根据另一个数据帧python pandas替换列值 - 更好的方法？

Question

注意:为了简单起见,我正在使用一个玩具示例,因为复制/粘贴数据帧很难在堆栈溢出(请告诉我是否有一个简单的方法来执行此操作).

有没有办法将一个数据框中的值合并到另一个数据框而不获取_X,_Y列？我希望一列上的值替换另一列的所有零值.

df1: 

Name   Nonprofit    Business    Education

X      1             1           0
Y      0             1           0   <- Y and Z have zero values for Nonprofit and Educ
Z      0             0           0
Y      0             1           0

df2:

Name   Nonprofit    Education
Y       1            1     <- this df has the correct values. 
Z       1            1



pd.merge(df1, df2, on='Name', how='outer')

Name   Nonprofit_X    Business    Education_X     Nonprofit_Y     Education_Y
Y       1                1          1                1               1
Y      1                 1          1                1               1
X      1                 1          0               nan             nan   
Z      1                 1          1                1               1

在上一篇文章中,我尝试了combine_First和dropna(),但这些都没有完成任务.

我想用df2中的值替换df1中的零.此外,我希望根据df2更改具有相同名称的所有行.

Name    Nonprofit     Business    Education
Y        1             1           1
Y        1             1           1 
X        1             1           0
Z        1             0           1

我现有的解决方案执行以下操作:我根据df2中存在的名称进行子集化,然后使用正确的值替换这些值.但是,我想要一个不那么黑客的方法来做到这一点.

pubunis_df = df2
sdf = df1 

regex = str_to_regex(', '.join(pubunis_df.ORGS))

pubunis = searchnamesre(sdf, 'ORGS', regex)

sdf.ix[pubunis.index, ['Education', 'Public']] = 1
searchnamesre(sdf, 'ORGS', regex)

Answer 1

注意：在最新版本的熊猫中，以上两个答案都不再适用：

KSD 的回答会引发错误：

df1 = pd.DataFrame([["X",1,1,0],
              ["Y",0,1,0],
              ["Z",0,0,0],
              ["Y",0,0,0]],columns=["Name","Nonprofit","Business", "Education"])    

df2 = pd.DataFrame([["Y",1,1],
              ["Z",1,1]],columns=["Name","Nonprofit", "Education"])   

df1.loc[df1.Name.isin(df2.Name), ['Nonprofit', 'Education']] = df2.loc[df2.Name.isin(df1.Name),['Nonprofit', 'Education']].values

df1.loc[df1.Name.isin(df2.Name), ['Nonprofit', 'Education']] = df2[['Nonprofit', 'Education']].values

Out[851]:
ValueError: shape mismatch: value array of shape (2,) could not be broadcast to indexing result of shape (3,)

而 EdChum 的回答会给我们错误的结果：

 df1.loc[df1.Name.isin(df2.Name), ['Nonprofit', 'Education']] = df2[['Nonprofit', 'Education']]

df1
Out[852]: 
  Name  Nonprofit  Business  Education
0    X        1.0         1        0.0
1    Y        1.0         1        1.0
2    Z        NaN         0        NaN
3    Y        NaN         1        NaN

好吧，只有当“名称”列中的值是唯一的并且在两个数据框中都排序时，它才会安全地工作。

这是我的回答：

方式一：

df1 = df1.merge(df2,on='Name',how="left")
df1['Nonprofit_y'] = df1['Nonprofit_y'].fillna(df1['Nonprofit_x'])
df1['Business_y'] = df1['Business_y'].fillna(df1['Business_x'])
df1.drop(["Business_x","Nonprofit_x"],inplace=True,axis=1)
df1.rename(columns={'Business_y':'Business','Nonprofit_y':'Nonprofit'},inplace=True)

方式二：

df1 = df1.set_index('Name')
df2 = df2.set_index('Name')
df1.update(df2)
df1.reset_index(inplace=True)

有关更新的更多指南。. 在“更新”之前，需要设置索引的两个数据框的列名不必相同。您可以尝试“Name1”和“Name2”。此外，即使 df2 中的其他不必要的行也不会更新 df1，它也能工作。换句话说，df2 不需要是 df1 的超集。

例子：

df1 = pd.DataFrame([["X",1,1,0],
              ["Y",0,1,0],
              ["Z",0,0,0],
              ["Y",0,1,0]],columns=["Name1","Nonprofit","Business", "Education"])    

df2 = pd.DataFrame([["Y",1,1],
              ["Z",1,1],
              ['U',1,3]],columns=["Name2","Nonprofit", "Education"])   

df1 = df1.set_index('Name1')
df2 = df2.set_index('Name2')


df1.update(df2)

结果：

      Nonprofit  Business  Education
Name1                                
X           1.0         1        0.0
Y           1.0         1        1.0
Z           1.0         0        1.0
Y           1.0         1        1.0

Answer 2

使用boolean mask from isin来过滤df并从rhs df中分配所需的行值:

In [27]:

df.loc[df.Name.isin(df1.Name), ['Nonprofit', 'Education']] = df1[['Nonprofit', 'Education']]
df
Out[27]:
  Name  Nonprofit  Business  Education
0    X          1         1          0
1    Y          1         1          1
2    Z          1         0          1
3    Y          1         1          1

[4 rows x 4 columns]

Answer 3

在[27]中:这是正确的.

df.loc[df.Name.isin(df1.Name), ['Nonprofit', 'Education']] = df1[['Nonprofit', 'Education']].values

df
Out[27]:

Name  Nonprofit  Business  Education

0    X          1         1          0
1    Y          1         1          1
2    Z          1         0          1
3    Y          1         1          1

[4行x 4列]

只有当df1中的所有行都存在于df中时,上述操作才有效.换句话说,df应该是df1的超级集合

如果你在df1中有一些不匹配的行到df,你应该按照下面的说法进行操作

换句话说,df不是df1的超集:

df.loc[df.Name.isin(df1.Name), ['Nonprofit', 'Education']] = 
df1.loc[df1.Name.isin(df.Name),['Nonprofit', 'Education']].values

Answer 4

df2.set_index('Name').combine_first(df1.set_index('Name')).reset_index()