Question

cow = {
    "data": [
        {
            "id": "1179640025492630",
            "indicator": "google-analytics|UA-97996598-1",
            "type": "API_KEY"
        },
        {
            "id": "1469013519830038",
            "indicator": "google-analytics|UA-96613605-2",
            "type": "API_KEY"
        },
        {
            "id": "1459958767410551",
            "indicator": "google-analytics|UA-86399386-2",
            "type": "API_KEY"
        },
        {
            "id": "1507839102569000",
            "indicator": "google-analytics|UA-89570367-2",
            "type": "API_KEY"
        },
        {
            "id": "1276736575767341",
            "indicator": "google-analytics|UA-69774312-4",
            "type": "API_KEY"
        },
        {
            "id": "1292251910882451",
            "indicator": "google-analytics|UA-93952538-3",
            "type": "API_KEY"
        }
    ],
    "paging": {
        "cursors": {
            "after": "NQZDZD",
            "before": "MAZDZD"
        }
    }
}

我想从数据字典中提取“id”：“1179640025492630”。

我试过使用for循环，如：

for i in cow['data']:
    for key,value in i:
        if(i == 'id'):
            print key,value
        else:
            pass

它给我一个值错误，如：

Traceback (most recent call last):
  File "./dict.py", line 14, in <module>
    for key,value  in ['data']:
  ValueError: too many values to unpack

如果我使用.iteritems，我会收到AttributeError：

Traceback (most recent call last):
  File "./dict.py", line 14, in <module>
    for key,value   in ['data'].iteritems:
  AttributeError: 'list' object has no attribute 'iteritems'

Answer 1

如果你想要一个使用嵌套for循环的更干净的解决方案，你可以使用python的map函数迭代data中的所有字典，只取id }值。

>>> list(map(lambda d: 'id: {}'.format(d['id']), cow['data']))
['id: 1179640025492630', 'id: 1469013519830038', 'id: 1459958767410551', 'id: 1507839102569000', 'id: 1276736575767341', 'id: 1292251910882451']

Answer 2

如果要提取 id 字段，请使用映射函数：

map(lambda item: ("id", item["id"]), cow["data"])

并且您将获得一个元组列表（请注意，py3地图是懒惰的，您将强制评估列表）。

Answer 3

或者你可以结合list和dictionary理解，这可能更容易处理。

result = [{k:v for k,v in dikt.items() if k == "id"} for dikt in cow['data']]

给你

[{'id'：'1179640025492630'}，{'id'：'1469013519830038'}，{'id'：'1459958767410551'}，{'id'：'1507839102569000'}，{'id'：' 1276736575767341'}，{'id'：'1292251910882451'}]

你可以这样打印出来：

print "\n".join([i['id'] for i in result])

Answer 4

如果您只想打印与id s相关联的所有值：

for thing in cow["data"]:
    print(thing["id"])

输出：

1179640025492630
1469013519830038
1459958767410551
1507839102569000
1276736575767341
1292251910882451

如何解析这个JSON dict来打印出一个键值对？

4 个答案: