基于另一个数据框 python pandas 替换列值 - 更好的方法？

Question

Note:for simplicity's sake, i'm using a toy example, because copy/pasting dataframes is difficult in stack overflow (please let me know if there's an easy way to do this).

Is there a way to merge the values from one dataframe onto another without getting the _X, _Y columns? I'd like the values on one column to replace all zero values of another column.

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

In a previous post, I tried combine_First and dropna(), but these don't do the job.

I want to replace zeros in df1 with the values in df2. Furthermore, I want all rows with the same Names to be changed according to df2.

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

(need to clarify: The value in 'Business' column where name = Z should 0.)

My existing solution does the following: I subset based on the names that exist in df2, and then replace those values with the correct value. However, I'd like a less hacky way to do this.

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

Attention: In latest version of pandas, both answers above doesn't work anymore:

KSD's answer will raise error:

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,)

and EdChum's answer will give us the wrong result:

 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

Well, it will work safely only if values in column 'Name' are unique and are sorted in both data frames.

Here is my answer:

Way 1:

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)

Way 2:

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

More guide about update. . The columns names of both data frames need to set index are not necessary same before 'update'. You could try 'Name1' and 'Name2'. Also, it works even if other unnecessary row in df2, which won't update df1. In other words, df2 doesn't need to be the super set of df1.

Example:

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)

result:

      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

Use the boolean mask from isin to filter the df and assign the desired row values from the rhs df:

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

In [27]: This is the correct one.

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 rows x 4 columns]

The above will work only when all rows in df1 exists in df . In other words df should be super set of df1

Incase if you have some non matching rows to df in df1,you should follow below

In other words df is not superset of df1 :

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()