Pythonic search of list of dictionaries

Skip Montanaro skip at
Tue Jan 4 19:14:06 CET 2005

    Bulba> I put those dictionaries into the list:

    Bulba>    oldl=[x for x in orig]  # where orig=csv.DictReader(ofile ...

    Bulba> ..and then search for matching source terms in two loops:

    Bulba>    for o in oldl:
    Bulba>        for n in newl:
    Bulba>            if n['English'] == o['English']:
    Bulba>            ...

    Bulba> Now, this works. However, not only this is very un-Pythonic, but
    Bulba> also very inefficient: the complexity is O(n**2), so it scales up
    Bulba> very badly.

How about using sets?

    oenglish = set([item['English'] for item in oldl])
    nenglish = set([item['English'] for item in newl])

    matching = oenglish & nenglish

Once you have those that match, you can constrain your outer loop to just
those cases where

    o['English'] in matching

If you're not using 2.4 yet, then get sets via:

    from sets import Set as set

That's still not all that Pythonic, but should be a bit faster.

You might want to sort your lists by the 'English' key.  I don't know how to
use the new key arg to list.sort(), but you can still do it the
old-fashioned way:

    oldl.sort(lambda a,b: cmp(a['English'], b['English']))
    newl.sort(lambda a,b: cmp(a['English'], b['English']))

Once sorted, you can then march through the lists in parallel, which should
give you an O(n) algorithm.  


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