迭代 3D 数组的 Pythonic 方式 [英] Pythonic way of iterating over 3D array

问题描述

我在 Python 中有一个 3D 数组，我需要遍历数组中的所有立方体.也就是说，对于数组维度中的所有 (x,y,z) 我需要访问多维数据集:

I have a 3D array in Python and I need to iterate over all the cubes in the array. That is, for all (x,y,z) in the array's dimensions I need to access the cube:

array[(x + 0, y + 0, z + 0)]
array[(x + 1, y + 0, z + 0)]
array[(x + 0, y + 1, z + 0)]
array[(x + 1, y + 1, z + 0)]
array[(x + 0, y + 0, z + 1)]
array[(x + 1, y + 0, z + 1)]
array[(x + 0, y + 1, z + 1)]
array[(x + 1, y + 1, z + 1)]

该数组是一个 Numpy 数组，尽管这并不是必需的.我刚刚发现使用 numpy.fromfile() 使用单行读取数据非常容易.

The array is a Numpy array, though that's not really necessary. I just found it very easy to read the data in with a one-liner using numpy.fromfile().

有没有比以下更多的 Pythonic 方法来迭代这些?这看起来就像使用 Python 语法的 C.

Is there any more Pythonic way to iterate over these than the following? That simply looks like C using Python syntax.

for x in range(x_dimension):
    for y in range(y_dimension):
        for z in range(z_dimension):
            work_with_cube(array[(x + 0, y + 0, z + 0)],
                           array[(x + 1, y + 0, z + 0)],
                           array[(x + 0, y + 1, z + 0)],
                           array[(x + 1, y + 1, z + 0)],
                           array[(x + 0, y + 0, z + 1)],
                           array[(x + 1, y + 0, z + 1)],
                           array[(x + 0, y + 1, z + 1)],
                           array[(x + 1, y + 1, z + 1)])

推荐答案

看看 itertools，尤其是 itertools.product.您可以使用

Have a look at itertools, especially itertools.product. You can compress the three loops into one with

import itertools

for x, y, z in itertools.product(*map(xrange, (x_dim, y_dim, z_dim)):
    ...

您也可以通过这种方式创建多维数据集:

You can also create the cube this way:

cube = numpy.array(list(itertools.product((0,1), (0,1), (0,1))))
print cube
array([[0, 0, 0],
       [0, 0, 1],
       [0, 1, 0],
       [0, 1, 1],
       [1, 0, 0],
       [1, 0, 1],
       [1, 1, 0],
       [1, 1, 1]])

并通过简单的加法添加偏移量

and add the offsets by a simple addition

print cube + (10,100,1000)
array([[  10,  100, 1000],
       [  10,  100, 1001],
       [  10,  101, 1000],
       [  10,  101, 1001],
       [  11,  100, 1000],
       [  11,  100, 1001],
       [  11,  101, 1000],
       [  11,  101, 1001]])

在您的情况下将转换为 cube + (x,y,z) .您的代码的非常紧凑的版本是

which would to translate to cube + (x,y,z) in your case. The very compact version of your code would be

import itertools, numpy

cube = numpy.array(list(itertools.product((0,1), (0,1), (0,1))))

x_dim = y_dim = z_dim = 10

for offset in itertools.product(*map(xrange, (x_dim, y_dim, z_dim))):
    work_with_cube(cube+offset)

编辑:itertools.product 根据不同的参数生成乘积，即 itertools.product(a,b,c)，所以我必须通过 map(xrange, ...) 和 *map(...)

Edit: itertools.product makes the product over the different arguments, i.e. itertools.product(a,b,c), so I have to pass map(xrange, ...) with as *map(...)

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