how do I sort dictionaries by value then by key?

In Python how do I sort a list of dictionaries by a certain value of the dictionary + alphabetically?

• Ok, here's what I'm trying to do... I know that itemgetter() sort could to alphabetical sort easy, but if I have something like this: [{'Name':'TOTAL', 'Rank':100}, {'Name':'Woo Company', 'Rank':15}, {'Name':'ABC Company', 'Rank':20}] And I want it sorted alphabetically (by Name) + include the condition that the one with Name:'TOTAL' should be listed last in the sequence, like this: [{'Name':'ABC Company', 'Rank':20}, {'Name':'Woo Company', 'Rank':15}, {'Name':'TOTAL', 'Rank':100}] How would I do that?

• Answer:

  The best approach here is to decorate the sort key... Python will sort a tuple by the tuple components in order, so build a tuple key with your sorting criteria: sorted(list_of_dicts, key=lambda d: (d['Name'] == 'TOTAL', d['Name'].lower())) This results in a sort key of: (True, 'total') for {'Name': 'TOTAL', 'Rank': 100} (False, 'woo company') for {'Name': 'Woo Company', 'Rank': 15} (False, 'abc company') for {'Name': 'ABC Company', 'Rank': 20} Since False sorts earlier than True, the ones whose names aren't TOTAL will end up together, then be sorted alphabetically, and TOTAL will end up at the end.