按值按字符排序字典按键Python3

Question

Let's say I have a dictionary like:

h=[('a',5), ('f', 3), ('b',3), ('c',3), ('d',1), ('e',4) ]

I want it sorted like:

[('a',5), ('e',4), ('b',3), ('c',3), ('f',3), ('d',1)]

I can solve this with Python 2 with something like this:

sortedList= sorted(h.iteritems(),key=lambda(k,v):(-v,k))

I can get really close in Python 3 with something like this:

import operator
sortedList =sorted(h.items(), key=operator.itemgetter(1,0) , reverse=True)

but it comes out like this

[('a',5), ('e',4), ('f',3), ('c',3), ('b',3),  ('d',1)]

How can I reverse the tiebreaker operation?

Answer 1

You can use this call to the sorted function in python 3:

sortedList = sorted(h, key=lambda k: (-k[1], k[0]))

This will give the same result as the python 2 sorting:

[('a',5), ('e',4), ('b',3), ('c',3), ('f',3), ('d',1)]