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Python尽可能高效地将字节转换为字符串字典_Python_String_Dictionary_Byte - Fatal编程技术网

Python尽可能高效地将字节转换为字符串字典

Python尽可能高效地将字节转换为字符串字典,python,string,dictionary,byte,Python,String,Dictionary,Byte,我正在尝试高效(快速)地将字节文件转换成字典。 到目前为止,我已经把它解码成str,去掉{}并把它编入字典。但是我仍然得到单字符串引号,所以我不能按键调用dict。我该怎么做 response = b'{"address":"david@gmail.com","username":"david","domain":"gmail.com","md5Hash":

我正在尝试高效(快速)地将字节文件转换成字典。 到目前为止,我已经把它解码成str,去掉{}并把它编入字典。但是我仍然得到单字符串引号,所以我不能按键调用dict。我该怎么做

response = b'{"address":"david@gmail.com","username":"david","domain":"gmail.com","md5Hash":"f3c52e5ef3d2b471d0ef51c66c21d10c","suggestion":"","validFormat":true,"deliverable":false,"fullInbox":false,"hostExists":true,"catchAll":false,"gravatar":false,"role":false,"disposable":false,"free":true}'
到目前为止,我采取了以下步骤:

print(response.decode("utf-8"))

{"address":"david@gmail.com","username":"david","domain":"gmail.com","md5Hash":"f3c52e5ef3d2b471d0ef51c66c21d10c","suggestion":"","validFormat":true,"deliverable":false,"fullInbox":false,"hostExists":true,"catchAll":false,"gravatar":false,"role":false,"disposable":false,"free":true}



print({response.decode("utf-8").replace("}","").replace("{","")})

{'"address":"david@gmail.com","username":"david","domain":"gmail.com","md5Hash":"f3c52e5ef3d2b471d0ef51c66c21d10c","suggestion":"","validFormat":true,"deliverable":false,"fullInbox":false,"hostExists":true,"catchAll":false,"gravatar":false,"role":false,"disposable":false,"free":true'}
但是我仍然无法通过按键调用dict,因为


我需要的是快速、低资源。

比如@juanpa.arrivillaga说只要使用
json
模块就可以了

在这里:

导入json
d=json.load(响应)
现在,
d
的值是您的解码dict


就像@juanpa.arrivillaga所说的,没有必要使用
.decode()
json.loads()
可以对字节进行操作。

这看起来像json?它甚至不表示有效的Python dict文本,例如
false
而不是
false
。使用
json
模块。这是什么来源?我的问题也是。你甚至不需要
。解码
json。加载
接受
字节
对象以及
str
对象