在python中的列表内交换元组/列表中的值? [英] Swap values in a tuple/list inside a list in python?

问题描述

我在这样的列表中有一个元组/列表:

I have a tuple/list inside a list like this:

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

python(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')]

当前使用:

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

有更好或更快速的方法吗???

Is there a better or faster way???

推荐答案

您可以使用地图:

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

或列表理解:

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

列表推导是首选，并且在需要lambda时，它应该比map快得多，但是请注意，列表推导具有严格的评估，即，如果担心内存，它将在绑定到变量后立即进行评估.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()

此处有更多详细信息: Python列表理解与.地图

There are some more details here: Python List Comprehension Vs. Map

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