问题描述
我需要在python中计算bspline曲线.我研究了scipy.interpolate.splprep和其他一些scipy模块，但找不到任何能满足我需求的东西.所以我在下面写了我自己的模块.代码可以正常工作，但是速度很慢(测试函数运行时间为0.03s，考虑到我只要求100个带有6个控制顶点的样本，这似乎很多.)
I need to compute bspline curves in python. I looked into scipy.interpolate.splprep and a few other scipy modules but couldn't find anything that readily gave me what I needed. So i wrote my own module below. The code works fine, but it is slow (test function runs in 0.03s, which seems like a lot considering i'm only asking for 100 samples with 6 control vertices).
是否有一种方法可以通过一些scipy模块调用来简化下面的代码，这大概可以加快速度?如果没有，我该如何对我的代码进行改进以提高其性能?
Is there a way to simplify the code below with a few scipy module calls, which presumably would speed it up? And if not, what could i do to my code to improve its performance?
import numpy as np
# cv = np.array of 3d control vertices
# n = number of samples (default: 100)
# d = curve degree (default: cubic)
# closed = is the curve closed (periodic) or open? (default: open)
def bspline(cv, n=100, d=3, closed=False):
# Create a range of u values
count = len(cv)
knots = None
u = None
if not closed:
u = np.arange(0,n,dtype='float')/(n-1) * (count-d)
knots = np.array([0]*d + range(count-d+1) + [count-d]*d,dtype='int')
else:
u = ((np.arange(0,n,dtype='float')/(n-1) * count) - (0.5 * (d-1))) % count # keep u=0 relative to 1st cv
knots = np.arange(0-d,count+d+d-1,dtype='int')
# Simple Cox - DeBoor recursion
def coxDeBoor(u, k, d):
# Test for end conditions
if (d == 0):
if (knots[k] <= u and u < knots[k+1]):
return 1
return 0
Den1 = knots[k+d] - knots[k]
Den2 = knots[k+d+1] - knots[k+1]
Eq1 = 0;
Eq2 = 0;
if Den1 > 0:
Eq1 = ((u-knots[k]) / Den1) * coxDeBoor(u,k,(d-1))
if Den2 > 0:
Eq2 = ((knots[k+d+1]-u) / Den2) * coxDeBoor(u,(k+1),(d-1))
return Eq1 + Eq2
# Sample the curve at each u value
samples = np.zeros((n,3))
for i in xrange(n):
if not closed:
if u[i] == count-d:
samples[i] = np.array(cv[-1])
else:
for k in xrange(count):
samples[i] += coxDeBoor(u[i],k,d) * cv[k]
else:
for k in xrange(count+d):
samples[i] += coxDeBoor(u[i],k,d) * cv[k%count]
return samples
if __name__ == "__main__":
import matplotlib.pyplot as plt
def test(closed):
cv = np.array([[ 50., 25., -0.],
[ 59., 12., -0.],
[ 50., 10., 0.],
[ 57., 2., 0.],
[ 40., 4., 0.],
[ 40., 14., -0.]])
p = bspline(cv,closed=closed)
x,y,z = p.T
cv = cv.T
plt.plot(cv[0],cv[1], 'o-', label='Control Points')
plt.plot(x,y,'k-',label='Curve')
plt.minorticks_on()
plt.legend()
plt.xlabel('x')
plt.ylabel('y')
plt.xlim(35, 70)
plt.ylim(0, 30)
plt.gca().set_aspect('equal', adjustable='box')
plt.show()
test(False)
下面的两个图像显示了在两个封闭条件下我的代码返回的结果:
The two images below shows what my code returns with both closed conditions:
推荐答案
因此，在对我的问题和研究进行了很多研究之后，我终于有了答案.一切都可以在scipy中找到，我将代码放在这里，希望其他人可以找到它.
So after obsessing a lot about my question, and much research, i finally have my answer. Everything is available in scipy , and i'm putting my code here so hopefully someone else can find this useful.
该函数接收N-d个点，一个曲线度，一个周期性状态(打开或关闭)的数组，并将沿着该曲线返回n个样本.有一些方法可以确保曲线样本是等距的，但是暂时我将重点关注这个问题，因为这全都与速度有关.
The function takes in an array of N-d points, a curve degree, a periodic state (opened or closed) and will return n samples along that curve. There are ways to make sure the curve samples are equidistant but for the time being i'll focus on this question, as it is all about speed.
值得一提的是:我似乎无法超越20度曲线.当然，这已经足够了，但是我认为这值得一提.
Worthy of note: I can't seem to be able to go beyond a curve of 20th degree. Granted, that's overkill already, but i figured it's worth mentioning.
还值得注意的是:在我的机器上，下面的代码可以在0.017s内计算100,000个样本
Also worthy of note: on my machine the code below can calculate 100,000 samples in 0.017s
import numpy as np
import scipy.interpolate as si
def bspline(cv, n=100, degree=3, periodic=False):
""" Calculate n samples on a bspline
cv : Array ov control vertices
n : Number of samples to return
degree: Curve degree
periodic: True - Curve is closed
False - Curve is open
"""
# If periodic, extend the point array by count+degree+1
cv = np.asarray(cv)
count = len(cv)
if periodic:
factor, fraction = divmod(count+degree+1, count)
cv = np.concatenate((cv,) * factor + (cv[:fraction],))
count = len(cv)
degree = np.clip(degree,1,degree)
# If opened, prevent degree from exceeding count-1
else:
degree = np.clip(degree,1,count-1)
# Calculate knot vector
kv = None
if periodic:
kv = np.arange(0-degree,count+degree+degree-1)
else:
kv = np.clip(np.arange(count+degree+1)-degree,0,count-degree)
# Calculate query range
u = np.linspace(periodic,(count-degree),n)
# Calculate result
return np.array(si.splev(u, (kv,cv.T,degree))).T
要对其进行测试:
import matplotlib.pyplot as plt
colors = ('b', 'g', 'r', 'c', 'm', 'y', 'k')
cv = np.array([[ 50., 25.],
[ 59., 12.],
[ 50., 10.],
[ 57., 2.],
[ 40., 4.],
[ 40., 14.]])
plt.plot(cv[:,0],cv[:,1], 'o-', label='Control Points')
for d in range(1,21):
p = bspline(cv,n=100,degree=d,periodic=True)
x,y = p.T
plt.plot(x,y,'k-',label='Degree %s'%d,color=colors[d%len(colors)])
plt.minorticks_on()
plt.legend()
plt.xlabel('x')
plt.ylabel('y')
plt.xlim(35, 70)
plt.ylim(0, 30)
plt.gca().set_aspect('equal', adjustable='box')
plt.show()
打开或周期性曲线的结果:
Results for both opened or periodic curves:
从scipy-0.19.0开始，有一个新的 scipy.interpolate.BSpline 函数.
As of scipy-0.19.0 there is a new scipy.interpolate.BSpline function that can be used.
import numpy as np
import scipy.interpolate as si
def scipy_bspline(cv, n=100, degree=3, periodic=False):
""" Calculate n samples on a bspline
cv : Array ov control vertices
n : Number of samples to return
degree: Curve degree
periodic: True - Curve is closed
"""
cv = np.asarray(cv)
count = cv.shape[0]
# Closed curve
if periodic:
kv = np.arange(-degree,count+degree+1)
factor, fraction = divmod(count+degree+1, count)
cv = np.roll(np.concatenate((cv,) * factor + (cv[:fraction],)),-1,axis=0)
degree = np.clip(degree,1,degree)
# Opened curve
else:
degree = np.clip(degree,1,count-1)
kv = np.clip(np.arange(count+degree+1)-degree,0,count-degree)
# Return samples
max_param = count - (degree * (1-periodic))
spl = si.BSpline(kv, cv, degree)
return spl(np.linspace(0,max_param,n))
进行等效性测试:
p1 = bspline(cv,n=10**6,degree=3,periodic=True) # 1 million samples: 0.0882 sec
p2 = scipy_bspline(cv,n=10**6,degree=3,periodic=True) # 1 million samples: 0.0789 sec
print np.allclose(p1,p2) # returns True
这篇关于具有numpy/scipy的快速b样条算法的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！