Sorting a dictionary by value then by key

Question

This seems like it has to be a dupe but my SO-searching-fu is poor today…

Say I have a dictionary of integer key/values, how can I sort the dictionary by the values descending, then by the key descending (for common values).

Input:

{12:2, 9:1,  14:2}
{100:1, 90:4, 99:3, 92:1, 101:1}

Output:

[(14,2), (12,2), (9,1)]  # output from print 
[(90,4), (99,3), (101,1), (100,1), (92,1)]

Asked By: Austin Salonen

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Source

Answer 1

In [62]: y={100:1, 90:4, 99:3, 92:1, 101:1}
In [63]: sorted(y.items(), key=lambda x: (x[1],x[0]), reverse=True)
Out[63]: [(90, 4), (99, 3), (101, 1), (100, 1), (92, 1)]

The key=lambda x: (x[1],x[0]) tells sorted that for each item x in y.items(), use (x[1],x[0]) as the proxy value to be sorted. Since x is of the form (key,value), (x[1],x[0]) yields (value,key). This causes sorted to sort by value first, then by key for tie-breakers.

reverse=True tells sorted to present the result in descending, rather than ascending order.

See this wiki page for a great tutorial on sorting in Python.

PS. I tried using key=reversed instead, but reversed(x) returns an iterator, which does not compare as needed here.

Answered By: unutbu

Answer 2

Try this:

>>> d={100:1, 90:4, 99:3, 92:1, 101:1}
>>> sorted(d.items(), lambda a,b:b[1]-a[1] or a[0]-b[0])

Answered By: Melug

Answer 3

Maybe this is more explicit:

>>> y = {100:1, 90:4, 99:3, 92:1, 101:1}
>>> reverse_comparison = lambda (a1, a2), (b1, b2):cmp((b2, b1), (a2, a1))
>>> sorted(y.items(), cmp=reverse_comparison)
[(90, 4), (99, 3), (101, 1), (100, 1), (92, 1)]

Answered By: Don

Sorting a dictionary by value then by key

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