blocks|key|4620028|text|我会看看numpy.ndarray.shape和itertools.product：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4620029|import+numpy,+itertools
A+=+numpy.ones((10,10,10,10))
B+=+numpy.zeros((10,10,10,10))

for+i,+j,+k,+l+in+itertools.product(*map(xrange,+A.shape)):
++++B[k,l,i,j]+=+A[i,j,k,l]|code-block|syntax|javascript|4620030|当然，我假设你所说的“不使用循环”是指“不使用嵌套循环”。除非有一些numpy内置的可以做到这一点，我认为这是你最好的选择。|4620031|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|$]]

I would look at numpy.ndarray.shape and itertools.product:

<pre><code>import numpy, itertools
A = numpy.ones((10,10,10,10))
B = numpy.zeros((10,10,10,10))

for i, j, k, l in itertools.product(*map(xrange, A.shape)):
 B[k,l,i,j] = A[i,j,k,l]
</code></pre>

By "without the use of loops" I'm assuming you mean "without the use of nested loops", of course. Unless there's some numpy built-in that does this, I think this is your best bet.

blocks|key|114250|text|您可以rollaxis两次：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|114251|>>>+A+=+np.random.random((2,4,3,5))
>>>+B+=+np.rollaxis(np.rollaxis(A,+2),+3,+1)
>>>+A.shape
(2,+4,+3,+5)
>>>+B.shape
(3,+5,+2,+4)
>>>+from+itertools+import+product
>>>+all(B[k,l,i,j]+==+A[i,j,k,l]+for+i,j,k,l+in+product(*map(range,+A.shape)))
True|code-block|syntax|javascript|114252|或者，swapaxes+to更容易理解：|114253|>>>+A+=+np.random.random((2,4,3,5))
>>>+C+=+A.swapaxes(0,+2).swapaxes(1,3)
>>>+C.shape
(3,+5,+2,+4)
>>>+all(C[k,l,i,j]+==+A[i,j,k,l]+for+i,j,k,l+in+product(*map(range,+A.shape)))
True|114254|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.rollaxis.html#numpy.rollaxis|1|http://docs.scipy.org/doc/numpy/reference/generated/numpy.swapaxes.html^0|3|8|3|8|0|0|0|3|8|3|8|1|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|F|3|G|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@$9|18|A|19|1|1A]]|E|$]]|$1|M|3|N|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]]]

You could <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.rollaxis.html#numpy.rollaxis"><code>rollaxis</code></a> twice:

<pre><code>&gt;&gt;&gt; A = np.random.random((2,4,3,5))
&gt;&gt;&gt; B = np.rollaxis(np.rollaxis(A, 2), 3, 1)
&gt;&gt;&gt; A.shape
(2, 4, 3, 5)
&gt;&gt;&gt; B.shape
(3, 5, 2, 4)
&gt;&gt;&gt; from itertools import product
&gt;&gt;&gt; all(B[k,l,i,j] == A[i,j,k,l] for i,j,k,l in product(*map(range, A.shape)))
True
</code></pre>

or maybe <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.swapaxes.html"><code>swapaxes</code></a> twice is easier to follow:

<pre><code>&gt;&gt;&gt; A = np.random.random((2,4,3,5))
&gt;&gt;&gt; C = A.swapaxes(0, 2).swapaxes(1,3)
&gt;&gt;&gt; C.shape
(3, 5, 2, 4)
&gt;&gt;&gt; all(C[k,l,i,j] == A[i,j,k,l] for i,j,k,l in product(*map(range, A.shape)))
True
</code></pre>

blocks|key|4620190|text|在numpy中执行此操作的规范方法是使用np.transpose的可选排列参数。在您的示例中，要从ijkl转到klij，排列是(2,+3,+0,+1)，例如：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4620191|In+[16]:+a+=+np.empty((2,+3,+4,+5))

In+[17]:+b+=+np.transpose(a,+(2,+3,+0,+1))

In+[18]:+b.shape
Out[18]:+(4,+5,+2,+3)|code-block|syntax|javascript|4620192|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.transpose.html^0|K|C|1D|4|1J|4|1R|C|K|C|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|F|3|G|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

The canonical way of doing this in numpy would be to use <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.transpose.html" rel="noreferrer"><code>np.transpose</code></a>'s optional permutation argument. In your case, to go from <code>ijkl</code> to <code>klij</code>, the permutation is <code>(2, 3, 0, 1)</code>, e.g.:

<pre><code>In [16]: a = np.empty((2, 3, 4, 5))

In [17]: b = np.transpose(a, (2, 3, 0, 1))

In [18]: b.shape
Out[18]: (4, 5, 2, 3)
</code></pre>

I would like to do the following:

<pre><code>for i in dimension1:
 for j in dimension2:
 for k in dimension3:
 for l in dimension4:
 B[k,l,i,j] = A[i,j,k,l]
</code></pre>

without the use of loops. In the end both A and B contain the same information but indexed
differently. 

I must point out that the dimension 1,2,3 and 4 can be the same or different. So a numpy.reshape() seems difficult.

Swapping the dimensions of a numpy array

我想做以下几件事：for i in dimension1:  for j in dimension2:    for k in dimension3:      for l in dimension4:        B[k,l,i,j] = A[i,j,k,l]而不使用循环。最后，A和B都包含相同的信息，但索引不同。...

问交换numpy数组的维数
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