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import itertools
a = itertools.combinations('ABCD',  2)
print(a)
# Es sollte eigentlich 'AB AC AD BC BD CD' herauskommen...

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for i in a:
    print i

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In [9]: itertools.combinations('ABCD', 2)
Out[9]: <itertools.combinations at 0x929189c>

In [10]: cs = itertools.combinations('ABCD', 2)

In [11]: next(cs)
Out[11]: ('A', 'B')

In [12]: next(cs)
Out[12]: ('A', 'C')

In [13]: next(cs)
Out[13]: ('A', 'D')

In [14]: next(cs)
Out[14]: ('B', 'C')

In [15]: next(cs)
Out[15]: ('B', 'D')

In [16]: next(cs)
Out[16]: ('C', 'D')

In [17]: next(cs)
---------------------------------------------------------------------------
StopIteration                             Traceback (most recent call last)
/home/bj/<ipython-input-17-2473a9b0b1c8> in <module>()
----> 1 next(cs)

StopIteration: 

In [18]: cs = itertools.combinations('ABCD', 2)

In [19]: for c in cs:
   ....:     print c
   ....: 
('A', 'B')
('A', 'C')
('A', 'D')
('B', 'C')
('B', 'D')
('C', 'D')

In [20]: cs = itertools.combinations('ABCD', 2)

In [21]: list(cs)
Out[21]: [('A', 'B'), ('A', 'C'), ('A', 'D'), ('B', 'C'), ('B', 'D'), ('C', 'D')]

Code: Alles auswählen

Python 3.3.0 (default, Sep 25 2013, 19:28:08) 
[GCC 4.7.2] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> from itertools import combinations
>>> pairs = map(''.join, combinations('ABCD', 2))
>>> print(*pairs)
AB AC AD BC BD CD