如果你不想失去一些速度,那就不要这样做。如果稍微慢一点没关系,你需要考虑使用普通的json.loads
并递归转换为str
可能更便宜,也许更快。
话虽如此,如果你真的非常想要一个返回字符串的loads
,以至于愿意接受通过扩展不打算使用的代码来实现,这里有一个可能的结果(大部分是通过复制粘贴进行扩展)这是荒谬的,感谢Lennart让我看到了光明(即,你只需要扩展JSONDecoder和一些技巧):
import json
from json import decoder, scanner
from json.scanner import make_scanner
from _json import scanstring as c_scanstring
_CONSTANTS = json.decoder._CONSTANTS
py_make_scanner = scanner.py_make_scanner
def str_scanstring(*args, **kwargs):
result = c_scanstring(*args, **kwargs)
return str(result[0]), result[1]
json.decoder.scanstring = str_scanstring
class StrJSONDecoder(decoder.JSONDecoder):
def __init__(self, encoding=None, object_hook=None, parse_float=None,
parse_int=None, parse_constant=None, strict=True,
object_pairs_hook=None):
self.encoding = encoding
self.object_hook = object_hook
self.object_pairs_hook = object_pairs_hook
self.parse_float = parse_float or float
self.parse_int = parse_int or int
self.parse_constant = parse_constant or _CONSTANTS.__getitem__
self.strict = strict
self.parse_object = decoder.JSONObject
self.parse_array = decoder.JSONArray
self.parse_string = str_scanstring
self.scan_once = py_make_scanner(self)
_default_decoder = StrJSONDecoder(encoding=None, object_hook=None,
object_pairs_hook=None)
json._default_decoder = _default_decoder
j = {1:'2', 1.1:[1,2,3], u'test': {12:12, 13:'o'}}
print json.loads(json.dumps(j))