Question

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)}

Answer 1

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, 'probability': p} for c,p in zip(d['class'], d['probability'])])
[{'class': 0.0, 'probability': 0.8488858872836712},
 {'class': 1.0, 'probability': 0.1511141127163287}]
>>>

Answer 2

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, 1.0], 'probability': [0.8488858872836712, 0.1511141127163287]}
result = [dict(zip(d.keys(), i)) for i in zip(*d.values())]

Answer 3

使用列表推导拆分json

dic = { 'class': [0.0, 1.0], 'probability': [0.8488858872836712, 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': 0.8488858872836712}, {'class': 1.0, 'probability': 0.1511141127163287}]

在字典中拆分列表值以分隔字典

3 个答案: