如何根据匹配的键：值对组合字典列表中的 N 个字典？答案

【问题标题】：how do I combine N dictionaries in list of dictionaries based on matching key:value pair?如何根据匹配的键：值对组合字典列表中的 N 个字典？
【发布时间】：2021-07-26 10:28:34
【问题描述】：

我想实现以下目标。它本质上是N个字典的组合或合并，从重复的id中累积所有数据，并在最终结果中附加来自多个数据源中所有字典的所有values(except id, updated_date)。

class A:
    def __init__(self):
        pass

    def run(self):
        return {"data":[{"id":"ID-2002-0201","updated_at":"2018-05-14T22:25:51Z","html_url":["https://github.com/ID-2002-0201"],"source":"github"},{"id":"ID-2002-0200","updated_at":"2018-05-14T21:49:15Z","html_url":["https://github.com/ID-2002-0200"],"source":"github"},{"id":"ID-2002-0348","updated_at":"2018-05-11T14:13:28Z","html_url":["https://github.com/ID-2002-0348"],"source":"github"}]}

class B:
    def __init__(self):
        pass

    def run(self):
        return {"data":[{"id":"ID-2002-0201","updated_at":"2006-03-28","html_url":["http://sample.com/files/1622"],"source":"sample"},{"id":"ID-2002-0200","updated_at":"2006-06-05","html_url":["http://sample.com/files/1880"],"source":"sample"},{"id":"ID-2002-0348","updated_at":"2007-03-09","html_url":["http://sample.com/files/3441"],"source":"sample"}]}
        
results = {}
data_sources = [A(),B()]
for data in data_sources:
    data_stream = data.run()
    for data in data_stream.get('data'):
        for key, value in data.items():
            if key in ['html_url']:
                results.setdefault(key, []).extend(value)
            elif key in ['source']:
                results.setdefault(key, []).append(value)
            else:
                results[key] = value
print(results)

想要的输出

[
    {
        "id":"ID-2002-0201",
        "updated_at":"2018-05-14T22:25:51Z",
        "html_url":[
            "https://github.com/ID-2002-0201",
            "https://github.com/ID-2002-0202",
            "https://github.com/ID-2002-0203",
            "https://github.com/ID-2002-0204"
        ],
        "source": [
            "github",
            "xxx",
            "22aas"
        ]
    },
]

【问题讨论】：

标签： python dictionary pipeline reduce

【解决方案1】：

我有点困惑，因为您提供的所需输出与您在代码中提供的示例类不匹配。但是，我想我得到了你想要的，如果我对你的问题的解释有误，请纠正我。

我使用您的结果数组，就像字典一样。外部字典包含所有唯一 ID 作为键，内部字典包含您在输出中想要的数据。在循环计算之后，我只返回 list(results.values()) 以获取 N 个组合的字典列表。

代码如下：

class A:
    def __init__(self):
        pass

    def run(self):
        return {"data":[{"id":"ID-2002-0201","updated_at":"2018-05-14T22:25:51Z","html_url":["https://github.com/ID-2002-0201"],"source":"github"},{"id":"ID-2002-0200","updated_at":"2018-05-14T21:49:15Z","html_url":["https://github.com/ID-2002-0200"],"source":"github"},{"id":"ID-2002-0348","updated_at":"2018-05-11T14:13:28Z","html_url":["https://github.com/ID-2002-0348"],"source":"github"}]}

class B:
    def __init__(self):
        pass

    def run(self):
        return {"data":[{"id":"ID-2002-0201","updated_at":"2006-03-28","html_url":["http://sample.com/files/1622"],"source":"sample"},{"id":"ID-2002-0200","updated_at":"2006-06-05","html_url":["http://sample.com/files/1880"],"source":"sample"},{"id":"ID-2002-0348","updated_at":"2007-03-09","html_url":["http://sample.com/files/3441"],"source":"sample"}]}
        
results = {}
data_sources = [A(),B()]
for data in data_sources:
    data_stream = data.run()
    for data in data_stream.get('data'):
        curr_id = data["id"]
        result = results.setdefault(curr_id, {})
        for key, value in data.items():
            if key in ['html_url']:
                result.setdefault(key, []).extend(value)
            elif key in ['source']:
                result.setdefault(key, []).append(value)
            else:
                result[key] = value
print(list(results.values()))

【讨论】：