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] > -- Mark's version without the print. Which says in a different way, Terry's point-2 : > 2. Separate interface code that gets input and presents output from the > function that processes the increasing sequence. The function should not > care whether the ints come from a user, file, or calculation. $ python3 Python 3.4.1 (default, Jun 9 2014, 10:28:44) [GCC 4.8.3] on linux Type "help", "copyright", "credits" or "license" for more information. >>> from operator import itemgetter >>> from itertools import groupby >>> data = [51, 53, 55, 67, 68, 91, 92, 93, 94, 99] >>> [group for k,g in groupby(enumerate(data), lambda t:t[0]-t[1]) ... for group in [list(map(itemgetter(1), g))] ... if len(group) > 1 ... ... ] [[67, 68], [91, 92, 93, 94]] >>> You can replace the outermost '[]' with '()' -- https://mail.python.org/mailman/listinfo/python-list
