熊猫数据栏中相似词的显示方法

Question

这是我的数据

Id   Column_1               Column_2
1    United States          United Tractor
2    Love of Fair           Tales of Grim
3    Hotel Marriot Jakarta  Marriot Hotel Jakarta 

这是我的预期输出

Id   Column_1               Column_2               Word
1    United States          United Tractor         united
2    Love of Fair           Tales of Grim          of
3    Hotel Marriot Jakarta  Marriot Hotel Jakarta  hotel marriot

数据：

{'Id': [1, 2, 3],
 'Column_1': ['United States', 'Love of Fair', 'Hotel Marriot Jakarta'],
 'Column_2': ['United Tractor', 'Tales of Grim', 'Marriot Hotel Jakarta']}

Answer 1

一种选择是在列表理解中使用set.intersection：

df['Word'] = [' '.join(set(x.lower().split()) & set(y.lower().split())) for x,y in zip(df['Column_1'], df['Column_2'])]

另一种选择是对列进行stack；然后在执行set.intersection的lambda上使用groupby.apply

df['Word'] = (df[['Column_1', 'Column_2']].stack().str.lower().str.split()
              .groupby(level=0).apply(lambda x: ' '.join(set(x.iat[0]) & set(x.iat[1]))))

输出：

   Id               Column_1                Column_2                   Word
0   1          United States          United Tractor                 united
1   2           Love of Fair           Tales of Grim                     of
2   3  Hotel Marriot Jakarta  Marriot Hotel Jakarta   hotel marriot jakarta

Answer 2

非常相似但有点不同的解决方案：

df['Word'] = (df[['Column_1', 'Column_2']].
              applymap(lambda x: set(x.lower().split())).
              apply(lambda x: ' '.join(x.Column_1 & x.Column_2),1))
>>> df
'''
   Id               Column_1               Column_2                   Word
0   1          United States         United Tractor                 united
1   2           Love of Fair          Tales of Grim                     of
2   3  Hotel Marriot Jakarta  Marriot Hotel Jakarta  hotel marriot jakarta