blocks|key|397043|text|我相信如果区段的数量不需要正确划分数组的大小，那么您正在寻找的是numpy.split，或者可能是numpy.array_split。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|397044|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.split.html|1|http://docs.scipy.org/doc/numpy/reference/generated/numpy.array_split.html#numpy.array_split^0|W|B|1D|H|W|B|0|1D|H|1|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@$9|U|A|V|1|W]|$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]|N|$5|I|J|K|E|$L|O]]]]

I believe that you're looking for <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.split.html" rel="noreferrer"><code>numpy.split</code></a> or possibly <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.array_split.html#numpy.array_split" rel="noreferrer"><code>numpy.array_split</code></a> if the number of sections doesn't need to divide the size of the array properly.

blocks|key|179218|text|试试numpy.array_split。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|179219|从文档中：|179220|>>>+x+=+np.arange(8.0)
>>>+np.array_split(x,+3)
++++[array([+0.,++1.,++2.]),+array([+3.,++4.,++5.]),+array([+6.,++7.])]|code-block|syntax|javascript|179221|与numpy.split相同，但如果组的长度不相等，则不会引发异常。|179222|如果块数>len(数组)，您将获得嵌套在其中的空数组，以解决这个问题-如果您的拆分数组保存在a中，则可以通过以下方式删除空数组：|179223|[x+for+x+in+a+if+x.size+>+0]|179224|如果您愿意，只需将其保存回a即可。|179225|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.array_split.html#numpy.array_split|1|http://docs.scipy.org/doc/numpy/reference/generated/numpy.split.html^0|2|H|2|H|0|0|0|0|1|B|1|B|1|0|1A|1|0|0|D|1|0^^$0|@$1|2|3|4|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@$9|17|A|18|1|19]]|E|$]]|$1|F|3|G|5|6|7|1A|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|1B|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1C|8|@$9|1D|A|1E|B|C]]|D|@$9|1F|A|1G|1|1H]]|E|$]]|$1|O|3|P|5|6|7|1I|8|@$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|J|7|1L|8|@]|D|@]|E|$K|L]]|$1|S|3|T|5|6|7|1M|8|@$9|1N|A|1O|B|C]]|D|@]|E|$]]|$1|U|3|-4|5|6|7|1P|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]|12|$5|X|Y|Z|E|$10|13]]]]

Try <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.array_split.html#numpy.array_split" rel="noreferrer"><code>numpy.array_split</code></a>.

From the documentation:

<pre><code>&gt;&gt;&gt; x = np.arange(8.0)
&gt;&gt;&gt; np.array_split(x, 3)
 [array([ 0., 1., 2.]), array([ 3., 4., 5.]), array([ 6., 7.])]
</code></pre>

Identical to <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.split.html" rel="noreferrer"><code>numpy.split</code></a>, but won't raise an exception if the groups aren't equal length.

If number of chunks > len(array) you get blank arrays nested inside, to address that - if your split array is saved in <code>a</code>, then you can remove empty arrays by:

<pre><code>[x for x in a if x.size &gt; 0]
</code></pre>

Just save that back in <code>a</code> if you wish.

blocks|key|397094|text|不完全是一个答案，而是一个很长的注释，与其他(正确)答案的代码格式很好。如果您尝试以下操作，您将看到您得到的是原始数组的视图，而不是副本，而您链接的问题中接受的答案并非如此。注意可能的副作用！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|397095|>>>+x+=+np.arange(9.0)
>>>+a,b,c+=+np.split(x,+3)
>>>+a
array([+0.,++1.,++2.])
>>>+a[1]+=+8
>>>+a
array([+0.,++8.,++2.])
>>>+x
array([+0.,++8.,++2.,++3.,++4.,++5.,++6.,++7.,++8.])
>>>+def+chunks(l,+n):
...+++++"""+Yield+successive+n-sized+chunks+from+l.
...+++++"""
...+++++for+i+in+xrange(0,+len(l),+n):
...+++++++++yield+l[i:i%2Bn]
...+
>>>+l+=+range(9)
>>>+a,b,c+=+chunks(l,+3)
>>>+a
[0,+1,+2]
>>>+a[1]+=+8
>>>+a
[0,+8,+2]
>>>+l
[0,+1,+2,+3,+4,+5,+6,+7,+8]|code-block|syntax|javascript|397096|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Not quite an answer, but a long comment with nice formatting of code to the other (correct) answers. If you try the following, you will see that what you are getting are views of the original array, not copies, and that was not the case for the accepted answer in the question you link. Be aware of the possible side effects!

<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; x = np.arange(9.0)
&gt;&gt;&gt; a,b,c = np.split(x, 3)
&gt;&gt;&gt; a
array([ 0., 1., 2.])
&gt;&gt;&gt; a[1] = 8
&gt;&gt;&gt; a
array([ 0., 8., 2.])
&gt;&gt;&gt; x
array([ 0., 8., 2., 3., 4., 5., 6., 7., 8.])
&gt;&gt;&gt; def chunks(l, n):
... """ Yield successive n-sized chunks from l.
... """
... for i in xrange(0, len(l), n):
... yield l[i:i+n]
... 
&gt;&gt;&gt; l = range(9)
&gt;&gt;&gt; a,b,c = chunks(l, 3)
&gt;&gt;&gt; a
[0, 1, 2]
&gt;&gt;&gt; a[1] = 8
&gt;&gt;&gt; a
[0, 8, 2]
&gt;&gt;&gt; l
[0, 1, 2, 3, 4, 5, 6, 7, 8]
</code></pre>

blocks|key|397128|text|这个怎么样？在这里，您使用所需的长度拆分数组。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|397129|a+=+np.random.randint(0,10,[4,4])

a
Out[27]:+
array([[1,+5,+8,+7],
+++++++[3,+2,+4,+0],
+++++++[7,+7,+6,+2],
+++++++[7,+4,+3,+0]])

a[0:2,:]
Out[28]:+
array([[1,+5,+8,+7],
+++++++[3,+2,+4,+0]])

a[2:4,:]
Out[29]:+
array([[7,+7,+6,+2],
+++++++[7,+4,+3,+0]])|code-block|syntax|javascript|397130|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

How about this? Here you split the array using the length you want to have.

<pre><code>a = np.random.randint(0,10,[4,4])

a
Out[27]: 
array([[1, 5, 8, 7],
 [3, 2, 4, 0],
 [7, 7, 6, 2],
 [7, 4, 3, 0]])

a[0:2,:]
Out[28]: 
array([[1, 5, 8, 7],
 [3, 2, 4, 0]])

a[2:4,:]
Out[29]: 
array([[7, 7, 6, 2],
 [7, 4, 3, 0]])
</code></pre>

blocks|key|179330|text|这可以使用numpy的as_strided来实现。我假设如果块大小不是总行数的一个因素，那么最后一批中的其余行将用零填充。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|179331|from+numpy.lib.stride_tricks+import+as_strided
def+batch_data(test,+chunk_count):
++m,n+=+test.shape
++S+=+test.itemsize
++if+not+chunk_count:
++++chunk_count+=+1
++batch_size+=+m//chunk_count
#+Batches+which+can+be+covered+fully
++test_batches+=+as_strided(test,+shape=(chunk_count,+batch_size,+n),+strides=(batch_size*n*S,n*S,S)).copy()
++covered+=+chunk_count*batch_size
++if+covered+<+m:
++++rest+=+test[covered:,:]
++++rm,+rn+=+rest.shape
++++mismatch+=+batch_size+-+rm
++++last_batch+=+np.vstack((rest,np.zeros((mismatch,rn)))).reshape(1,-1,n)
++++return+np.vstack((test_batches,last_batch))
++return+test_batches|code-block|syntax|javascript|179332|这是基于我的答案https://stackoverflow.com/a/68238815/5462372。|179333|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/68238815/5462372^0|B|A|0|0|8|18|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|@$9|Z|A|10|1|11]]|E|$]]|$1|M|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

This can be achieved using <code>as_strided</code> of numpy. I have put a spin to answer by assuming that if chunk size is not a factor of total number of rows, then rest of the rows in the last batch will be filled with zeros.
<pre><code>from numpy.lib.stride_tricks import as_strided
def batch_data(test, chunk_count):
 m,n = test.shape
 S = test.itemsize
 if not chunk_count:
 chunk_count = 1
 batch_size = m//chunk_count
# Batches which can be covered fully
 test_batches = as_strided(test, shape=(chunk_count, batch_size, n), strides=(batch_size*n*S,n*S,S)).copy()
 covered = chunk_count*batch_size
 if covered &lt; m:
 rest = test[covered:,:]
 rm, rn = rest.shape
 mismatch = batch_size - rm
 last_batch = np.vstack((rest,np.zeros((mismatch,rn)))).reshape(1,-1,n)
 return np.vstack((test_batches,last_batch))
 return test_batches
</code></pre>
This is based on my answer <a href="https://stackoverflow.com/a/68238815/5462372">https://stackoverflow.com/a/68238815/5462372</a>.

There is this <a href="https://stackoverflow.com/questions/312443/how-do-you-split-a-list-into-evenly-sized-chunks-in-python">How do you split a list into evenly sized chunks?</a>
for splitting an array into chunks. Is there anyway to do this more efficiently for giant arrays using Numpy?

Partition array into N chunks with Numpy

有一个用于将数组分成块的。对于使用Numpy的巨型数组，有没有更有效的方法来做到这一点？

问使用Numpy将数组划分为N个块
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