Reputation: 12608
I am comparing 2 CSV files with pandas python and all is working correctly. It is matching on the employee_id column and outputting the results to a csv file
df1 = pd.read_csv('input1.csv', sep=',\s+', delimiter=',', encoding="utf-8")
df2 = pd.read_csv('input2.csv', sep=',\s,', delimiter=',', encoding="utf-8")
df3 = pd.merge(df1,df2, on='employee_id', how='right')
df3.to_csv('output.csv', encoding='utf-8', index=False)
When it does not find a match it returns a blank result, I would like it to instead return not_found
Is this possible with Pandas or should I do some processing afterwards?
Upvotes: 3
Views: 1344
Reputation: 863031
I think you can use fillna
if no NaN
in df1
:
df1 = pd.DataFrame({'employee_id':[1,2,3],
'B':[4,5,6],
'C':[7,8,9]})
print (df1)
B C employee_id
0 4 7 1
1 5 8 2
2 6 9 3
df2 = pd.DataFrame({'employee_id':[1,4,6],
'D':[4,5,6],
'E':[7,8,9]})
print (df2)
D E employee_id
0 4 7 1
1 5 8 4
2 6 9 6
df3 = pd.merge(df1,df2, on='employee_id', how='right')
df3[df1.columns] = df3[df1.columns].fillna('not_found')
print (df3)
B C employee_id D E
0 4 7 1 4 7
1 not_found not_found 4 5 8
2 not_found not_found 6 6 9
But if NaN
in df1
is necessary create mask for identify missing values from right
join - with parameter indicator=True
in merge
or with isin
and negate mask by ~
:
df1 = pd.DataFrame({'employee_id':[1,2,3],
'B':[np.nan,5,6],
'C':[7,8,9]})
print (df1)
B C employee_id
0 NaN 7 1
1 5.0 8 2
2 6.0 9 3
df3 = pd.merge(df1,df2, on='employee_id', how='right', indicator=True)
mask = df3['_merge'] == 'right_only'
df3.loc[mask, df1.columns.difference(['employee_id'])] =
df3.loc[mask,df1.columns.difference(['employee_id'])].fillna('not_found')
df3 = df3.drop('_merge', axis=1)
print (df3)
B C employee_id D E
0 NaN 7 1 4 7
1 not_found not_found 4 5 8
2 not_found not_found 6 6 9
df3 = pd.merge(df1,df2, on='employee_id', how='right')
mask = ~df2['employee_id'].isin(df1['employee_id'])
df3.loc[mask, df1.columns.difference(['employee_id'])] = \
df3.loc[mask,df1.columns.difference(['employee_id'])].fillna('not_found')
print (df3)
B C employee_id D E
0 NaN 7 1 4 7
1 not_found not_found 4 5 8
2 not_found not_found 6 6 9
Upvotes: 2