Question

我将2个CSV文件与pandas python进行比较，一切正常。它匹配 employee_id 列并将结果输出到csv文件

    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)

如果找不到匹配项，则会返回空白结果，我希望它返回 not_found

Pandas可以这样做，或者之后我应该做一些处理吗？

Answer 1

我认为如果NaN中没有df1，您可以使用fillna：

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

但是，如果NaN中df1是必需的，请创建掩码以识别right加入中的缺失值 - merge中的参数indicator=True或{{3}通过~否定掩码：

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

Python Pandas - 未找到匹配时返回默认值

1 个答案: