比较python中字典列表中的值

此代码将在线性时间内解决您的问题,即O(n),其中n是字典长度的增长顺序。只考虑那些 Id 两者兼而有之 title 和 location 并将忽略其余部分。

from collections import Counter

data = [{"Id": 1, "title":"example1"},
        {"Id": 2, "title":"example2"},
        {"Id": 3, "title":"example3"},
        {"Id": 4, "title":"example4"},
        {"Id": 1, "location":"city1"},
        {"Id": 2, "location":"city2"},
        {"Id": 4, "location":"city4"},
        {"Id": 5, "location":"city5"}]

paired_ids = set([key for key, val in dict(Counter([item["Id"] for item in data])).items() if val == 2]) # O(n)

def combine_dict(data):
    result = {key: [] for key in paired_ids} # O(m), m: number of paired ids (m <= n/2)
    for item in data: # O(n)
        items = list(item.items())
        id, tl, val = items[0][1], items[1][0], items[1][1]

        if id in paired_ids: # O(1), as paired_ids is a set lookup takes O(1)
            result[id].append({tl: val})

    return [{"Id": id, "title": lst[0]["title"], "location": lst[1]["location"]} for id, lst in result.items()] # O(n)


print(*combine_dict(data), sep="\n")

输出:

{'Id': 1, 'title': 'example1', 'location': 'city1'}
{'Id': 2, 'title': 'example2', 'location': 'city2'}
{'Id': 4, 'title': 'example4', 'location': 'city4'}

以下代码应该可以工作:

def resultHandler(extractedResult):
  jsonList = []
  for i in range(len(extractedResult) // 2):
    jsonList.append({"Id": i})
  for i in range(len(extractedResult)):
    for j in range(len(jsonList)):
      if jsonList[j]["Id"] == extractedResult[i]["Id"]:
        if "title" in extractedResult[i]:
          jsonList[j]["title"] = extractedResult[i]["title"];
        else:
          jsonList[j]["location"] = extractedResult[i]["location"];
  return jsonList;

extractedResult = [{"Id": 0, "title":"example1"}, {"Id": 1, "title":"example2"}, {"Id": 0, "location":"example3"}, {"Id": 1, "location":"example4"}]

jsonList = resultHandler(extractedResult)

print(jsonList)

输出:

[{'Id': 0, 'title': 'example1', 'location': 'example3'}, {'Id': 1, 'title': 'example2', 'location': 'example4'}]

此代码首先填充 jsonList Id值从0到长度的一半 extractedResult (ID的数量也是如此)。

然后,对于每一本字典 提取结果 ,我们在字典里找到了 jsonList 和匹配的ID。如果 提取结果 包含一个密钥, "title" ,然后我们在 jsonList .这同样适用于 "location" .

我希望这有助于回答你的问题!如果您需要任何进一步的澄清或详细信息,请告诉我:)

此函数有一个嵌套循环。外部循环遍历字典列表。内环 再次遍历字典列表,检查当前字典的id是否正确 已经在字典列表中了。如果不是,它会将字典添加到 字典。如果是,它会使用以下内容更新字典列表中的字典: 当前的字典。

lst = [
    {"id": 1, "fname": "John"},
    {"id": 2, "name": "Bob"},
    {"id": 1, "lname": "Mary"},
]
def combine_dicts(lst):
    res = []
    for d in lst:
        if d.get("id") not in [x.get("id") for x in res]:
            res.append(d)
        else:
            for r in res:
                if r.get("id") == d.get("id"):
                    r.update(d)
    return res


print(combine_dicts(last))
# output: [{'id': 1, 'fname': 'John', 'lname': 'Mary'}, {'id': 2, 'name': 'Bob'}]

更实用(但效率稍低)的方法:

from itertools import groupby
from functools import reduce
from operator import itemgetter


new_data = []
for _, g in groupby(sorted(data, key=itemgetter("Id")), key=itemgetter("Id")):
    new_data.append(reduce(lambda d1, d2: {**d1, **d2}, g))

按ID对DICT进行分组,然后合并每组。

from collections import defaultdict

def merge_dicts(dicts):
    grouped = defaultdict(list)
    for d in dicts:
        grouped[d['id']].append(d)

    merged = []
    for ds in grouped.values():
        m = {}
        for d in ds:
            m |= d        # If Python < 3.9 : m = {**m, **d}
        merged.append(m)

    return merged