blocks|key|4253734|text|像这样的吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4253735|>>>+data+=+np.array([[1,2,3],[4,5,6],[7,8,9]])
>>>+for+layer+in+[data[:,i]+for+i+in+range(3)]:
...+++++print+layer
...+
[1+4+7]
[2+5+8]
[3+6+9]|code-block|syntax|javascript|4253736|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|$]]

something like that?

<pre><code>&gt;&gt;&gt; data = np.array([[1,2,3],[4,5,6],[7,8,9]])
&gt;&gt;&gt; for layer in [data[:,i] for i in range(3)]:
... print layer
... 
[1 4 7]
[2 5 8]
[3 6 9]
</code></pre>

blocks|key|2127498|text|如果我错了，请纠正我，但在我看来，您的3D数组将如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2127499|>>>+my_array.shape
++++(3,N)|code-block|syntax|javascript|2127500|其中N是数组的大小。所以如果你想迭代一个维度，你可以这样做：|2127501|>>>+for+item+in+my_array[1,:]:|2127502|这将在第二个维度上迭代。|2127503|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Correct me if I am wrong but it seems to me that your 3D array is gonna look like :

<pre><code>&gt;&gt;&gt; my_array.shape
 (3,N)
</code></pre>

Where N is the size of your array. So if you want to iterate over one dimension, you can just do :

<pre><code>&gt;&gt;&gt; for item in my_array[1,:]:
</code></pre>

This will iterate on the second dimension.

blocks|key|2127699|text|这可能比这更优雅地解决，但如果你事先知道dim+(例如2)，一种方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2127700|for+i+in+range(data.shape[dim]):
++++layer+=+data[:,:,i]|code-block|syntax|javascript|2127701|或者如果是dim=0|2127702|for+i+in+range(data.shape[dim]):
++++layer+=+data[i,:,:]|2127703|等。|2127704|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

This could probably be solved more elegantly than this, but one way of doing it if you know dim beforehand(e.g. 2) is:

<pre><code>for i in range(data.shape[dim]):
 layer = data[:,:,i]
</code></pre>

or if dim=0

<pre><code>for i in range(data.shape[dim]):
 layer = data[i,:,:]
</code></pre>

etc.

blocks|key|2127938|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2127939|>>>+data+=+np.arange(24).reshape(2,+3,+4)
>>>+for+dim_0_slice+in+data:+#+the+first+dimension+is+easy
...+++++print+dim_0_slice
...+
[[+0++1++2++3]
+[+4++5++6++7]
+[+8++9+10+11]]
[[12+13+14+15]
+[16+17+18+19]
+[20+21+22+23]]
>>>+for+dim_1_slice+in+np.rollaxis(data,+1):+#+for+the+others,+roll+it+to+the+front
...+++++print+dim_1_slice
...+
[[+0++1++2++3]
+[12+13+14+15]]
[[+4++5++6++7]
+[16+17+18+19]]
[[+8++9+10+11]
+[20+21+22+23]]
>>>+for+dim_2_slice+in+np.rollaxis(data,+2):
...+++++print+dim_2_slice
...+
[[+0++4++8]
+[12+16+20]]
[[+1++5++9]
+[13+17+21]]
[[+2++6+10]
+[14+18+22]]
[[+3++7+11]
+[15+19+23]]|code-block|syntax|javascript|2127940|2127941|编辑一些计时，以比较不同的方法对较大的数组：|offset|length|style|BOLD|2127942|In+[7]:+a+=+np.arange(200*100*300).reshape(200,+100,+300)

In+[8]:+%25timeit+for+j+in+xrange(100):+a[:,+j]
10000+loops,+best+of+3:+60.2+us+per+loop

In+[9]:+%25timeit+for+j+in+xrange(100):+a[:,+j,+:]
10000+loops,+best+of+3:+82.8+us+per+loop

In+[10]:+%25timeit+for+j+in+np.rollaxis(a,+1):+j
10000+loops,+best+of+3:+28.2+us+per+loop

In+[11]:+%25timeit+for+j+in+np.swapaxes(a,+0,+1):+j
10000+loops,+best+of+3:+26.7+us+per+loop|2127943|entityMap^0|0|0|0|0|2|0|0^^$0|@$1|2|3|-4|4|5|6|Q|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|R|7|@]|8|@]|9|$D|E]]|$1|F|3|-4|4|5|6|S|7|@]|8|@]|9|$]]|$1|G|3|H|4|5|6|T|7|@$I|U|J|V|K|L]]|8|@]|9|$]]|$1|M|3|N|4|C|6|W|7|@]|8|@]|9|$D|E]]|$1|O|3|-4|4|5|6|X|7|@]|8|@]|9|$]]]|P|$]]

Iterating over the first dimension is very easy, see below. To iterate over the others, roll that dimension to the front and do the same:

<pre><code>&gt;&gt;&gt; data = np.arange(24).reshape(2, 3, 4)
&gt;&gt;&gt; for dim_0_slice in data: # the first dimension is easy
... print dim_0_slice
... 
[[ 0 1 2 3]
 [ 4 5 6 7]
 [ 8 9 10 11]]
[[12 13 14 15]
 [16 17 18 19]
 [20 21 22 23]]
&gt;&gt;&gt; for dim_1_slice in np.rollaxis(data, 1): # for the others, roll it to the front
... print dim_1_slice
... 
[[ 0 1 2 3]
 [12 13 14 15]]
[[ 4 5 6 7]
 [16 17 18 19]]
[[ 8 9 10 11]
 [20 21 22 23]]
&gt;&gt;&gt; for dim_2_slice in np.rollaxis(data, 2):
... print dim_2_slice
... 
[[ 0 4 8]
 [12 16 20]]
[[ 1 5 9]
 [13 17 21]]
[[ 2 6 10]
 [14 18 22]]
[[ 3 7 11]
 [15 19 23]]
</code></pre>

<hr>

EDIT Some timings, to compare different methods for largish arrays:

<pre><code>In [7]: a = np.arange(200*100*300).reshape(200, 100, 300)

In [8]: %timeit for j in xrange(100): a[:, j]
10000 loops, best of 3: 60.2 us per loop

In [9]: %timeit for j in xrange(100): a[:, j, :]
10000 loops, best of 3: 82.8 us per loop

In [10]: %timeit for j in np.rollaxis(a, 1): j
10000 loops, best of 3: 28.2 us per loop

In [11]: %timeit for j in np.swapaxes(a, 0, 1): j
10000 loops, best of 3: 26.7 us per loop
</code></pre>

blocks|key|745356|text|我认为原来的问题是模棱两可的，标题也是如此：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|745357|745358|在x[k,:,:,...]上迭代所有k会产生和x.shape[0]一样多的项目，所以我称之为在一个维度上迭代；|unordered-list-item|offset|length|style|CODE|745359|相反，在我的脑海中，在一个切片上迭代意味着对所有i，j，k迭代。例如，在ndarray中迭代所有列。|745360|745361|尽管这不是OP所要求的(在澄清之后)，但寻找后者的解决方案的人可能会发现以下代码很有用：|745362|from+itertools+import+product

def+iterslice(x,axis=0):
++++sub+=+[+range(s)+for+s+in+x.shape+]
++++sub[axis]+=+(slice(None),)
++++for+p+in+product(*sub):
++++++++yield+x[p]|code-block|syntax|javascript|745363|entityMap^0|0|0|1|C|N|A|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|W|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|X|8|@$F|Y|G|Z|H|I]|$F|10|G|11|H|I]]|9|@]|A|$]]|$1|J|3|K|5|E|7|12|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|13|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|14|8|@]|9|@]|A|$]]|$1|O|3|P|5|Q|7|15|8|@]|9|@]|A|$R|S]]|$1|T|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|U|$]]

I think the original question was ambiguous, and so is the title:
<ul>
<li>Iterating over <code>x[k,:,:,...]</code> for all k yields as many items as <code>x.shape[0]</code>, so I would call this iterating over a dimension;</li>
<li>In contrast, iterating over a slice in my mind means iterating e.g. over <code>x[:,i,j,k,...]</code> for all i,j,k. For example, iterating over all columns in an ndarray.</li>
</ul>
Although that's not what the OP asked (after clarification), people looking for a solution to the latter might find the following code useful:
<pre><code>from itertools import product

def iterslice(x,axis=0):
 sub = [ range(s) for s in x.shape ]
 sub[axis] = (slice(None),)
 for p in product(*sub):
 yield x[p]
</code></pre>

Say I have a 3D numpy.array, e.g. with dimensions x y z, is there a way to iterate over slices along a particular axis? Something like:

<pre><code>for layer in data.slices(dim=2):
 # do something with layer
</code></pre>

Edit:
To clarify, the example is a dim=3 array, i.e. shape=(len_x, len_y, len_z). Elazar and equivalently kamjagin's solutions work, but aren't that general - you have to construct the <code>[:, :, i]</code> by hand, which means you need to know the dimensions, and the code isn't general enough to handle arrays of arbitrary dimensions. You can fill missing dimension by using something like <code>[..., :]</code>, but again you still have to construct this yourself.

Sorry, should have been clearer, the example was a bit too simple!

iterate over slices of an ndarray

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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假设我有一个3D numpy.array，例如维度为x，y，z的，有没有一种方法可以沿着特定的轴迭代切片？类似于：for layer in data.slices(dim=2):    # do something with layer编辑:为了清楚起见，示例是一个dim=3数组，即shape=(len_x，len_y...

问在ndarray的切片上迭代
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问在ndarray的切片上迭代EN