How do I sort a list of dictionaries by a value of

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

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 2

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 3

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 4

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 5

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 6

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.