Question

轻松过滤的最佳包装是什么？像django-filter，但我在django对象之外。

例如

>>dict = {
{'name':'apple','amount':10},
{'name':'apple','amount':5},
{'name':'banana','amount':10},
{'name':'green-apple','amount':20},
}

>>filter(dict,name='apple')
{'name':'apple','amount':10},{'name':'apple','amount':10}

 >>filter(dict,name='apple',amount=10)
{'name':'apple','amount':10}

 >>filter(dict,amount_gt=5)
{'name':'apple','amount':10},{'name':'banana','amount':10}

 >>filter(dict,amount_gte=5)
{'name':'apple','amount':10},{'name':'apple','amount':5},{'name':'banana','amount':10}

 >>filter(dict,amount_lt=10)
{'name':'apple','amount':5}

 >>filter(dict,name_regex=".*apple")
{'name':'apple','amount':5},{'name':'apple','amount':10},{'name':'green-apple','amount':20}

Answer 1

您可以使用以下功能：

def filter_dicts(dict_list, **kwargs):
    for item in dict_list:
        if all(item[k] == v for k, v in kwargs.items()):
            yield item

用法：

dict_list = [
    {"name": "apple", "amount": 10},
    {"name": "apple", "amount": 15},
    {"name": "orange", "amount": 5}
]

for d in filter_dicts(dict_list, name="apple", amount=10):
    print(d)

或者，您也可以这样做：

filtered_dicts = [item for item in dict_list if item["amount"] > 10 and item["name"] == "apple"]

Answer 2

您可以使用items将dict转换为列表并在该列表上使用过滤器，即

filter(lambda x : x[0] == 'amount', dict.items())

你也可以将它转换回来

什么是dict过滤器的最佳包

2 个答案: