创建3D的numpy的阵列从三个一维数组坐标 [英] Creating a numpy array of 3D coordinates from three 1D arrays

问题描述

假设我有三乱一维数组，例如：

Suppose I have three arbitrary 1D arrays, for example:

x_p = np.array((1.0, 2.0, 3.0, 4.0, 5.0))
y_p = np.array((2.0, 3.0, 4.0))
z_p = np.array((8.0, 9.0))

这三个阵列重新在一个三维网格present采样间隔，我想构建三维向量的一维数组的所有路口，像

These three arrays represent sampling intervals in a 3D grid, and I want to construct a 1D array of three-dimensional vectors for all intersections, something like

points = np.array([[1.0, 2.0, 8.0],
                   [1.0, 2.0, 9.0],
                   [1.0, 3.0, 8.0],
                   ...
                   [5.0, 4.0, 9.0]])

顺序并不实际上很重要这一点。最显而易见的方法来生成它们：

Order doesn't actually matter for this. The obvious way to generate them:

npoints = len(x_p) * len(y_p) * len(z_p)
points = np.zeros((npoints, 3))
i = 0
for x in x_p:
    for y in y_p:
        for z in z_p:
            points[i, :] = (x, y, z)
            i += 1

所以，问题是...有没有更快的方法？我已经看了，但没有找到（可能只是没有找到合适的谷歌关键字）。

So the question is... is there a faster way? I have looked but not found (possibly just failed to find the right Google keywords).

我目前使用的：

npoints = len(x_p) * len(y_p) * len(z_p)
points = np.zeros((npoints, 3))
i = 0
nz = len(z_p)
for x in x_p:
    for y in y_p:
        points[i:i+nz, 0] = x
        points[i:i+nz, 1] = y
        points[i:i+nz, 2] = z_p
        i += nz

但我觉得我缺少一些聪明的奇特方式numpy的？

but I feel like I am missing some clever fancy Numpy way?

推荐答案

要使用numpy的网格在上面的例子中下面的工作：

To use numpy mesh grid on the above example the following will work:

np.vstack(np.meshgrid(x_p,y_p,z_p)).reshape(3,-1).T

numpy的meshgrid了两个多维度的电网需要numpy的1.7。为了规避这一点，从源$ C ​​$ C 。

def ndmesh(*xi,**kwargs):
    if len(xi) < 2:
        msg = 'meshgrid() takes 2 or more arguments (%d given)' % int(len(xi) > 0)
        raise ValueError(msg)

    args = np.atleast_1d(*xi)
    ndim = len(args)
    copy_ = kwargs.get('copy', True)

    s0 = (1,) * ndim
    output = [x.reshape(s0[:i] + (-1,) + s0[i + 1::]) for i, x in enumerate(args)]

    shape = [x.size for x in output]

    # Return the full N-D matrix (not only the 1-D vector)
    if copy_:
        mult_fact = np.ones(shape, dtype=int)
        return [x * mult_fact for x in output]
    else:
        return np.broadcast_arrays(*output)

检查结果是：

print np.vstack((ndmesh(x_p,y_p,z_p))).reshape(3,-1).T

[[ 1.  2.  8.]
 [ 1.  2.  9.]
 [ 1.  3.  8.]
 ....
 [ 5.  3.  9.]
 [ 5.  4.  8.]
 [ 5.  4.  9.]]

有关上面的例子：

%timeit sol2()
10000 loops, best of 3: 56.1 us per loop

%timeit np.vstack((ndmesh(x_p,y_p,z_p))).reshape(3,-1).T
10000 loops, best of 3: 55.1 us per loop

有关当每个尺寸为100：

For when each dimension is 100:

%timeit sol2()
1 loops, best of 3: 655 ms per loop
In [10]:

%timeit points = np.vstack((ndmesh(x_p,y_p,z_p))).reshape(3,-1).T
10 loops, best of 3: 21.8 ms per loop

根据你想要的数据做什么，你可以返回一个视图：

Depending on what you want to do with the data, you can return a view:

%timeit np.vstack((ndmesh(x_p,y_p,z_p,copy=False))).reshape(3,-1).T
100 loops, best of 3: 8.16 ms per loop

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