How do I sort a list of dictionaries by values of the dictionary in Python?

By : masi
Source: Stackoverflow.com
Question!

I got a list of dictionaries and want that to be sorted by a value of that dictionary.

This

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

sorted by name, should become

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


Answers

Using the pandas package is a fairly lightweight and quick method:

import pandas as pd

listOfDicts = [{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}]
df = pd.DataFrame(listOfDicts)
df = df.sort_values('name')
sorted_listOfDicts = df.T.to_dict().values()

Here are some benchmark values for a tiny list and a large (100k+) list of dicts, pandas outperforms on the large list:

setup_large = "listOfDicts = [];\
[listOfDicts.extend(({'name':'Homer', 'age':39}, {'name':'Bart', 'age':10})) for _ in range(50000)];\
from operator import itemgetter;import pandas as pd;\
df = pd.DataFrame(listOfDicts);"

setup_small = "listOfDicts = [];\
listOfDicts.extend(({'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}));\
from operator import itemgetter;import pandas as pd;\
df = pd.DataFrame(listOfDicts);"

method1 = "newlist = sorted(listOfDicts, key=lambda k: k['name'])"
method2 = "newlist = sorted(listOfDicts, key=itemgetter('name')) "
method3 = "df = df.sort_values('name');\
sorted_listOfDicts = df.T.to_dict().values()"

import timeit
t = timeit.Timer(method1, setup_small)
print('Small Method LC: ' + str(t.timeit(100)))
t = timeit.Timer(method2, setup_small)
print('Small Method LC2: ' + str(t.timeit(100)))
t = timeit.Timer(method3, setup_small)
print('Small Method Pandas: ' + str(t.timeit(100)))

t = timeit.Timer(method1, setup_large)
print('Large Method LC: ' + str(t.timeit(100)))
t = timeit.Timer(method2, setup_large)
print('Large Method LC2: ' + str(t.timeit(100)))
t = timeit.Timer(method3, setup_small)
print('Large Method Pandas: ' + str(t.timeit(100)))

#Small Method LC: 0.000163078308105
#Small Method LC2: 0.000134944915771
#Small Method Pandas: 0.0712950229645
#Large Method LC: 3.24144887924
#Large Method LC2: 1.78758502007
#Large Method Pandas: 0.0736699104309
By : abby sobh


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

my_list.sort(lambda x,y : cmp(x['name'], y['name']))

my_list will now be what you want.

(3 years later) Edited to add:

The new key argument is more efficient and neater. A better answer now looks like:

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

...the lambda is, IMO, easier to understand than operator.itemgetter, but YMMV.

By : pjz


Here is the alternative general solution - it sorts elements of dict by keys and values. The advantage of it - no need to specify keys, and it would still work if some keys are missing in some of dictionaries.

def sort_key_func(item):
    """ helper function used to sort list of dicts

    :param item: dict
    :return: sorted list of tuples (k, v)
    """
    pairs = []
    for k, v in item.items():
        pairs.append((k, v))
    return sorted(pairs)
sorted(A, key=sort_key_func)


This video can help you solving your question :)
By: admin