在python列表中交换元组/列表中的值?

2022-01-19 00:00:00 python swap list tuples

问题描述

我有一个这样的元组列表:

I have a list of tuples like this:

[('foo','bar'),('foo1','bar1'),('foofoo','barbar')]

在 python 中(在非常低的 cpu/ram 机器上运行)交换这样的值的最快方法是什么...

What is the fastest way in python (running on a very low cpu/ram machine) to swap values like this...

[('bar','foo'),('bar1','foo1'),('barbar','foofoo')]

我目前正在使用:

for x in mylist:
    self.my_new_list.append(((x[1]),(x[0])))

有没有更好更快的方法???

Is there a better or faster way???


解决方案

你可以使用map:

map (lambda t: (t[1], t[0]), mylist)

或列表理解:

[(t[1], t[0]) for t in mylist]

当需要 lambda 时,列表推导式是首选并且据说比 map 快得多,但是请注意,列表推导式有一个严格的评估,也就是说,如果您担心内存,它将在绑定到变量时立即进行评估消费使用 generator 代替:

List comprehensions are preferred and supposedly much faster than map when lambda is needed, however note that list comprehension has a strict evaluation, that is it will be evaluated as soon as it gets bound to variable, if you're worried about memory consumption use a generator instead:

g = ((t[1], t[0]) for t in mylist)
#call when you need a value
g.next()

这里有更多详细信息:Python 列表理解 Vs.地图

相关文章