space-efficient top-N algorithm

[John Machin]

b_mcerlean at yahoo.com (Brian McErlean) wrote in message news:<f040c4f9.0302092300.79757ffd at posting.google.com>...

> I take it by the references to sorting and descending order that you're
> getting the top 50 elements by sorting the dict?  ie. something like:

I hope the OP is not doing this.

> l = [(d[url],url) for url in d]
> l.sort()

# l is now in ascending order by count

> top50 = l[:50]

# and you now take the first 50, which are the ones with the smaller counts.

> for count,url in top50:
>     print url, count