Python-未找到匹配项时返回默认值
我将2个CSV文件与python进行比较,所有文件都正常工作。它在employee_id列上进行匹配,并将结果输出到csv文件Python-未找到匹配项时返回默认值,python,csv,pandas,Python,Csv,Pandas,我将2个CSV文件与python进行比较,所有文件都正常工作。它在employee_id列上进行匹配,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,d
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)
当它没有找到匹配项时,会返回一个空白结果,我希望它返回未找到
熊猫可以这样做吗?或者我应该在以后做一些处理吗?我认为您可以在df1
中使用if noNaN
:
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,如果有必要在df1
中创建掩码,以识别右侧的缺失值,则在或中使用参数连接-indicator=True
,并通过~/code>取消掩码:
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
非常感谢,它们是很好的例子。省去了我事后处理文件的麻烦。Grad可以帮助您;)天气真好!
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