如何简化Pandas账户名称与机构部门代码一致性校验函数

现有数据集包含ACCOUNT_NAME（账户名称）和INSTITUTIONAL_SECTOR_CODE（机构部门代码）两列，业务校验规则如下：

• 若账户名称包含list1 = ['limited','Hotel','Company','Restaurant','Sacco']中的元素，对应的机构部门代码需匹配list2 = ['NFCPVT','OLENT','MENT','SENT','MFI']中的对应值
• list1前4个元素对应的账户可匹配list2前4个任意代码
• list1最后一个元素Sacco仅可匹配list2的MFI值
• 最终需要输出不在对应允许值内的机构部门代码列表或计数统计

当前采用硬编码方式实现，现有代码如下：

import pandas as pd
import numpy as np

def acc_name_and_inst_sec_codes_consistency(df):
    
    #account name
    Comp_acc_name = ['ltdt','limited','Hotel','Company','Restaurant','Consultant','technologies','Service','atelier','quincaillerie']
    Sacco_acc_name = ['SACCO']
    inst_sector_code_comp_acc_name = ['NFCPVT','OLENT','MENT','SENT','NFCPUB']
    inst_sector_code_Sacco_acc_name = ['MFI']

    Filtered_comp_acc_name = df[df['ACCOUNT_NAME'].str.contains('|'.join(Comp_acc_name), case=False)]
    Filtered_sacco_acc_name = df[df['ACCOUNT_NAME'].str.contains('|'.join(Sacco_acc_name), case=False)]

    print('Below are institutional sector codes that shouldnot be included for the Account name with',\
      Comp_acc_name,'in it\n\n',Filtered_comp_acc_name[~Filtered_comp_acc_name['INSTITUTIONAL_SECTOR_CODE'].\
      isin(inst_sector_code_comp_acc_name)]['INSTITUTIONAL_SECTOR_CODE_DESC'].value_counts().\
      rename_axis('INSTITUTIONAL_SECTOR_CODE_DESC').reset_index(name='counts').sort_values('counts', ascending = False).\
      to_string(index=False),'\n\n' 'Below are institutional sector codes that shouldnot be included for the Account name with', Sacco_acc_name,'in it\n\n',Filtered_sacco_acc_name[~Filtered_sacco_acc_name['INSTITUTIONAL_SECTOR_CODE'].\
    isin(inst_sector_code_Sacco_acc_name)]['INSTITUTIONAL_SECTOR_CODE_DESC'].value_counts().\
      rename_axis('INSTITUTIONAL_SECTOR_CODE_DESC').reset_index(name='counts').sort_values('counts', ascending = False).\
      to_string(index=False))

样例输出如下：

Below are institutional sector codes that shouldnot be included for the Account name with ['ltdt', 'limited', 'Hotel', 'Company', 'Restaurant', 'Consultant', 'technologies', 'Service', 'atelier', 'quincaillerie'] in it
 INSTITUTIONAL_SECTOR_CODE_DESC  counts
                Not Applicable   11012
                   Individuals    1537
                     Insurance       9
    MFI/SACCO (Deposit Taking)       3 
Below are institutional sector codes that shouldnot be included for the Account name with ['SACCO'] in it
     INSTITUTIONAL_SECTOR_CODE_DESC  counts
                    Not Applicable    1001
                       Individuals      28
 Non Financial Companies - Private       1 

请问有什么方法可以简化上述函数？

核心优化思路是把重复的统计逻辑抽离，同时将匹配规则配置化，避免硬编码带来的维护问题，简化后代码如下：

import pandas as pd
import numpy as np

# 抽离重复的非法值统计逻辑
def get_invalid_count_str(filtered_df, allowed_codes):
    invalid_df = filtered_df[~filtered_df['INSTITUTIONAL_SECTOR_CODE'].isin(allowed_codes)]
    count_res = invalid_df['INSTITUTIONAL_SECTOR_CODE_DESC'].value_counts()\
                .rename_axis('INSTITUTIONAL_SECTOR_CODE_DESC')\
                .reset_index(name='counts')\
                .sort_values('counts', ascending=False)
    return count_res.to_string(index=False)

def acc_name_and_inst_sec_codes_consistency(df):
    # 规则配置：每个元素为(关键词列表, 允许的代码列表, 规则描述)
    RULES = [
        (
            ['ltdt','limited','Hotel','Company','Restaurant','Consultant','technologies','Service','atelier','quincaillerie'],
            ['NFCPVT','OLENT','MENT','SENT','NFCPUB'],
            '普通企业类账户关键词'
        ),
        (
            ['SACCO'],
            ['MFI'],
            'Sacco类账户关键词'
        )
    ]

    for keywords, allowed_codes, rule_desc in RULES:
        filtered_df = df[df['ACCOUNT_NAME'].str.contains('|'.join(keywords), case=False)]
        count_str = get_invalid_count_str(filtered_df, allowed_codes)
        print(f"以下是账户名称包含{rule_desc}时不允许出现的机构部门代码统计：\n\n{count_str}\n")

优化点说明 #

• 重复的统计逻辑统一封装为工具函数，后续调整统计格式、排序规则只需要修改一处即可
• 匹配规则用结构化的配置管理，新增校验规则不需要修改主逻辑，直接在RULES列表里加对应条目即可，可维护性大幅提升
• 去掉冗余的代码拼接逻辑，用f-string格式化输出，可读性更强
• 输出内容改为中文，更符合使用习惯，如果需要保留英文输出只需要修改print的模板即可

内容的提问来源于stack exchange，提问作者VMO