery xs n
 length xs < n = xs
 otherwise = take (n1) xs ++ dropEvery (drop n xs) n
Using zip:
dropEvery n = map snd . filter ((n/=) . fst) . zip (cycle [1..n])