当值搜索匹配多个键（python）时，只需要打印dict中的高优先级键答案

【问题标题】：Need to print only the high priority key in dict, when a values search matches multiple keys ( python)当值搜索匹配多个键（python）时，只需要打印dict中的高优先级键
【发布时间】：2020-08-24 12:39:59
【问题描述】：

当搜索匹配多个dict值时，只需要打印最高优先级的KEY。

ERROR_CATEGORIES_TO_SEARCH_1 = {
                "VEHICAL_1" : ['CAR'],
                "VEHICAL_2": ['BIKE'],
        "VEHICAL_3"  : ["TRUCK"],
                "VEHICAL_4" : ['AEROPLANE'],
                "VEHICAL_5" : ['SHIP'],
        "VEHICAL_6"   : ['BOAT'],
                "VEHICAL_7" : ['CART'],
                "VEHICAL_8" : ['CYCLE']
            }
  
 
  prio_list = ['VEHICAL_1', 'VEHICAL_2' , 'VEHICAL_3' , 'VEHICAL_4', 'VEHICAL_5', 'VEHICAL_6','VEHICAL_7', 'VEHICAL_8']
 
 
  res = ERROR_CATEGORIES_TO_SEARCH_1.get(prio_list[0], ERROR_CATEGORIES_TO_SEARCH_1.get(prio_list[1], 
                                  ERROR_CATEGORIES_TO_SEARCH_1.get(prio_list[2], ERROR_CATEGORIES_TO_SEARCH_1.get(prio_list[3], ERROR_CATEGORIES_TO_SEARCH_1.get(prio_list[4], ERROR_CATEGORIES_TO_SEARCH_1.get(prio_list[5] , ERROR_CATEGORIES_TO_SEARCH_1.get(prio_list[6], ERROR_CATEGORIES_TO_SEARCH_1.get(prio_list[7]))))))))

MESSAGE_A  = "I bought a new CAR and a BIKE"
MESSAGE_B  = "WOW That SHIP is so huge and beautiful"
MESSAGE_C  = "only mode to travel through hill station is by TRUCK"
MESSAGE_D  = "First will ride on a CYCLE and then by BOAT"

resultant_key = [key for (key, lst) in ERROR_CATEGORIES_TO_SEARCH_1.items() for ele in lst if ele in MESSAGE_A]
//OUTPUT:: VEHICAL_1, VEHICAL_2

resultant_key = [key for (key, lst) in ERROR_CATEGORIES_TO_SEARCH_1.items() for ele in lst if ele in MESSAGE_B]
//OUTPUT:: VEHICAL_5

resultant_key = [key for (key, lst) in ERROR_CATEGORIES_TO_SEARCH_1.items() for ele in lst if ele in MESSAGE_c]
//OUTPUT:: VEHICAL_3

resultant_key = [key for (key, lst) in ERROR_CATEGORIES_TO_SEARCH_1.items() for ele in lst if ele in MESSAGE_D]
//OUTPUT :: VEHICAL_6, VEHICAL_8

由于我只需要打印要打印的优先键，以下是预期输出：

resultant_key = [key for (key, lst) in ERROR_CATEGORIES_TO_SEARCH_1.items() for ele in lst if ele in MESSAGE_A]
//OUTPUT:: VEHICAL_1


resultant_key = [key for (key, lst) in ERROR_CATEGORIES_TO_SEARCH_1.items() for ele in lst if ele in MESSAGE_D]
//OUTPUT :: VEHICAL_6

对此的任何建议都会有所帮助

【问题讨论】：

标签： python

【解决方案1】：

您可以将逻辑放在函数中，并在循环中使用 return 语句，仅返回第一个（即最高优先级）键。

def find_highest(message):
    for key in prio_list:
        for word in ERROR_CATEGORIES_TO_SEARCH_1.get(key, []):
            if word in message:
                return key

>>> for message in (MESSAGE_A, MESSAGE_B, MESSAGE_C, MESSAGE_D):
...     print(find_highest(message))
... 
VEHICAL_1
VEHICAL_5
VEHICAL_3
VEHICAL_6

【讨论】：

我是 python 新手，我能解释一下吗:: for word in ERROR_CATEGORIES_TO_SEARCH_1.get(key, []):
dict.get(...) 的第二个参数是默认值，如果找不到密钥将返回（通常为：None）。但是对于默认的空列表，迭代 for word in ... 仍然有效（迭代步数 = 0，而不是迭代失败，因为您无法迭代 None）。它与您在巨大的 res = ... 声明中使用的相同，您根本不需要，在此答案中进行了更改，另请参阅 docs.python.org/3/library/stdtypes.html#dict.get
谢谢，这让人困惑 :: key, []