Split a dictionary in half?

If you use python +3.3, and want your splitted dictionaries to be the same across different python invocations, do not use .items, since the hash-values of the keys, which determines the order of .items() will change between python invocations. See Hash randomization

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Here's a way to do it using an iterator over the items in the dictionary and itertools.islice:

import itertools

def splitDict(d):
    n = len(d) // 2          # length of smaller half
    i = iter(d.items())      # alternatively, i = d.iteritems() works in Python 2

    d1 = dict(itertools.islice(i, n))   # grab first n items
    d2 = dict(i)                        # grab the rest

    return d1, d2

This would work, although I didn't test edge-cases:

>>> d = {'key1': 1, 'key2': 2, 'key3': 3, 'key4': 4, 'key5': 5}
>>> d1 = dict(d.items()[len(d)/2:])
>>> d2 = dict(d.items()[:len(d)/2])
>>> print d1
{'key1': 1, 'key5': 5, 'key4': 4}
>>> print d2
{'key3': 3, 'key2': 2}

In python3:

d = {'key1': 1, 'key2': 2, 'key3': 3, 'key4': 4, 'key5': 5}
d1 = dict(list(d.items())[len(d)//2:])
d2 = dict(list(d.items())[:len(d)//2])

Also note that order of items is not guaranteed

d1 = {key: value for i, (key, value) in enumerate(d.viewitems()) if i % 2 == 0}
d2 = {key: value for i, (key, value) in enumerate(d.viewitems()) if i % 2 == 1}