Am 02.01.2011 19:19, schrieb Emile van Sebille:
On 1/2/2011 8:31 AM Stefan Sonnenberg-Carstens said...A last one: l = [1, 2, 3, 4, 5, 6, 7, 'a', 8, 'b'] dict((x[1],x[0]) for x in ((l.pop(),l.pop()) for x in xrange(len(l)/2)))This also works: l = [1, 2, 3, 4, 5, 6, 7, 'a', 8, 'b'] pop=l.pop dict([(pop(),pop()) for i in l])
No, it does not:
>>> l = [1, 2, 3, 4, 5, 6, 7, 'a', 8, 'b']
>>>
>>> pop=l.pop
>>>
>>> dict([(pop(),pop()) for i in l])
{'a': 7, 'b': 8, 4: 3, 6: 5}
>>>
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