から辞書が欲しい
>>> page_detail_string = urllib2.urlopen("http://graph.facebook.com/Ideas4India").read()
次のような文字列を返します
>>> page_detail_string
'{"about":"Ideas for development of India","category":"Community","description":"Platform where you can discuss and share your ideas which you think might help in betterment of our country.\\nPlease respect other community members and do not talk politics here.","is_published":true,"talking_about_count":0,"username":"Ideas4India","were_here_count":0,"id":"250014455083430","name":"Ideas 4 India","link":"http:\\/\\/www.facebook.com\\/Ideas4India","likes":23}'
今、私はそれを辞書に変換したいのですが、これはast.literal_evalを使用して簡単に行うことができます
>>> import ast
>>> dict_page = ast.literal_eval(page_detail_string)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/python2.7/ast.py", line 80, in literal_eval
return _convert(node_or_string)
File "/usr/lib/python2.7/ast.py", line 63, in _convert
in zip(node.keys, node.values))
File "/usr/lib/python2.7/ast.py", line 62, in <genexpr>
return dict((_convert(k), _convert(v)) for k, v
File "/usr/lib/python2.7/ast.py", line 79, in _convert
raise ValueError('malformed string')
しかし、私はそれが原因でこのエラーをスローすると思います
"is_published":true
上記のキーと値( "is_published":true)を削除して、辞書に変換する方法はありますか?
ありがとう