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To: python-list@python.org
From: Mark Lawrence <breamoreboy@yahoo.co.uk>
Subject: Re: Help me write better Code
Date: Wed, 09 Jul 2014 16:44:01 +0100
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On 09/07/2014 15:27, sssdevelop wrote:
> Hello,
>
> I have working code - but looking for better/improved code. Better coding practices, better algorithm :)
>
> Problem: Given sequence of increasing integers, print blocks of consecutive integers.
>
> Example:
>
> Input: [10, 11, 12, 15]
> Output: [10, 11, 12]
>
> Input: [51, 53, 55, 67, 68, 91, 92, 93, 94, 99]
> Outout: [67, 68], [91, 92, 93, 94]
>
> My code looks as below:
> -----------------------------
> #!/usr/bin/python
> a = [51, 53, 55, 67, 68, 91, 92, 93, 94, 99]
> #a = []
> #a = [10]
> #a = [10, 11, 12, 15]
> print "Input: "
> print  a
>
> prev = 0
> blocks = []
> tmp = []
> last = 0
> for element in a:
>     if prev == 0:
>        prev = element
>        next
>     if element == prev + 1:
>         if tmp:
>             pass
>         else:
>             tmp.append(prev)
>         tmp.append(element)
>     else:
>         if tmp:
>            blocks.append(tmp)
>         tmp = []
>
>     prev = element
>
> if tmp:
>      blocks.append(tmp)
>
> if blocks:
>      #print "I have repeated elements and those are:"
>      for b in blocks:
>          print b
>
> -----------------------
>
> thank you in advance!
>

Adopted from here https://docs.python.org/3.0/library/itertools.html

data = [51, 53, 55, 67, 68, 91, 92, 93, 94, 99]
for k, g in groupby(enumerate(data), lambda t:t[0]-t[1]):
     group = list(map(operator.itemgetter(1), g))
     if len(group) > 1:
         print(group)

 >>>
[67, 68]
[91, 92, 93, 94]
 >>>

-- 
My fellow Pythonistas, ask not what our language can do for you, ask 
what you can do for our language.

Mark Lawrence

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