迭代ndarray的切片

Question

假设我有一个3D numpy.array,例如尺寸为xyz,有没有办法沿特定轴迭代切片？就像是:

for layer in data.slices(dim=2):
    # do something with layer

编辑:为了澄清,该示例是一个dim = 3数组,即shape =(len_x,len_y,len_z).Elazar和等效的kamjagin解决方案有效,但不是那么通用 - 你必须[:, :, i]手工构建,这意味着你需要知道尺寸,而且代码不够通用,无法处理任意尺寸的数组.您可以通过使用类似的东西来填充缺失的维度[..., :],但是您仍然需要自己构建它.

对不起,应该更清楚,这个例子有点太简单了!

Answer 1

迭代第一维很容易,见下文.要迭代其他维度,请将该维度滚动到前面并执行相同操作:

>>> 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]]

编辑一些时间,比较大型阵列的不同方法:

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