blocks|key|5528889|text|看起来你基本上是从2D输入数组的开头开始切片，直到2+rows和4+columns，然后拆分每一行。您可以使用c[:2,:4]进行切片，然后使用np.vsplit拆分行，从而获得一个单行解决方案，如下所示-|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5528890|res_out+=+np.vsplit(c[:2,:4],2)|code-block|syntax|javascript|5528891|示例运行-|5528892|In+[10]:+c
Out[10]:+
array([[0,+2,+5,+1,+0],
+++++++[1,+5,+5,+0,+3],
+++++++[0,+1,+0,+6,+6],
+++++++[2,+6,+2,+3,+3]])

In+[11]:+indices
Out[11]:+[[0,+slice(0,+4,+None)],+[1,+slice(0,+4,+None)]]

In+[12]:+#+desired+result+using+a+for+loop
++++...:+res+=+[]
++++...:+for+idx+in+indices:
++++...:+++++res.append(c[idx])
++++...:+++++

In+[13]:+res
Out[13]:+[array([0,+2,+5,+1]),+array([1,+5,+5,+0])]

In+[14]:+np.vsplit(c[:2,:4],2)
Out[14]:+[array([[0,+2,+5,+1]]),+array([[1,+5,+5,+0]])]|5528893|请注意，np.vsplit的输出将是二维数组的列表，而不是问题中发布的代码中的一维数组列表。|5528894|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.vsplit.html^0|P|6|W|9|1J|8|20|9|20|9|0|0|0|0|0|4|9|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@$9|17|A|18|1|19]]|E|$]]|$1|F|3|G|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1B|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1C|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1D|8|@$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

It seems that you are basically slicing until <code>2 rows</code> and <code>4 columns</code> from the start of the 2D input array and then splitting each row. You can do the slicing with <code>c[:2,:4]</code> and then split rows with <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.vsplit.html" rel="nofollow"><code>np.vsplit</code></a> to have a one-liner solution like so -

<pre><code>res_out = np.vsplit(c[:2,:4],2)
</code></pre>

Sample run -

<pre><code>In [10]: c
Out[10]: 
array([[0, 2, 5, 1, 0],
 [1, 5, 5, 0, 3],
 [0, 1, 0, 6, 6],
 [2, 6, 2, 3, 3]])

In [11]: indices
Out[11]: [[0, slice(0, 4, None)], [1, slice(0, 4, None)]]

In [12]: # desired result using a for loop
 ...: res = []
 ...: for idx in indices:
 ...: res.append(c[idx])
 ...: 

In [13]: res
Out[13]: [array([0, 2, 5, 1]), array([1, 5, 5, 0])]

In [14]: np.vsplit(c[:2,:4],2)
Out[14]: [array([[0, 2, 5, 1]]), array([[1, 5, 5, 0]])]
</code></pre>

Please note that the output from <code>np.vsplit</code> would be a list of 2D arrays, rather than a list of 1D arrays as with the posted code in the question.

blocks|key|1195652|text|您的示例可以重写为列表理解：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1195653|In+[121]:+[c[idx]+for+idx+in+indices]
Out[121]:+
[array([4,+2,+1,+2]),
+array([3,+2,+2,+3]),
+array([3,+2,+2,+3]),
+array([0,+3,+4,+4])]|code-block|syntax|javascript|1195654|它可以变成一个漂亮的二维数组：|1195655|In+[122]:+np.array([c[idx]+for+idx+in+indices])
Out[122]:+
array([[4,+2,+1,+2],
+++++++[3,+2,+2,+3],
+++++++[3,+2,+2,+3],
+++++++[0,+3,+4,+4]])|1195656|在这里，np.array()是一种连接形式，沿着新的轴连接数组。|offset|length|style|CODE|1195657|由于第2个索引对于所有行都是相同的(slice(4))，因此此索引也适用：|1195658|In+[123]:+c[[0,1,1,2],slice(4)]++#+or+[...,:4]
Out[123]:+
array([[4,+2,+1,+2],
+++++++[3,+2,+2,+3],
+++++++[3,+2,+2,+3],
+++++++[0,+3,+4,+4]])|1195659|在第一轴上重复是不成问题的。在第二个不同的切片需要更多的操作。除了这种特殊的:4情况，您必须将切片转换为范围。没有办法用多个切片来索引一个维度。|1195660|1195661|切片的长度相同，但'start‘值不同的情况与https://stackoverflow.com/a/28007256/901925+access-multiple-elements-of-an-array中讨论的情况类似。|1195662|In+[135]:+c.flat[[i*c.shape[1]%2Bnp.arange(j.start,j.stop)+for+i,j+in+indices]]
Out[135]:+
array([[4,+2,+1,+2],
+++++++[3,+2,+2,+3],
+++++++[3,+2,+2,+3],
+++++++[0,+3,+4,+4]])|1195663|我以这种方式生成的索引是：|1195664|In+[136]:+[i*c.shape[1]%2Bnp.arange(j.start,j.stop)+for+i,j+in+indices]
Out[136]:+
[array([0,+1,+2,+3]),
+array([5,+6,+7,+8]),
+array([5,+6,+7,+8]),
+array([10,+11,+12,+13])]|1195665|如果indices有点不规则，它工作得很好：indices1+=+[[0,+slice(0,+3)],+[1,+slice(2,+5)],+[1,+slice(1,+4)],+[2,+slice(0,+3)]]|1195666|我之前的回答看了一些其他的索引方式。但是，即使考虑到生成索引数组所需的计算，在扁平数组上进行索引通常也是最快的。|1195667|如果片段的长度不同，那么您就不得不生成一个数组列表，或者生成这样一个列表的hstack：|1195668|In+[158]:+indices2+=+[[0,+slice(0,+2)],+[1,+slice(2,+5)],+
++++[1,+slice(0,+4)],+[2,+slice(0,+5)]]

In+[159]:+c.flat[np.hstack([i*c.shape[1]%2Bnp.arange(j.start,j.stop)+
+++++for+i,j+in+indices2])]
Out[159]:+array([4,+2,+2,+3,+1,+3,+2,+2,+3,+0,+3,+4,+4,+3])

In+[160]:+[c.flat[i*c.shape[1]%2Bnp.arange(j.start,j.stop)]+for+i,j+in+indices2]
Out[160]:+[array([4,+2]),+array([2,+3,+1]),+array([3,+2,+2,+3]),
++++array([0,+3,+4,+4,+3])]

In+[161]:+np.hstack(_)
Out[161]:+array([4,+2,+2,+3,+1,+3,+2,+2,+3,+0,+3,+4,+4,+3])|1195669|1195670|更多关于不同但长度相等的切片：|1195671|In+[190]:+indices1+=+[[0,+slice(0,+3)],+[1,+slice(2,+5)],+[1,+slice(1,+4)],+[2,+slice(0,+3)]]

In+[191]:+c.flat[[i*c.shape[1]%2Bnp.arange(j.start,j.stop)+for+i,j+in+indices1]]Out[191]:+
array([[4,+2,+1],
+++++++[2,+3,+1],
+++++++[2,+2,+3],
+++++++[0,+3,+4]])

In+[193]:+rows+=+[[i]+for+i,j+in+indices1]
In+[200]:+cols=[np.arange(j.start,j.stop)+for+i,j+in+indices1]

In+[201]:+c[rows,cols]
Out[201]:+
array([[4,+2,+1],
+++++++[2,+3,+1],
+++++++[2,+2,+3],
+++++++[0,+3,+4]])|1195672|在这种情况下，rows是可以用cols广播的垂直列表。|1195673|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/28007256/901925^0|0|0|0|0|4|A|0|I|8|0|0|12|2|0|0|1V|10|N|17|0|0|0|0|0|2|7|M|2B|0|0|11|6|0|0|0|0|0|7|4|F|4|0^^$0|@$1|2|3|4|5|6|7|1S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|1V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|1W|8|@$M|1X|N|1Y|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1Z|8|@$M|20|N|21|O|P]]|9|@]|A|$]]|$1|S|3|T|5|D|7|22|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|23|8|@$M|24|N|25|O|P]]|9|@]|A|$]]|$1|W|3|-4|5|6|7|26|8|@]|9|@]|A|$]]|$1|X|3|Y|5|6|7|27|8|@$M|28|N|29|O|P]]|9|@$M|2A|N|2B|1|2C]]|A|$]]|$1|Z|3|10|5|D|7|2D|8|@]|9|@]|A|$E|F]]|$1|11|3|12|5|6|7|2E|8|@]|9|@]|A|$]]|$1|13|3|14|5|D|7|2F|8|@]|9|@]|A|$E|F]]|$1|15|3|16|5|6|7|2G|8|@$M|2H|N|2I|O|P]|$M|2J|N|2K|O|P]]|9|@]|A|$]]|$1|17|3|18|5|6|7|2L|8|@]|9|@]|A|$]]|$1|19|3|1A|5|6|7|2M|8|@$M|2N|N|2O|O|P]]|9|@]|A|$]]|$1|1B|3|1C|5|D|7|2P|8|@]|9|@]|A|$E|F]]|$1|1D|3|-4|5|6|7|2Q|8|@]|9|@]|A|$]]|$1|1E|3|1F|5|6|7|2R|8|@]|9|@]|A|$]]|$1|1G|3|1H|5|D|7|2S|8|@]|9|@]|A|$E|F]]|$1|1I|3|1J|5|6|7|2T|8|@$M|2U|N|2V|O|P]|$M|2W|N|2X|O|P]]|9|@]|A|$]]|$1|1K|3|-4|5|6|7|2Y|8|@]|9|@]|A|$]]]|1L|$1M|$5|1N|1O|1P|A|$1Q|1R]]]]

Your example can be rewritten as a list comprehension:

<pre><code>In [121]: [c[idx] for idx in indices]
Out[121]: 
[array([4, 2, 1, 2]),
 array([3, 2, 2, 3]),
 array([3, 2, 2, 3]),
 array([0, 3, 4, 4])]
</code></pre>

which can be turned into a nice 2d array:

<pre><code>In [122]: np.array([c[idx] for idx in indices])
Out[122]: 
array([[4, 2, 1, 2],
 [3, 2, 2, 3],
 [3, 2, 2, 3],
 [0, 3, 4, 4]])
</code></pre>

Here <code>np.array()</code> is a form of concatenation, joining the arrays along a new axis.

Since the 2nd index is the same for all rows (<code>slice(4)</code>), this indexing also works:

<pre><code>In [123]: c[[0,1,1,2],slice(4)] # or [...,:4]
Out[123]: 
array([[4, 2, 1, 2],
 [3, 2, 2, 3],
 [3, 2, 2, 3],
 [0, 3, 4, 4]])
</code></pre>

Repetition on the 1st axis is not a problem. Differing slices in the 2nd take some more manipulation. Except for this special <code>:4</code> case, you will have to turn the slices in to ranges. There's no way of indexing one dimension with multiple slices.

<hr>

The case where the slices all have same length, but different 'start' values, is similar to the one discussed in <a href="https://stackoverflow.com/a/28007256/901925">https://stackoverflow.com/a/28007256/901925</a> <code>access-multiple-elements-of-an-array</code>. 

<pre><code>In [135]: c.flat[[i*c.shape[1]+np.arange(j.start,j.stop) for i,j in indices]]
Out[135]: 
array([[4, 2, 1, 2],
 [3, 2, 2, 3],
 [3, 2, 2, 3],
 [0, 3, 4, 4]])
</code></pre>

The indices that I generate this way are:

<pre><code>In [136]: [i*c.shape[1]+np.arange(j.start,j.stop) for i,j in indices]
Out[136]: 
[array([0, 1, 2, 3]),
 array([5, 6, 7, 8]),
 array([5, 6, 7, 8]),
 array([10, 11, 12, 13])]
</code></pre>

It works fine if <code>indices</code> is somewhat irregular: <code>indices1 = [[0, slice(0, 3)], [1, slice(2, 5)], [1, slice(1, 4)], [2, slice(0, 3)]]</code>

My earlier answer looks at some other ways indexing. But often indexing on a flatten array is fastest, even if you take into account the calculation required to generate the index array.

If the slices vary in length, then you are stuck with generating a list of arrays, or an <code>hstack</code> of such a list:

<pre><code>In [158]: indices2 = [[0, slice(0, 2)], [1, slice(2, 5)], 
 [1, slice(0, 4)], [2, slice(0, 5)]]

In [159]: c.flat[np.hstack([i*c.shape[1]+np.arange(j.start,j.stop) 
 for i,j in indices2])]
Out[159]: array([4, 2, 2, 3, 1, 3, 2, 2, 3, 0, 3, 4, 4, 3])

In [160]: [c.flat[i*c.shape[1]+np.arange(j.start,j.stop)] for i,j in indices2]
Out[160]: [array([4, 2]), array([2, 3, 1]), array([3, 2, 2, 3]),
 array([0, 3, 4, 4, 3])]

In [161]: np.hstack(_)
Out[161]: array([4, 2, 2, 3, 1, 3, 2, 2, 3, 0, 3, 4, 4, 3])
</code></pre>

<hr>

more on the varying, but equal length slices:

<pre><code>In [190]: indices1 = [[0, slice(0, 3)], [1, slice(2, 5)], [1, slice(1, 4)], [2, slice(0, 3)]]

In [191]: c.flat[[i*c.shape[1]+np.arange(j.start,j.stop) for i,j in indices1]]Out[191]: 
array([[4, 2, 1],
 [2, 3, 1],
 [2, 2, 3],
 [0, 3, 4]])

In [193]: rows = [[i] for i,j in indices1]
In [200]: cols=[np.arange(j.start,j.stop) for i,j in indices1]

In [201]: c[rows,cols]
Out[201]: 
array([[4, 2, 1],
 [2, 3, 1],
 [2, 2, 3],
 [0, 3, 4]])
</code></pre>

In this case <code>rows</code> is a vertical list that can be broadcasted with <code>cols</code>.

I want to get the result of a list (or array) of indices from a numpy array, in the shape: ( len(indices), (shape of one indexing operation) ).

Is there any way to use a list of indices directly, without using a for loop, like I used in the mininal example, shown below?

<pre><code>c = np.random.randint(0, 5, size=(4, 5))
indices = [[0, slice(0, 4)], [1, slice(0, 4)], [1, slice(0, 4)], [2, slice(0, 4)]]

# desired result using a for loop
res = []
for idx in indices:
 res.append(c[idx])
</code></pre>

It should be noted, that the indices list is not representative of my problem, it serves as an example, in general it is generated during runtime.
However, each index operation returns the same shape

numpy, return of array of indices in shape of

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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我希望从numpy数组中获得索引列表(或数组)的结果，其形式为：(len(索引)，(一个索引操作的形状))。有没有办法直接使用索引列表，而不使用for循环，就像我在mininal示例中使用的那样，如下所示？c = np.random.randint(0, 5, size=(4, 5))indices = [[0, sl...

问形的索引数组的返回值
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问形的索引数组的返回值EN