StackOverflow2013

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PO
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Id
12876438
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0
AnswerCount
0
ClosedDate
CommentCount
2
CommunityOwnedDate
CreationDate
2012-10-13T19:53:50.620
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0
LastActivityDate
2012-10-13T20:08:39.817
LastEditDate
2012-10-13T20:08:39.817
LastEditorUserId
371834
OwnerUserId
371834
ParentId
12876384
PostTypeId
2
Score
37
ViewCount
0
LastEditorDisplayName
text
Body
A quick search on <a href="http://www.haskell.org/hoogle/?hoogle=Int+-%3E+%5Ba%5D+-%3E+%5B%5Ba%5D%5D" rel="noreferrer">Hoogle</a> showed that there is no such function. On the other hand, it was replied that there is one in the <code>split</code> package, called <code>chunksOf</code>. However, you can do it on your own <pre><code>group :: Int -> [a] -> [[a]] group _ [] = [] group n l | n > 0 = (take n l) : (group n (drop n l)) | otherwise = error "Negative n" </code></pre> Of course, some parentheses can be removed, I left there here for understanding what the code does: The base case is simple: whenever the list is empty, simply return the empty list. The recursive case tests first if <code>n</code> is positive. If <code>n</code> is <code>0</code> or lower we would enter an infinite loop and we don't want that. Then we split the list into two parts using <a href="http://hackage.haskell.org/packages/archive/base/latest/doc/html/Prelude.html#v:take" rel="noreferrer"><code>take</code></a> and <a href="http://hackage.haskell.org/packages/archive/base/latest/doc/html/Prelude.html#v:drop" rel="noreferrer"><code>drop</code></a>: <code>take</code> gives back the first <code>n</code> elements while <code>drop</code> returns the other ones. Then, we add the first <code>n</code> elements to the list obtained by applying our function to the other elements in the original list. <img src="https://i.stack.imgur.com/voD4C.png" alt="enter image description here">
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1. POGrouping a list into lists of n elements in Haskell
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1. PTAnswer
UserLastEditorUserId
1. USMihai Maruseac
UserOwnerUserId
1. USMihai Maruseac
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1. POGrouping a list into lists of n elements in Haskell
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1. COGreat diagram! +1
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 USAndrewC

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