Python中的自定义JSON sort_keys顺序

Question

在Python 2.6中有没有办法为JSON的sort_keys提供自定义键或cmp函数？

我有一个来自JSON的dicts列表,如下所示:

[
  {
    "key": "numberpuzzles1",
    "url": "number-puzzle-i.html",
    "title": "Number Puzzle I",
    "category": "nestedloops",
    "points": "60",
    "n": "087"
  },
  {
     "key": "gettingindividualdigits",
     "url": "getting-individual-digits.html",
     "title": "Getting Individual Digits",
     "category": "nestedloops",
     "points": "80",
     "n": "088"
  }
]

...我已经存储到list变量中了assigndb.我希望能够加载JSON,修改它,并使用dumps(或其他)将其序列化,保持密钥的顺序完好无损.

到目前为止,我尝试过这样的事情:

ordering = {'key': 0, 'url': 1, 'title': 2, 'category': 3,
             'flags': 4, 'points': 5, 'n': 6}

def key_func(k):
    return ordering[k]

# renumber assignments sequentially
for (i, a) in enumerate(assigndb):
    a["n"] = "%03d" % (i+1)

s = json.dumps(assigndb, indent=2, sort_keys=True, key=key_func)

...但当然dumps不支持自定义键list.sort().JSONEncoder也许有定制的东西？我似乎无法得到它.

Answer 1

一个想法(用2.7测试):

import json
import collections
json.encoder.c_make_encoder = None
d = collections.OrderedDict([("b", 2), ("a", 1)])
json.dumps(d)
# '{"b": 2, "a": 1}'

请参阅:OrderedDict + issue6105.在c_make_encoder似乎只需要破解的Python 2.x的 不是直接的解决方案,因为您必须dict为OrderedDicts 更改s,但它仍然可用.我检查了json库(encode.py),并且有序是硬编码的:

if _sort_keys:
    items = sorted(dct.items(), key=lambda kv: kv[0])

Answer 2

这有点丑陋，但如果托克兰的解决方案不适合你：

data = [{'category': 'nestedloops', 'title': 'Number Puzzle I', 'url': 'number-puzzle-i.html', 'n': '087', 'points': '60', 'key': 'numberpuzzles1'}, {'category': 'nestedloops', 'title': 'Getting Individual Digits', 'url': 'getting-individual-digits.html', 'n': '088', 'points': '80', 'key': 'gettingindividualdigits'}]
ordering = {'key': 0, 'url': 1, 'title': 2, 'category': 3,
            'flags': 4, 'points': 5, 'n': 6}
outlist = []
for d in data:
    outlist.append([])
    for k in sorted(d.keys(), key=lambda k: ordering[k]):
        outlist[-1].append(json.dumps({k: d[k]}))

for i, l in enumerate(outlist):
    outlist[i] = "{" + ",".join((s[1:-1] for s in outlist[i])) + "}"

s = "[" + ",".join(outlist) + "]"