删除段落中的引号和逗号

Question

x =['Jeff Bezos (chairman', ' president and CEO)', 'Werner Vogels (CTO)', '']
x ="".join(x)

我希望去掉括号内的引号，然后退出

x =['Jeff Bezos (chairman,president and CEO)', 'Werner Vogels (CTO)', '']

Answer 1

您的示例输入太小，无法创建一个可信的模式，但假设您希望在遇到右大括号时加入列表元素，您可以做一些简单的事情，如：

x = ['Jeff Bezos (chairman', ' president and CEO)', 'Werner Vogels (CTO)', '']
y = ['']
for element in x:
    y[-1] += element
    if element.endswith(')'):
        y.append('')
print(y)
# ['Jeff Bezos (chairman president and CEO)', 'Werner Vogels (CTO)', '']

但要真正基于左大括号进行连续连接，并为每个连接添加逗号，您需要记录左大括号的前一个状态，然后等待结束状态，然后使用逗号将所有内容连接在一起-类似于：

x = ['Jeff Bezos (chairman', ' president and CEO)', 'Werner Vogels (CTO)', '']
y = []
cont = False
for element in x:
    if not cont:
        y.append('')
    y[-1] += (',' if cont else '') + element
    cont = element.rfind('(') > element.rfind(')')
print(y)
# ['Jeff Bezos (chairman, president and CEO)', 'Werner Vogels (CTO)', '']

请注意，如果在同一元素中有多个大括号，即使这样也容易导致糟糕的连接，但问题是您愿意深入到多深以确保覆盖每个边缘情况。

Answer 2

import re

x = ['Jeff Bezos (chairman', ' president and CEO)', 'Werner Vogels (CTO)', '']

m=re.search('(\(.*?\))', str(x)) #finding the pattern starting with'('ending with')'
srt=m.group(0).replace("'", "")  # replacing single quotes within brackets.
x = re.sub('(\(.*?\))', srt, str(x),1)  #replacing the updated string in list
print '\n',x

输出:

['Jeff Bezos (chairman,  president and CEO)', 'Werner Vogels (CTO)', '']