正如json.loads简单地返回一个字典一样，您可以使用适用于字典的运算符：

>>> jdata = json.load('{"uri": "http:", "foo", "bar"}')
>>> 'uri' in jdata       # Check if 'uri' is in jdata's keys
True
>>> jdata['uri']         # Will return the value belonging to the key 'uri'
u'http:'

编辑：要给出有关如何遍历数据的想法，请考虑以下示例：

>>> import json
>>> jdata = json.loads(open ('bookmarks.json').read())
>>> for c in jdata['children'][0]['children']:
...     print 'Title: {}, URI: {}'.format(c.get('title', 'No title'),
                                          c.get('uri', 'No uri'))
...
Title: Recently Bookmarked, URI: place:folder=BOOKMARKS_MENU(...)
Title: Recent Tags, URI: place:sort=14&type=6&maxResults=10&queryType=1
Title: , URI: No uri
Title: Mozilla Firefox, URI: No uri

检查jdata数据结构将使您可以根据需要进行导航。pprint您已经拥有的电话就是一个很好的起点。

Edit2：另一个尝试。这将获得您在词典列表中提到的文件。这样，我认为您应该能够使其适应您的需求。

>>> def build_structure(data, d=[]):
...     if 'children' in data:
...         for c in data['children']:
...             d.append({'title': c.get('title', 'No title'),
...                                      'uri': c.get('uri', None)})
...             build_structure(c, d)
...     return d
...
>>> pprint.pprint(build_structure(jdata))
[{'title': u'Bookmarks Menu', 'uri': None},
 {'title': u'Recently Bookmarked',
  'uri':   u'place:folder=BOOKMARKS_MENU&folder=UNFILED_BOOKMARKS&(...)'},
 {'title': u'Recent Tags',
  'uri':   u'place:sort=14&type=6&maxResults=10&queryType=1'},
 {'title': u'', 'uri': None},
 {'title': u'Mozilla Firefox', 'uri': None},
 {'title': u'Help and Tutorials',
  'uri':   u'http://www.mozilla.com/en-US/firefox/help/'},
 (...)
}]

然后 “在其中搜索u'uri': u'http:'”，执行以下操作：

for c in build_structure(jdata):
    if c['uri'].startswith('http:'):
        print 'Started with http'