在 Python 中对嵌套词典列表进行排序-解网

问：

如何对嵌套词典列表进行排序？字典的实际列表是

[
   {
      "Name":"21_12",
      "Details":[
         {
            "name":"Cat",
            "Data":[
               {
                  "status":"Passed",
                  "id":3,
                  "loop_count":1
               },
               {
                  "status":"Passed",
                  "id":5,
                  "loop_count":1
               }
            ]
         },
         {
            "name":"Dog",
            "Data":[
               {
                  "status":"Passed",
                  "id":1,
                  "loop_count":1
               },
               {
                  "status":"Passed",
                  "id":2,
                  "loop_count":1
               }
            ]
         }
      ]
   }
]

其中，即使在“数据”和“详细信息”中，字典的内部列表也应使用“id”进行排序

所需的输出：

[
   {
      "Name":"21_12",
      "Details":[
         {
            "name":"Dog",
            "Data":[
               {
                  "status":"Passed",
                  "id":1,
                  "loop_count":1
               },
               {
                  "status":"Passed",
                  "id":2,
                  "loop_count":1
               }
            ]
         },
         {
            "name":"Cat",
            "Data":[
               {
                  "status":"Passed",
                  "id":3,
                  "loop_count":1
               },
               {
                  "status":"Passed",
                  "id":5,
                  "loop_count":1
               }
            ]
         }
      ]
   }
]

尝试排序内置函数。没有按预期工作

python-3.x 字典排序嵌套列表

def order_structure(structure):
    for details in structure:
        sorted_detail = sorted(details["Details"], key=lambda x: max(x["Data"], key= lambda y: y["id"])["id"])
        details["Details"] = sorted_detail
        for data in details["Details"]:
            sorted_data = sorted(data["Data"], key= lambda x: x["id"])
            data["Data"] = sorted_data
    

order_structure(structure)

如果您不想就地订购它，但想生成订购的副本，只需在structure.copy()

def get_ordered_structure(structure):
    structure_copy = structure.copy()
    for details in structure_copy:
        sorted_detail = sorted(details["Details"], key=lambda x: max(x["Data"], key= lambda y: y["id"])["id"])
        details["Details"] = sorted_detail
        for data in details["Details"]:
            sorted_data = sorted(data["Data"], key= lambda x: x["id"])
            data["Data"] = sorted_data
    return structure_copy

ordered_structure = get_ordered_structure(structure)

这样，原始结构就不会改变

即使主列表中有多个元素，如果它们遵循相同的结构，该函数也会起作用。

0赞 Sagar Madutha 12/31/2022 #2

你可以用python pandas来解决这个问题。

myDictionary = [
   {
      "Name":"21_12",
      "Details":[
         {
            "name":"Cat",
            "Data":[
               {
                  "status":"Passed",
                  "id":3,
                  "loop_count":1
               },
               {
                  "status":"Passed",
                  "id":5,
                  "loop_count":1
               }
            ]
         },
         {
            "name":"Dog",
            "Data":[
               {
                  "status":"Passed",
                  "id":1,
                  "loop_count":1
               },
               {
                  "status":"Passed",
                  "id":2,
                  "loop_count":1
               }
            ]
         }
      ]
   }
]

import pandas as pd
for i in range(len(myDictionary[0]["Details"])):
    df = pd.DataFrame(myDictionary[0]["Details"][i]["Data"])
    df.sort_values("id", inplace=True)
    myDictionary[0]["Details"][i]["Data"] = df.to_dict(orient="records")
    print(myDictionary)

0赞 tomerar 1/1/2023 #3

您可以将 Python 排序函数与正确的键函数一起使用

sorted_list = sorted(list_of_dicts, key=lambda x: x['Details']['Data']['id'])

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在 Python 中对嵌套词典列表进行排序

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