python - 如何合并包含相同键值的字典对象，并输出一个列表？

Question

我想添加包含相同键值的字典对象，并得到一个列表。我有一个功能，我这样做:

def arizakestirimi_func(self):
    arizatahmin = []
    ...
    record = {} //create a new dict
    record["isyeri"]=key
    record["mekanikarizatahminsuresi"]= arizatahminsüresi
    arizatahmin.append(record);
    ...
    record = {} //create a new dict
    record["isyeri"]=key //key can be same 
    record["elektrikarizatahminsuresi"]= arizatahminsüresi //this is different key value
    arizatahmin.append(record);
    ...
    record = {} //create a new dict
    record["isyeri"]=key //key can be same 
    record["kesintilerarizatahminsuresi"]= arizatahminsüresi //this is different key value

    arizatahmin.append(record);

    #Return value as a HttpResponse.
    return HttpResponse(json.dumps({'arizaTahminSuresi': arizatahmin},indent=2), content_type="application/json") 

我的输出是这样的json:

{
  "arizaTahminSuresi": [
    {
      "isyeri": "15400005",//The same key
      "mekanikarizatahminsuresi": 38.95637287225691
    },

    {
      "isyeri": "15400005",//The same key
      "elektrikarizatahminsuresi": 427.9069449086019
    },

      "isyeri": "15400005",//The same key
      "kesintilerarizatahminsuresi": 4.882768280126584
    }

  ]
}

我希望输出如下。我该怎么做？

{
  "arizaTahminSuresi": [
    {
      "isyeri": "15400005",
      "mekanikarizatahminsuresi": 38.95637287225691
      "elektrikarizatahminsuresi": 427.9069449086019
      "kesintilerarizatahminsuresi": 4.882768280126584
    }   
  ]
}

Answer 1

您可以使用 groupby 获取所有匹配的字典，然后使用 update 取消它们，如下所示:

from itertools import groupby
from operator import itemgetter
from pprint import pprint

original = {
    "arizaTahminSuresi": [
        {
            "isyeri": "15400005",
            "mekanikarizatahminsuresi": 38.95637287225691
        },
        {
            "isyeri": "15400003",
            "mekanikarizatahminsuresi": 312.05942081303965
        },
        {
            "isyeri": "15400002",
            "mekanikarizatahminsuresi": 773.600387779049
        },
        {
            "isyeri": "15400004",
            "mekanikarizatahminsuresi": 275.25620818317475
        },
        {
            "isyeri": "15400001",
            "mekanikarizatahminsuresi": 9.032135504089638
        },
        {
            "isyeri": "15400004",
            "elektrikarizatahminsuresi": 105.3605215202133
        },
        {
            "isyeri": "15400003",
            "elektrikarizatahminsuresi": 463.8990083193239
        },
        {
            "isyeri": "15400001",
            "elektrikarizatahminsuresi": 49.71052726257949
        },
        {
            "isyeri": "15400005",
            "elektrikarizatahminsuresi": 427.9069449086019
        },
        {
            "isyeri": "15400002",
            "elektrikarizatahminsuresi": 25.78649617892279
        },
        {
            "isyeri": "15400004",
            "kesintilerarizatahminsuresi": 67.41996630760787
        },
        {
            "isyeri": "15400005",
            "kesintilerarizatahminsuresi": 4.882768280126584
        },
        {
            "isyeri": "15400002",
            "kesintilerarizatahminsuresi": 58.96635620693996
        },
        {
            "isyeri": "15400001",
            "kesintilerarizatahminsuresi": 357.28675445627414
        },
        {
            "isyeri": "15400003",
            "kesintilerarizatahminsuresi": 434.16131082294305
        }

    ]
}

grouped_subdicts = groupby(sorted(original["arizaTahminSuresi"], key=itemgetter("isyeri")), itemgetter("isyeri"))

result = []
for k,g in grouped_subdicts:
    d = {}
    for sub_d in g:
        d.update(sub_d)
    result.append(d)

original["arizaTahminSuresi"] = result

pprint(original)