将迭代器转换为列表的最快方法 [英] Fastest way to convert an iterator to a list

问题描述

有一个迭代器对象，是否有比列表解析更快，更好或更正确的东西来获取迭代器返回的对象列表？

Having an iterator object, is there something faster, better or more correct than a list comprehension to get a list of the objects returned by the iterator?

user_list = [user for user in user_iterator]

推荐答案

list(your_iterator)

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