Merge two different dataframes on different column names [duplicate]

Answer 1

You can also use join with default left join or merge, last if necessary remove rows with NaNs by dropna:

print (df1.join(df2.set_index('A'), on='A').dropna())
    A    B   C  BB   CC  DD
0  A1  123  K0  B0  121  D0
1  A1  345  K1  B0  121  D0
3  A3  146  K1  B3  345  D1

---

print (pd.merge(df1, df2, on='A', how='left').dropna())
    A    B   C  BB   CC  DD
0  A1  123  K0  B0  121  D0
1  A1  345  K1  B0  121  D0
3  A3  146  K1  B3  345  D1

EDIT:

I think you need inner join (by default, so on='inner' can be omit):

Both_DFs = pd.merge(df1,df2, left_on=['A','B'],right_on=['A','CC'])
print (Both_DFs)
    A    B   C  BB   CC  DD
0  A1  121  K0  B0  121  D0

Answer 2

I don't know if your example show exactly your problem but,

If we try to merge with MultiIndex, we need to have the 2 index matching.

df1['A'] == df2['A'] && df1['B'] == df2['CC']

Here we haven't any row that match the 2 index.

If we merge just by df1['A'], we got something like this : Both_DFs=pd.merge(df1, df2, how='left', left_on=['A'], right_on=['A'])

    A    B   C   BB   CC   DD
0  A1  123  K0   B0  121   D0
1  A1  345  K1   B0  121   D0
2  A2  121  K0  NaN  NaN  NaN
3  A3  146  K1   B3  345   D1

If you wan't remove line row that not in df2 try to change 'how' method to inner.

Both_DFs=pd.merge(df1, df2, how='left', left_on=['A'], right_on=['A'])
   A    B   C   BB   CC   DD
0  A1  123  K0   B0  121   D0
1  A1  345  K1   B0  121   D0
2  A3  146  K1   B3  345   D1

Did this approach of what you're looking for ?

Answer 3

Well, if you declare column A as index, it works:

Both_DFs = pd.merge(df1.set_index('A', drop=True),df2.set_index('A', drop=True), how='left',left_on=['B'],right_on=['CC'], left_index=True, right_index=True).dropna().reset_index()

This results in:

    A    B   C  BB   CC  DD
0  A1  123  K0  B0  121  D0
1  A1  345  K1  B0  121  D0
2  A3  146  K1  B3  345  D1

EDIT

You just needed:

Both_DFs = pd.merge(df1,df2, how='left',left_on=['A','B'],right_on=['A','CC']).dropna()

Which gives:

    A    B   C  BB   CC  DD
0  A1  121  K0  B0  121  D0