Question

我有这个代码输出（使用键盘模块）：

[KeyboardEvent(enter up), KeyboardEvent(h down), KeyboardEvent(h up), KeyboardEvent(e down), KeyboardEvent(e up), KeyboardEvent(y down), KeyboardEvent(y up)]

如何从此列表中删除每个'KeyboardEvent'？

Answer 1

如何使用KeyboardEvent.name：

newList = [event.name for event in myList]

要获得更好的结果，您可以将其与KeyboardEvent.event_type结合使用：

newList = [event.name + ' ' + event.event_type for event in myList]

演示：

>>> myList
[KeyboardEvent(enter up), KeyboardEvent(h down), KeyboardEvent(h up), KeyboardEvent(e down), KeyboardEvent(e up), KeyboardEvent(y down)]

>>> [event.name for event in myList]
['enter', 'h', 'h', 'e', 'e', 'y']

>>> [event.name + ' ' + event.event_type for event in myList]
['enter up', 'h down', 'h up', 'e down', 'e up', 'y down']

Answer 2

a=[KeyboardEvent(enter up), KeyboardEvent(h down), KeyboardEvent(h up), KeyboardEvent(e down), KeyboardEvent(e up), KeyboardEvent(y down), KeyboardEvent(y up)]
a=[elem for elem in a if not isinstance(a, KeyboardEvent)]

此列表补偿应该有效

Answer 3

我会尝试正则表达式

import re

Foo = [KeyboardEvent(enter up), KeyboardEvent(h down), KeyboardEvent(h up), KeyboardEvent(e down), KeyboardEvent(e up), KeyboardEvent(y down), KeyboardEvent(y up)]

strList = []

for item in Foo:
  bar = re.sub('KeyboardEvent(\(.*?)', '', str(item))
  bar = re.sub('\)', '', bar)
  strList.append(bar)

print strList

Answer 4

尝试使用循环删除它：

list = [KeyboardEvent(enter up), KeyboardEvent(h down), KeyboardEvent(h up), KeyboardEvent(e down), KeyboardEvent(e up), KeyboardEvent(y down), KeyboardEvent(y up)]

for x in list:
    del list[str(x)]

Answer 5

或者您可以尝试将其实际删除KeyBoard事件作为字符串：

a=[KeyboardEvent(enter up), KeyboardEvent(h down), KeyboardEvent(h up), KeyboardEvent(e down), KeyboardEvent(e up), KeyboardEvent(y down), KeyboardEvent(y up)]
a=[str(elem).strip('KeyboardEvent') for elem in a]

如何删除python列表中的部分字符串？

5 个答案: