Question

我有一个字典列表，我想用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"}]

过滤器可以具有多个键。可能的键是（name，last_name，number）。基本上，我想要的是浏览字典列表，并检查每个字典是否该字典包含给定过滤器中的键，如果包含，则检查键值是否匹配。如果没有，请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。因此，从orig_list中删除字典不是强制性的。也可以将字典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"}，您将获得：

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

Answer 2

如果您确定所有字典词典的键都存在于其他词典中，则可以这样编写搜索（换句话说，缺少的键将被视为匹配）：

filterDict = {"name":"Peter"}
result = [ d for d in orig_list if {**d,**filterDict} == d ]

如果缺少的键不匹配，则可以执行以下操作：

result = [d for d in orig_list if {*filterDict.items()}<={*d.items()}]

如果键和值与其他字典不匹配，则从字典列表中删除字典

2 个答案: