Sorting by arbitrary lambda

Question

How can I sort a list by a key described by an arbitrary function? For example, if I have:

mylist = [[\"quux\", 1, \"a\"], [\"bar\", 0, \"b\"]]

Answer 1

Solution of your question is: sorted_list = sorted(mylist, key=lambda value:value[1])

Solution for dictionary of list is:

mylist = [{'name':'kk', 'age':21},{'name':'bk', 'age':21}]

sorted_list = sorted(mylist, key=lambda k: k['key_name'])

Answer 2

The answer is to use "sorted", i.e.

sorted(mylist, key=lambda x: x[1])

Answer 3

This is such a common need that support for it has been added to the standard library, in the form of operator.itemgetter:

from operator import itemgetter
mylist = [["quux", 1, "a"], ["bar", 0, "b"]]
mylist.sort(key=itemgetter(1)) # or sorted(mylist, key=...)

Answer 4

You basically have it already:

>>> mylist = [["quux", 1, "a"], ["bar", 0, "b"]]
>>> mylist.sort(key=lambda x: x[1])
>>> print mylist

gives:

[['bar', 0, 'b'], ['quux', 1, 'a']]

That will sort mylist in place.

[this para edited thanks to @Daniel's correction.] sorted will return a new list that is sorted rather than actually changing the input, as described in http://wiki.python.org/moin/HowTo/Sorting/.

Answer 5

You have two options, very close to what you described, actually:

mylist.sort(key=lambda x: x[1]) # In place sort
new_list = sorted(mylist, key=lambda x: x[1])

Answer 6

Sort and itemgetter is the fastest.

>>> import operator
>>> import timeit

>>> mylist = [["quux", 1, "a"], ["bar", 0, "b"]]
>>> t1 = timeit.Timer(lambda: mylist.sort(key=lambda x: x[1]))
>>> t1.timeit()
1.6330803055632404

>>> t2 = timeit.Timer(lambda: mylist.sort(key=operator.itemgetter(1)))
>>> t2.timeit()
1.3985503043467773

>>> t3 = timeit.Timer(lambda: sorted(mylist, key=operator.itemgetter(1)))
>>> t3.timeit()
2.6329514733833292

>>> t4 = timeit.Timer(lambda: sorted(mylist, key=lambda x: x[1]))
>>> t4.timeit()
2.9197154810598533