In Python, when we want to sort a list of tuples or lists, we may want to sort it based on certain element in each sub-list, for example, sort the list based on the first element in each sub-list.
There are two different ways to achieve this, one is using lambda and the
other is using
itemgetter.
# using lambda
my_list.sort(key=lambda x:x[0])
# using itemgetter
from operator import itemgetter
my_list.sort(key=itemgetter(0))According to my benchmark on a list of 1000 tuples, using itemgetter is
almost twice as quick as the plain lambda method. The following is my code:
In [1]: a = list(range(1000))
In [2]: b = list(range(1000))
In [3]: import random
In [4]: random.shuffle(a)
In [5]: random.shuffle(b)
In [6]: c = list(zip(a, b))
In [7]: %timeit c.sort(key=lambda x: x[1])
81.4 µs ± 433 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [8]: random.shuffle(c)
In [9]: from operator import itemgetter
In [10]: %timeit c.sort(key=itemgetter(1))
47 µs ± 202 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)I have also tested the performance (run time in $µs$) of this two method for various list size. See the table below for a comparison:
+-----------+--------+------------+
| List size | lambda | itemgetter |
+-----------+--------+------------+
| 100 | 8.19 | 5.09 |
+-----------+--------+------------+
| 1000 | 81.4 | 47 |
+-----------+--------+------------+
| 10000 | 855 | 498 |
+-----------+--------+------------+
| 100000 | 14600 | 10100 |
+-----------+--------+------------+
| 1000000 | 172000 | 131000 |
+-----------+--------+------------+
(The code producing the above image can be found here)
From the above image, we can see that itemgetter is consistently faster than
lambda regardless of the list size. Combined with its conciseness to select
multiple elements from a list, itemgetter is clearly the winner to use in
sort method.
Note: this post is based on my answer on StackOverflow.
此内容由惯性聚合(RSS阅读器)自动聚合整理，仅供阅读参考。 原文来自 — 版权归原作者所有。