Simon Mullis napisał(a): > Something like: > > dict_of_counts = dict([(v[0:3], "count") for v in l]) > > I can't seem to figure out how to get "count", as I cannot do x += 1 > or x++ as x may or may not yet exist, and I haven't found a way to > create default values.
It seems to me that the "count" you're looking for is the number of elements from l whose first 3 characters are the same as the v[0:3] thing. So you may try: >>> dict_of_counts = dict((v[0:3], sum(1 for x in l if x[:3] == v[:3])) for v >>> in l) But this isn't particularly efficient. The 'canonical way' to construct such histograms/frequency counts in python is probably by using defaultdict: >>> dict_of_counts = collections.defaultdict(int) >>> for x in l: >>> dict_of_counts[x[:3]] += 1 Regards, Marek -- http://mail.python.org/mailman/listinfo/python-list
