python - 如果键和值与其他字典不匹配，则从字典列表中删除字典

Question

我有一个字典列表，我想按其他字典中的值过滤它。

orig_list = [{"name":"Peter","last_name":"Wick","mail":"Peter@mail.com","number":"111"},
{"name":"John","last_name":"Hen","mail":"John@mail.com","number":"222"},
{"name":"Jack","last_name":"Malm","mail":"Jack@mail.com","number":"542"},
{"name":"Anna","last_name":"Hedge","mail":"Anna@mail.com"},
{"name":"Peter","last_name":"Roesner","mail":"Peter2@mail.com","number":"445"},
{"name":"Tino","last_name":"Tes","mail":"Tino@mail.com","number":"985"},]

预期结果示例 1:

filter = {"name":"Peter"}

orig_list[{"name":"Peter","last_name":"Wick","mail":"Peter@mail.com","number":"111"},
{"name":"Peter","last_name":"Roesner","mail":"Peter2@mail.com","number":"445"}]

预期结果示例 2:

filter = {"name":"Peter","number":"445"}

orig_list[
{"name":"Peter","last_name":"Roesner","mail":"Peter2@mail.com","number":"445"}]

过滤器可以有多个键。可能的键是(姓名、姓氏、号码)。 基本上我想要的是遍历字典列表并检查每个字典是否包含来自给定过滤器的键，如果包含，则检查键值是否匹配。如果他们不这样做，则从字典列表中删除整个字典。

最终列表不必是 orig_list。它可以是一个新列表。所以从 orig_list 中删除字典不是强制性的。字典也可以复制到新的字典列表中。

Answer 1

您可以使用列表理解:

orig_list = [{"name":"Peter","last_name":"Wick","mail":"Peter@mail.com","number":"111"},
{"name":"John","last_name":"Hen","mail":"John@mail.com","number":"222"},
{"name":"Jack","last_name":"Malm","mail":"Jack@mail.com","number":"542"},
{"name":"Anna","last_name":"Hedge","mail":"Anna@mail.com"},
{"name":"Peter","last_name":"Roesner","mail":"Peter2@mail.com","number":"445"},
{"name":"Tino","last_name":"Tes","mail":"Tino@mail.com","number":"985"},]

filter_by = {"name":"Peter"}


result = [dic for dic in orig_list if all(key in dic and dic[key] == val for key, val in filter_by.items())]
print(result)

输出:

[
  {
    "name": "Peter",
    "last_name": "Wick",
    "mail": "Peter@mail.com",
    "number": "111"
  },
  {
    "name": "Peter",
    "last_name": "Roesner",
    "mail": "Peter2@mail.com",
    "number": "445"
  }
]

对于 filter_by = {"name":"Peter","number":"445"} 你得到:

[
  {
    "name": "Peter",
    "last_name": "Roesner",
    "mail": "Peter2@mail.com",
    "number": "445"
  }
]