Implement traditional group-by as found in most other languages like clojure/scala/ruby/etc

[sbinq]

This default group-by logic is different from what is called groupby within itertools.

[aavezel]

from itertools import groupby
lst = ['1', '12', 'a', '123', 'ab']
keyfunc = lambda x: len(x)
for k, g in groupby(sorted(lst, key=keyfunc), key=keyfunc):
    print k
    print list(g)

[kachayev]

@aavezel The main problem with given solution is that we need to calculate keyfunc(item) twice of each item in lst (it can be problematic if keyfunc is something more sophisticated than len).

[sbinq]

@aavezel @kachayev yes, agree with @kachayev arguments against itertools.groupby. Also don't like the idea of having to write such boilerplate (well, not much - but still a little annoying) and convert manually to dict.

[aavezel]

@kachayev : you are mistake. itertools.groupby return iterators, and not calculate keys if it not needed. It is more suitable for principle of lazy calculation.

[kachayev]

@aavezel

1. sorted function returns a list, not iterable. So there is no room for lazy calculation.
2. In your code keyfunc will be evaluated for each element by sorted function (cause it's necessary to compare items on sorting) and then it be evaluated by groupby function to return you k value and find out the boundaries of each group.

[sbinq]

@kachayev updated code according to your review - styling issues and more explicit test

[kachayev]

@sbinq Thanks for your contribution!