blocks|key|2920298|text|高级索引要求您为每个维度提供所有索引。您为第一个索引提供了3个索引，对于第二个索引只提供了2个索引，因此出现了错误。你想做这样的事|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2920299|>>>+a[[[0,+0],+[1,+1],+[3,+3]],+[[0,2],+[0,2],+[0,+2]]]
array([[+0,++2],
+++++++[+4,++6],
+++++++[12,+14]])|code-block|syntax|javascript|2920300|这当然是一个痛苦的写作，所以你可以让广播帮助你：|2920301|>>>+a[[[0],+[1],+[3]],+[0,+2]]
array([[+0,++2],
+++++++[+4,++6],
+++++++[12,+14]])|2920302|如果使用数组(而不是列表)进行索引，这要简单得多：|2920303|>>>+row_idx+=+np.array([0,+1,+3])
>>>+col_idx+=+np.array([0,+2])
>>>+a[row_idx[:,+None],+col_idx]
array([[+0,++2],
+++++++[+4,++6],
+++++++[12,+14]])|2920304|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|$]]

Fancy indexing requires you to provide all indices for each dimension. You are providing 3 indices for the first one, and only 2 for the second one, hence the error. You want to do something like this:

<pre><code>&gt;&gt;&gt; a[[[0, 0], [1, 1], [3, 3]], [[0,2], [0,2], [0, 2]]]
array([[ 0, 2],
 [ 4, 6],
 [12, 14]])
</code></pre>

That is of course a pain to write, so you can let broadcasting help you:

<pre><code>&gt;&gt;&gt; a[[[0], [1], [3]], [0, 2]]
array([[ 0, 2],
 [ 4, 6],
 [12, 14]])
</code></pre>

This is much simpler to do if you index with arrays, not lists:

<pre><code>&gt;&gt;&gt; row_idx = np.array([0, 1, 3])
&gt;&gt;&gt; col_idx = np.array([0, 2])
&gt;&gt;&gt; a[row_idx[:, None], col_idx]
array([[ 0, 2],
 [ 4, 6],
 [12, 14]])
</code></pre>

blocks|key|511211|text|使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|511212|+>>>+a[[0,1,3]][:,[0,2]]
array([[+0,++2],
+++[+4,++6],
+++[12,+14]])|code-block|syntax|javascript|511213|或：|511214|>>>+a[[0,1,3],::2]
array([[+0,++2],
+++[+4,++6],
+++[12,+14]])|511215|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|$]]

USE:

<pre><code> &gt;&gt;&gt; a[[0,1,3]][:,[0,2]]
array([[ 0, 2],
 [ 4, 6],
 [12, 14]])
</code></pre>

OR:

<pre><code>&gt;&gt;&gt; a[[0,1,3],::2]
array([[ 0, 2],
 [ 4, 6],
 [12, 14]])
</code></pre>

blocks|key|2920339|text|正如Toan所建议的那样，一个简单的黑客就是先选择行，然后再选择上面的列。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2920340|>>>+a[[0,1,3],+:]++++++++++++#+Returns+the+rows+you+want
array([[+0,++1,++2,++3],
+++++++[+4,++5,++6,++7],
+++++++[12,+13,+14,+15]])
>>>+a[[0,1,3],+:][:,+[0,2]]++#+Selects+the+columns+you+want+as+well
array([[+0,++2],
+++++++[+4,++6],
+++++++[12,+14]])|code-block|syntax|javascript|2920341|编辑内置方法：np.ix_|offset|length|style|CODE|2920342|我最近发现，numpy为您提供了一个内置的一行程序来执行@Jaime建议的操作，但不需要使用广播语法(因为缺乏可读性)。从医生那里：|2920343|使用ix_可以快速构造索引数组，从而对交叉产品进行索引。a[np.ix_([1,3],[2,5])]返回数组[[a[1,2]+a[1,5]],+[a[3,2]+a[3,5]]]。
|blockquote|2920344|2920345|所以你就这样用它：|2920346|>>>+a+=+np.arange(20).reshape((5,4))
>>>+a[np.ix_([0,1,3],+[0,2])]
array([[+0,++2],
+++++++[+4,++6],
+++++++[12,+14]])|2920347|它的工作方式是按照Jaime建议的方式对齐阵列，这样广播就能正常地进行：|2920348|>>>+np.ix_([0,1,3],+[0,2])
(array([[0],
++++++++[1],
++++++++[3]]),+array([[0,+2]]))|2920349|此外，正如MikeC在评论中所说，np.ix_的优势在于返回一个视图，而我的第一个(预编辑)答案没有。这意味着您现在可以分配给索引数组：|2920350|>>>+a[np.ix_([0,1,3],+[0,2])]+=+-1
>>>+a++++
array([[-1,++1,+-1,++3],
+++++++[-1,++5,+-1,++7],
+++++++[+8,++9,+10,+11],
+++++++[-1,+13,+-1,+15],
+++++++[16,+17,+18,+19]])|2920351|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy-1.10.0/reference/generated/numpy.ix_.html^0|0|0|7|6|7|6|0|0|0|S|M|1I|Y|0|0|0|0|0|0|H|6|0|0^^$0|@$1|2|3|4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1D|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1E|8|@$I|1F|J|1G|K|L]]|9|@$I|1H|J|1I|1|1J]]|A|$]]|$1|M|3|N|5|6|7|1K|8|@]|9|@]|A|$]]|$1|O|3|P|5|Q|7|1L|8|@$I|1M|J|1N|K|L]|$I|1O|J|1P|K|L]]|9|@]|A|$]]|$1|R|3|-4|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1R|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|1S|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1T|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|D|7|1U|8|@]|9|@]|A|$E|F]]|$1|10|3|11|5|6|7|1V|8|@$I|1W|J|1X|K|L]]|9|@]|A|$]]|$1|12|3|13|5|D|7|1Y|8|@]|9|@]|A|$E|F]]|$1|14|3|-4|5|6|7|1Z|8|@]|9|@]|A|$]]]|15|$16|$5|17|18|19|A|$1A|1B]]]]

As Toan suggests, a simple hack would be to just select the rows first, and then select the columns over that.

<pre><code>&gt;&gt;&gt; a[[0,1,3], :] # Returns the rows you want
array([[ 0, 1, 2, 3],
 [ 4, 5, 6, 7],
 [12, 13, 14, 15]])
&gt;&gt;&gt; a[[0,1,3], :][:, [0,2]] # Selects the columns you want as well
array([[ 0, 2],
 [ 4, 6],
 [12, 14]])
</code></pre>

<h3>[Edit] The built-in method: <a href="http://docs.scipy.org/doc/numpy-1.10.0/reference/generated/numpy.ix_.html" rel="noreferrer"><code>np.ix_</code></a></h3>

I recently discovered that numpy gives you an in-built one-liner to doing exactly what @Jaime suggested, but without having to use broadcasting syntax (which suffers from lack of readability). From the docs:

<blockquote>
 Using ix_ one can quickly construct index arrays that will index the
 cross product. <code>a[np.ix_([1,3],[2,5])]</code> returns the array <code>[[a[1,2] a[1,5]], [a[3,2] a[3,5]]]</code>.
</blockquote>

So you use it like this:

<pre><code>&gt;&gt;&gt; a = np.arange(20).reshape((5,4))
&gt;&gt;&gt; a[np.ix_([0,1,3], [0,2])]
array([[ 0, 2],
 [ 4, 6],
 [12, 14]])
</code></pre>

And the way it works is that it takes care of aligning arrays the way Jaime suggested, so that broadcasting happens properly:

<pre><code>&gt;&gt;&gt; np.ix_([0,1,3], [0,2])
(array([[0],
 [1],
 [3]]), array([[0, 2]]))
</code></pre>

Also, as MikeC says in a comment, <code>np.ix_</code> has the advantage of returning a view, which my first (pre-edit) answer did not. This means you can now assign to the indexed array:

<pre><code>&gt;&gt;&gt; a[np.ix_([0,1,3], [0,2])] = -1
&gt;&gt;&gt; a 
array([[-1, 1, -1, 3],
 [-1, 5, -1, 7],
 [ 8, 9, 10, 11],
 [-1, 13, -1, 15],
 [16, 17, 18, 19]])
</code></pre>

blocks|key|3462547|text|使用np.ix_是最方便的方法(正如其他人所回答的)，但也可以这样做：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3462548|>>>+rows+=+[0,+1,+3]
>>>+cols+=+[0,+2]

>>>+(a[rows].T)[cols].T

array([[+0,++2],
+++++++[+4,++6],
+++++++[12,+14]])|code-block|syntax|javascript|3462549|entityMap^0|2|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Using <code>np.ix_</code> is the most convenient way to do it (as answered by others), but it also can be done as follows:
<pre><code>&gt;&gt;&gt; rows = [0, 1, 3]
&gt;&gt;&gt; cols = [0, 2]

&gt;&gt;&gt; (a[rows].T)[cols].T

array([[ 0, 2],
 [ 4, 6],
 [12, 14]])
</code></pre>

I've been going crazy trying to figure out what stupid thing I'm doing wrong here.

I'm using NumPy, and I have specific row indices and specific column indices that I want to select from. Here's the gist of my problem:

<pre><code>import numpy as np

a = np.arange(20).reshape((5,4))
# array([[ 0, 1, 2, 3],
# [ 4, 5, 6, 7],
# [ 8, 9, 10, 11],
# [12, 13, 14, 15],
# [16, 17, 18, 19]])

# If I select certain rows, it works
print a[[0, 1, 3], :]
# array([[ 0, 1, 2, 3],
# [ 4, 5, 6, 7],
# [12, 13, 14, 15]])

# If I select certain rows and a single column, it works
print a[[0, 1, 3], 2]
# array([ 2, 6, 14])

# But if I select certain rows AND certain columns, it fails
print a[[0,1,3], [0,2]]
# Traceback (most recent call last):
# File "&lt;stdin&gt;", line 1, in &lt;module&gt;
# ValueError: shape mismatch: objects cannot be broadcast to a single shape
</code></pre>

Why is this happening? Surely I should be able to select the 1st, 2nd, and 4th rows, and 1st and 3rd columns? The result I'm expecting is:

<pre><code>a[[0,1,3], [0,2]] =&gt; [[0, 2],
 [4, 6],
 [12, 14]]
</code></pre>

Selecting specific rows and columns from NumPy array

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

我已经疯了，想弄清楚我在这里做了什么蠢事。我使用的是NumPy，我有特定的行索引和特定的列索引，我想从中选择。以下是我问题的要点：import numpy as npa = np.arange(20).reshape((5,4))# array([[ 0,  1,  2,  3],#        [ 4,  5,  ...

问从NumPy数组中选择特定的行和列
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

问从NumPy数组中选择特定的行和列EN