按值范围划分字典的交集

Question

我有一本字典，它们的形式是：

"id_1" : { 
    "_id" : ObjectId("5cefddaae40ed94ea713f7c8"), 
    "derived_from" : ObjectId("5cefddaae40ed94ea713f7c7"), 
    "epsg" : "32723", 
    "active" : True, 
    "name" : "DPA-Time", 
    "range_time_depth" : [ 4648.91338, 5641.545770000001 ], 
    "type" : "Polygon",
    "iline": 6998 
}

我有一本像数据库查询一样工作的字典：

{
    '_id': { '$in': ['5cefddaae40ed94ea713f7c8','5cefddaae40ed94ea713f7c8'] },
    'iline': {'$gte': 6998, '$lte': 7000}
}

所以，我想检索词典字典中所有 _id 等于 '5cefddaae40ed94ea713f7c8' 或 '5cefddaae40ed94ea713f7c8' 和 'iline' 在 6998 和 7000 之间的词典。我知道交集方法有助于匹配字典值，但是当我想将值与值的 range 匹配时，我该怎么做？

我做了什么：

def check_correspondence(obj, query):
    for _, value in obj.items():

        if value.keys() & query.keys():
            return True
        else: return False

我循环遍历提到的字典并应用 & 运算符。但是，此运算符似乎不适用于范围比较，只能用于严格相等。当我尝试时，我收到 TypeError: unhashable type: 'list'

Answer 1

您不需要检查所有密钥，因为您只想检查 ID 和 iline。

注意，& 和 and 在 python 中是不一样的。 & 是 bit-wise operator.

objs = {
    "id_1" : { 
        "_id" : "5cefddaae40ed94ea713f7c8", 
        "derived_from" : "5cefddaae40ed94ea713f7c7", 
        "epsg" : "32723", 
        "active" : True, 
        "name" : "DPA-Time", 
        "range_time_depth" : [ 4648.91338, 5641.545770000001 ], 
        "type" : "Polygon",
        "iline": 6998 
    },
    "id_2" : { 
        "_id" : "5cefddaae40ed94ea713f7c9", 
        "derived_from" : "5cefddaae40ed94ea713f7c7", 
        "epsg" : "32723", 
        "active" : True, 
        "name" : "DPA-Time", 
        "range_time_depth" : [ 4648.91338, 5641.545770000001 ], 
        "type" : "Polygon",
        "iline": 7002
    } 
}

query_ = {
    '_id': { '$in': ['5cefddaae40ed94ea713f7c8','5cefddaae40ed94ea713f7c9'] },
    'iline': {'$gte': 6998, '$lte': 7000}
}

def check_correspondence(obj, query):
    is_in = obj['_id'] in query['_id']['$in']
    is_iline = query['iline']['$lte'] >= obj['iline'] >= query['iline']['$gte']
    
    return is_in and is_iline

for obj_name, obj_ in objs.items():
    print(obj_name, check_correspondence(obj_, query_))
    
# id_1 True
# id_2 False - because 7002 > 7000

按值范围划分字典的交集

Intersection of dictionaries by value ranges

python

search

dictionary

intersection

python-3.x