Python 我需要比较两个df'；s对于匹配和不匹配，我还需要确定在不匹配的情况下哪个答案来自主df

我在python中有两个数据帧，希望比较这两个数据帧以查找匹配和不匹配。但重要的是，我可以在不匹配中确定哪个答案来自主答案表，哪个答案来自用户答案

我决定使用pandas df.where函数来实现这一点，除了能够识别哪个答案来自主答案表，以及在出现不匹配的情况下哪些答案来自用户答案之外，它还起到了作用

# I have a DataFrame called df_master (master answer sheet)

import pandas as pd

df_master = pd.DataFrame({'B0': [1, 0, 0, 0, 0, 1],
            'B1': [0, 0, 0, 0, 1, 0],
            'B2': [0, 1, 0, 0, 0, 0],
            'B3': [0, 0, 1, 0, 0, 0],
            'B4': [0, 0, 0, 1, 0, 0]})
print(df_master)

#    B0  B1  B2  B3  B4
# 0   1   0   0   0   0
# 1   0   0   1   0   0
# 2   0   0   0   1   0
# 3   0   0   0   0   1
# 4   0   1   0   0   0
# 5   1   0   0   0   0

# I also have a DataFrame called df_answers (users answers)

df_answers = pd.DataFrame({'B0': [0, 0, 0, 0, 0, 1],
            'B1': [1, 0, 0, 0, 1, 0],
            'B2': [0, 0, 0, 0, 0, 0],
            'B3': [0, 1, 1, 0, 0, 0],
            'B4': [0, 0, 0, 1, 0, 0]})

print(df_answers)

#    B0  B1  B2  B3  B4
# 0   0   1   0   0   0
# 1   0   0   0   1   0
# 2   0   0   0   1   0
# 3   0   0   0   0   1
# 4   0   1   0   0   0
# 5   1   0   0   0   0

# when I compare the the two df's, for each match, matches correctly, where there
# is no match I have used other=2.  However this is a problem as I cannot see which is
# the correct answer.  It would be great if there was a way to work the code to reflect
# the master as a 3 and the incorrect answer from the users to stay 2?

comparison = df_master.where(df_master.values==df_answers.values, other=2)

print(comparison)

# My Results

#    B0  B1  B2  B3  B4
# 0   2   2   0   0   0
# 1   0   0   2   2   0
# 2   0   0   0   1   0
# 3   0   0   0   0   1
# 4   0   1   0   0   0
# 5   1   0   0   0   0

# Expected Results

#    B0  B1  B2  B3  B4
# 0   3   2   0   0   0
# 1   0   0   3   2   0
# 2   0   0   0   1   0
# 3   0   0   0   0   1
# 4   0   1   0   0   0
# 5   1   0   0   0   0

---

在您的示例中，在str sum之后使用

replace

，ps：您可以自己定义映射，如

{'00'：'tware failed'，'01'：'master failed'…}

(df_answers.astype(str)+df_master.astype(str)).replace({'00':0,'01':3,'10':2,'11':1})
Out[129]: 
   B0  B1  B2  B3  B4
0   3   2   0   0   0
1   0   0   3   2   0
2   0   0   0   1   0
3   0   0   0   0   1
4   0   1   0   0   0
5   1   0   0   0   0

---

谢谢@WeNYoBen，你的回答非常有效，我花了一点时间才理解，因为我是python新手，但现在我了解了它的工作原理，我也接受了你的建议，使用标签名称，例如：“01”：“master failed”。再次感谢您快速准确的回复。