是否有一种优雅的合并方式来通过一个公共键合并一个指令列表？

Question

我正在寻找一个更优雅的方法，将这个指令列表变成一个单一的合并指令，其中键'SKU'作为合并的键。这些列表实际上都是很容易被转化为指令的模型。我没有找到一个更优雅的解决方案使用Pydantic。

Dicts/PydanticInstances列表

[a for a in m]
[
    PackMapPydantic(id=60, pack_sku='PACKFAC01', sku='FAC01', prop_name='Facebook Username'),
    PackMapPydantic(id=61, pack_sku='PACKFAC01', sku='FAC02', prop_name='Facebook Username'),
    PackMapPydantic(id=62, pack_sku='PACKFAC01', sku='FAC03', prop_name='Facebook Username'),
    PackMapPydantic(id=63, pack_sku='PACKFAC01', sku='FAC05', prop_name='Facebook Username'),
    PackMapPydantic(id=64, pack_sku='PACKFAC01', sku='FAC06', prop_name='Facebook Username'),
    PackMapPydantic(id=70, pack_sku='PACKFAC01', sku='FAC01', prop_name='Channel Name'),
    PackMapPydantic(id=71, pack_sku='PACKFAC01', sku='FAC02', prop_name='Channel Name'),
    PackMapPydantic(id=72, pack_sku='PACKFAC01', sku='FAC03', prop_name='Channel Name'),
    PackMapPydantic(id=73, pack_sku='PACKFAC01', sku='FAC05', prop_name='Channel Name'),
    PackMapPydantic(id=74, pack_sku='PACKFAC01', sku='FAC06', prop_name='Channel Name'),
    PackMapPydantic(id=87, pack_sku='PACKFAC01', sku='FAC01', prop_name='__uk_avatar (direct link)'),
    PackMapPydantic(id=88, pack_sku='PACKFAC01', sku='FAC02', prop_name='__uk_avatar (direct link)'),
    PackMapPydantic(id=89, pack_sku='PACKFAC01', sku='FAC03', prop_name='__uk_avatar (direct link)'),
    PackMapPydantic(id=90, pack_sku='PACKFAC01', sku='FAC06', prop_name='__uk_avatar (direct link)')
]

所需输出：

[
  {
      'sku': 'FAC05', 
      'prop_name': [
          'Username', 
          'Channel'
      ]
  },
  {
      'sku': 'FAC01',
      'prop_name': [
          'Username',
          'Channel',
          '__uk_avatar (direct link)'
      ]
  },
  {
      'sku': 'FAC06',
      'prop_name': [
          'Username',
          'Channel',
          '__uk_avatar (direct link)'
      ]
  },
  {
      'sku': 'FAC02',
      'prop_name': [
          'Username',
          '__uk_avatar (direct link)'
      ]
  },
  {
      'sku': 'FAC03',
      'prop_name': [
          'Username',
          'Channel',
          '__uk_avatar (direct link)'
      ]
  }
]

当前解决方案：

for sku in skus_distinct:
    this_assets = []
    for item in items_listed:
        if item['sku'] == sku:
            # item = list(item)
            this_assets.append(item['prop_name'])

    now = {'sku': sku,
           'prop_name': this_assets }

    print(now)

Answer 1

使用pandas（假设dict1是您的dict的初始列表）-

import pandas as pd
df  = pd.DataFrame(dict1)[['sku','prop_name']]
df  = df.groupby('sku').agg({'prop_name': list}).reset_index()
result_dict = df.to_dict(orient='records')

输出-

[{'sku': 'FAC01',
  'prop_name': ['Facebook Username',
   'Channel Name',
   '__uk_avatar {direct link}']},
 {'sku': 'FAC02',
  'prop_name': ['Facebook Username',
   'Channel Name',
   '__uk_avatar {direct link}']},
 {'sku': 'FAC03',
  'prop_name': ['Facebook Username',
   'Channel Name',
   '__uk_avatar {direct link}']},
 {'sku': 'FAC05', 'prop_name': ['Facebook Username', 'Channel Name']},
 {'sku': 'FAC06',
  'prop_name': ['Facebook Username',
   'Channel Name',
   '__uk_avatar {direct link}']}]