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}]

Solution 1

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)

Solution 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'))

Solution 3

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.

Solution 4

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).

Solution 5

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']) 

View more solutions

Comments

[Claudiu]

Reading the answer and looking on operator.itemgetter. Can I sort on multiple value in the same process (for example we have [{'name':'Bart', 'age':10, 'note':3},{'name':'Homer','age':10,'note':2},{'name':'Vasile‌​','age':20,'note':3}‌​] And to use: from operator import itemgetter newlist = sorted(old_list, key=itemgetter(-'note','name') EDIT: Tested, and it is working but I don't know how to make note DESC and name ASC.

[jfs]

Using key is not only cleaner but more effecient too.

[Mario F]

The fastest way would be to add a newlist.reverse() statement. Otherwise you can define a comparison like cmp=lambda x,y: - cmp(x['name'],y['name']).

[Philluminati]

if the sort value is a number you could say: lambda k: (k['age'] * -1) to get a reverse sort

[monojohnny]

Anyway to combine name and age ? (like in SQL ORDER BY name,age ?)

[radicand]

This also applies to a list of tuples, if you use itemgetter(i) where i is the index of the tuple element to sort on.

[Bakuriu]

itemgetter accepts more than one argument: itemgetter(1,2,3) is a function that return a tuple like obj[1], obj[2], obj[3], so you can use it to do complex sorts.

[njzk2]

sorted using timsort which is stable, you can call sorted several times to have a sort on several criteria

[Permafacture]

njzk2's comment wasn't immediately clear to me so I found the following. You can just sort twice as njzk2 suggests, or pass multiple arguments to operator.itemgetter in the top answer. Link: stackoverflow.com/questions/5212870/…

[Claudiu]

@monojohnny: yes, just have the key return a tuple, key=lambda k: (k['name'], k['age']). (or key=itemgetter('name', 'age')). tuple's cmp will compare each element in turn. it's bloody brilliant.

[Winston Ewert]

No need to convert to string. Just return a tuple as the key.

[TTT]

In the documentation (docs.python.org/2/tutorial/datastructures.html) the optional key argument for list.sort() is not described. Any idea where to find that?

[Antti Haapala -- Слава Україні]

Python has supported the key= for .sort since 2.4, that is year 2004, it does the Schwartzian transform within the sorting code, in C; thus this method is useful only on Pythons 2.0-2.3. all of which are more than 12 years old.

[Kevin]

@TTT: See the library documentation for list and friends.

[wouter bolsterlee]

Sorting multiple times is the easiest generic solution without hacks: stackoverflow.com/a/29849371/1805397

[clp2]

I ran your code and found a mistake in the the timeit.Timer args for Large Method Pandas: you specify "setup_small" where it should be "setup_large". Changing that arg caused the program to run without finishing, and I stopped it after more than 5 minutes. When I ran it with "timeit(1)", the Large Method Pandas finished in 7.3 sec, much worse than LC or LC2.

[abby sobh]

You're quite right, that was quite an oversight on my part. I no longer recommend it for large cases! I have edited the answer to simply allow it as a possibility, the use case is still up for debate.

[webenformasyon]

in case of unicode strings import locale / locale.setlocale(locale.LC_ALL, "") / newlist = sorted(list_to_be_sorted, key=lambda k: locale.strxfrm(k['name'])) works inspired from stackoverflow.com/questions/1097908/…

[Peter Mortensen]

This relies too much on the link. Can you provide a more complete answer?

[Peter Mortensen]

Why do we need to use lower() in this case?

[Peter Mortensen]

What do you mean by "sorts elements of a dict by keys and values"? In what way is it sorting? Where do the values come in?

[G.J]

Proper anwers are already provided by other contributors as well. Feel free to either keep the link, or delete the answer.

[Sam]

what could be done if the key is unknown and keeps changing?I mean list of dicts with just one key and value but the key and value could not be defined as they keep changing.

[Notre]

I'd need more of an example to look at. Try submitting a possible solution on the codereview stackexchange and asking if there's a better way.

[sattva_venu]

what if we don't know the value of key?

[Notre]

@Sam if you want to sort by the value of the single key in the dict, even if you don't know the key, you can do key=lambda k: list(k.values())[0]

[Andrew]

operator.itemgetter worked for me, the first option didn't work or maybe I didn't fill in the right info... lambda? d: ?

[reza_khalafi]

list indices must be integers or slices, not str

[ciurlaro]

@Bakuriu is there any way to do complex sorts with both ascending and descending order? e.g. key=itemgetter("speed", "color", "cost"), reverse=(True, False, True)

[Bakuriu]

@ciurlaro Not that I know of. In order to make that work you'd first have to map the values in a way that make the fields you want sort in the opposite direction, then use a plain sort by those fields and at the end "un-wrap" the values. Obviously instead of using itemgetter you could use a custom function and do whatever you want there.

[Bakuriu]

@ciurlaro Note that in python sorting is stable, so say you want to achieve the equivalent of key=itemgetter('A', 'B', 'C', 'D', 'E'), reverse=(True, True, False, True, True) you could do that by sorting 3 times: first on D, E with reverse=True, then on C with reverse=False and finally on A,B with reverse=True.

[ciurlaro]

@Bakuriu I found this answer stackoverflow.com/a/56842689/3004162 that worked to achieve what I wanted. Still far from a beauty of a builtin function unfortunately, but that's the best I found