删除嵌套列表中的字符串的一部分python

Question

我想只保留我的嵌套列表中的字符串的特定部分。我的代码如下：

answers = [ 
    ['person1', ' 2 1": ["answer 1"], "', ' 3 1": ["answer 0"], "', ...] 
    ['person2', ' 2 1": ["answer 3"], "', ' 3 1": ["answer 1"], "', ...]] 

我想删除该字符串的部分，使得只有一块剩下的是：

answers = [ 
    ['person1', 'answer 1', 'answer 0', ...] 
    ['person 2', 'answer 3', 'answer 1', ...]] 

Answer 1

此代码：

answers = [ 
      ['person1', ' 2 1": ["answer 1"], "', ' 3 1": ["answer 0"], "'], 
      ['person2', ' 2 1": ["answer 3"], "', ' 3 1": ["answer 1"], "']] 

result = [] 
for person_ans in answers: 
    list = [] 
    for val in person_ans: 
     if "[" in val: 
      temp = val.split("[") 
      output = temp[1].split("]")[0] 
      list.append(output) 
     else: 
      list.append(val) 
    result.append(list) 
print result 

输出：

[['person1', '"answer 1"', '"answer 0"'], ['person2', '"answer 3"', '"answer 1"']] 

Answer 2

鉴于：

answers = [ 
    ['person1', ' 2 1": ["answer 1"], "', ' 3 1": ["answer 0"], "',], 
    ['person2', ' 2 1": ["answer 3"], "', ' 3 1": ["answer 1"], "',] 
] 

可以使用正则表达式来解析每个字符串为所需图案p：

import re 


p = re.compile(r'(?!.*\[)(\w+\s*\d)') 
new_answers = [] 
for lst in answers: 
    new_answers.append([re.search(p, s).groups(0)[0] for s in lst]) 
new_answers 
# [['person1', 'answer 1', 'answer 0'], ['person2', 'answer 3', 'answer 1']] 

此特定pattern忽略所述第一支架的后面的字符“[”和找到的字母具有零个或多个空格和数字。