Python新手在这里，试图创建一个布尔掩码子集数据集。我将感谢任何指导，如何使这个面具的工作和我做错了什么。谢谢你，谢谢
我的df看起来像这样：

import pandas as pd

df= pd.DataFrame({'A': ['A-dog', "B-dog","C-cat" , "E-snake", "F-hamser"],
                  'B': ['F-dog', "B-parrot","C-snake" , "E-cat", "F-bird"],
                  'C': [1, 2, 3, 4, 5],
                  'D': [22,23,24,25,26],
                  'E': ['A-snake', "B-dog","C-snake" , "E-snake", "F-snake"],
                  'Flag': [0,0,0,0,0]})

df

我想评估列A，B和E，并将以“狗”和“猫”结尾的单元格替换为“”，并将我进行替换的行的“标志”列更改为1。
我想创建一个布尔掩码，这样我就可以替换字符串并将“Flag”更改为1，但我的掩码不起作用。
这就是我所尝试的：

cols=['A','B','E']
mask=df[cols].apply(lambda x: 'dog'  or 'cat' in x[-3:])
# x[-3:] to select the last three characters of the string.
# If the mask were working, I would change the flag variable in this way
df.loc(mask.any(axis=1),'Flag')=1

我得到的df看起来像这样：

res= pd.DataFrame({'A': ['A-', "B-","C-" , "E-snake", "F-hamser"],
                  'B': ['F-', "B-parrot","C-snake" , "E-", "F-bird"],
                  'C': [1, 2, 3, 4, 5],
                  'D': [22,23,24,25,26],
                  'E': ['A-snake', "B-","C-snake" , "E-snake", "F-snake"],
                  'Flag': [1,1,1,1,0]})

res