vector addition

[Dan Stromberg]

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 <python at 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
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://mail.python.org/pipermail/python-list/attachments/20100607/e9967751/attachment-0001.html>