什么是最有效的（pythonic）方式来忽略列表中有偏执的单词？

Question

我有以下清单

x = ['Accara building model (ABM)','tri-com model (tcm)']

使用re，我可以忽略括号中的单词。就像下面一样

import re 
x = ['Accara building model (ABM)','tri-com model (tcm)']
for i in x:
    ko= list(re.sub("[\(\[].*?[\)\]]", "", i))
    print (ko)

但我得到的输出格式如下

['A', 'c', 'c', 'a', 'r', 'a', ' ', 'b', 'u', 'i', 'l', 'd', 'i', 'n', 'g', ' ', 'm', 'o', 'd', 'e', 'l', ' ']
['t', 'r', 'i', '-', 'c', 'o', 'm', ' ', 'm', 'o', 'd', 'e', 'l', ' ']

理想情况下，我想要的是尽可能少的代码行。（我知道我的代码目前效率低下）

需要理想输出

['Accara building model', 'tri-com model']

Answer 1

import re
x = ['Accara building model (ABM)','tri-com model (tcm)']

print([ "".join(list(re.sub("[\(\[].*?[\)\]]", "", i))) for i in x ])


python test
['Accara building model ', 'tri-com model ']