在python中将Dictionary转换为JSON
she*_*158 270 python json dictionary python-2.7
r = {'is_claimed': 'True', 'rating': 3.5}
r = json.dumps(r)
file.write(str(r['rating']))
我无法在json中访问我的数据.我究竟做错了什么？
TypeError: string indices must be integers, not str
Ima*_*deh 429
json.dumps()将字典转换为str对象,而不是json(dict)对象!所以你必须将你的str加载到dict中才能通过使用str方法来使用它
请参阅json(dict)保存方法和str检索方法.
这是代码示例,可以帮助您更好地理解它:
import json
r = {'is_claimed': 'True', 'rating': 3.5}
r = json.dumps(r)
loaded_r = json.loads(r)
loaded_r['rating'] #Output 3.5
type(r) #Output str
type(loaded_r) #Output dict
- 难道你不认为将json加载到一个单独的变量`loaded_r`中是有点低效的,考虑到你已经在`r`中使用它了吗？ (8认同)
- 我怀疑@TimCastelijns误解了伊姆兰的意图。将JSON加载到新变量中并不是因为它是新信息，而是为了表明新字典中的值（即在JSON编码和解码往返之后）的值是相同的。 (5认同)
- @SchrodingersStat 嗨。我认为这不会对性能产生太大影响。您可以使用“timeit”功能进行检查。如果你真的想要一个快速的 json 加载/转储，你可以使用 'ujson' 库！ (3认同)
Tim*_*Tim 36
json.dumps()返回python dict的JSON字符串表示形式.查看文档
你不能这样做,r['rating']因为r是一个字符串,而不是一个字典
也许你的意思是
r = {'is_claimed': 'True', 'rating': 3.5}
json = json.dumps(r) # note i gave it a different name
file.write(str(r['rating']))
Mil*_*vić 12
json.dumps() 用于解码 JSON 数据
json.loads将一个字符串作为输入并返回一个字典作为输出。json.dumps以字典作为输入并返回一个字符串作为输出。
import json
# initialize different data
str_data = 'normal string'
int_data = 1
float_data = 1.50
list_data = [str_data, int_data, float_data]
nested_list = [int_data, float_data, list_data]
dictionary = {
'int': int_data,
'str': str_data,
'float': float_data,
'list': list_data,
'nested list': nested_list
}
# convert them to JSON data and then print it
print('String :', json.dumps(str_data))
print('Integer :', json.dumps(int_data))
print('Float :', json.dumps(float_data))
print('List :', json.dumps(list_data))
print('Nested List :', json.dumps(nested_list, indent=4))
print('Dictionary :', json.dumps(dictionary, indent=4)) # the json data will be indented
输出：
String : "normal string"
Integer : 1
Float : 1.5
List : ["normal string", 1, 1.5]
Nested List : [
1,
1.5,
[
"normal string",
1,
1.5
]
]
Dictionary : {
"int": 1,
"str": "normal string",
"float": 1.5,
"list": [
"normal string",
1,
1.5
],
"nested list": [
1,
1.5,
[
"normal string",
1,
1.5
]
]
}
- Python 对象到 JSON 数据的转换
| Python | JSON |
|:--------------------------------------:|:------:|
| dict | object |
| list, tuple | array |
| str | string |
| int, float, int- & float-derived Enums | number |
| True | true |
| False | false |
| None | null |
更新
在 JSON 文件中
nested_dictionary = {
'one': nested_list,
'two': dictionary,
}
json_dict = {'Nested Dictionary': nested_dictionary,
'Multiple':[nested_dictionary, nested_dictionary, nested_dictionary]
}
with open("test_nested.json", "w") as outfile:
json.dump(json_dict, outfile, indent=4, sort_keys=False)
图表反应
输出到 test_nested.json
{
"Nested Dictionary": {
"one": [
1,
1.5,
[
"normal string",
1,
1.5
]
],
"two": {
"int": 1,
"str": "normal string",
"float": 1.5,
"list": [
"normal string",
1,
1.5
],
"nested list": [
1,
1.5,
[
"normal string",
1,
1.5
]
]
}
},
"Multiple": [
{
"one": [
1,
1.5,
[
"normal string",
1,
1.5
]
],
"two": {
"int": 1,
"str": "normal string",
"float": 1.5,
"list": [
"normal string",
1,
1.5
],
"nested list": [
1,
1.5,
[
"normal string",
1,
1.5
]
]
}
},
{
"one": [
1,
1.5,
[
"normal string",
1,
1.5
]
],
"two": {
"int": 1,
"str": "normal string",
"float": 1.5,
"list": [
"normal string",
1,
1.5
],
"nested list": [
1,
1.5,
[
"normal string",
1,
1.5
]
]
}
},
{
"one": [
1,
1.5,
[
"normal string",
1,
1.5
]
],
"two": {
"int": 1,
"str": "normal string",
"float": 1.5,
"list": [
"normal string",
1,
1.5
],
"nested list": [
1,
1.5,
[
"normal string",
1,
1.5
]
]
}
}
]
}
class 实例到 JSON
- 一个简单的解决方案：
class Foo(object):
def __init__(
self,
data_str,
data_int,
data_float,
data_list,
data_n_list,
data_dict,
data_n_dict):
self.str_data = data_str
self.int_data = data_int
self.float_data = data_float
self.list_data = data_list
self.nested_list = data_n_list
self.dictionary = data_dict
self.nested_dictionary = data_n_dict
foo = Foo(
str_data,
int_data,
float_data,
list_data,
nested_list,
dictionary,
nested_dictionary)
# Because the JSON object is a Python dictionary.
result = json.dumps(foo.__dict__, indent=4)
# See table above.
# or with built-in function that accesses .__dict__ for you, called vars()
# result = json.dumps(vars(foo), indent=4)
print(result) # same as before
- 更简单
class Bar:
def toJSON(self):
return json.dumps(self, default=lambda o: o.__dict__,
sort_keys=False, indent=4)
bar = Bar()
bar.web = "Stackoverflow"
bar.type = "Knowledge"
bar.is_the_best = True
bar.user = Bar()
bar.user.name = "Milovan"
bar.user.age = 34
print(bar.toJSON())
图表反应
输出：
{
"web": "Stackoverflow",
"type": "Knowledge",
"is_the_best": true,
"user": {
"name": "Milovan",
"age": 34
}
}
- 认为它必须被接受的答案！ (2认同)
小智 6
您可以通过在默认字典中声明新字典来在上面的示例中创建嵌套字典。
import json
dictionary = {
'fruit':{"Grapes": "10","color": "green"},
'vegetable':{"chilli": "4","color": "red"},
}
result = json.dumps(dictionary, indent = 3)
打印（结果）
在这里，我使用了 indent=3
参考： https: //favtutor.com/blogs/dict-to-json-python