如何在python中处理json数据？

Question

# This code searches the query from googlecustomsearch api and returns data in json format   
import pprint
import json

from googleapiclient.discovery import build


my_api_key = "**************************************"
my_cse_id = "*************************************"

def google_search(search_term, api_key, cse_id, **kwargs):
    service = build("customsearch", "v1", developerKey=api_key)
    res = service.cse().list(q=search_term, cx=cse_id, **kwargs).execute()
    return res['items']


results = google_search(
    'Roshan Patel', my_api_key, my_cse_id, num=10)
for result in results:
    pprint.pprint(result)
    result_dict = json.loads(result)
    print result_dict['formattedUrl'] 

这是我得到的输出：

我想只获取例如：u'formattedUrl'的url部分并将其存储在列表中，该怎么做？

新错误

Traceback (most recent call last):

  File "<ipython-input-38-eb898c8de239>", line 1, in <module>
    runfile('C:/Users/abc/untitled9.py', wdir='C:/Users/abc')

  File "C:\Users\abc\Anaconda2\lib\site-packages\spyder\utils\site\sitecustomize.py", line 880, in runfile
    execfile(filename, namespace)

  File "C:\Users\abc\Anaconda2\lib\site-packages\spyder\utils\site\sitecustomize.py", line 87, in execfile
    exec(compile(scripttext, filename, 'exec'), glob, loc)

  File "C:/Users/abc/untitled9.py", line 28, in <module>
    result_dict = json.loads(result)

  File "C:\Users\abc\Anaconda2\lib\json\__init__.py", line 339, in loads
    return _default_decoder.decode(s)

  File "C:\Users\abc\Anaconda2\lib\json\decoder.py", line 364, in decode
    obj, end = self.raw_decode(s, idx=_w(s, 0).end())

TypeError: expected string or buffer

Answer 1

你是否需要首先将结果转换为Python对象，比如dict，使用json模块：

result_dict = json.loads(result)

现在你可以过滤你的词典：

result_dict['formattedUrl']

Answer 2

result不是JSON，而是实际的Python字典，所以只需：

print result[u'formattedUrl']