Inefficient summing

[Chris Rebert]

I personally would probably do:

from collections import defaultdict

label2sum = defaultdict(lambda: 0)
for r in rec:
    for key, value in r.iteritems():
        label2sum[key] += value

ratio = label2sum["F1"] / label2sum["F2"]

This iterates through each 'r' only once, and (imho) is pretty
readable provided you know how defaultdicts work. Not everything has
to unnecessarily be made a one-liner. Coding is about readability
first, optimization second. And optimized code should not be
abbreviated, which would make it even harder to understand.

I probably would have gone with your second solution if performance
was no object.

Cheers,
Chris
-- 
Follow the path of the Iguana...
http://rebertia.com

On Wed, Oct 8, 2008 at 1:23 PM, beginner <zyzhu2000 at gmail.com> wrote:
> Hi All,
>
> I have a list of records like below:
>
> rec=[{"F1":1, "F2":2}, {"F1":3, "F2":4} ]
>
> Now I want to write code to find out the ratio of the sums of the two
> fields.
>
> One thing I can do is:
>
> sum(r["F1"] for r in rec)/sum(r["F2"] for r in rec)
>
> But this is slow because I have to iterate through the list twice.
> Also, in the case where rec is an iterator, it does not work.
>
> I can also do this:
>
> sum1, sum2= reduce(lambda x, y: (x[0]+y[0], x[1]+y[1]), ((r["F1"],
> r["F2"]) for r in rec))
> sum1/sum2
>
> This loops through the list only once, and is probably more efficient,
> but it is less readable.
>
> I can of course use an old-fashioned loop. This is more readable, but
> also more verbose.
>
> What is the best way, I wonder?
>
>
> -a new python programmer
> --
> http://mail.python.org/mailman/listinfo/python-list
>