blocks|key|1950807|text|假设输出存储在d中，您可以这样做|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1950808|[{'class':+c,+'probability':+p}+for+c,p+in+zip(d['class'],+d['probability'])]|code-block|syntax|javascript|1950809|这将导致：|1950810|[{'class':+0.0,+'probability':+0.8488858872836712},
+{'class':+1.0,+'probability':+0.1511141127163287}]|1950811|这里有一个概念的证明：|1950812|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}]
>>>+|1950813|entityMap^0|7|1|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|T|$]]

Assuming the output stored in <code>d</code>, you can do 

<pre><code>[{'class': c, 'probability': p} for c,p in zip(d['class'], d['probability'])]
</code></pre>

That will result in:

<pre><code>[{'class': 0.0, 'probability': 0.8488858872836712},
 {'class': 1.0, 'probability': 0.1511141127163287}]
</code></pre>

Here is a proof of concept:

<pre><code>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.
&gt;&gt;&gt; from pprint import pprint
&gt;&gt;&gt; d={'class': [0.0, 1.0],
... 'probability': [0.8488858872836712, 0.1511141127163287]}
&gt;&gt;&gt; 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}]
&gt;&gt;&gt; 
</code></pre>

blocks|key|1950820|text|一个更通用的解决方案(如果您不提前知道关键名称)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1950821|d+=+{'class':+[0.0,+1.0],+'probability':+[0.8488858872836712,+0.1511141127163287]}
result+=+[dict(zip(d.keys(),+i))+for+i+in+zip(*d.values())]|code-block|syntax|javascript|1950822|输出：|1950823|[{'class':+0.0,+'probability':+0.8488858872836712},+{'class':+1.0,+'probability':+0.1511141127163287}]|1950824|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

A more generic solution (if you do not know the key names ahead of time):

<pre><code>d = {'class': [0.0, 1.0], 'probability': [0.8488858872836712, 0.1511141127163287]}
result = [dict(zip(d.keys(), i)) for i in zip(*d.values())]
</code></pre>

Output:

<pre><code>[{'class': 0.0, 'probability': 0.8488858872836712}, {'class': 1.0, 'probability': 0.1511141127163287}]
</code></pre>

blocks|key|1700783|text|使用列表理解拆分json|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1700784|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)|code-block|syntax|javascript|1700785|产出：-|1700786|[{'class':+0.0,+'probability':+0.8488858872836712},+{'class':+1.0,+'probability':+0.1511141127163287}]|1700787|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Split json using list comprehension

<pre><code>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)
</code></pre>

Output:-

<pre><code>[{'class': 0.0, 'probability': 0.8488858872836712}, {'class': 1.0, 'probability': 0.1511141127163287}]
</code></pre>

I have the following json response from a flask application and am wondering how I can split it out into multiple "rows"/dicts. 

Output:

<pre><code>{'class': [0.0, 1.0],
 'probability': [0.8488858872836712, 0.1511141127163287]}
</code></pre>

What I desire is:

<pre><code>[{"class": 0.0, "probability": 0.8488858872836712},{"class": 1.0, "probability": 0.1511141127163287}]
</code></pre>

I've tried something like the following but am not sure how to get both keys:

<pre><code>{k: v for e in zip(model.classes_, probabilities[0]) for k, v in zip(('class', 'probability'), e)}
</code></pre>

Split list values inside dictionary to separate dictionaries

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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下面是来自一个烧瓶应用程序的json响应，我想知道如何将它拆分成多个“行”/dicts。输出：{'class': [0.0, 1.0], 'probability': [0.8488858872836712, 0.1511141127163287]}我想要的是：[{"class": 0.0, "probability"...

问将字典中的列表值拆分为单独的字典
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