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如果一个字符串列包含在 Pandas 的另一列中,则合并两个数据框

[英]Merge two dataframe if one string column is contained in another column in Pandas

I need to merge the following df1 and df2 , based on condition: if address in df1 contains state in df2 .我需要根据条件合并以下df1df2 :如果df1中的address包含df2中的state

df1: df1:

                                                                           address  \
0      Cecilia Chapman 711-2880 Nulla St. Mankato Mississippi 96522 (257) 563-7401   
1  Iris Watson P.O. Box 283 8562 Fusce Rd. Frederick Nebraska 20620 (372) 587-2335   
2    Celeste Slater 606-3727 Ullamcorper. Street Roseville NH 11523 (786) 713-8616   
3            Theodore Lowe Ap #867-859 Sit Rd. Azusa New York 39531 (793) 151-6230   
4                 Calista Wise 7292 Dictum Av. San Antonio MI 47096 (492) 709-6392   

   quantity  price  
0         2     20  
1         3     13  
2         5     23  
3         3     32  
4         5     45  

df2: df2:

   id        state
0   1  Mississippi
1   2     Nebraska
2   3     New York

My expected output will like this.我的预期输出会像这样。 How could I do that?我怎么能那样做? Thank you.谢谢你。

                                                                           address  \
0      Cecilia Chapman 711-2880 Nulla St. Mankato Mississippi 96522 (257) 563-7401   
1  Iris Watson P.O. Box 283 8562 Fusce Rd. Frederick Nebraska 20620 (372) 587-2335   
2    Celeste Slater 606-3727 Ullamcorper. Street Roseville NH 11523 (786) 713-8616   
3            Theodore Lowe Ap #867-859 Sit Rd. Azusa New York 39531 (793) 151-6230   
4                 Calista Wise 7292 Dictum Av. San Antonio MI 47096 (492) 709-6392   

   quantity  price   id        state  
0         2     20  1.0  Mississippi  
1         3     13  2.0     Nebraska  
2         5     23  NaN          NaN  
3         3     32  3.0     New York  
4         5     45  NaN          NaN  

Update: the output of pat = '|'.join(r"\\b{}\\b".format(x) for x in df2['state']); print(df1['address'].str.extract('('+ pat + ')', expand=False))更新: pat = '|'.join(r"\\b{}\\b".format(x) for x in df2['state']); print(df1['address'].str.extract('('+ pat + ')', expand=False)) pat = '|'.join(r"\\b{}\\b".format(x) for x in df2['state']); print(df1['address'].str.extract('('+ pat + ')', expand=False))

      0    1    2    3    4    5    6    7    8    9  ...    40   41   42  \
0    NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN ...   NaN  NaN  NaN   
1    NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN ...   NaN  NaN  NaN   
2    NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN ...   NaN  NaN  NaN   
3    NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN ...   NaN  NaN  NaN    
..   ...  ...  ...  ...  ...  ...  ...  ...  ...  ... ...   ...  ...  ...  
158  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN ...   NaN  NaN  NaN   
159  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN ...   NaN  NaN  NaN  

You can extract all possible states by Series.str.extract with \\b\\b for words boundaries to new column and then merge with left join:您可以通过Series.str.extract提取所有可能的状态,使用\\b\\b将单词边界提取到新列,然后与左连接merge

pat = '|'.join(r"\b{}\b".format(x) for x in df2['state'])
df1['state']= df1['address'].str.extract('('+ pat + ')', expand=False)
print (df1)
                                             address  quantity  price  \
0  Cecilia Chapman 711-2880 Nulla St. Mankato Mis...         2     20   
1  Iris Watson P.O. Box 283 8562 Fusce Rd. Freder...         3     13   
2  Celeste Slater 606-3727 Ullamcorper. Street Ro...         5     23   
3  Theodore Lowe Ap #867-859 Sit Rd. Azusa New Yo...         3     32   
4  Calista Wise 7292 Dictum Av. San Antonio MI 47...         5     45   

         state  
0  Mississippi  
1     Nebraska  
2          NaN  
3     New York  
4          NaN  

df = df1.merge(df2, on='state', how='left')
print (df)
                                             address  quantity  price  \
0  Cecilia Chapman 711-2880 Nulla St. Mankato Mis...         2     20   
1  Iris Watson P.O. Box 283 8562 Fusce Rd. Freder...         3     13   
2  Celeste Slater 606-3727 Ullamcorper. Street Ro...         5     23   
3  Theodore Lowe Ap #867-859 Sit Rd. Azusa New Yo...         3     32   
4  Calista Wise 7292 Dictum Av. San Antonio MI 47...         5     45   

         state   id  
0  Mississippi  1.0  
1     Nebraska  2.0  
2          NaN  NaN  
3     New York  3.0  
4          NaN  NaN  

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