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

Question:

How do I sort a list of dictionaries by a specific key’s value? Given:

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

When sorted by name, it should become:

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

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Answers:

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

Answered By: Matej

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']))
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.

Or better:

Since Python 2.4, there’s a key argument is both more efficient and neater:

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

…the lambda is, IMO, easier to understand than operator.itemgetter, but your mileage may vary.

Answered By: pjz

The sorted() function takes a key= parameter

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

Alternatively, you can use operator.itemgetter instead of defining the function yourself

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

For completeness, add reverse=True to sort in descending order

newlist = sorted(list_to_be_sorted, key=itemgetter('name'), reverse=True)
Answered By: Mario F
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.

Answered By: efotinis

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'))
Answered By: Owen
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'))
Answered By: vemury

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).

Answered By: Dologan

Using the 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 the 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.

Answered By: kiriloff

Let’s say I have a dictionary D with the elements below. To sort, just use the key argument in sorted to pass a 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

Check this out.

Answered By: Shank_Transformer

Here is the alternative general solution – it sorts elements of a 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)
Answered By: vvladymyrov

Using the Pandas package is another method, though its 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
Answered By: abby sobh
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']) 
Answered By: forzagreen

Sometimes 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}]
Answered By: uingtea

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}]
Answered By: srikavineehari

I have been a big fan of a filter with lambda. However, it is not best option if you consider time complexity.

First option

sorted_list = sorted(list_to_sort, key= lambda x: x['name'])
# Returns list of values

Second option

list_to_sort.sort(key=operator.itemgetter('name'))
# Edits the list, and does not return a new list

Fast comparison of execution times

# First option
python3.6 -m timeit -s "list_to_sort = [{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}, {'name':'Faaa', 'age':57}, {'name':'Errr', 'age':20}]" -s "sorted_l=[]" "sorted_l = sorted(list_to_sort, key=lambda e: e['name'])"

1000000 loops, best of 3: 0.736 µsec per loop

# Second option
python3.6 -m timeit -s "list_to_sort = [{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}, {'name':'Faaa', 'age':57}, {'name':'Errr', 'age':20}]" -s "sorted_l=[]" -s "import operator" "list_to_sort.sort(key=operator.itemgetter('name'))"

1000000 loops, best of 3: 0.438 µsec per loop

Answered By: Bejür

If performance is a concern, I would use operator.itemgetter instead of lambda as built-in functions perform faster than hand-crafted functions. The itemgetter function seems to perform approximately 20% faster than lambda based on my testing.

From https://wiki.python.org/moin/PythonSpeed:

Likewise, the builtin functions run faster than hand-built equivalents. For example, map(operator.add, v1, v2) is faster than map(lambda x,y: x+y, v1, v2).

Here is a comparison of sorting speed using lambda vs itemgetter.

import random
import operator

# Create a list of 100 dicts with random 8-letter names and random ages from 0 to 100.
l = [{'name': ''.join(random.choices(string.ascii_lowercase, k=8)), 'age': random.randint(0, 100)} for i in range(100)]

# Test the performance with a lambda function sorting on name
%timeit sorted(l, key=lambda x: x['name'])
13 µs ± 388 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

# Test the performance with itemgetter sorting on name
%timeit sorted(l, key=operator.itemgetter('name'))
10.7 µs ± 38.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

# Check that each technique produces the same sort order
sorted(l, key=lambda x: x['name']) == sorted(l, key=operator.itemgetter('name'))
True

Both techniques sort the list in the same order (verified by execution of the final statement in the code block), but the first one is a little faster.

Answered By: swac

As indicated by @Claudiu to @monojohnny in comment section of this answer,
given:

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

to sort the list of dictionaries by key 'age', 'name'

(like in SQL statement ORDER BY age, name), you can use:

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

or, likewise

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

print(newlist)

[{‘name’: ‘Bart’, ‘age’: 10},
{‘name’: ‘Milhouse’, ‘age’: 10},
{‘name’: ‘Homer’, ‘age’: 39}]

Answered By: Tms91

sorting by multiple columns, while in descending order on some of them:
the cmps array is global to the cmp function, containing field names and inv == -1 for desc 1 for asc

def cmpfun(a, b):
    for (name, inv) in cmps:
        res = cmp(a[name], b[name])
        if res != 0:
            return res * inv
    return 0

data = [
    dict(name='alice', age=10), 
    dict(name='baruch', age=9), 
    dict(name='alice', age=11),
]

all_cmps = [
    [('name', 1), ('age', -1)], 
    [('name', 1), ('age', 1)], 
    [('name', -1), ('age', 1)],]

print 'data:', data
for cmps in all_cmps: print 'sort:', cmps; print sorted(data, cmpfun)
Answered By: alex