迭代时如何从列表中删除项目？

问题描述：

我正在迭代Python中的元组列表，并且如果它们符合某些条件，我会尝试删除它们。

I'm iterating over a list of tuples in Python, and am attempting to remove them if they meet certain criteria.

for tup in somelist:
    if determine(tup):
         code_to_remove_tup

我应该用什么代替 code_to_remove_tup ？我无法弄清楚如何以这种方式删除项目。

What should I use in place of code_to_remove_tup? I can't figure out how to remove the item in this fashion.

答

您可以使用列表推导来创建新的列表只包含您不想删除的元素：

You can use a list comprehension to create a new list containing only the elements you don't want to remove:

somelist = [x for x in somelist if not determine(x)]

或者，通过分配给切片 somelist [：] ，您可以改变现有列表以仅包含您想要的项目：

Or, by assigning to the slice somelist[:], you can mutate the existing list to contain only the items you want:

somelist[:] = [x for x in somelist if not determine(x)]

如果还有其他方法，此方法可能很有用引用 somelist 需要反映更改。

This approach could be useful if there are other references to somelist that need to reflect the changes.

除了理解，您还可以使用 itertools 。在Python 2中：

Instead of a comprehension, you could also use itertools. In Python 2:

from itertools import ifilterfalse
somelist[:] = ifilterfalse(determine, somelist)

或者在Python 3中：

Or in Python 3:

from itertools import filterfalse
somelist[:] = filterfalse(determine, somelist)