Hi, The problem that I wish to solve is to divide a (sored) list of integers into sublists such that each sublist contains numbers in consecutive sequence. For example, *Input:* [1,2,3,7,8,10,11,12] *Output:* [[1,2,3],[7,8],[10,11,12]] I have written the following code and it does the trick. -- Take a list and divide it at first point of non-consecutive number encounter divide :: [Int] -> [Int] -> ([Int], [Int]) divide first [] = (first, []) divide first second = if (last first) /= firstSecond - 1 then (first, second) else divide (first ++ [firstSecond]) (tail second) where firstSecond = head second -- Helper for breaking a list of numbers into consecutive sublists breakIntoConsecsHelper :: [Int] -> [[Int]] -> [[Int]] breakIntoConsecsHelper [] [[]] = [[]] breakIntoConsecsHelper lst ans = if two == [] then ans ++ [one] else ans ++ [one] ++ breakIntoConsecsHelper two [] where firstElem = head lst remaining = tail lst (one, two) = divide [firstElem] remaining -- Break the list into sublists of consective numbers breakIntoConsecs :: [Int] -> [[Int]] breakIntoConsecs lst = breakIntoConsecsHelper lst [[]] -- Take the tail of the result given by the function above to get the required list of lists. However, I was wondering if there was a better way of doing this. Any help would be highly appreciated. Thank you. Best Regards, Saqib Shamsi
On Fri, Jan 13, 2017 at 09:35:54PM +0530, Saqib Shamsi wrote:
The problem that I wish to solve is to divide a (sored) list of integers into sublists such that each sublist contains numbers in consecutive sequence.
For example, *Input:* [1,2,3,7,8,10,11,12] *Output:* [[1,2,3],[7,8],[10,11,12]]
[...]
However, I was wondering if there was a better way of doing this. Any help would be highly appreciated.
Hello Saquib, you could try using a 'trick' like this: λ> zipWith (-) [1,2,3,7,8,10,11,12] (enumFrom 1) [0,0,0,3,3,4,4,4] Now you have an 'helper' list which can be glued to the first one with zip λ> zip [1,2,3,7,8,10,11,12] it [(1,0),(2,0),(3,0),(7,3),(8,3),(10,4),(11,4),(12,4)] and now grouped by using `groupBy` in Data.List. Does that help?
The "helper list" technique is ingenious, but seems very specific to Int as
the type within the list.
I have had other tasks that seem to fit a more generalized problem
statement, of which the original question appears to me as special case:
Given a criterion of type a -> a -> Bool and a list [a], produce a list of
lists [[a]] in which sub-lists are made up of consecutive elements from the
original list that satisfy the criterion.
It appears to me that List.groupBy may meet that need, but I'm not able to
verify that at the moment (and would be glad of feedback).
-jn-
On Fri, Jan 13, 2017 at 10:55 AM, Francesco Ariis
On Fri, Jan 13, 2017 at 09:35:54PM +0530, Saqib Shamsi wrote:
The problem that I wish to solve is to divide a (sored) list of integers into sublists such that each sublist contains numbers in consecutive sequence.
For example, *Input:* [1,2,3,7,8,10,11,12] *Output:* [[1,2,3],[7,8],[10,11,12]]
[...]
However, I was wondering if there was a better way of doing this. Any help would be highly appreciated.
Hello Saquib, you could try using a 'trick' like this:
λ> zipWith (-) [1,2,3,7,8,10,11,12] (enumFrom 1) [0,0,0,3,3,4,4,4]
Now you have an 'helper' list which can be glued to the first one with zip
λ> zip [1,2,3,7,8,10,11,12] it [(1,0),(2,0),(3,0),(7,3),(8,3),(10,4),(11,4),(12,4)]
and now grouped by using `groupBy` in Data.List.
Does that help? _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
-- Beauty of style and harmony and grace and good rhythm depend on simplicity. - Plato
Had a chance to chat with ghci, so earlier conjecture not confirmed:
Prelude Data.List> groupBy (\x y -> x == y-1) [1,2,3,7,8,10,11,12]
[[1,2],[3],[7,8],[10,11],[12]]
So close but no cigar.
On Fri, Jan 13, 2017 at 10:05 AM, Saqib Shamsi
Hi,
The problem that I wish to solve is to divide a (sored) list of integers into sublists such that each sublist contains numbers in consecutive sequence.
For example, *Input:* [1,2,3,7,8,10,11,12] *Output:* [[1,2,3],[7,8],[10,11,12]]
I have written the following code and it does the trick.
-- Take a list and divide it at first point of non-consecutive number encounter divide :: [Int] -> [Int] -> ([Int], [Int]) divide first [] = (first, []) divide first second = if (last first) /= firstSecond - 1 then (first, second) else divide (first ++ [firstSecond]) (tail second) where firstSecond = head second
-- Helper for breaking a list of numbers into consecutive sublists breakIntoConsecsHelper :: [Int] -> [[Int]] -> [[Int]] breakIntoConsecsHelper [] [[]] = [[]] breakIntoConsecsHelper lst ans = if two == [] then ans ++ [one] else ans ++ [one] ++ breakIntoConsecsHelper two [] where firstElem = head lst remaining = tail lst (one, two) = divide [firstElem] remaining
-- Break the list into sublists of consective numbers breakIntoConsecs :: [Int] -> [[Int]] breakIntoConsecs lst = breakIntoConsecsHelper lst [[]]
-- Take the tail of the result given by the function above to get the required list of lists.
However, I was wondering if there was a better way of doing this. Any help would be highly appreciated.
Thank you. Best Regards, Saqib Shamsi
_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
-- Beauty of style and harmony and grace and good rhythm depend on simplicity. - Plato
Here's one that does what you want, doesn't require the list to be sorted, and groups together consecutive and equal terms: groupConsecutive :: (Enum a,Eq a) => [a] -> [[a]] groupConsecutive = foldr go [] where go x ls@(hd@(y:_):yss) | x == y || x == pred y = (x:hd):yss | otherwise = [x]:ls go x [] = [[x]] go x ([]:yss) = [x]:yss Then
groupConsecutive [1,2,3,7,8,10,11,12] [[1,2,3],[7,8],[10,11,12]]
groupConsecutive [1,2,2,3,2,3] [[1,2,2,3],[2,3]]
and
groupConsecutive "bookkeeper understudy" ["b","oo","kk","ee","p","e","r"," ","u","n","de","rstu","d","y"]
The third case of go will never be reached. If you use a type that is also an instance of Bounded, and if your list contains the minimum element of the type, you'll get a runtime error on the use of pred. For example:
groupConsecutive [True,False,True] *** Exception: Prelude.Enum.Bool.pred: bad argument
Graham On 13-Jan-2017 11:05 AM, Saqib Shamsi wrote:
Hi,
The problem that I wish to solve is to divide a (sored) list of integers into sublists such that each sublist contains numbers in consecutive sequence.
For example, *Input:* [1,2,3,7,8,10,11,12] *Output:* [[1,2,3],[7,8],[10,11,12]]
I have written the following code and it does the trick.
-- Take a list and divide it at first point of non-consecutive number encounter divide :: [Int] -> [Int] -> ([Int], [Int]) divide first [] = (first, []) divide first second = if (last first) /= firstSecond - 1 then (first, second) else divide (first ++ [firstSecond]) (tail second) where firstSecond = head second
-- Helper for breaking a list of numbers into consecutive sublists breakIntoConsecsHelper :: [Int] -> [[Int]] -> [[Int]] breakIntoConsecsHelper [] [[]] = [[]] breakIntoConsecsHelper lst ans = if two == [] then ans ++ [one] else ans ++ [one] ++ breakIntoConsecsHelper two [] where firstElem = head lst remaining = tail lst (one, two) = divide [firstElem] remaining
-- Break the list into sublists of consective numbers breakIntoConsecs :: [Int] -> [[Int]] breakIntoConsecs lst = breakIntoConsecsHelper lst [[]]
-- Take the tail of the result given by the function above to get the required list of lists.
However, I was wondering if there was a better way of doing this. Any help would be highly appreciated.
Thank you. Best Regards, Saqib Shamsi
_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
Sorry, I mean, "groups together consecutive and equal terms, that occur sequentially". Examples speak louder than general descriptions! On 13-Jan-2017 6:43 PM, Graham Gill wrote:
Here's one that does what you want, doesn't require the list to be sorted, and groups together consecutive and equal terms:
You could zip the original list to its own tail.
On Fri, Jan 13, 2017 at 4:09 PM, Graham Gill
Sorry, I mean, "groups together consecutive and equal terms, that occur sequentially". Examples speak louder than general descriptions!
On 13-Jan-2017 6:43 PM, Graham Gill wrote:
Here's one that does what you want, doesn't require the list to be sorted, and groups together consecutive and equal terms:
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participants (5)
-
Francesco Ariis -
Graham Gill -
Jeffrey Brown -
Joel Neely -
Saqib Shamsi