遍历嵌套的字典值并根据字典的值返回一个有序列表答案

【问题标题】：Iterate through nested dictionary values and return an ordered list based on dict's values遍历嵌套的字典值并根据字典的值返回一个有序列表
【发布时间】：2021-10-21 09:46:38
【问题描述】：

我有一个名为“transportation_costs”的嵌套字典，其中包含与每个设施-客户组合相关的运输成本。我需要遍历字典中的每个客户 (key2) 并生成五个有序列表（每个客户一个），其中包含根据嵌套字典中的值对给定客户从最便宜到最昂贵的设施排名。

*** 当我说“最便宜”时，我的意思是它的运输成本更低。

 transportation_cost=
                     {'Fac-1' : {"Apple":4,"Samsung":5,"Huawei":6,"Nokia":8,"Motorolla":10},
                      'Fac-2' : {"Apple":6,"Samsung":4,"Huawei":3,"Nokia":5,"Motorolla":8},
                      'Fac-3' : {"Apple":9,"Samsung":7,"Huawei":4,"Nokia":3,"Motorolla":4},
                      'Fac-4' : {"Apple":3,"Samsung":4,"Huawei":8,"Nokia":4,"Motorolla":4},
                      'Fac-5' : {"Apple":4,"Samsung":7,"Huawei":5,"Nokia":3,"Motorolla":2}}

我需要最终的输出是这样的：

         "Apple" = ['FAC-4', 'FAC-5', 'FAC-3', 'FAC-2','FAC-1']
         "Samsung" = ['FAC-2', 'FAC-4', 'FAC-3', 'FAC-5','Fac-1']
         "Huawei"=  ['FAC-3', 'FAC-5', 'FAC-1', 'FAC-4','FAC-2']
         "Nokia" = ['FAC-5', 'FAC-3', 'FAC-2', 'FAC-4','FAC-1']
         "Motorolla" = ['FAC-5', 'FAC-1', 'FAC-3', 'FAC-2', 'FAC-4']

【问题讨论】：

到目前为止你尝试了什么，你在哪里卡住了？

标签： python dictionary for-loop nested

【解决方案1】：

试试：

all_keys = set(k for v in transportation_cost.values() for k in v.keys())
out = {
    k: [
        *sorted(
            transportation_cost.keys(), key=lambda x: transportation_cost[x][k]
        )
    ]
    for k in all_keys
}
print(out)

打印：

{
    "Huawei": ["Fac-2", "Fac-3", "Fac-5", "Fac-1", "Fac-4"],
    "Motorolla": ["Fac-5", "Fac-3", "Fac-4", "Fac-2", "Fac-1"],
    "Apple": ["Fac-4", "Fac-1", "Fac-5", "Fac-2", "Fac-3"],
    "Nokia": ["Fac-3", "Fac-5", "Fac-4", "Fac-2", "Fac-1"],
    "Samsung": ["Fac-2", "Fac-4", "Fac-1", "Fac-3", "Fac-5"],
}

【讨论】：