python有没有什么技巧可以将字典展平并同时对其进行排序?

for annotation in sorted(s, key = lambda tup: tup['points'][0]['start']):
    print(annotation["label"],annotation["points"][0]["start"],annotation["points"][0]["end"])

或者,只需要识别 start 按姓名一次:

for annotation in sorted([(x["label"][0],x["points"][0]["start"],x["points"][0]["end"]) for x in s], key = lambda tup: tup[1]):
    print(annotation)

诀窍是打开包装 第一 .

(item['label'], item['points'][0]['start'], item['points'][0]['end'])

我们如何将这样的规则应用于整个列表?使用表达式生成器或表达式列表:

(
    (item['label'], item['points'][0]['start'], item['points'][0]['end'])
    for item in s
)

我们如何对这些结果进行排序?使用 sorted 分类

sorted(
    (
        (item['label'], item['points'][0]['start'], item['points'][0]['end'])
        for item in s
    ),
    key = lambda filtered: filtered[1]
)

假设字典键始终按所示顺序排列,则可以将字典列表转换为重新排列的元组列表(以“开始”键/值对开头)。然后对元组进行排序,并从中重新形成字典。

s = [{"label":["College Name"],
     "points":[{"start":939,"end":956,"text":"Kendriya Vidyalaya"}]},
     {"label":["College Name"],
      "points":[{"start":883,"end":904,"text":"Woodbine modern school"}]}]

r = [*map(dict,sorted( ((*pd.items(),(lb,lv)) for d in s   
                       for (lb,(lv,*_)),(_,(pd,*_)) in [(*d.items(),)])))]

print(r)

[{'start': 883, 'end': 904, 'text': 'Woodbine modern school',
  'label': 'College Name'},
 {'start': 939, 'end': 956, 'text': 'Kendriya Vidyalaya', 
  'label': 'College Name'}]

r = sorted(({k:v for k,(v,*_) in d.items()} for d in s),
            key=lambda d:d["points"]["start"])

print(r)

[{'label': 'College Name', 
  'points': {'start': 883, 'end': 904, 'text': 'Woodbine modern school'}}, 
 {'label': 'College Name', 
  'points': {'start': 939, 'end': 956, 'text': 'Kendriya Vidyalaya'}}]