如何在Python字典中匹配指定短语并输出对应键？代码纠错

需求：在给定的dict_ep字典中，找到指定短语（如'FRAUDED BILL'）对应的字典键（例如匹配该短语时输出EP1_2）。

原代码 #

dict_ep = {'EP1_2':['FRAUDED BILL','IMPROPER BILLING - FRAUDED CARD (CARDS)','EMBEZZLEMENT','FRAUD'], 
            'EP1_4':['2nd COPY OF CONTRACT (CONSIGNEE)','ACCIDENT WITH DISPOSED VEHICLE'],
            'EP1_6':['BANK STRIKE'],
            'EP1_8':['ACCESS TO BALANCE AND CARD LIMIT','PAYMENT AGREEMENT']}

problem = ('frauded bill').upper()


for i in dict_ep:
    if problem == dict_ep.keys():
        print('EP found')
    else:
        print('EP no exist, try again!')

错误运行结果 #

EP no exist, try again!
EP no exist, try again!
EP no exist, try again!
EP no exist, try again!

错误排查 #

• 判断逻辑完全错误：代码中用problem == dict_ep.keys()将目标字符串和字典的键集合做相等比较，类型和内容均不匹配，永远不会成立。
• 未访问对应的值列表：循环遍历字典键时，没有取出键对应的短语列表，无法检查目标短语是否存在。
• 输出逻辑错误：每次循环都输出结果，而非找到匹配项时输出对应键、未找到时仅输出一次提示。

修正后的代码 #

精确匹配版本（完全匹配短语） #

dict_ep = {'EP1_2':['FRAUDED BILL','IMPROPER BILLING - FRAUDED CARD (CARDS)','EMBEZZLEMENT','FRAUD'], 
            'EP1_4':['2nd COPY OF CONTRACT (CONSIGNEE)','ACCIDENT WITH DISPOSED VEHICLE'],
            'EP1_6':['BANK STRIKE'],
            'EP1_8':['ACCESS TO BALANCE AND CARD LIMIT','PAYMENT AGREEMENT']}

problem = ('frauded bill').upper()

found = False
for ep_key, phrases in dict_ep.items():
    if problem in phrases:
        print(f"匹配到的EP键：{ep_key}")
        found = True
        break  # 找到第一个匹配项后停止循环，需所有匹配可删除此句

if not found:
    print('EP no exist, try again!')

模糊匹配版本（短语包含目标关键词） #

如果需要支持模糊匹配（比如目标短语是'FRAUD'时匹配所有包含该词的项），可修改判断逻辑：

dict_ep = {'EP1_2':['FRAUDED BILL','IMPROPER BILLING - FRAUDED CARD (CARDS)','EMBEZZLEMENT','FRAUD'], 
            'EP1_4':['2nd COPY OF CONTRACT (CONSIGNEE)','ACCIDENT WITH DISPOSED VEHICLE'],
            'EP1_6':['BANK STRIKE'],
            'EP1_8':['ACCESS TO BALANCE AND CARD LIMIT','PAYMENT AGREEMENT']}

problem = ('fraud').upper()

found = False
for ep_key, phrases in dict_ep.items():
    for phrase in phrases:
        if problem in phrase:
            print(f"匹配到的EP键：{ep_key}")
            found = True
            break  # 找到当前EP下的匹配后停止检查该EP的其他短语
    if found:
        break  # 找到第一个匹配的EP后停止循环，需所有匹配可删除此句

if not found:
    print('EP no exist, try again!')

内容的提问来源于stack exchange，提问作者William Soares