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)
Videos
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)
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'))
There are lots of ways to write this. Before you're comfortable with lambdas, the best thing to do is write it out explicitly:
def sort_key(d):
return d['c']
sorted(list_of_dict, key=sort_key, reverse=True)
A trained eye will see that this is the same as the following one-liner:
sorted(list_of_dict, key=lambda d: d['c'], reverse=True)
Or, possibly:
from operator import itemgetter
sorted(list_of_dict, key=itemgetter('c'), reverse=True)
import operator
list_of_dicts.sort(key=operator.itemgetter('c'), reverse=True)
Python 3.7+ or CPython 3.6
Dicts preserve insertion order in Python 3.7+. Same in CPython 3.6, but it's an implementation detail.
>>> x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
>>> {k: v for k, v in sorted(x.items(), key=lambda item: item[1])}
{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}
or
>>> dict(sorted(x.items(), key=lambda item: item[1]))
{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}
Older Python
It is not possible to sort a dictionary, only to get a representation of a dictionary that is sorted. Dictionaries are inherently orderless, but other types, such as lists and tuples, are not. So you need an ordered data type to represent sorted values, which will be a list—probably a list of tuples.
For instance,
import operator
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=operator.itemgetter(1))
sorted_x will be a list of tuples sorted by the second element in each tuple. dict(sorted_x) == x.
And for those wishing to sort on keys instead of values:
import operator
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=operator.itemgetter(0))
In Python3 since unpacking is not allowed we can use
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=lambda kv: kv[1])
If you want the output as a dict, you can use collections.OrderedDict:
import collections
sorted_dict = collections.OrderedDict(sorted_x)
As simple as: sorted(dict1, key=dict1.get)
Well, it is actually possible to do a "sort by dictionary values". Recently I had to do that in a Code Golf (Stack Overflow question Code golf: Word frequency chart). Abridged, the problem was of the kind: given a text, count how often each word is encountered and display a list of the top words, sorted by decreasing frequency.
If you construct a dictionary with the words as keys and the number of occurrences of each word as value, simplified here as:
from collections import defaultdict
d = defaultdict(int)
for w in text.split():
d[w] += 1
then you can get a list of the words, ordered by frequency of use with sorted(d, key=d.get) - the sort iterates over the dictionary keys, using the number of word occurrences as a sort key.
for w in sorted(d, key=d.get, reverse=True):
print(w, d[w])
or, if we want a dictionary back (since Python 3.6+ preserves insertion order):
{w: d[w] for w in sorted(d, key=d.get, reverse=True)}
I am writing this detailed explanation to illustrate what people often mean by "I can easily sort a dictionary by key, but how do I sort by value" - and I think the original post was trying to address such an issue. And the solution is to do sort of list of the keys, based on the values, as shown above.