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primarykey
Id
16861883
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0
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2013-05-31T16:30:42.130
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2013-05-31T16:43:23.097
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2013-05-31T16:43:23.097
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2407038
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1
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text
Body
From what I understand, you want to take a list of type <code>[[(Int, Int)]]</code> and sum the first element of any tuple that has the same second element. You can completely disregard that it is a list of lists and just <code>concat</code> right away, unless the structure of the lists can help you determine which elements should be added (but I have not seem any evidence that would indicate that). Then you get the minimum <code>snd</code>, sort the list by the <code>snd</code> values, and recurse through the list, calling <code>span :: (a -> Bool) -> [a] -> ([a], [a])</code> which takes a predicate and splits a list at the last element which satisfies the predicate. Since the list is sorted, this will capture all of the tuples with same <code>snd</code>. Then we fold over that list of tuples. <pre><code>add :: [[(Int, Int)]] -> [(Int, Int)] add list = addT (minBound :: Int) $ sortBy sortF (concat list) where sortF (_,x) (_,y) | x < y = LT | x > y = GT | x == y = EQ addT n list = sumFunc $ span (sndIs n) list where sndIs n (_,y) = y == n sumFunc ([],b)= addT (n+1) b sumFunc (a,[])= [(foldr (\(x,y) (w,v) -> (x+w,y)) (0,0) a)] sumFunc (a,b) = (foldr (\(x,y) (w,v) -> (x+w,y)) (0,0) a):(addT (n+1) b) </code></pre> then : <pre><code>> add [[(1,2), (3,0)], [(3,1), (7,0)]] > [(10,0),(3,1),(1,2)] </code></pre> The only problem is the complexity is very bad (try <code>add [[(43, minBound :: Int), (10, maxBound :: Int)]]</code>). If you want to make it faster, you should get all the <code>snd</code>s first and only call addT with those, if you are likely to be using this function on values that differ greatly.
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1. POAdding the first element lists of tuples
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1. PTAnswer
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1. USuser2407038
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1. USuser2407038
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