如果找到匹配，则比较两个不同数据框中的列，将电子邮件从 df2 复制到 df1

Question

我有两个列名不同的数据框，每个有 10 行。我要做的是比较列值，如果它们匹配，则将电子邮件地址从 df2 复制到 df1。我看过这个例子，但我的列名不同How to join (merge) data frames (inner, outer, left, right)?。我已经看到this example 以及np.where 使用了多个条件，但是当我这样做时，它给了我以下错误：

ValueError: Wrong number of items passed 2, placement implies 1

我想做的事：

我想要做的是将 df1 的第一行 2 列（first，last_huge）与 df2 列的所有行（first_small，last_small）进行比较，如果找到匹配项，则从 df2 中的特定列获取电子邮件地址并分配它到 df1 中的一个新列。任何人都可以帮我解决这个问题我只复制了下面的相关代码并且它不能完全工作只是向 new_email 添加 5 条新记录。

最初我所做的是将 df1['first'] 与 df2['first'] 进行比较

data1 = {"first":["alice", "bob", "carol"],
         "last_huge":["foo", "bar", "baz"],
         "street_huge": ["Jaifo Road", "Wetib Ridge", "Ucagi View"],
         "city_huge": ["Egviniw", "Manbaali", "Ismazdan"],
         "age_huge": ["23", "30", "36"],
         "state_huge": ["MA", "LA", "CA"],
         "zip_huge": ["89899", "78788", "58999"]}

df1 = pd.DataFrame(data1)

data2 = {"first_small":["alice", "bob", "carol"],
         "last_small":["foo", "bar", "baz"],
         "street_small": ["Jsdffo Road", "sdf Ridge", "sdfff View"],
         "city_huge": ["paris", "london", "rome"],
         "age_huge": ["28", "40", "56"],
         "state_huge": ["GA", "EA", "BA"],
         "zip_huge": ["89859", "78728", "56999"],
         "email_small":["alice@xyz.com", "bob@abc.com", "carol@jkl.com"],
         "dob": ["31051989", "31051980", "31051981"],
         "country": ["UK", "US", "IT"],
         "company": ["microsoft", "apple", "google"],
         "source": ["bing", "yahoo", "google"]}

df2 = pd.DataFrame(data2)

df1['new_email'] = np.where((df1[['first']] == df2[['first_small']]), df2[['email_small']], np.nan)

现在只向 new_email 添加 5 条记录，其余为 nan。并向我显示此错误：

ValueError: Can only compare identically-labeled Series objects

Answer 1

试试merge:

(df1.merge(df2[["first_small", "last_small", "email_small"]], 
           how="left", 
           left_on=["first", "last_huge"], 
           right_on=["first_small", "last_small"])
    .drop(['first_small','last_small'], 1))

例子：

data1 = {"first":["alice", "bob", "carol"], 
         "last_huge":["foo", "bar", "baz"]}
df1 = pd.DataFrame(data1)

data2 = {"first_small":["alice", "bob", "carol"], 
         "last_small":["foo", "bar", "baz"],
         "email_small":["alice@xyz.com", "bob@abc.com", "carol@jkl.com"]}
df2 = pd.DataFrame(data2)

(df1.merge(df2[["first_small", "last_small", "email_small"]], 
           how="left", 
           left_on=["first", "last_huge"], 
           right_on=["first_small", "last_small"])
    .drop(['first_small','last_small'], 1))

输出：

   first last_huge    email_small
0  alice       foo  alice@xyz.com
1    bob       bar    bob@abc.com
2  carol       baz  carol@jkl.com

Answer 2

通过使用 andrew_reece 的示例数据 :-) pd.concat

pd.concat([df1.set_index(["first", "last_huge"]),df2.set_index(["first_small", "last_small"])['email_small']],axis=1).reset_index().dropna()
Out[23]: 
   first last_huge    email_small
0  alice       foo  alice@xyz.com
1    bob       bar    bob@abc.com
2  carol       baz  carol@jkl.com

通过使用您的数据

pd.concat([df1.set_index(["first", "last_huge"]),df2.set_index(["first_small", "last_small"])['email_small']],axis=1).reset_index()
Out[97]: 
   first last_huge age_huge city_huge state_huge  street_huge zip_huge  \
0  alice       foo       23   Egviniw         MA   Jaifo Road    89899   
1    bob       bar       30  Manbaali         LA  Wetib Ridge    78788   
2  carol       baz       36  Ismazdan         CA   Ucagi View    58999   
     email_small  
0  alice@xyz.com  
1    bob@abc.com  
2  carol@jkl.com  

使用map更新

df1['email_small']=(df1['first']+df1['last_huge']).map(df2.set_index(df2['first_small']+df2['last_small'])['email_small'])
df1
Out[115]: 
  age_huge city_huge  first last_huge state_huge  street_huge zip_huge  \
0       23   Egviniw  alice       foo         MA   Jaifo Road    89899   
1       30  Manbaali    bob       bar         LA  Wetib Ridge    78788   
2       36  Ismazdan  carol       baz         CA   Ucagi View    58999   
     email_small  
0  alice@xyz.com  
1    bob@abc.com  
2  carol@jkl.com