How do I sort a list of dictionaries by a value of the dictionary?

Question

I have a list of dictionaries and want each item to be sorted by a specific property values.

Take into consideration the array below,

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

When sorted by name, should become

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

Answer 1

It may look cleaner using a key instead a cmp:

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

or as J.F.Sebastian and others suggested,

from operator import itemgetter
newlist = sorted(list_to_be_sorted, key=itemgetter('name')) 

For completeness (as pointed out in comments by fitzgeraldsteele), add reverse=True to sort descending

newlist = sorted(l, key=itemgetter('name'), reverse=True)

Answer 2

import operator

To sort the list of dictionaries by key='name':

list_of_dicts.sort(key=operator.itemgetter('name'))

To sort the list of dictionaries by key='age':

list_of_dicts.sort(key=operator.itemgetter('age'))

Answer 3

If you want to sort the list by multiple keys you can do the following:

my_list = [{'name':'Homer', 'age':39}, {'name':'Milhouse', 'age':10}, {'name':'Bart', 'age':10} ]
sortedlist = sorted(my_list , key=lambda elem: "%02d %s" % (elem['age'], elem['name']))

It is rather hackish, since it relies on converting the values into a single string representation for comparison, but it works as expected for numbers including negative ones (although you will need to format your string appropriately with zero paddings if you are using numbers)

Answer 4

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.

Answer 5

import operator
a_list_of_dicts.sort(key=operator.itemgetter('name'))

'key' is used to sort by an arbitrary value and 'itemgetter' sets that value to each item's 'name' attribute.

Answer 6

a = [{'name':'Homer', 'age':39}, ...]

# This changes the list a
a.sort(key=lambda k : k['name'])

# This returns a new list (a is not modified)
sorted(a, key=lambda k : k['name']) 

Answer 7

I guess you've meant:

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

This would be sorted like this:

sorted(l,cmp=lambda x,y: cmp(x['name'],y['name']))

Answer 8

Using Schwartzian transform from Perl,

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

do

sort_on = "name"
decorated = [(dict_[sort_on], dict_) for dict_ in py]
decorated.sort()
result = [dict_ for (key, dict_) in decorated]

gives

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

More on Perl Schwartzian transform

In computer science, the Schwartzian transform is a Perl programming idiom used to improve the efficiency of sorting a list of items. This idiom is appropriate for comparison-based sorting when the ordering is actually based on the ordering of a certain property (the key) of the elements, where computing that property is an intensive operation that should be performed a minimal number of times. The Schwartzian Transform is notable in that it does not use named temporary arrays.

Answer 9

You could use a custom comparison function, or you could pass in a function that calculates a custom sort key. That's usually more efficient as the key is only calculated once per item, while the comparison function would be called many more times.

You could do it this way:

def mykey(adict): return adict['name']
x = [{'name': 'Homer', 'age': 39}, {'name': 'Bart', 'age':10}]
sorted(x, key=mykey)

But the standard library contains a generic routine for getting items of arbitrary objects: itemgetter. So try this instead:

from operator import itemgetter
x = [{'name': 'Homer', 'age': 39}, {'name': 'Bart', 'age':10}]
sorted(x, key=itemgetter('name'))

Answer 10

You have to implement your own comparison function that will compare the dictionaries by values of name keys. See Sorting Mini-HOW TO from PythonInfo Wiki

Answer 11

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)

Answer 12

sometime we need to use lower() for example

lists = [{'name':'Homer', 'age':39},
  {'name':'Bart', 'age':10},
  {'name':'abby', 'age':9}]

lists = sorted(lists, key=lambda k: k['name'])
print(lists)
# [{'name':'Bart', 'age':10}, {'name':'Homer', 'age':39}, {'name':'abby', 'age':9}]

lists = sorted(lists, key=lambda k: k['name'].lower())
print(lists)
# [ {'name':'abby', 'age':9}, {'name':'Bart', 'age':10}, {'name':'Homer', 'age':39}]

Answer 13

I tried something like this:

my_list.sort(key=lambda x: x['name'])

It worked for integers as well.

Answer 14

Using the pandas package is another method, though it's runtime at large scale is much slower than the more traditional methods proposed by others:

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:

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_large)
print('Large Method Pandas: ' + str(t.timeit(1)))

#Small Method LC: 0.000163078308105
#Small Method LC2: 0.000134944915771
#Small Method Pandas: 0.0712950229645
#Large Method LC: 0.0321750640869
#Large Method LC2: 0.0206089019775
#Large Method Pandas: 5.81405615807

Answer 15

Lets Say I h'v a Dictionary D with elements below. To sort just use key argument in sorted to pass custom function as below

D = {'eggs': 3, 'ham': 1, 'spam': 2}

def get_count(tuple):
    return tuple[1]

sorted(D.items(), key = get_count, reverse=True)
or
sorted(D.items(), key = lambda x: x[1], reverse=True)  avoiding get_count function call

https://wiki.python.org/moin/HowTo/Sorting/#Key_Functions

Answer 16

Here is my answer to a related question on sorting by multiple columns. It also works for the degenerate case where the number of columns is only one.

Answer 17

If you do not need the original list of dictionaries, you could modify it in-place with sort() method using a custom key function.

Key function:

def get_name(d):
    """ Return the value of a key in a dictionary. """

    return d["name"]

The list to be sorted:

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

Sorting it in-place:

data_one.sort(key=get_name)

If you need the original list, call the sorted() function passing it the list and the key function, then assign the returned sorted list to a new variable:

data_two = [{'name': 'Homer', 'age': 39}, {'name': 'Bart', 'age': 10}]
new_data = sorted(data_two, key=get_name)

Printing data_one and new_data.

>>> print(data_one)
[{'name': 'Bart', 'age': 10}, {'name': 'Homer', 'age': 39}]
>>> print(new_data)
[{'name': 'Bart', 'age': 10}, {'name': 'Homer', 'age': 39}]

Answer 18

You can use itemgetter ,if you want to consider the performance. itemgetter typically runs bit faster than lambda.

from operator import itemgetter
result = sorted(data, key=itemgetter('age'))  # this will sort list by property order 'age'.

Answer 19

You may use the following code

sorted_dct = sorted(dct_name.items(), key = lambda x : x[1])