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Today's Topics:
1. Better Code (Saqib Shamsi)
2. Re: Better Code (Francesco Ariis)
3. Re: Better Code (Joel Neely)
4. Re: Better Code (Joel Neely)
----------------------------------------------------------------------
Message: 1
Date: Fri, 13 Jan 2017 21:35:54 +0530
From: Saqib Shamsi <shamsi.sa...@gmail.com>
To: beginners@haskell.org
Subject: [Haskell-beginners] Better Code
Message-ID:
<CAKpYsCGKeWoHttqpZAF7jAkhe++PqLSfwZA820tA=je8hfi...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
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
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Message: 2
Date: Fri, 13 Jan 2017 17:55:16 +0100
From: Francesco Ariis <fa...@ariis.it>
To: beginners@haskell.org
Subject: Re: [Haskell-beginners] Better Code
Message-ID: <20170113165516.gb23...@casa.casa>
Content-Type: text/plain; charset=utf-8
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?
------------------------------
Message: 3
Date: Fri, 13 Jan 2017 11:40:57 -0600
From: Joel Neely <joel.ne...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Better Code
Message-ID:
<CAEEzXAjPaZ+F3-+3FDMy+DRdXX-Qvrqvw6=hwghijsreubw...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
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 <fa...@ariis.it> wrote:
> 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
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Message: 4
Date: Fri, 13 Jan 2017 13:14:56 -0600
From: Joel Neely <joel.ne...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Better Code
Message-ID:
<CAEEzXAg=a2-xJm+vu0cGA3ETkViZP=pzx8o31oocaamf5fh...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
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 <shamsi.sa...@gmail.com>
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
>
>
--
Beauty of style and harmony and grace and good rhythm depend on simplicity.
- Plato
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