我从flask应用程序收到以下json响应，想知道如何将其拆分为多个“行” / dict。
输出：
{'class': [0.0,1.0],'probability': [0.8488858872836712,0.1511141127163287]}
我想要的是：
[{"class": 0.0,"probability": 0.8488858872836712},{"class": 1.0,"probability": 0.1511141127163287}]
我尝试了以下操作，但不确定如何获取这两个密钥：
{k: v for e in zip(model.classes_,probabilities[0]) for k,v in zip(('class','probability'),e)}
假设输出存储在d中，您可以完成
[{'class': c,'probability': p} for c,p in zip(d['class'],d['probability'])]
这将导致：
[{'class': 0.0,'probability': 0.8488858872836712},{'class': 1.0,'probability': 0.1511141127163287}]
这里是概念证明：
Python 3.7.5 (default,Oct 17 2019,12:16:48)
[GCC 9.2.1 20190827 (Red Hat 9.2.1-1)] on linux
Type "help","copyright","credits" or "license" for more information.
>>> from pprint import pprint
>>> d={'class': [0.0,1.0],... 'probability': [0.8488858872836712,0.1511141127163287]}
>>> pprint([{'class': c,d['probability'])])
[{'class': 0.0,'probability': 0.1511141127163287}]
>>>
更通用的解决方案（如果您不提前知道密钥名称）：
implicit val orderer = Ordering.Double.TotalOrdering
myMap.toList.sortBy(_._2)
// res1: List[(String,Double)] = List((4,0.3),(2,2.1),(3,3.4),(1,4.6))
输出：
d = {'class': [0.0,'probability': [0.8488858872836712,0.1511141127163287]}
result = [dict(zip(d.keys(),i)) for i in zip(*d.values())]
使用列表推导拆分json
dic = { 'class': [0.0,0.1511141127163287] }
split_value = [{'class':i,'probability':j} for i,j in zip(dic['class'],dic['probability'])]
print(split_value)
输出：-
[{'class': 0.0,'probability': 0.1511141127163287}]