如何按多个元素对字典列表进行排序

Question

This thread How do I sort a list of dictionaries by values of the dictionary in Python? explains very clearly how to sort a list of dictionaries. In summary for,

[{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}]

You do:

newlist = sorted(list_to_be_sorted, key=lambda k: k['name']) 

However, I have much more elements in my dictionary and I need to sort by four of them - not just one. Any tips how to do this?

Note: I read further into the thread and this suggestion:

sortedlist = sorted(input, key=lambda elem: "%02d %s" % (elem['age'], elem['name']))

does not work. It gives error:

TypeError: list indices must be intergers, not str

I try various versions of this and to no avail.

Any help appreciated,

Answer 1

Simply change the key function to return a tuple containing the elements you wish to sort by:

newlist = sorted(input, key=lambda k: (k['age'], k['name'])) 

For example:

In [13]: input = [{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}, {'name':'Milhouse', 'age':10}, {'name': 'Mr. Burns', 'age': 104}]

In [14]: sorted(input, key=lambda k: (k['age'], k['name']))
Out[14]: 
[{'age': 10, 'name': 'Bart'},
 {'age': 10, 'name': 'Milhouse'},
 {'age': 39, 'name': 'Homer'},
 {'age': 104, 'name': 'Mr. Burns'}]

Your approach with string formatting could also be made to work, but the number of digits in the format would need to be increased to accommodate the likes of Cornelius Chapman . ;-)

Having said that, I would recommend using a tuple, as this tackles the problem more directly, is more efficient and is less prone to coding errors.

edit: As suggested by @JBernardo in the comments, the above can also be phrased using operator.itemgetter() :

In [19]: sorted(input, key=operator.itemgetter('age','name'))
Out[19]: 
[{'age': 10, 'name': 'Bart'},
 {'age': 10, 'name': 'Milhouse'},
 {'age': 39, 'name': 'Homer'},
 {'age': 104, 'name': 'Mr. Burns'}]