如果value在另一个数据框的另一列中，则添加列

Question

我的一个数据框架是:

name    value
0    Harry     a
1    Kenny     b
2    Zoey      h

另一个原因是:

list                    topic
0    Jame, Harry, Noah      topic1
1    lee, zee               topic2  

我想如果dataframe1的任何名称在dataframe2的列表中，它应该在dataframe1中添加一个名称列'present'，其值为各自的主题。

name    value   present
0    Harry     a      topic1
1    Kenny     b       none
2    Zoey      h       none

更新查询

df1

name        value
0    Harry Lee     a
1    Kenny         b
2    Zoey          h

df2相同，期望结果为

name    value   present
0 Harry Lee    a      topic1 topic2
1    Kenny     b       none
2    Zoey      h       none

Answer 1

我们需要用explode修剪df1然后我们可以做map

df1['list'] = df1['list'].str.split(',')
s = df1.explode('list')
df['present'] = df.name.map(dict(zip(s['list'],s['topic'])))
df
Out[550]: 
name value present
0  Harry     a  topic1
1  Kenny     b     NaN
2   Zoey     h     NaN

Answer 2

import pandas as pd
import numpy as np
df1 = pd.DataFrame({"name":['Harry', 'Kenny', 'Zoey'], "value":["a", "b", "h"]})
df2 = pd.DataFrame({"list": ["Jame, Harry, Noah", "lee, zee"], "topic": ["topic1", "topic2"]})
def add_column(x):
try:
present = df2[df2['list'].str.contains(x)].iloc[0,1]
except IndexError:
present = np.NAN
return present
df1['present'] = df1['name'].apply(add_column)

Answer 3

您可以爆炸第二个数据框的list列，然后将其与name和list列合并回left连接上的第一个数据框，然后使用None删除list列和fillna:

df1.merge(df2.explode('list'), left_on='name', right_on='list', how='left').drop(columns='list').fillna('None')
name value   topic
0  Harry     a  topic1
1  Kenny     b    None
2   Zoey     h    None