如果键和值与其他字典不匹配，则从字典列表中删除字典答案

【问题标题】：Remove dict from list of dict if key and value doesnt match with other dict如果键和值与其他字典不匹配，则从字典列表中删除字典
【发布时间】：2019-10-26 01:49:57
【问题描述】：

我有一个字典列表，我想按其他字典中的值对其进行过滤。

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（名称，姓氏，数字）。基本上我想要的是通过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，请从 dict 列表中删除整个 dict。

最终列表不一定是 orig_list。它可以是一个新列表。因此，从 orig_list 中删除 dicts 不是强制性的。字典也可以复制到新的字典列表中。

【问题讨论】：

你尝试过什么，它到底有什么问题？

标签： python python-3.x algorithm dictionary

【解决方案1】：

您可以使用列表推导：

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"}，你会得到：

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

【讨论】：

我不知道我可以这样使用if all ;)
@DSC all 只是一个函数，它不是关键字：docs.python.org/3/library/functions.html#all
不要真正使用如此复杂的理解，所以还不需要它
哇，非常酷的解决方案！我试图写一些列表理解，但他们总是失败。感谢您的解决方案。它为我节省了很多时间来尝试正确地编写理解。

【解决方案2】：

如果您确定过滤器字典的所有键都存在于其他字典中，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，您可以这样做：

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

【讨论】：