Question

输入解释： 我有两个数据框df1和df2，它们分别包含如下所述的列。

df1

Description   Col1    Col2
AAA           1.2     2.5
BBB           1.3     2.0
CCC           1.1     2.3

df2

Description   Col1    Col2
AAA           1.2     1.3
BBB           1.3     2.0

场景： 当两者相等时，必须比较df1['Description']和df2['Description']，然后必须将df1['Col1']与df2['Col1']和df1['Col2']与df2['Col2']比较，并按预期产生结果在下面。

预期输出：

Description    Col1    Col2   Col1_Result           Col2_Result
AAA            1.2     2.5    Pass                  Fail
BBB            1.3     2.0    Pass                  Pass
CCC            1.1     2.3    Not found in df2      Not found in df2

尝试代码： 已经针对上述情况尝试了以下提到的代码行，但是不起作用。通过错误“ ValueError：只能比较标记相同的Series对象”

df1['Col1_Result'] = np.where(df1['Description']== df2['Description'],np.where(df1['Col1'] == df2['Col1'], 'Pass', 'Fail'),'Not found in df2')
df1['Col2_Result'] = np.where(df1['Description']== df2['Description'],np.where(df1['Col2'] == df2['Col2'], 'Pass', 'Fail'),'Not found in df2')

预先感谢！

Answer 1

或者，下面的代码适用于给定的示例。如果存在边缘情况，可以根据需要进行修改。

# Import libraries
import pandas as pd

# Create DataFrame
df1 = pd.DataFrame({
    'Description':['AaA', 'BBB','CCC'],
    'Col1': [1.2,1.3,1.1],
    'Col2':[2.5,2.0,2.3]
})
df2 = pd.DataFrame({
    'Description': ['AAA', 'BBB'],
    'Col1': [1.2, 1.3],
    'Col2': [1.3, 2.0]
})

# Convert to lower case
df1['Description'] = df1['Description'].str.lower()
df2['Description'] = df2['Description'].str.lower()

# Merge df
df = df1.merge(df2, on='Description', how='left')


# Compare
df['Col1_result'] = df.apply(lambda x: 'Not found in df2' if (pd.isna(x['Col1_y'])) else
                                        'Pass' if x['Col1_x']==x['Col1_y'] else
                                        'Fail', axis=1)
df['Col2_result'] = df.apply(lambda x: 'Not found in df2' if (pd.isna(x['Col2_y'])) else
                                        'Pass' if x['Col2_x']==x['Col2_y'] else
                                        'Fail', axis=1)

# Keep only columns from df1
df = df.drop(['Col1_y', 'Col2_y'], axis=1)
# Remove '_x' from column names
df.columns = df.columns.str.replace(r'_x$', '')

# Change to upper case
df['Description'] = df['Description'].str.upper()

输出

df

  Description  Col1  Col2       Col1_result       Col2_result
0         AAA   1.2   2.5              Pass              Fail
1         BBB   1.3   2.0              Pass              Pass
2         CCC   1.1   2.3  Not found in df2  Not found in df2

Answer 2

将DataFrame.merge与左联接一起用于输出DataFrame，然后按DataFrame.filter选择添加的列，并通过首先比较缺失值的值来创建输出，然后在numpy.select中彼此列：

df1['desc'] = df1['Description'].str.lower() 
df2['desc'] = df2['Description'].str.lower()
df = (df1.merge(df2, on='desc', suffixes=['', '_Result'], how='left')
         .drop(['Description_Result','desc'], axis=1))

df3 = df.filter(like='_Result')
new = df3.rename(columns=lambda x: x.replace('_Result',''))

df[df3.columns] = np.select([new.isna(), 
                             df[new.columns].eq(new)], 
                            ['Not found in df2', 'Pass'], 'Fail')
print (df)
  Description  Col1  Col2       Col1_Result       Col2_Result
0         AAA   1.2   2.5              Pass              Fail
1         BBB   1.3   2.0              Pass              Pass
2         CCC   1.1   2.3  Not found in df2  Not found in df2

详细信息：

print (df3)
   Col1_Result  Col2_Result
0          1.2          1.3
1          1.3          2.0
2          NaN          NaN

print (new)
   Col1  Col2
0   1.2   1.3
1   1.3   2.0
2   NaN   NaN

根据比较其他列将数据框“ df1”的字符串列与数据框“ df2”的另一个字符串列进行比较

2 个答案: