Question

Suppose I have the following list:

 ls = ['a', 'b', 'c', 'd']

I get a combination using

 list(itertools.combinations(iterable, 2))  

 >>> [('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]

What I’d like to do is break this combination into subsets, such that the first member of each tuple in the subset is the same:

 subset1: [('a', 'b'), ('a', 'c'), ('a', 'd')]
 subset2: [('b', 'c'), ('b', 'd'), 
 subset3: [('c', 'd')]

Any ideas?

Answer 1

>>> import itertools as it
>>> ls = ['a', 'b', 'c', 'd']
>>> ii=it.groupby( it.combinations(ls, 2), lambda x: x[0] )
>>> for key, iterator in ii:
...     print key, list(iterator)
... 
a [('a', 'b'), ('a', 'c'), ('a', 'd')]
b [('b', 'c'), ('b', 'd')]
c [('c', 'd')]

If you don’t like lambda, you could use operator.itemgetter(0) instead of lambda x: x[0].