如果您不愿意损失一些速度,则不会。如果稍微慢一点是可以的,你必须考虑使用普通json.loads
并递归地转换为str
可能更便宜而且更快。
话虽如此,如果您确实想要loads
返回字符串的程度足以接受扩展本不应该的代码,这是一个可能的结果(主要是通过复制粘贴扩展)这是愚蠢的,感谢Lennart让我看到了轻量级(即,您只需要扩展 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
# Convert from unicode to str
def str_scanstring(*args, **kwargs):
result = c_scanstring(*args, **kwargs)
return str(result[0]), result[1]
# Little dirty trick here
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)
# And another little dirty trick there
_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))