基于 Python 中的另一个数据帧更新数据帧

Updating a dataframe based on another dataframe in Python

我有一个 DataFrame ,比如说 df1,它的所有列都是正确的,除了 'Employee' 列。还有另一个 DataFrame ,比如 df2,它有正确的员工姓名,但存储在 'Staff' 列中。我正在尝试根据来自各自数据帧的 'key_df1' 和 'key_df2' 更新 df1。需要一些关于如何解决这个问题的帮助。 (请参阅下面图像中的预期输出)

data1=[['NYC-URBAN','JON','00','yes','BANKING','AC32456'],['WDC-RURAL','XING','00','Yes','FINANCE','AD45678'],['LONDON-URBAN','EDWARDS','00','No','IT','DE43216'],
     ['SINGAPORE-URBAN','WOLF','00','No','SPORTS','RT45327'],['MUMBAI-RURAL','NEMBIAR','00','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,仅针对 Key_df1 和 Key_df2 两者中的关系的行数据框。请参阅下面的第二个预期输出。

您可以使用布尔索引,例如:

mask = df1.Key_df1 == df2.Key_df1.reindex(df1.index)
df1.loc[mask, 'Employee'] = df2.Staff

输出:

              City Employee Income Travelling Industry   Key_df1
0        NYC-URBAN      JON  00        yes  BANKING   AC32456
1        WDC-RURAL    ALPHA  00        Yes  FINANCE   AD45678
2     LONDON-URBAN     BETA  00         No       IT   DE43216
3  SINGAPORE-URBAN     WOLF  00         No   SPORTS   RT45327
4     MUMBAI-RURAL  NEMBIAR  00         No       IT  Rs454457

您还可以在以下位置使用 numpy:

import numpy as np

df1['Employee'] = np.where(df1['Key_df1'] == df2['Key_df1'], df2['Staff'], df1['Employee'])

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 和索引更新(就地)它的 Employee 列 设置为 Key_df2,仅占用 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 并将其移动到之前的位置:

result = wrk.reset_index().reindex(columns=List1)

结果是:

              City Employee Income Travelling Industry   Key_df1
0        NYC-URBAN      JON  00        yes  BANKING   AC32456
1        WDC-RURAL    ALPHA  00        Yes  FINANCE   AD45678
2     LONDON-URBAN     BETA  00         No       IT   DE43216
3  SINGAPORE-URBAN     WOLF  00         No   SPORTS   RT45327
4     MUMBAI-RURAL  NEMBIAR  00         No       IT  Rs454457

根据有关 旅行 专栏

的评论进行编辑

现在只是 更新 是不够的,必须以其他方式解决任务。

从加入 df1df2.Staff 开始(与 set_index 正确加入):

result = df1.join(df2.set_index('Key_df2').Staff, on='Key_df1')

第二步(真正的更新)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)

最后的结果是:

              City   Employee Income Travelling Industry   Key_df1
0        NYC-URBAN        JON  00        yes  BANKING   AC32456
1        WDC-RURAL  ALPHA_Yes  00        Yes  FINANCE   AD45678
2     LONDON-URBAN    BETA_No  00         No       IT   DE43216
3  SINGAPORE-URBAN       WOLF  00         No   SPORTS   RT45327
4     MUMBAI-RURAL    NEMBIAR  00         No       IT  Rs454457