python将所有键转换为字符串
Clo*_*ave 2 python json dictionary
我有嵌套的字典，有一些数字键。我需要将此字典存储为 JSON，并且由于这些键是数字，因此我无法将它们存储为 JSON。我写了下面的代码，但它给了我一个错误，说字典的长度已经改变（RuntimeError: dictionary changed size during iteration）。
def convert_to_str(dictionary):
for key in dictionary:
print (key)
found = False
non_str_keys = []
if not isinstance(key, str):
print(key, 'is not a string')
dictionary[str(key)] = dictionary[key]
non_str_keys.append(key)
if isinstance(dictionary[str(key)], dict):
dictionary[str(key)] = convert_to_str(dictionary[str(key)])
non_str_keys.append(key)
if non_str_keys:
for each_non_str_key in non_str_keys:
del dictionary[each_non_str_key]
return dictionary
我如何避免这种情况？我的字典是——
a = {
"age": {
1: 25.0,
2: 50.25,
3: 50.0,
4: 75.0,
5: 14.580906789680968,
6: [
25.0,
30.0,
34.800000000000004,
40.0,
46.60000000000001,
50.0,
56.0,
61.0,
65.0,
69.0,
75.0
],
"quartiles": [
38.0,
64.0
],
"decile_event_rate": [
0.8125,
0.7142857142857143,
0.65625,
0.42857142857142855,
0.45161290322580644,
0.4857142857142857,
0.5925925925925926,
0.5,
0.5142857142857142,
0.375
]
},
"income": {
"min": 10198.0,
"mean": 55621.78666666667,
"median": 52880.0,
"max": 99783.0,
"std": 24846.911384024643,
"deciles": [
10198.0,
25269.4,
31325.800000000003,
37857.0,
43721.8,
52880.0,
63996.0,
72526.9,
82388.2,
89765.90000000001,
99783.0
],
"quartiles": [
35088.5,
78687.25
],
"decile_event_rate": [
0.6666666666666666,
0.6,
0.5333333333333333,
0.5666666666666667,
0.5,
0.6451612903225806,
0.4827586206896552,
0.5,
0.5666666666666667,
0.5
]
},
"edu_yrs": {
"min": 0.0,
"mean": 12.73,
"median": 13.0,
"max": 25.0,
"std": 7.86234623342895,
"deciles": [
0.0,
2.0,
4.0,
7.0,
9.600000000000009,
13.0,
16.0,
18.0,
21.200000000000017,
23.0,
25.0
],
"quartiles": [
6.0,
20.0
],
"decile_event_rate": [
0.5384615384615384,
0.6521739130434783,
0.5151515151515151,
0.48,
0.6111111111111112,
0.5,
0.5,
0.6071428571428571,
0.5151515151515151,
0.6666666666666666
]
},
"yrs_since_exercise": {
"min": 0.0,
"mean": 18.566666666666666,
"median": 16.0,
"max": 60.0,
"std": 14.417527732194037,
"deciles": [
0.0,
3.0,
5.0,
8.0,
12.0,
16.0,
20.0,
25.0,
31.0,
41.0,
60.0
],
"quartiles": [
6.0,
27.0
],
"decile_event_rate": [
1.0,
1.0,
1.0,
0.9629629629629629,
0.75,
0.4857142857142857,
0.15384615384615385,
0.06666666666666667,
0.0,
0.0
]
},
"security_label": {
"event_rate": {
"A": {
"1.0": 0.6,
"0.0": 0.4
},
"B": {
"1.0": 0.57,
"0.0": 0.43
},
"C": {
"0.0": 0.5,
"1.0": 0.5
}
},
"freq": {
"A": 100,
"B": 100,
"C": 100
},
"var_type": "categorical"
}
}
编辑
json.dump(self.entity_data, open(path, 'w'), indent=2, cls=CustomEncoder)
File "/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/json/__init__.py", line 179, in dump
for chunk in iterable:
File "/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/json/encoder.py", line 430, in _iterencode
yield from _iterencode_dict(o, _current_indent_level)
File "/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/json/encoder.py", line 404, in _iterencode_dict
yield from chunks
File "/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/json/encoder.py", line 404, in _iterencode_dict
yield from chunks
File "/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/json/encoder.py", line 404, in _iterencode_dict
yield from chunks
File "/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/json/encoder.py", line 376, in _iterencode_dict
raise TypeError("key " + repr(key) + " is not a string")
TypeError: key 0 is not a string
添加错误图像
编辑-2
我在使用 numpy 对象时遇到了序列化错误 ebfore。所以我开始，使用这个编码器将它们转换为 python 对象。
class CustomEncoder(json.JSONEncoder):
def default(self, obj):
if isinstance(obj, np.integer):
return int(obj)
elif isinstance(obj, np.floating):
return float(obj)
elif isinstance(obj, np.ndarray):
return obj.tolist()
else:
return super(CustomEncoder, self).default(obj)
我一直json.dump在使用cls = CustomEncoder. 这是我用过的命令
json.dump(self.entity_data, open(path, 'w'), indent=2, cls=CustomEncoder)
您需要递归转换所有键；生成具有 dict 理解的新字典，这比就地更改键容易得多。您不能在正在迭代的字典中添加字符串键和删除非字符串键，因为这会改变哈希表，这很容易改变字典键的列出顺序，所以这是不允许的。
你不应该忘记处理列表；它们也可以包含更多的字典。
每当我需要转换这样的嵌套结构时，我都会使用@functools.singledispatch装饰器将不同容器类型的处理拆分为不同的功能：
from functools import singledispatch
@singledispatch
def keys_to_strings(ob):
return ob
@keys_to_strings.register
def _handle_dict(ob: dict):
return {str(k): keys_to_strings(v) for k, v in ob.items()}
@keys_to_strings.register
def _handle_list(ob: list):
return [keys_to_strings(v) for v in ob]
然后 JSON 编码的结果keys_to_string()：
json.dumps(keys_to_string(a))
并不是说这都是需要的。json.dumps()本机接受整数键，将它们转换为字符串。您的输入示例无需转换即可工作：
json.dumps(a)
注意： JSON 的键/值对中的键始终是 类型
str。当字典转换为 JSON 时，字典的所有键都被强制转换为字符串。因此，如果将字典转换为 JSON，然后再转换回字典，则该字典可能不等于原始字典。也就是说，loads(dumps(x)) != x如果x有非字符串键。
这仅适用于JSON 本来可以处理的类型，所以None，布尔值float和int对象。对于其他任何事情，你仍然会得到你的例外。您可能有一个表示为的对象0，但它不是 Python int0：
>>> json.dumps({0: 'works'})
'{"0": "works"}'
>>> import numpy
>>> numpy.int32()
0
>>> json.dumps({numpy.int32(): 'fails'})
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/Users/mjpieters/Development/Library/buildout.python/parts/opt/lib/python3.6/json/__init__.py", line 231, in dumps
return _default_encoder.encode(obj)
File "/Users/mjpieters/Development/Library/buildout.python/parts/opt/lib/python3.6/json/encoder.py", line 199, in encode
chunks = self.iterencode(o, _one_shot=True)
File "/Users/mjpieters/Development/Library/buildout.python/parts/opt/lib/python3.6/json/encoder.py", line 257, in iterencode
return _iterencode(o, 0)
TypeError: keys must be a string
我选择了一个numpy整数类型，因为这是一个常见的混淆整数值，而不是 Python int。
您添加到帖子中的自定义编码器不会用于密钥；这只适用于字典中的值，所以如果你有非标准的键对象，那么你确实仍然需要使用上面的递归解决方案。