Question

我有一个列表列表，例如

l =  ["{{{1star}}} Do not bother with this  restaurant.",  "{{{1star}}}
The food is good but the service is awful."]

我想将此列表转换为其他列表。新列表将包含一个元组和另一个列表。元组的第一个元素将是{{}}内的字符串，而元组的第二个元素将包含其余文本作为列表。我期待以下输出：

output =  [("{{{1star}}}", ["Do not bother with this  restaurant."]), 
("{{{1star}}}", ["The food is good but the service is awful."])]

谢谢！

Answer 1

我尝试通过一些字符串操作来获得所需的内容

l =  ["{{{1star}}} Do not bother with this  restaurant.",  "{{{1star}}} The food is good but the service is awful."]

out = []

for strings in l : 
  s = strings.split()
  first = s[0] 
  second = strings.replace(s[0],'')
  tuple = ( first , second ) 
  out.append(tuple)

print(out)

或使用正则表达式

import re

l =  ["{{{1star}}} Do not bother with this  restaurant.",  "{{{1star}}} The food is good but the service is awful."]

out = []

for strings in l : 
  s = re.match( r'(.*)\}(.*?) .*', strings, re.M|re.I)
  print(s)
  if s : 
    first = s.group(1) + '}'
    second = strings.replace(first,'')
    tuple = ( first , second ) 
    out.append(tuple)

print(out)

如何将列表列表转换为元组列表+列表

1 个答案: