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Groups > comp.lang.python > #74309
| Newsgroups | comp.lang.python |
|---|---|
| Date | 2014-07-10 07:39 -0700 |
| References | <3a608dd2-d8bf-429e-af21-ce5ac8f18272@googlegroups.com> <mailman.11695.1404920643.18130.python-list@python.org> |
| Message-ID | <63a40b77-f3d5-45b4-bd11-ff9c099bdb67@googlegroups.com> (permalink) |
| Subject | Re: Help me write better Code |
| From | sssdevelop <sssdevelop@gmail.com> |
Mark - thank you so much. You have suggested be new best tool/module. It's going to help me many places. Was not aware of such powerful tool. thank you, On Wednesday, July 9, 2014 9:14:01 PM UTC+5:30, Mark Lawrence wrote: > 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 > > > > --- > > This email is free from viruses and malware because avast! Antivirus protection is active. > > http://www.avast.com
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Help me write better Code sssdevelop <sssdevelop@gmail.com> - 2014-07-09 07:27 -0700
Re: Help me write better Code Mark Lawrence <breamoreboy@yahoo.co.uk> - 2014-07-09 16:44 +0100
Re: Help me write better Code sssdevelop <sssdevelop@gmail.com> - 2014-07-10 07:39 -0700
Re: Help me write better Code Mark Lawrence <breamoreboy@yahoo.co.uk> - 2014-07-10 15:51 +0100
Re: Help me write better Code Rustom Mody <rustompmody@gmail.com> - 2014-07-10 11:38 -0700
Re: Help me write better Code Ian Kelly <ian.g.kelly@gmail.com> - 2014-07-09 12:16 -0600
Re: Help me write better Code sssdevelop <sssdevelop@gmail.com> - 2014-07-10 07:38 -0700
Re: Help me write better Code Terry Reedy <tjreedy@udel.edu> - 2014-07-09 14:46 -0400
Re: Help me write better Code sssdevelop <sssdevelop@gmail.com> - 2014-07-10 07:38 -0700
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