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
如果你想对整个事物进行矢量化,生成一个方形格子然后剪切.然后把你的盒子外面的边缘切断.
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