On 06/07/2010 07:45 PM, Dan Stromberg wrote: > > 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 this line is shouting to be replaced with #!/usr/bin/env 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 > <mailto: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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