Answers for "python merge using or statement"

0

python merge using or statement

dupes = (df.B_on_A_match_1 == df.B_on_A_match_2) # also could remove A_on_B_match
df.loc[~dupes]

     A  A_on_B_match_1  A_on_B_match_2    B  B_on_A_match_1  B_on_A_match_2
0  1.0             NaN             NaN  NaN             9.0             4.0
0  NaN             2.0             6.0  8.0             NaN             NaN
1  NaN             4.0             4.0  5.0             NaN             NaN
Posted by: Guest on January-04-2020
0

python merge using or statement

suff_A = ['_on_A_match_1', '_on_A_match_2']
suff_B = ['_on_B_match_1', '_on_B_match_2']

pd.concat([df1.merge(df2, on='A', suffixes=suff_A), 
           df1.merge(df2, on='B', suffixes=suff_B)])

     A  A_on_B_match_1  A_on_B_match_2    B  B_on_A_match_1  B_on_A_match_2
0  1.0             NaN             NaN  NaN             9.0             4.0
1  4.0             NaN             NaN  NaN             5.0             5.0
0  NaN             2.0             6.0  8.0             NaN             NaN
1  NaN             4.0             4.0  5.0             NaN             NaN
Posted by: Guest on January-04-2020
0

python merge using or statement

dupes = (df.B_on_A_match_1 == df.B_on_A_match_2) # also could remove A_on_B_match
df.loc[~dupes]

     A  A_on_B_match_1  A_on_B_match_2    B  B_on_A_match_1  B_on_A_match_2
0  1.0             NaN             NaN  NaN             9.0             4.0
0  NaN             2.0             6.0  8.0             NaN             NaN
1  NaN             4.0             4.0  5.0             NaN             NaN
Posted by: Guest on January-04-2020
0

python merge using or statement

suff_A = ['_on_A_match_1', '_on_A_match_2']
suff_B = ['_on_B_match_1', '_on_B_match_2']

pd.concat([df1.merge(df2, on='A', suffixes=suff_A), 
           df1.merge(df2, on='B', suffixes=suff_B)])

     A  A_on_B_match_1  A_on_B_match_2    B  B_on_A_match_1  B_on_A_match_2
0  1.0             NaN             NaN  NaN             9.0             4.0
1  4.0             NaN             NaN  NaN             5.0             5.0
0  NaN             2.0             6.0  8.0             NaN             NaN
1  NaN             4.0             4.0  5.0             NaN             NaN
Posted by: Guest on January-04-2020

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