On Oct 17, 10:06 am, "Diez B. Roggisch" <[EMAIL PROTECTED]> wrote: > Diez B. Roggisch wrote: > > Abandoned wrote: > > >> Hi.. > >> I have a dictionary like these: > >> a={'a': '1000', 'b': '18000', 'c':'40', 'd': '600'} ...... 100.000 > >> element > >> I want to sort this by value and i want to first 100 element.. > >> Result must be: > >> [b, a, d, c .....] ( first 100 element) > > >> I done this using FOR and ITERATOR but it tooks 1 second and this is > >> very big time to my project. > >> I want to learn the fastest method.. > > > That is the fastest method, unless you write your own ordered dict > > implementation that sorts by value. But that will most probably lose the > > time when holding up the invariant when inserting key/values into the > > dictionary. > > Actually, I somehow read the FOR and ITERATOR above as something like this: > > entries = sorted(a.items(), key=lambda v: v[1])[:100] > > The gist of my statement above is nontheless the same: if you want sorted > results, you need to sort... > > Diez
If you want the top 100 out of 100K, heapq.nlargest is more than an order of magnitude faster. George -- http://mail.python.org/mailman/listinfo/python-list