blocks|key|1449350|text|你可以使用字典理解和切片。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1449351|lst=+[[country,usa,uk,india,japan],
[city,tokyo,berlin,moscow,abudhabi],
[planet,earth,mars,venus]]

result+=+[{l[0]:l[1:]}+for+l+in+lst]|code-block|syntax|javascript|1449352|在这里，我假设lst是一个列表，您想要从它创建一个字典列表。这在你的问题中并不是百分之百清楚。|1449353|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You can use dictionary comprehension and slicing for that
<pre class="lang-py prettyprint-override"><code>lst= [[country,usa,uk,india,japan],
[city,tokyo,berlin,moscow,abudhabi],
[planet,earth,mars,venus]]

result = [{l[0]:l[1:]} for l in lst]
</code></pre>
here I have assumed that lst is a list of lists you want to create a list of dictionaries from it. This was not 100% clear from you question.

blocks|key|2991734|text|假设这个有效的python输入：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2991735|lst+=+[['country','usa','uk','india','japan'],
+++++++['city','tokyo','berlin','moscow','abudhabi'],
+++++++['planet','earth','mars','venus']]|code-block|syntax|javascript|2991736|使用zip和dict构造函数：|offset|length|style|CODE|2991737|keys,+*vals+=+zip(*lst)
d+=+dict(zip(keys,+map(list,vals)))|2991738|输出：|2991739|{'country':+['usa',+'uk',+'india',+'japan'],
+'city':+['tokyo',+'berlin',+'moscow',+'abudhabi'],
+'planet':+['earth',+'mars',+'venus']}|2991740|entityMap^0|0|0|2|3|6|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]|$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|T|$]]

Assuming this valid python input:
<pre><code>lst = [['country','usa','uk','india','japan'],
 ['city','tokyo','berlin','moscow','abudhabi'],
 ['planet','earth','mars','venus']]
</code></pre>
Use <code>zip</code>, and the <code>dict</code> constructor:
<pre><code>keys, *vals = zip(*lst)
d = dict(zip(keys, map(list,vals)))
</code></pre>
Output:
<pre><code>{'country': ['usa', 'uk', 'india', 'japan'],
 'city': ['tokyo', 'berlin', 'moscow', 'abudhabi'],
 'planet': ['earth', 'mars', 'venus']}
</code></pre>

blocks|key|2682147|text|我假设您没有使用字符串，并且您提供的列表不是嵌套列表，因此我根据对代码的理解修改了代码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2682148|列表是由|2682149|lst=+[['country','usa','uk','india','japan'],
['city','tokyo','berlin','moscow','abudhabi'],
['planet','earth','mars','venus']]|code-block|syntax|javascript|2682150|把它转换成我用过的字典|2682151|dict={lst[0][0]:lst[0][1:5],
++++++lst[1][0]:lst[1][1:5],
++++++lst[2][0]:lst[2][1:4]}|2682152|这就是印刷字典的样子|2682153|{'city':+['tokyo',+'berlin',+'moscow',+'abudhabi'],
+'country':+['usa',+'uk',+'india',+'japan'],
+'planet':+['earth',+'mars',+'venus']}|2682154|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

I am assuming that you did not use strings and the list you gave was not a nested list so I modified the code based on my understanding of your code.
The list is defined by
<pre><code>lst= [['country','usa','uk','india','japan'],
['city','tokyo','berlin','moscow','abudhabi'],
['planet','earth','mars','venus']]
</code></pre>
to convert it to a dictionary I used
<pre><code>dict={lst[0][0]:lst[0][1:5],
 lst[1][0]:lst[1][1:5],
 lst[2][0]:lst[2][1:4]}
</code></pre>
and this is what printing the dictionary looks like
<pre><code>{'city': ['tokyo', 'berlin', 'moscow', 'abudhabi'],
 'country': ['usa', 'uk', 'india', 'japan'],
 'planet': ['earth', 'mars', 'venus']}
</code></pre>

blocks|key|2991822|text|假设这是您想要的输入：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2991823|lst+=+[['country','usa','uk','india','japan'],
+++++++['city','tokyo','berlin','moscow','abudhabi'],
+++++++['planet','earth','mars','venus']]|code-block|syntax|javascript|2991824|输出是这样的一个字典：|2991825|{'country':+['usa',+'uk',+'india',+'japan'],+'city':+['tokyo',+'berlin',+'moscow',+'abudhabi'],+'planet':+['earth',+'mars',+'venus']}|2991826|您可以使用dict理解和列表解压来获得以下结果：|2991827|mydict+=+{k:+vals+for+(k,+*vals)+in+lst}|2991828|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Assuming this is the input you want:
<pre><code>lst = [['country','usa','uk','india','japan'],
 ['city','tokyo','berlin','moscow','abudhabi'],
 ['planet','earth','mars','venus']]
</code></pre>
and the output is a single dictionary like this:
<pre><code>{'country': ['usa', 'uk', 'india', 'japan'], 'city': ['tokyo', 'berlin', 'moscow', 'abudhabi'], 'planet': ['earth', 'mars', 'venus']}
</code></pre>
You can use dict comprehension with list unpacking to get the result:
<pre><code>mydict = {k: vals for (k, *vals) in lst}
</code></pre>

Lets say we have a set of input lists:
<pre><code>lst= [country,usa,uk,india,japan],
[city,tokyo,berlin,moscow,abudhabi],
[planet,earth,mars,venus]
</code></pre>
We want to create a dictionary in python for each of these lists, so that the output looks like:
<pre><code>mydict= {country:[usa,uk,india,japan]},
{city:[tokyo,berlin,moscow,abudhabi]},
{planet:[earth,mars,venus]}
</code></pre>

Convert the given lists into dictionary using Python?

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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假设我们有一组输入列表：lst= [country,usa,uk,india,japan],[city,tokyo,berlin,moscow,abudhabi],[planet,earth,mars,venus]我们希望在python中为每个列表创建一个字典，以便输出如下所示：mydict= {country:[us...

问使用Python将给定的列表转换为字典？
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问使用Python将给定的列表转换为字典？EN