Sorting dictionary containing lists

Question

Assuming I've the following list of lists:

dict1 = [['Jeremy', 25, 120000], ['Paul', 23, 75000], ['Mike', 32, 80000]]

I can easily sort the lists on index 2 as follows:

from operator import itemgetter

sorted_dict = sorted(dict1, key=itemgetter(1))
print(sorted_dict)
>>>
[['Paul', 23, 75000], ['Jeremy', 25, 120000], ['Mike', 32, 80000]]

Things get a little more complicated with a dictionary of lists. Assuming I've the following:

dict2 = {'employee1':['Paul', 23, 75000], 
         'employee2':['Mike', 32, 80000], 
         'employee3':['Jeremy', 25, 120000]}

I can approximate a sort on index 2 as follows:

from operator import itemgetter

#First, extract lists
result1 = dict2.values()
#Second, sort list by index 2
result2 = sorted(result1, key=itemgetter(1))
#Finally, use for loop to sort dictionary
for a in result2:
    for b in dict2.keys():
        if a == dict2[b]:
            print("{0}:{1}".format(b,a))

>>>
   employee1:['Paul', 23, 75000]
   employee3:['Jeremy', 25, 120000]
   employee2:['Mike', 32, 80000]

I would like a better way to perform a sort on the dictionary. I found a community wiki post on sorting dictionaries (here) but the dictionary in that post has constant key. In my dictionary above, each list has a unique key.

Thanks.

Using Python 3.4.1

mgilson · Accepted Answer · 2014-08-18T06:11:15.667

3

First, you can't sort a dict because it doesn't have order ¹. You can sort it's items however:

sorted(d.items(), key=lambda t: t[1][1])

Should do the trick.

notes

t[1] # => the "value", t[0] is the "key"
t[1][1] # => The second element in the value.

You'll get a list of 2-tuples (where the first element is the key, and the second element is the value). Of course, you could always pass this list of 2-tuples directly to collections.OrderedDict if you wanted to construct a dict which has an ordering after the fact ...

^{¹More correctly, the order is arbitrary and could change based on key insertions, deletions, python implementation or version ... (which is why it's easier to say that dict's are unordered)}

edited Aug 18 '14 at 06:11

answered Aug 18 '14 at 06:03

mgilson

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Perfecto! Aside, please could you refer me to the best resource on lambda. It always looks Greek to me. – sedeh Aug 18 '14 at 13:23
@sedeh -- `lamda arguments: expression` simply creates a function which evaluates the expression with the given arguments. I think I used this one a good bit back when I was first familiarizing myself with them: http://www.diveintopython.net/power_of_introspection/lambda_functions.html – mgilson Aug 18 '14 at 16:43

xecgr · Answer 2 · 2014-08-18T06:16:18.550

1

You can sort the keys and create another dict with ordered keys

>>>>from collections import OrderedDict
>>>>dict2 = {'employee1':['Paul', 23, 75000], 
     'employee2':['Mike', 32, 80000], 
     'employee3':['Jeremy', 25, 120000]}

>>>>OrderedDict([
        #new item: ordered key, it's value form original dict
        ( x, dict2[x]) 
        #sort keys of original dict
        for x in sorted(dict2, key=lambda x: dict2[x][1])
])
OrderedDict([('employee1', ['Paul', 23, 75000]), ('employee3', ['Jeremy', 25, 120000]), ('employee2', ['Mike', 32, 80000])])

edited Aug 18 '14 at 06:16

answered Aug 18 '14 at 06:06

xecgr

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I'm not sure what you're going for here -- `dict(...)` will destroy any order that you got from the initial sort. – mgilson Aug 18 '14 at 06:07
you are right, I've change default dict to OrderedDict which lets you define key's order. This scripts output is the same that question's one – xecgr Aug 18 '14 at 06:12

Sorting dictionary containing lists

2 Answers2