python-3.x 如何从两个字典列表中获取特定对象的特定键值？

avwztpqn 于 2022-12-05 发布在 Python

关注(0)|答案(2)|浏览(147)

我有两本字典：

timing = [
    {"day_name": "sunday"},
    {"day_name": "monday"},
    {"day_name": "tuesday"},
    {"day_name": "wednesday"},
    {"day_name": "thursday"},
    {"day_name": "friday"},
    {"day_name": "saturday"},
]

hours_detail = [
    {"day_name": "sunday", "peak_hour": False},
    {"day_name": "monday", "peak_hour": False},
    {"day_name": "tuesday", "peak_hour": False},
    {"day_name": "wednesday", "peak_hour": False},
    {"day_name": "thursday", "peak_hour": False},
    {"day_name": "friday", "peak_hour": False},
    {"day_name": "saturday", "peak_hour": False},

    {"day_name": "saturday", "peak_hour": True},
    {"day_name": "friday", "peak_hour": True},
    {"day_name": "thursday", "peak_hour": True},
]

我想创建另一个字典列表，看起来像下面的一个。我基本上是结合这两个列表，并根据日期名称重新排列。

final_data_object = [
    {
        "timing": {"day_name": "saturday"},
        "hour_detail": [
            {"day_name": "saturday", "peak_hour": False},
            {"day_name": "saturday", "peak_hour": True},
        ]
    },
    {
        "timing": {"day_name": "friday"},
        "hour_detail": [
            {"day_name": "friday", "peak_hour": False},
            {"day_name": "friday", "peak_hour": True},
        ]
    },
    soon on...
]

我试过这个，但没有用：

data = []
for time_instance in timing:
    obj = {
        "timing": time_instance
    }
    for hour_instance in hour_detail:
        if time_instance["day_name"] == hour_instance["day_name"]:
            obj["hour_detail"] = hour_instance
            data.append(obj)

return data

python-3.x

来源：https://stackoverflow.com/questions/74629179/how-to-get-specific-objects-from-two-list-of-dictionaries-on-a-specific-key-valu

2条答案

按热度按时间

jc3wubiy1#

如果pricing = hour_detail，则obj["pricing"]必须是列表，因为它可以有多个值。
你必须在循环for time_instance in timing:中创建一个新的列表，并在每次time_instance["day_name"] == pricing_instance["day_name"]时追加。
示例：

data = []

for time_instance  in timing:
    current_hour_detail = []
    for line in hours_detail:
        if line["day_name"] == time_instance["day_name"]:
            current_hour_detail.append(line)
            
    data.append({
        "timing": time_instance,
        "pricing": current_hour_detail
    })

或更小的方式

data = []

for time_instance  in timing:
    current_hour_detail = [line for line in hours_detail if line["day_name"] == time_instance["day_name"]]
    data.append({
        "timing": time_instance,
        "pricing": current_hour_detail
    })

O（n*m）的时间复杂度，首先通过哈希表{day_name：时间复杂度为O（n + m），所以时间复杂度为O（n+m）

赞(0）回复(0）举报 2022-12-05

xxhby3vn2#

这应该可以回答您的问题：

...
data = []
for time_instance in timing:
    obj = {"timing": time_instance}
    # use a list to store all your pricing
    obj["hour_detail"] = []
    for pricing_instance in pricing:
        if time_instance["day_name"] == pricing_instance["day_name"]:
            obj["hour_detail"].append(pricing_instance)
    # add 'obj' after the loop
    data.append(obj)
print(data)

赞(0）回复(0）举报 2022-12-05

我来回答

python-3.x 如何从两个字典列表中获取特定对象的特定键值？

2条答案

相关问题

热门标签

最新问答