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"]]
I’d like to sort “mylist” by the second element of each member, e.g.
sort(mylist, key=lambda x: x[1])
how can I do this?
Answers:
The answer is to use “sorted”, i.e.
sorted(mylist, key=lambda x: x[1])
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])
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/.
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=...)
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
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'])
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"]]
I’d like to sort “mylist” by the second element of each member, e.g.
sort(mylist, key=lambda x: x[1])
how can I do this?
The answer is to use “sorted”, i.e.
sorted(mylist, key=lambda x: x[1])
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])
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/.
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=...)
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
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'])