EDIT：首先将列转换为列表：

import ast

cols = ['CAR1','CAR2']
df[cols] = df[cols].apply(ast.literal_eval)

在列表解析中使用set.intersection，并转换为布尔值和整数，以实现True,False到1/0的Map：

df['Flag'] = [int(bool(set(x).intersection(y))) for x,y in zip(df['CAR1'], df['CAR2'])]

替代解决方案：

df['Flag'] = [1 if set(x).intersection(y) else 0 for x,y in zip(df['CAR1'], df['CAR2'])]

print (df)
              CAR1               CAR2  Flag
0  [ford, hyundai]    [ford, hyundai]     1
1  [ford, hyundai]  [hyundai, nissan]     1
2  [ford, hyundai]        [bmw, audi]     0

您可以在列表解析中使用set运算（如果集合重叠，isdisjoint返回False，使用1-x将其反转并转换为整数）：

df['Flag'] = [1-set(s1).isdisjoint(s2) for s1, s2 in zip(df['CAR1'], df['CAR2'])]

• 注意：isdisjoint的速度非常快，因为它不需要读取完整的集合，只要找到一个公共项就返回False。*

输出：

CAR1               CAR2  Flag
0  [ford, hyundai]    [ford, hyundai]     1
1  [ford, hyundai]  [hyundai, nissan]     1
2  [ford, hyundai]        [bmw, audi]     0

从字符串

from ast import literal_eval

df['Flag'] = [1-set(s1).isdisjoint(s2) for s1, s2 in
               zip(df['CAR1'].apply(literal_eval),
                   df['CAR2'].apply(literal_eval))]