python - 检查列内的字符串错误行为 - Pandas

Question

我有以下数据框:

import pandas as pd 

df = pd.DataFrame ({
    'Name':['Unable host=test1, status=c0000192','Unable host=test2, status=22228001','Unable host=test3, status=c000018d'],
    'Error':['NaN','NaN','NaN']
    })


print(df)
                                   Name Error
0  Unable host=test1, status=c0000192   NaN
1  Unable host=test2, status=22228001   NaN
2  Unable host=test3, status=c000018d   NaN

代码应遵循以下 if 语句:

import pandas as pd 

df = pd.DataFrame ({
    'Name':['Unable host=test1, status=c0000192','Unable host=test2, status=22228001','Unable host=test3, status=c000018d'],
    'Error':['NaN','NaN','NaN']
    })

if df['Name'].str.contains('status=c000018d').any():
    df['Error'] = 'Trust Failure'

elif df['Name'].str.contains('status=22228001').any():
    df['Error'] = 'No creds'

else:
    df['Error'] = 'Other'

print(df)

输出应该是:

                                   Name          Error
0  Unable host=test1, status=c0000192  Other
1  Unable host=test2, status=22228001  No creds
2  Unable host=test3, status=c000018d  Trust Failure

相反，我得到:

                                   Name          Error
0  Unable host=test1, status=c0000192  Trust Failure
1  Unable host=test2, status=22228001  Trust Failure
2  Unable host=test3, status=c000018d  Trust Failure

为什么会这样？这就像第一个 if 语句匹配所有内容。

Answer 1

我认为你在这里混合了一些东西，试试这个方法:

df.loc[df['Name'].str.contains('status=c000018d'), 'Error'] = 'Trust Failure'
df.loc[df['Name'].str.contains('status=22228001'), 'Error'] = 'No creds'
df.loc[df['Name'] == 'NaN', 'Error'] = 'Other'

您正在执行的操作在第一个 if 语句中返回 true，并用“信任失败”填充所有值。