Question

基于此答案， Avoiding KeyError in dataframe，我可以进行验证。但是我需要跟踪由于哪种验证条件而导致哪一行失败。

是否可以添加新列并提供失败消息？

我的代码-

valid_dict = {'name': 'WI 80 INDEMNITY 18 OPTION 1 SILVER RX $10/45/90/25%',
                          'issuer_id': 484,
                          'service_area_id': 1,
                          'plan_year': 2018,
                          'network_url': np.nan,
                          'formulary_url': np.nan,
                          'sbc_download_url': np.nan,
                          'treatment_cost_calculator_url': np.nan,
                          'promotional_label': np.nan,
                          'hios_plan_identifier': '99806CAAUSJ-TMP',
                          'type': 'MetalPlan',
                          'price_period': 'Monthly',
                          'is_age_29_plan': False,
                          'sort_rank_override': np.nan,
                          'composite_rating': False,
                          }

            data_obj = DataService()
            hios_issuer_identifer_list = data_obj.get_hios_issuer_identifer(df)

            d1 = {k: v for k, v in valid_dict.items() if k in set(valid_dict.keys()) - set(df.columns)}
            df1 = df.assign(**d1)
            cols_url = df.columns.intersection(['network_url', 'formulary_url', 'sbc_download_url', 'treatment_cost_calculator_url'])
            m1 = (df1[['name', 'issuer_id', 'service_area_id']].notnull().all(axis=1))
            m2 = (df1[['promotional_label']].astype(str).apply(lambda x: (x.str.len <= 65) | x.isin(['nan'])).all(axis=1))
            m3 = (df1[cols_url].astype(str).apply(lambda x: (x.str.contains('\A(https?:\/\/)([a-zA-Z0-9\-_])*(\.)*([a-zA-Z0-9\-]+)\.([a-zA-Z\.]{2,5})(\.*.*)?\Z')) | x.isin(['nan'])).all(axis=1))
            m4 = ((df1['plan_year'].notnull()) & (df['plan_year'].astype(str).str.isdigit()) & (df['plan_year'].astype(str).str.len() == 4))
            m5 = ((df1['hios_plan_identifier'].notnull()) & (df['hios_plan_identifier'].str.len() >= 10) & (df['hios_plan_identifier'].str.contains('\A(\d{5}[A-Z]{2}[a-zA-Z0-9]{3,7}-TMP|\d{5}[A-Z]{2}\d{3,7}(\-?\d{2})*)\Z')))
            m6 = (df1['type'].isin(['MetalPlan', 'MedicarePlan', 'BasicHealthPlan', 'DualPlan', 'MedicaidPlan', 'ChipPlan']))
            m7 = (df1['price_period'].isin(['Monthly', 'Yearly']))
            m8 = (df1['is_age_29_plan'].astype(str).isin(['True', 'False', 'nan']))
            m9 = (df1[['sort_rank_override']].astype(str).apply(lambda x: (x.str.isdigit()) | x.isin(['nan'])).all(axis=1))
            m10 = (df1['composite_rating'].astype(str).isin(['True', 'False']))
            m11 = (df1['hios_plan_identifier'].astype(str).str[:5].isin(hios_issuer_identifer_list))

            df1 = df1[m1 & m2 & m3 & m4 & m5 & m6 & m7 & m8 & m9 & m10 & m11].drop(d1.keys(), axis=1)

            merged =  df.merge(df1.drop_duplicates(), how='outer', indicator=True)
            merged[merged['_merge'] == 'left_only'].to_csv('logs/invalid_plan_data.csv')

            return df1

类似下面的内容

 wellthie_issuer_identifier  issuer_name    ...     service_area_id     _error
0                   UHC99806  Fake Humana    ...                   1  failed on plan_year

Answer 1

使用df1 = df1[m1 & m2 & m3 & m4 & m5 & m6 & m7 & m8 & m9 & m10 & m11].drop(d1.keys(), axis=1)，您将选择所有条件均未失败的行。显然，您在这里不会有想要的东西，这没关系，因为这是经过验证的部分，应该没有错误。

在删除失败的行之前，可以通过再次选择来得到错误：

df_error = df1.copy()
df_error['error_message'] = ~m1
...

如果列中有错误，则可以定义一些错误文本以显示在表中：

df_error['failed_on_name'] = pd.where(m1, your_message_here)

如果要在日志中显示错误，则可以遍历错误表并输出消息（考虑第一个版本的列中具有布尔值）

for _, row in df_error.iterrows():
    print (error_message(dict(row)))

因此，您将能够使用如下功能处理行：

def error_message(row):
    row_desc = []
    error_msg = []
    for k, v in row.items():
        if isinstance(v, bool):
            if v:
                error_msg.append(k)
        else:
            row_desc.append(v)
    return 'Row ' + ' '.join(row_desc) + ' failed with errors: ' + ' '.join(error_msg)

在熊猫数据框中添加错误日志消息行

1 个答案: