I'm trying to get a sorted dictionary. The dictionary should be sorted according to the following two criteria:

1. According to the values (integers)
2. If two keys (str) have the same value, sort alphabetically

So far, I have written the following line in my code:

sorted_word_values_dict = dict(sorted(word_values_dict.items(), key=operator.itemgetter(1)))

This code sorts correctly according to the first criteria, but if two keys (str) have the same value (e.g. both have a value of 2), they need to be sorted alphabetically. How can I include this extra criterium in my code?

Thanks in advance for any help!

CodePudding user response：

Given this dictionary:

d = {'c': 1251, 'b': 2372, 'f': 5172, 'a': 2372}

If you want to sort the values in ascending order and keys lexicographically:

out = dict(sorted(d.items(), key=lambda x: (x[1],x[0])))

Output:

{'c': 1251, 'a': 2372, 'b': 2372, 'f': 5172}

If however, you want to sort values in descending order (and keys lexicographically):

out = dict(sorted(d.items(), key=lambda x: (-x[1],x[0])))

Output:

{'f': 5172, 'a': 2372, 'b': 2372, 'c': 1251}

CodePudding user response：

You can use lambda as a key function in sorted call

Following should get you the new dict in sorted order -

sorted_word_values_dict = dict(sorted(word_values_dict.items(), key=lambda x: (x[1], x[0])))

The key argument in sorted function can take a function which returns multiple values, in case there is a tie between the first value, the comparison would move to next value in the key.

In this example, x[1](dict value) would be compared first, if it is identical, we would compare the x[0](dict key).