Trimming dictionaries

Question

I'd like to know if there is a more simple/pythonic way to trim the "size/length" of a dictionary in python. If I have a dict with 10 key-value-pairs(elements) and I'd like to restrict the size to be 5. Is deleting elements in a loop the best solution (order/identity does not matter)

def _trim_search_results(self):

    MAX_RESULTS = 5

    # a big dict
    results_to_be_trimmed = {result_1 ... result_n}

    # dict with length MAX_RESULTS
    trimmed_results = {}

    for index, key in enumerate(results_to_be_trimmed):
        if index == MAX_RESULTS:
            break
        else:
            trimmed_results[key] = results_to_be_trimmed[key]

    results_to_be_trimmed = trimmed_results

I think there must be a better solution ...

Answer 1

You could try

d = dict(d.items()[:MAX_RESULTS])

In Python 3:

d = dict(list(d.items())[:MAX_RESULTS])

Answer 2

You can use itertools.islice on dict.iteritems.

dict.iteritems() returns an iterator in py2.x, you can slice that iterator using itertools.islice and pass it to dict() to get the new dict.

Demo:

>>> from itertools import islice
>>> d = dict.fromkeys(range(10))
>>> dict(islice(d.iteritems(), 5))
{0: None, 1: None, 2: None, 3: None, 4: None}

Timings:

>>> d = dict.fromkeys(range(100))
>>> %timeit from itertools import islice;dict(islice(d.iteritems(), 10)) #winner
10000 loops, best of 3: 10.7 us per loop
>>> %timeit dict(d.items()[0: 10])
100000 loops, best of 3: 10.9 us per loop

>>> d = dict.fromkeys(range(10**5))
>>> %timeit from itertools import islice;dict(islice(d.iteritems(), 1000)) #winner
1000 loops, best of 3: 106 us per loop
>>> %timeit dict(d.items()[0: 1000])
100 loops, best of 3: 20 ms per loop