blocks|key|1660698|text|使用两个索引，一个用于所需的嵌套列表，一个用于所需嵌套列表的哪个元素。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1660699|因此，在本例中，您需要第二个列表的第二个元素：|1660700|NL[1][1]='X'|code-block|syntax|javascript|1660701|输出：|1660702|[[1,+2,+3],+[4,+'X',+6],+[7,+8,+9]]|1660703|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Use 2 indexes, 1 for the what nested list you want and one for what element of the nested list you want.
So in this case you want the 2nd list's 2nd element:
<pre><code>NL[1][1]='X'
</code></pre>
Output:
<pre><code>[[1, 2, 3], [4, 'X', 6], [7, 8, 9]]
</code></pre>

blocks|key|3384113|text|假设您需要找到元素5，并希望用10替换它。我们遍历外部列表，然后遍历每个内部列表的元素。一旦我们找到我们寻找的元素，我们就可以用索引来替换它。一旦找到匹配的元素，我们就使用枚举来拥有索引。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3384114|下面的代码替换所有匹配的元素(所有出现的5)。|3384115|NL+=+[[1,2,3],+[4,5,6],+[7,8,9]]

print(NL)++#+prints:+[[1,+2,+3],+[4,+5,+6],+[7,+8,+9]]

for+i,+sublist+in+enumerate(NL):
++++for+y,+element+in+enumerate(sublist):
++++++++if+element+==+5:
++++++++++++NL[i][y]+=+10

print(NL)++#+prints:+[[1,+2,+3],+[4,+10,+6],+[7,+8,+9]]+++++++++++|code-block|syntax|javascript|3384116|entityMap^0|9|1|F|2|0|K|1|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@$9|U|A|V|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

Let's say you need to find the element <code>5</code> and want to replace it with <code>10</code>.
We iterate through the outer list and then each inner-list's elements. Once we find the element we look for, we can replace it by the indexes. We use enumerate to have the indexes once we find a matching element.
The following code replaces ALL matching elements (all occurences of <code>5</code>).
<pre><code>NL = [[1,2,3], [4,5,6], [7,8,9]]

print(NL) # prints: [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

for i, sublist in enumerate(NL):
 for y, element in enumerate(sublist):
 if element == 5:
 NL[i][y] = 10

print(NL) # prints: [[1, 2, 3], [4, 10, 6], [7, 8, 9]] 
</code></pre>

blocks|key|2513214|text|使用numpy|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2513215|NL+=+np.array(NL)

mask+=+np.where(NL+==+5)
NL[mask]+=+10|code-block|syntax|javascript|2513216|2513217|array([[+1,++2,++3],
+++++++[+4,+10,++6],
+++++++[+7,++8,++9]])|2513218|2513219|Solution2:|BOLD|2513220|def+get_index(num,+List):
++++for+row,+i+in+enumerate(List):
++++++++if+num+in+i:
++++++++++++return+row,+i.index(num)
++++return+-1

idx+=+get_index(5,NL)
if+idx>0:
++++NL[idx[0]][idx[1]]+=+7|2513221|2513222|[[1,+2,+3],+[4,+7,+6],+[7,+8,+9]]|2513223|entityMap^0|2|5|0|0|0|0|0|0|A|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|12|8|@]|D|@]|E|$]]|$1|L|3|M|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|N|3|-4|5|6|7|14|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|15|8|@$9|16|A|17|B|Q]]|D|@]|E|$]]|$1|R|3|S|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|T|3|-4|5|6|7|19|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|W|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|X|$]]

Using <code>numpy</code>:
<pre><code>NL = np.array(NL)

mask = np.where(NL == 5)
NL[mask] = 10
</code></pre>
<hr />
<pre><code>array([[ 1, 2, 3],
 [ 4, 10, 6],
 [ 7, 8, 9]])
</code></pre>
<hr />
Solution2:
<pre><code>def get_index(num, List):
 for row, i in enumerate(List):
 if num in i:
 return row, i.index(num)
 return -1

idx = get_index(5,NL)
if idx&gt;0:
 NL[idx[0]][idx[1]] = 7
</code></pre>
<hr />
<pre><code>[[1, 2, 3], [4, 7, 6], [7, 8, 9]]
</code></pre>

blocks|key|2513308|text|您应该按索引访问元素。您有2D列表(数组)，因此应该使用两个索引：NL[1][1]+=+"X"。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2513309|完整代码：|BOLD|2513310|NL+=+[[1,+2,+3],
++++++[4,+5,+6],
++++++[7,+8,+9]]

print("Original:+{}".format(NL))

NL[1][1]+=+"X"

print("New:+{}".format(NL))|code-block|syntax|javascript|2513311|输出：|2513312|>>>+python3+test.py
Original:+[[1,+2,+3],+[4,+5,+6],+[7,+8,+9]]
New:+[[1,+2,+3],+[4,+'X',+6],+[7,+8,+9]]|2513313|entityMap^0|X|E|0|0|5|0|0|0|3|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@$9|U|A|V|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|W|8|@$9|X|A|Y|B|H]]|D|@]|E|$]]|$1|I|3|J|5|K|7|Z|8|@]|D|@]|E|$L|M]]|$1|N|3|O|5|6|7|10|8|@$9|11|A|12|B|H]]|D|@]|E|$]]|$1|P|3|Q|5|K|7|13|8|@]|D|@]|E|$L|M]]|$1|R|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|S|$]]

You should access to element by indexes. You have 2D list (array) so you should use 2 indexes: <code>NL[1][1] = &quot;X&quot;</code>.
Complete code:
<pre><code>NL = [[1, 2, 3],
 [4, 5, 6],
 [7, 8, 9]]

print(&quot;Original: {}&quot;.format(NL))

NL[1][1] = &quot;X&quot;

print(&quot;New: {}&quot;.format(NL))
</code></pre>
Output:
<pre><code>&gt;&gt;&gt; python3 test.py
Original: [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
New: [[1, 2, 3], [4, 'X', 6], [7, 8, 9]]
</code></pre>

blocks|key|3384247|text|只需使用NL1+=+'X‘|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3384248|然后打印(NL)|3384249|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

just use NL[1][1] = 'X'
then print(NL)

blocks|key|1659498|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1659499|，我很难找到元素的位置，然后改变它。|blockquote|1659500|1659501|1659502|这里的大多数答案似乎都错过了这部分，并假设你有这个位置。|1659503|1659504|1659505|您可以使用嵌套列表理解：|1659506|NL+=+[[1,2,3],
++++++[4,5,6],
++++++[7,8,9]]

NL+=+[['X'+if+i==5+else+i+for+i+in+j]+for+j+in+NL]

print(NL)|code-block|syntax|javascript|1659507|输出：|1659508|[[1,+2,+3],
+[4,'X',6],
+[7,+8,+9]]|1659509|entityMap^0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|-4|4|5|6|W|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|X|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|Y|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|Z|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|10|7|@]|8|@]|9|$]]|$1|H|3|-4|4|5|6|11|7|@]|8|@]|9|$]]|$1|I|3|-4|4|5|6|12|7|@]|8|@]|9|$]]|$1|J|3|K|4|5|6|13|7|@]|8|@]|9|$]]|$1|L|3|M|4|N|6|14|7|@]|8|@]|9|$O|P]]|$1|Q|3|R|4|5|6|15|7|@]|8|@]|9|$]]|$1|S|3|T|4|N|6|16|7|@]|8|@]|9|$O|P]]|$1|U|3|-4|4|5|6|17|7|@]|8|@]|9|$]]]|V|$]]

<blockquote>
I am having trouble trying to find the position of the element and then changing it.
</blockquote>
Most of the answers here seem to have missed that part, and assumed you had the position.
<hr />
<hr />
You can use a nested list comprehension:
<pre><code>NL = [[1,2,3],
 [4,5,6],
 [7,8,9]]

NL = [['X' if i==5 else i for i in j] for j in NL]

print(NL)
</code></pre>
Output:
<pre><code>[[1, 2, 3],
 [4,'X',6],
 [7, 8, 9]]
</code></pre>

I have a nested list and I am trying to replace a certain element of the list with something else.
<pre><code> NL = [[1,2,3],
 [4,5,6],
 [7,8,9]];
</code></pre>
Now, I need to update the list, let's say the user wants to change element at NL[1][1] (i.e. 5) to 'X'.
NL will be updated as
<pre><code> NL = [[1,2,3],
 [4,'X',6],
 [7,8,9]];`
</code></pre>
I am having trouble trying to find the position of the element and then changing it. Any help is much appreciated.
Thanks

Finding an element in nested python list and then replacing it

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。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我有一个嵌套的列表，我试图用其他的东西替换列表中的某个元素。  NL = [[1,2,3],       [4,5,6],       [7,8,9]];现在，我需要更新列表，假设用户希望将NL1上的元素(即5)更改为“X”。NL将更新为  NL = [[1,2,3],       [4,'X',6],       [...

问在嵌套python列表中查找元素，然后替换它
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问在嵌套python列表中查找元素，然后替换它EN