Question

我和朋友一起为这个项目提供了这些数据：

arr = [{'Key':'Key 1', 'SecondKey':'SecondKey 1', 'ThirdKey':'Thirdkey 1', 'Value':100},
   {'Key':'Key 1', 'SecondKey':'SecondKey 1', 'ThirdKey':'ThirdKey 2', 'Value':130},
   {'Key':'Key 1', 'SecondKey':'SecondKey 2', 'ThirdKey':'ThirdKey 1', 'Value':230},
   {'Key':'Key 1', 'SecondKey':'SecondKey 2', 'ThirdKey':'ThirdKey 2', 'Value':300},
   {'Key':'Key 2', 'SecondKey':'SecondKey 4', 'ThirdKey':'ThirdKey 1', 'Value':111},
   {'Key':'Key 2', 'SecondKey':'SecondKey 2', 'ThirdKey':'ThirdKey 2', 'Value':400},
   {'Key':'Key 2', 'SecondKey':'SecondKey 1', 'ThirdKey':'ThirdKey 2', 'Value':230}
   ]

我尝试过这样的功能：

def array_to_dict(arr, classification):
    dct = tree()
    for d in arr:
        dct[d["Key"]]=(d)
    pprint.pprint(dct)

def tree():
    return defaultdict(tree)

def array_to_dict_recursive(arr, classification):
    result = defaultdict(arr)
    for v, k in arr:
        result[k].append(v)
    final_result = [{'type': k, 'items': v} for k, v in result.items()]
    print (str(final_result))

def array_cool(arr):
    #pprint.pprint(arr)
    arr.sort(key=operator.itemgetter('Key'))
    pprint.pprint(arr)
    list1= []
    print("")
    for key, items in itertools.groupby(arr, operator.itemgetter('Key')):
        list1.append(list(items))
    pprint.pprint(list1)

我希望它以这种方式显示为JSON文件：

{
"Key 1": {
    "SecondKey 2": {
        "ThirdKey 2": [
            {
                "Key": "Key 1",
                "Value": 300,
                "SecondKey": "SecondKey 2",
                "ThirdKey": "ThirdKey 2"
            }
        ],
        "ThirdKey 1": [
            {
                "Key": "Key 1",
                "Value": 230,
                "SecondKey": "SecondKey 2",
                "ThirdKey": "ThirdKey 1"
            }
        ]
    },
    "SecondKey 1": {
        "ThirdKey 2": [
            {
                "Key": "Key 1",
                "Value": 130,
                "SecondKey": "SecondKey 1",
                "ThirdKey": "ThirdKey 2"
            }
        ],
        "ThirdKey 1": [
            {
                "Key": "Key 1",
                "Value": 100,
                "SecondKey": "SecondKey 1",
                "ThirdKey": "ThirdKey 1"
            }
        ]
    }
},
"Key 2": {
    "SecondKey 4": {
        "ThirdKey 1": [
            {
                "Key": "Key 2",
                "Value": 111,
                "SecondKey": "SecondKey 4",
                "ThirdKey": "ThirdKey 1"
            }
        ]
    },
    "SecondKey 2": {
        "ThirdKey 2": [
            {
                "Key": "Key 2",
                "Value": 400,
                "SecondKey": "SecondKey 2",
                "ThirdKey": "ThirdKey 2"
            }
        ]
    },
    "SecondKey 1": {
        "ThirdKey 2": [
            {
                "Key": "Key 2",
                "Value": 230,
                "SecondKey": "SecondKey 1",
                "ThirdKey": "ThirdKey 2"
            }
        ]
    }
}

}

我尝试了排序，但是排序将Key置于第二位并且混乱了下一个排序过程，并且它无法正常工作。

Answer 1

尝试

print json.dumps(arr, indent=4, sort_keys=True)

有关详细信息，请参阅json.dumps文档：

json.dumps（obj，skipkeys = False，ensure_ascii = True，check_circular = True，allow_nan = True，cls = None，indent = None，separators = None，encoding =“utf-8”，default = None，sort_keys =假，** kw）

Answer 2

Python dicts未分类。你有几个选择。

使用将排序应用于标准字典的工具，例如：

import json
print json.dumps(thedict, indent=4, sort_keys=True)

使用维护dict条目创建顺序的类型：

from collections import OrderedDict
d = OrderedDict()
d[KEY] = VALUE
d[KEY][SUBKEY] = VALUE
etc.

或者最简单的方法，只需按照您想要的顺序将它们添加到数组/列表中。

Answer 3

这是你想要的第一个刺。它会让你开始，然后你可以简化它并使它漂亮。要点：

json.dumps()按照看到的顺序输出字典键，因此显式排序的OrderedDict可能是您获得该字典的唯一方法。
您的结果与输入的数据结构不同，因此您必须构建它
Python和JSON中的字典不已订购密钥。因此，当您将结果导入其他内容时，订单可能会丢失。
这段代码效率很低，可能会对大型数据集造成严重影响。

您的数据结构：

为每个Key做出决定，按顺序排序
- 每个SecondKey的dict，反向排序
  - 每个ThirdKey的dict，反向排序
    - 包含每个值的dict的列表（没有值的排序）以及这些排序的键和值
      - Key1：来自外部循环的Key值
      - 值：当前项的值
      - SecondKey：来自外部循环的SecondKey值
      - ThirdKey：来自外部循环的ThirdKey值

from collections import OrderedDict
import json

arr = [{'Key':'Key 1', 'SecondKey':'SecondKey 1', 'ThirdKey':'ThirdKey 1', 'Value':100},
    {'Key':'Key 1', 'SecondKey':'SecondKey 1', 'ThirdKey':'ThirdKey 2', 'Value':130},
    {'Key':'Key 1', 'SecondKey':'SecondKey 2', 'ThirdKey':'ThirdKey 1', 'Value':230},
    {'Key':'Key 1', 'SecondKey':'SecondKey 2', 'ThirdKey':'ThirdKey 2', 'Value':300},
    {'Key':'Key 2', 'SecondKey':'SecondKey 4', 'ThirdKey':'ThirdKey 1', 'Value':111},
    {'Key':'Key 2', 'SecondKey':'SecondKey 2', 'ThirdKey':'ThirdKey 2', 'Value':400},
    {'Key':'Key 2', 'SecondKey':'SecondKey 1', 'ThirdKey':'ThirdKey 2', 'Value':230}
    ]

def arr_to_ordereddict():
    d = OrderedDict()
    keys = set(a['Key'] for a in arr)
    #print(sorted(keys))
    for k1 in sorted(keys):
        k1_dict = OrderedDict()
        d[k1] = k1_dict
        arr2 = [a for a in arr if a['Key'] == k1]
        keys2 = set(a['SecondKey'] for a in arr2)
        #print('\t', k1, sorted(keys2, reverse=True))
        for k2 in sorted(keys2, reverse=True):
            k2_dict = OrderedDict()
            k1_dict[k2] = k2_dict
            arr3 = [a for a in arr if a['SecondKey'] == k2]
            keys3 = set(a['ThirdKey'] for a in arr3)
            #print('\t\t', k1, k2, sorted(keys3, reverse=True))
            for k3 in sorted(keys3, reverse=True):
                k3_list = []
                k2_dict[k3] = k3_list
                for item in (a for a in arr3 if a['ThirdKey'] == k3):
                    value_dict = OrderedDict()
                    value_dict['Key'] = k1
                    value_dict['Value'] = item['Value']
                    value_dict['SecondKey'] = k2
                    value_dict['ThirdKey'] = k3
                    k3_list.append(value_dict)
    return d

print(json.dumps(arr_to_ordereddict(), indent=4))

从Python中的字典列表创建JSON

3 个答案: