从三个一维数组创建一个 3D 坐标的 numpy 数组 [英] Creating a numpy array of 3D coordinates from three 1D arrays

问题描述

假设我有三个任意的一维数组，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))

这三个数组代表3D网格中的采样间隔，我想为所有交叉点构造一个三维向量的一维数组，类似于

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]])

订单实际上并不重要.生成它们的明显方法:

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?我查看过但没有找到(可能只是没找到正确的 Google 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).

我目前正在使用这个:

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

但我觉得我错过了一些聪明的奇特 Numpy 方式?

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

推荐答案

要在上面的示例中使用 numpy 网格，以下将起作用:

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

用于二维以上网格的 Numpy meshgrid 需要 numpy 1.7.为了规避这一点并从源代码中提取相关数据.

Numpy meshgrid for grids of more then two dimensions require numpy 1.7. To circumvent this and pulling the relevant data from the source code.

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:

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.]]

对于上面的例子:

%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

当每个维度为 100 时:

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

根据你想对数据做什么，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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