[英]Updating a dataframe based on another dataframe in Python
我有一个 DataFrame,比如说 df1,它的所有列都正确,除了“员工”列。 还有另一个 DataFrame,比如说 df2,它有正确的员工姓名,但存储在“员工”列中。 我正在尝试根据相应 DataFrame 中的 'key_df1' 和 'key_df2' 更新 df1。 需要一些帮助来解决这个问题。 (请看下图中预期的output)
data1=[['NYC-URBAN','JON','$5000','yes','BANKING','AC32456'],['WDC-RURAL','XING','$4500','Yes','FINANCE','AD45678'],['LONDON-URBAN','EDWARDS','$3500','No','IT','DE43216'],
['SINGAPORE-URBAN','WOLF','$5000','No','SPORTS','RT45327'],['MUMBAI-RURAL','NEMBIAR','$2500','No','IT','Rs454457']]
data2=[['NYC','MIKE','BANKING','BIKING','AH56245'],['WDC','ALPHA','FINANCE','TREKKING','AD45678'],
['LONDON-URBAN','BETA','FINANCE','SLEEPING','DE43216'],['SINGAPORE','WOLF','SPORTS','DANCING','RT45307'],
['MUMBAI','NEMBIAR','IT','ZUDO','RS454453']]
List1=['City','Employee', 'Income','Travelling','Industry', 'Key_df1']
List2=['City','Staff','Industry','Hobby', 'Key_df1']
df1=pd.DataFrame(data1,columns=List1)
df2=pd.DataFrame(data2,columns=List2)
预期输出:
编辑(附加查询):
感谢您的回复。 除了上述问题,我想将 'Employee' 列的值与 df1 中的 'Travelling' 列连接起来,仅用于 Key_df1 和 Key_df2 在两个 DataFrame 中关联的行。 请参阅下面的第二个预期 output。
首先将df1中的索引设置为Key_df1 ,并保存为临时DataFrame:
wrk = df1.set_index('Key_df1')
然后使用df2更新(就地)其Employee列,索引设置为Key_df2 ,仅采用Staff列:
wrk.Employee.update(df2.set_index('Key_df2').Staff)
最后一个操作是将索引更改为“常规”列并将其移动到上一个位置:
result = wrk.reset_index().reindex(columns=List1)
结果是:
City Employee Income Travelling Industry Key_df1
0 NYC-URBAN JON $5000 yes BANKING AC32456
1 WDC-RURAL ALPHA $4500 Yes FINANCE AD45678
2 LONDON-URBAN BETA $3500 No IT DE43216
3 SINGAPORE-URBAN WOLF $5000 No SPORTS RT45327
4 MUMBAI-RURAL NEMBIAR $2500 No IT Rs454457
现在仅仅更新是不够的,任务必须以另一种方式解决。
从加入df1和df2.Staff开始(使用set_index正确加入):
result = df1.join(df2.set_index('Key_df2').Staff, on='Key_df1')
第二步(实际更新)是:
result.Employee.where(result.Staff.isna(), result.Staff + '_' + result.Travelling,
inplace=True)
最后一步是删除Staff列(不再需要):
result.drop(columns=['Staff'], inplace=True)
最终结果是:
City Employee Income Travelling Industry Key_df1
0 NYC-URBAN JON $5000 yes BANKING AC32456
1 WDC-RURAL ALPHA_Yes $4500 Yes FINANCE AD45678
2 LONDON-URBAN BETA_No $3500 No IT DE43216
3 SINGAPORE-URBAN WOLF $5000 No SPORTS RT45327
4 MUMBAI-RURAL NEMBIAR $2500 No IT Rs454457
您可以使用 Boolean 索引,例如:
mask = df1.Key_df1 == df2.Key_df1.reindex(df1.index)
df1.loc[mask, 'Employee'] = df2.Staff
Output:
City Employee Income Travelling Industry Key_df1
0 NYC-URBAN JON $5000 yes BANKING AC32456
1 WDC-RURAL ALPHA $4500 Yes FINANCE AD45678
2 LONDON-URBAN BETA $3500 No IT DE43216
3 SINGAPORE-URBAN WOLF $5000 No SPORTS RT45327
4 MUMBAI-RURAL NEMBIAR $2500 No IT Rs454457
您还可以使用 numpy 其中:
import numpy as np
df1['Employee'] = np.where(df1['Key_df1'] == df2['Key_df1'], df2['Staff'], df1['Employee'])
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