订购dict默认？

Question

是否有可能使默认的dict文字创建有序的dicts而不是无序的？

我想键入复杂的嵌套配置,如:

config = {
   'train': {
       'speed': 0.001,
       'initial_values': [1, 2, 3]
   },
   'model': {
...
   }
}

并且想要写一堆括号

config = OrderedDict([(
    'train', OrderedDict([(
       'speed', 0.001), (
       'initial_values', [1, 2, 3])]),(
    'model', OrderedDict([(
...

绝对是无法解决的.

关于为什么我的愿望不好,请不要有哲学.

---

好的,目前我会写一些像:

def od(*args):
   return OrderedDict([(args[i], args[i+1]) for i in range(0, len(args), 2)])

config = od(
    'train', od(
        'speed', 0.001,
        'initial_values', [1, 2, 3]
     ),
     'model', od(
    ...
     )
)

Answer 1

不,你不能改变Python语法,不能不改变CPython源代码和重新编译,但它不再是Python了.

您可以做的最好是 升级到Python 3.6或更高版本,其中默认情况下字典保留插入顺序.如果您必须具有完整的OrderedDict功能集(重新排序,反转,带有排序的dict视图),那么OrderedDict在事实之后将这些常规词典转换为对象:

from collections import OrderedDict
from functools import singledispatch

@singledispatch
def to_ordered_dict(d):
    """Convert dictionaries to OrderedDict objects, recursively

    Assumes you want to use current dictionary iteration order; in Python 3.7
    and newer that's the same as insertion order (or earlier if on PyPy, or 
    when using CPython, 3.6 and newer).

    """
    return d

@to_ordered_dict.register(dict)
def _dict(d):
    return OrderedDict(
        (to_ordered_dict(k), to_ordered_dict(v))
        for k, v in d.items()
    )

@to_ordered_dict.register(list)
@to_ordered_dict.register(tuple)
def _sequence(s):
    return type(s)(map(to_ordered_dict, s))

# add additional type registrations for other compound or container types

然后坚持使用最后一行的{...}符号config = to_ordered_dict(config).