5 maximum values in a python dictionary

Solution 1:

No need to use iteritems and itemgetter. The dict's own get method works fine.

max(A, key=A.get)

Similarly for sorting:

sorted(A, key=A.get, reverse=True)[:5]

Finally, if the dict size is unbounded, using a heap will eventually be faster than a full sort.

import heapq
heapq.nlargest(5, A, key=A.get)

For more information, have a look at the heapq documentation.

Solution 2:

You are close. You can sort the list using sorted ^[docs] and take the first five elements:

newA = dict(sorted(A.iteritems(), key=operator.itemgetter(1), reverse=True)[:5])

See also: Python Sorting HowTo

Solution 3:

You could use collections.Counter here:

dict(Counter(A).most_common(5))

Example:

>>> from collections import Counter
>>> A = {'a' : 1, 'b' : 3, 'c' : 2, 'd' : 4, 'e' : 0, 'f' :5}
>>> dict(Counter(A).most_common(5))
{'a': 1, 'c': 2, 'b': 3, 'd': 4, 'f': 5}

Solution 4:

For Python 3

import operator
dict(sorted(A.items(), key=operator.itemgetter(1), reverse=True)[:5])