如何創建字典的優先級列表

Question

• 清單如下
• 偏好字典在下面
• 如果除type之外的所有鍵和值都相同，那么..
• 需要比較偏好字典中最高順序的每個列表中的type
• Output 是最高級別的字典列表

list_ = [
  {
    "id": "11",
    "name": "son",
    "email": "n@network.com",
    "type": "Owner"
  },
      {
    "id": "11",
    "name": "son",
    "email": "n@network.com",
    "type": "Manager"
  },
{
    "id": "21",
    "name": "abc",
    "email": "abc@network.com",
    "type": "Employ"
  },
{
    "id": "21",
    "name": "abc",
    "email": "abc@network.com",
    "type": "Manager"
  }
]

偏好字典 = {'Owner': 1, 'Manager':2, 'employ':3, 'HR': 4 }

我預期的 output 字典如下

[{'id': '11', 'name': 'son', 'email': 'n@network.com', 'type': 'Owner'},
{'id':'21','name': 'abc','email': 'abc@network.com','type': 'Manager'}]

new_list = []
for each in list_:
    if each['type'] in priority.keys():
        if each['id'] not in new_list:
            new_list.append(each)

Answer 1

您可以簡單地執行src.sort(key = lambda x: preference[x["type"]])並且您的列表將被排序。

Answer 2

此解決方案按id對所有元素進行分組，並根據偏好對組進行排序（因此Owner是第一個， HR是最后一個），然后從每個組中選擇第一個：

from collections import defaultdict

src = [
  {
    "id": "11",
    "name": "son",
    "email": "n@network.com",
    "type": "Owner"
  },
      {
    "id": "11",
    "name": "son",
    "email": "n@network.com",
    "type": "Manager"
  },
{
    "id": "21",
    "name": "abc",
    "email": "abc@network.com",
    "type": "Employ"
  },
{
    "id": "21",
    "name": "abc",
    "email": "abc@network.com",
    "type": "Manager"
  }
]
preference = {'Owner': 1, 'Manager':2, 'Employ':3, 'HR': 4 }

d = defaultdict(list)
# group all the records by id
for item in src:
    d[item['id']].append(item)

# sort each group by the preference
for item in d.values():
    item.sort(key=lambda x: preference[x['type']])

# select only the first from each group
result = [item[0] for item in d.values()]

print(result)

Output：

[{'id': '11', 'name': 'son', 'email': 'n@network.com', 'type': 'Owner'}, {'id': '21', 'name': 'abc', 'email': 'abc@network.com', 'type': 'Manager'}]

Answer 3

您可以創建一個優先級隊列：

from queue import PriorityQueue

priority = {'Owner': 1, 'Manager':2, 'employ':3, 'HR': 4 }

q = PriorityQueue()
for elem in list_:
    p = priority[elem['type']]
    q.put((p, id(elem), elem))

或者您也可以根據type對列表進行排序：

priority_list = sorted(list_, key=lambda x: priority[x['type']], reverse=True)

Answer 4

好吧，這是我的鏡頭！

它並不漂亮，但似乎有效。

list_ = [
  {
    "id": "11",
    "name": "son",
    "email": "n@network.com",
    "type": "Owner"
  },
      {
    "id": "11",
    "name": "son",
    "email": "n@network.com",
    "type": "Manager"
  },
{
    "id": "21",
    "name": "abc",
    "email": "abc@network.com",
    "type": "Employ"
  },
{
    "id": "21",
    "name": "abc",
    "email": "abc@network.com",
    "type": "Manager"
  }
]

new = dict( Owner = 0, Manager = 0, Employ = 0, HR = 0 )

for a in list_ :
    type_ = a[ 'type' ]
    if type_ == 'Owner':
        new[ 'Owner' ] += 1
    if type_ == 'Manager':
        new[ 'Manager' ] += 1
    if type_ in [ 'Employ', 'Manager' ]:
        new[ 'Employ' ] += 1
    new[ 'HR' ] += 1

print( new )