在python中有效地生成点的格子

帮助我的代码更快：我的 python代码需要生成一个二维格点的落在一个有界的矩形内.我汇集了一些生成这个格子的代码(如下所示).然而,这个功能被称为很多次,并且已经成为我的应用程序的严重瓶颈.

我确定有一个更快的方式来做到这一点,可能涉及到numpy数组而不是列表.有什么建议,以更快,更优雅的方式做到这一点？

功能描述：
我有两个2D矢量v1和v2.这些矢量define a lattice.在我的情况下,我的矢量定义了一个格子,几乎是六边形,但不完全相同.我想在这个格子上生成一些在一些边界矩形中的所有2D点的集合.在我的情况下,矩形的一个角是(0,0),其他角是正坐标.

例：
如果我的边框的最远的角落在(3,3),我的格子矢量是：

v1 = (1.2,0.1)
v2 = (0.2,1.1)

我想要我的功能返回点数：

(1.2,0.1) #v1
(2.4,0.2) #2*v1
(0.2,1.1) #v2
(0.4,2.2) #2*v2
(1.4,1.2) #v1 + v2
(2.6,1.3) #2*v1 + v2
(1.6,2.3) #v1 + 2*v2
(2.8,2.4) #2*v1 + 2*v2

我不关心边缘病例;例如,函数返回(0,0)并不重要.

我目前做的这个缓慢的方式：

import numpy,pylab

def generate_lattice( #Help me speed up this function,please!
    image_shape,lattice_vectors,center_pix='image',edge_buffer=2):

    ##Preprocessing. Not much of a bottleneck:
    if center_pix == 'image':
        center_pix = numpy.array(image_shape) // 2
    else: ##Express the center pixel in terms of the lattice vectors
        center_pix = numpy.array(center_pix) - (numpy.array(image_shape) // 2)
        lattice_components = numpy.linalg.solve(
            numpy.vstack(lattice_vectors[:2]).T,center_pix)
        lattice_components -= lattice_components // 1
        center_pix = (lattice_vectors[0] * lattice_components[0] +
                      lattice_vectors[1] * lattice_components[1] +
                      numpy.array(image_shape)//2)
    num_vectors = int( ##Estimate how many lattice points we need
        max(image_shape) / numpy.sqrt(lattice_vectors[0]**2).sum())
    lattice_points = []
    lower_bounds = numpy.array((edge_buffer,edge_buffer))
    upper_bounds = numpy.array(image_shape) - edge_buffer

    ##SLOW LOOP HERE. 'num_vectors' is often quite large.
    for i in range(-num_vectors,num_vectors):
        for j in range(-num_vectors,num_vectors):
            lp = i * lattice_vectors[0] + j * lattice_vectors[1] + center_pix
            if all(lower_bounds < lp) and all(lp < upper_bounds):
                lattice_points.append(lp)
    return lattice_points


##Test the function and display the output.
##No optimization needed past this point.
lattice_vectors = [
    numpy.array([-40.,-1.]),numpy.array([ 18.,-37.])]
image_shape = (1000,1000)
spots = generate_lattice(image_shape,lattice_vectors)

fig=pylab.figure()
pylab.plot([p[1] for p in spots],[p[0] for p in spots],'.')
pylab.axis('equal')
fig.show()

解决方法

这是我想出来的还有很多改进可以做,但这是基本的想法.

def generate_lattice(image_shape,lattice_vectors) :
    center_pix = numpy.array(image_shape) // 2
    # Get the lower limit on the cell size.
    dx_cell = max(abs(lattice_vectors[0][0]),abs(lattice_vectors[1][0]))
    dy_cell = max(abs(lattice_vectors[0][1]),abs(lattice_vectors[1][1]))
    # Get an over estimate of how many cells across and up.
    nx = image_shape[0]//dx_cell
    ny = image_shape[1]//dy_cell
    # Generate a square lattice,with too many points.
    # Here I generate a factor of 4 more points than I need,which ensures 
    # coverage for highly sheared lattices.  If your lattice is not highly
    # sheared,than you can generate fewer points.
    x_sq = np.arange(-nx,nx,dtype=float)
    y_sq = np.arange(-ny,dtype=float)
    x_sq.shape = x_sq.shape + (1,)
    y_sq.shape = (1,) + y_sq.shape
    # Now shear the whole thing using the lattice vectors
    x_lattice = lattice_vectors[0][0]*x_sq + lattice_vectors[1][0]*y_sq
    y_lattice = lattice_vectors[0][1]*x_sq + lattice_vectors[1][1]*y_sq
    # Trim to fit in box.
    mask = ((x_lattice < image_shape[0]/2.0)
             & (x_lattice > -image_shape[0]/2.0))
    mask = mask & ((y_lattice < image_shape[1]/2.0)
                    & (y_lattice > -image_shape[1]/2.0))
    x_lattice = x_lattice[mask]
    y_lattice = y_lattice[mask]
    # Translate to the centre pix.
    x_lattice += center_pix[0]
    y_lattice += center_pix[1]
    # Make output compatible with original version.
    out = np.empty((len(x_lattice),2),dtype=float)
    out[:,0] = y_lattice
    out[:,1] = x_lattice
    return out

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