我有
k= (('answer ', ' Answer the call for a channel(answer )'), ('att_xfer ', ' Attended Transfer(att_xfer )'), ('bind_digit_action ', ' Bind a key sequence or regex to an action.(bind_digit_action )'))
我想去掉所有多余的空格。我怎样才能做到这一点
我有
k= (('answer ', ' Answer the call for a channel(answer )'), ('att_xfer ', ' Attended Transfer(att_xfer )'), ('bind_digit_action ', ' Bind a key sequence or regex to an action.(bind_digit_action )'))
我想去掉所有多余的空格。我怎样才能做到这一点
如果您的意思是“额外空格”,则每个字符串开头和结尾的空格:
k = tuple(tuple(b.strip() for b in a) for a in k)
如果要删除字符串中的一些其他“额外空格”(例如(answer )
=> (answer)
),则必须定义更多规则。
如果要删除所有空格:
tuple(tuple("".join(i.split()) for i in a) for a in k)
出去:
(('answer', 'Answerthecallforachannel(answer)'),
('att_xfer', 'AttendedTransfer(att_xfer)'),
('bind_digit_action',
'Bindakeysequenceorregextoanaction.(bind_digit_action)'))
或者如果您因此不需要元组:
from itertools import chain
["".join(i.split()) for i in chain.from_iterable(k)]
出去:
['answer',
'Answerthecallforachannel(answer)',
'att_xfer',
'AttendedTransfer(att_xfer)',
'bind_digit_action',
'Bindakeysequenceorregextoanaction.(bind_digit_action)']
另一种方法是:
tuple(map(lambda x:tuple(map(lambda y:y.strip(),x)),k))