在python列表中交换元组/列表中的值?
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
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