Question

我有这个json，我想只将follower_ids w.r.t project_id提取到列表中：

followers_ids_for_01 = [123,124,...]
followers_ids_for_02 = [125,126,...]

Json List：

{
    "data": [{
        "id": 01,
        "followers": [{
            "id": 123,
            "name": "Shivkumar"
        }, {
            "id": 124,
            "name": "Sneha"
        },
        {
            .....
        }],
    }, {
        "id": 02,
        "followers": [{
            "id": 125,
            "name": "Teja"
        }, {
            "id": 126,
            "name": "Harsha"
        },
        {
            ......
        }],
    }]
}

我尝试了什么：

data  =json_list
projects = data['data']

followers_ids =[]
for project in projects:
    for i in project['followers']:
       followers_ids.append( i['id'])

print followers_ids

输出：

[123, 124, 125, 126]

有人可以为＆＃34; followers_ids_for_{project_id}＆＃34;建议如何生成列表并在其中添加其各自的关注者ID？

修改/更新

我需要为空白列表生成命名变量

datalists = dict()
for project in projects:
    datalists["followers_ids_for_{}".format(project['id'])] = []

print datalists
#{'followers_ids_for_1': [], 'followers_ids_for_2': []}

Answer 1

您可以使用列表理解

[follower['id'] for project in projects for follower in project['followers']]

如果您只对特定项目ID感兴趣

[follower['id'] for project in projects for follower in project['followers'] if project['id] == 'some id']

Answer 2

您可以使用nest dict comprehension 表达式来创建dict对象：

>>> {val['id']: [item['id'] for item in val['followers']] for val in data['data']}
{1: [123, 124], 2: [125, 126]}

让我们说上面的dict存储在变量result中。然后，您可以访问与每个所谓的项目ID 对应的dict中的元素：

>>> result[1]    # For id '1'
[123, 124]

>>> result[2]    # For id '2'
[125, 126]

PS：上述字典理解中的data变量保存了JSON对象的值，即：

{
    "data": [{
        "id": 1,
        "followers": [{
            "id": 123,
            "name": "Shivkumar"
        }, {
            "id": 124,
            "name": "Sneha"
        }],
    }, {
        "id": 2,
        "followers": [{
            "id": 125,
            "name": "Teja"
        }, {
            "id": 126,
            "name": "Harsha"
        }]
    }]
}

Answer 3

您要做的第一件事是以您想要的格式构建字典：

In [514]: f = {'followers_ids_for_%s' %d['id'] : [x['id'] for x in d['followers']] for d in data['data'] }; f
Out[514]: {'followers_ids_for_01': [123, 124], 'followers_ids_for_02': [125, 126]}

我建议你在这里停止。但是，这是第二步， 高度不推荐的 步骤（来自您的hard requirement）：更新globals()。< / p>

In [516]: globals().update(f)

In [517]: followers_ids_for_01
Out[517]: [123, 124]

现在您可以将这些变量作为列表进行访问。再一次，这很糟糕，因为它会污染你的全局。

通过使用python对字典中的所有ID进行分组来生成列表

3 个答案: