GZ
Guest
Posts: n/a

 06-06-2010
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.

Chris Rebert
Guest
Posts: n/a

 06-06-2010
On Sat, Jun 5, 2010 at 6:20 PM, GZ <(E-Mail Removed)> 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 don't really get how that relates to vectors or why you'd use that
representation, and it looks like you're forgotten that Python uses
0-based indexing, but anyway, here's my crack at something more
efficient:

from collections import defaultdict

letter2count = defaultdict(int)
for item in l:
initial = item[0]
letter2count[initial] += 1

count_a = letter2count['a']
count_b = letter2count['b']
count_c = letter2count['c']

Cheers,
Chris
--
http://blog.rebertia.com

MRAB
Guest
Posts: n/a

 06-06-2010
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.)