Call me strange, but I regard this as a good place to use a functional style - IE, to use reduce, and furthermore I regard this as a good example of why reduce is useful for more than just summing numbers:
#!/disc/gx/sdfw/dans/python26/bin/python import collections def count_first_letters(dictionary, string_): dictionary[string_[0]] += 1 return dictionary print reduce( count_first_letters, ['abc', 'def', 'air', 'ghi'], collections.defaultdict(int), ) On Sat, Jun 5, 2010 at 7:01 PM, MRAB <pyt...@mrabarnett.plus.com> wrote: > GZ wrote: > >> Hi, >> >> I am looking for a fast internal vector representation so that >> (a1,b2,c1)+(a2,b2,c2)=(a1+a2,b1+b2,c1+c2). >> >> So I have a list >> >> l = ['a'a,'bb','ca','de'...] >> >> I want to count all items that start with an 'a', 'b', and 'c'. >> >> What I can do is: >> >> count_a = sum(int(x[1]=='a') for x in l) >> count_b = sum(int(x[1]=='b') for x in l) >> count_c = sum(int(x[1]=='c') for x in l) >> >> But this loops through the list three times, which can be slow. >> >> I'd like to have something like this: >> count_a, count_b, count_c = >> sum( (int(x[1]=='a',int(x[1]=='b',int(x[1]=='c') for x in l) >> >> I hesitate to use numpy array, because that will literally create and >> destroy a ton of the arrays, and is likely to be slow. >> >> If you want to do vector addition then numpy is the way to go. However, > first you could try: > > from collections import defaultdict > counts = defaultdict(int) > for x in l: > counts[x[0]] += 1 > > (Note that in Python indexes are zero-based.) > > -- > http://mail.python.org/mailman/listinfo/python-list >
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