如果鍵和值與其他字典不匹配，則從字典列表中刪除字典

Question

我有一個字典列表，我想用其他字典中的值過濾它。

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"},]

預期結果示例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"}]

預期結果示例2：

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

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

篩選器可以具有多個鍵。 可能的鍵是（name，last_name，number）。 基本上，我想要的是瀏覽字典列表，並檢查每個字典是否該字典包含給定過濾器中的鍵，如果包含，則檢查鍵值是否匹配。 如果他們不這樣做，請從字典列表中刪除整個字典。

最終列表不必是orig_list。 它可以是一個新列表。 因此，從orig_list中刪除字典不是強制性的。 字典也可以復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)

輸出：

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

對於filter_by = {"name":"Peter","number":"445"}您將獲得：

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

Answer 2

如果確定所有字典詞典的鍵都存在於其他字典中，則可以這樣編寫搜索（換句話說，缺少的鍵將被認為是匹配的）：

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

如果缺少的鍵不匹配，則可以執行以下操作：

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