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Python-B-spline-examples

Examples in Python about evaluating and interpolating a B-spline curve and their comparaison using Numpy, Scipy and Matplotlib.

Requirements

  • Python 2.7.x or 3.x.x,
  • Numpy
  • Scipy
  • Matplotlib

Getting hands dirty

Assuming you already know a little about B-spline curves, otherwise google it or check this out.

Let consider a 9 points list named plist :

plist= [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),]

We will do two things :

  • A. Drawing a cubic B-Spline curve where plist is the control polygon.

  • B. Find(interpolate) and draw the B-spline curve that go trough plist points and or in other words a curve fitting using a cubic B-spline curve.

As we will work with Numpy , let's create a numpy array named ctr from plist ,and then split it into x and y arrays.

ctr=np.array(plist)
x=ctr[:,0]
y=ctr[:,1]

A. Drawing the B-spline Curve

We will draw a cubic B-spline which degree k=3 as it's the most used one. We need now to define the knots vector, with k ending equal knots:

l=len(x)
t=np.linspace(0,1,l-2,endpoint=True)
t=np.append([0,0,0],t)
t=np.append(t,[1,1,1])

Now we have defined all requirement to draw it. we can construct a sequence of length 3 containing the knots, coefficients, and degree of the spline to pass it as the tck argument to scipy.interpolate.splev , the function that will evaluate our curve:

tck=[t,[x,y],3]

And as we will evaluate the B-spline curve, we need a second parameter__u__ , an array of points :

u3=np.linspace(0,1,(max(l*2,70)),endpoint=True)

And we evaluate it :

out=interpolate.splev(u3,tck) 

Drawing it using matplotlib :

Cubic B-spline curve drawing

With this code :

plt.plot(x,y,'k--',label='Control polygon',marker='o',markerfacecolor='red')
plt.plot(out[0],out[1],'b',linewidth=2.0,label='B-spline curve')
plt.legend(loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('Cubic B-spline curve evaluation')
plt.show()

B. interpolate the B-spline Curve

We will interpolate now the B-spline cruve that should go through n points. As we done before, a numpy array must be created using plist and split into x and y arrays :

plist= [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),]
ctr=np.array(plist)
x=ctr[:,0]
y=ctr[:,1]

To interpolate the B-spline that go through this points. To do so, we will be using scipy.interpolate.splprep, to get the interpolated S-pline parameters :

tck,_=interpolate.splprep([x,y],k=3,s=0)

And we use them to evaluate it using scipy.interpolate.splev, we still have a to redefine u variable to get a clear plot :

u=np.linspace(0,1,num=50,endpoint=True)
out=interpolate.splev(u,tck)

and we finaly draw it : Cubic B-spline curve interplation

With this code :

plt.figure()
plt.plot(x, y, 'ro', out[0], out[1], 'b')
plt.legend(['Points', 'Interpolated B-spline', 'True'],loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('B-Spline interpolation')
plt.show()

Full code :

B-Spline evalutation

importnumpyasnpfromscipyimportinterpolateimportmatplotlib.pyplotaspltctr=np.array( [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),
(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),])
x=ctr[:,0]
y=ctr[:,1]
# uncomment both lines for a closed curve#x=np.append(x,[x[0]]) #y=np.append(y,[y[0]])l=len(x) t=np.linspace(0,1,l-2,endpoint=True)
t=np.append([0,0,0],t)
t=np.append(t,[1,1,1])
tck=[t,[x,y],3]
u3=np.linspace(0,1,(max(l*2,70)),endpoint=True)
out=interpolate.splev(u3,tck)
plt.plot(x,y,'k--',label='Control polygon',marker='o',markerfacecolor='red')
#plt.plot(x,y,'ro',label='Control points only')plt.plot(out[0],out[1],'b',linewidth=2.0,label='B-spline curve')
plt.legend(loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('Cubic B-spline curve evaluation')
plt.show()

B-Spline interpolation

importnumpyasnpfromscipyimportinterpolateimportmatplotlib.pyplotasplt#x = np.arange(0, 2*np.pi+np.pi/4, 2*np.pi/8)#y = np.sin(x)ctr=np.array( [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),
(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1)])
x=ctr[:,0]
y=ctr[:,1]
#x=np.append(x,x[0])#y=np.append(y,y[0])tck,u=interpolate.splprep([x,y],k=3,s=0)
u=np.linspace(0,1,num=50,endpoint=True)
out=interpolate.splev(u,tck)
plt.figure()
plt.plot(x, y, 'ro', out[0], out[1], 'b')
plt.legend(['Points', 'Interpolated B-spline', 'True'],loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('B-Spline interpolation')
plt.show()

You can comment about it here

Please help improving this code.

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Examples in Python about plotting and interpolating a B-spline curve and their comparison using Numpy, Scipy and Matplotlib.

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GitHub - kawache/Python-B-spline-examples: Examples in Python about plotting and interpolating a B-spline curve and their comparison using Numpy, Scipy and Matplotlib. · GitHub
Skip to content

Repository files navigation

Python-B-spline-examples

Examples in Python about evaluating and interpolating a B-spline curve and their comparaison using Numpy, Scipy and Matplotlib.

Requirements

  • Python 2.7.x or 3.x.x,
  • Numpy
  • Scipy
  • Matplotlib

Getting hands dirty

Assuming you already know a little about B-spline curves, otherwise google it or check this out.

Let consider a 9 points list named plist :

plist= [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),]

We will do two things :

  • A. Drawing a cubic B-Spline curve where plist is the control polygon.

  • B. Find(interpolate) and draw the B-spline curve that go trough plist points and or in other words a curve fitting using a cubic B-spline curve.

As we will work with Numpy , let's create a numpy array named ctr from plist ,and then split it into x and y arrays.

ctr=np.array(plist)
x=ctr[:,0]
y=ctr[:,1]

A. Drawing the B-spline Curve

We will draw a cubic B-spline which degree k=3 as it's the most used one. We need now to define the knots vector, with k ending equal knots:

l=len(x)
t=np.linspace(0,1,l-2,endpoint=True)
t=np.append([0,0,0],t)
t=np.append(t,[1,1,1])

Now we have defined all requirement to draw it. we can construct a sequence of length 3 containing the knots, coefficients, and degree of the spline to pass it as the tck argument to scipy.interpolate.splev , the function that will evaluate our curve:

tck=[t,[x,y],3]

And as we will evaluate the B-spline curve, we need a second parameter__u__ , an array of points :

u3=np.linspace(0,1,(max(l*2,70)),endpoint=True)

And we evaluate it :

out=interpolate.splev(u3,tck) 

Drawing it using matplotlib :

Cubic B-spline curve drawing

With this code :

plt.plot(x,y,'k--',label='Control polygon',marker='o',markerfacecolor='red')
plt.plot(out[0],out[1],'b',linewidth=2.0,label='B-spline curve')
plt.legend(loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('Cubic B-spline curve evaluation')
plt.show()

B. interpolate the B-spline Curve

We will interpolate now the B-spline cruve that should go through n points. As we done before, a numpy array must be created using plist and split into x and y arrays :

plist= [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),]
ctr=np.array(plist)
x=ctr[:,0]
y=ctr[:,1]

To interpolate the B-spline that go through this points. To do so, we will be using scipy.interpolate.splprep, to get the interpolated S-pline parameters :

tck,_=interpolate.splprep([x,y],k=3,s=0)

And we use them to evaluate it using scipy.interpolate.splev, we still have a to redefine u variable to get a clear plot :

u=np.linspace(0,1,num=50,endpoint=True)
out=interpolate.splev(u,tck)

and we finaly draw it : Cubic B-spline curve interplation

With this code :

plt.figure()
plt.plot(x, y, 'ro', out[0], out[1], 'b')
plt.legend(['Points', 'Interpolated B-spline', 'True'],loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('B-Spline interpolation')
plt.show()

Full code :

B-Spline evalutation

importnumpyasnpfromscipyimportinterpolateimportmatplotlib.pyplotaspltctr=np.array( [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),
(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),])
x=ctr[:,0]
y=ctr[:,1]
# uncomment both lines for a closed curve#x=np.append(x,[x[0]]) #y=np.append(y,[y[0]])l=len(x) t=np.linspace(0,1,l-2,endpoint=True)
t=np.append([0,0,0],t)
t=np.append(t,[1,1,1])
tck=[t,[x,y],3]
u3=np.linspace(0,1,(max(l*2,70)),endpoint=True)
out=interpolate.splev(u3,tck)
plt.plot(x,y,'k--',label='Control polygon',marker='o',markerfacecolor='red')
#plt.plot(x,y,'ro',label='Control points only')plt.plot(out[0],out[1],'b',linewidth=2.0,label='B-spline curve')
plt.legend(loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('Cubic B-spline curve evaluation')
plt.show()

B-Spline interpolation

importnumpyasnpfromscipyimportinterpolateimportmatplotlib.pyplotasplt#x = np.arange(0, 2*np.pi+np.pi/4, 2*np.pi/8)#y = np.sin(x)ctr=np.array( [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),
(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1)])
x=ctr[:,0]
y=ctr[:,1]
#x=np.append(x,x[0])#y=np.append(y,y[0])tck,u=interpolate.splprep([x,y],k=3,s=0)
u=np.linspace(0,1,num=50,endpoint=True)
out=interpolate.splev(u,tck)
plt.figure()
plt.plot(x, y, 'ro', out[0], out[1], 'b')
plt.legend(['Points', 'Interpolated B-spline', 'True'],loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('B-Spline interpolation')
plt.show()

You can comment about it here

Please help improving this code.

About

Examples in Python about plotting and interpolating a B-spline curve and their comparison using Numpy, Scipy and Matplotlib.

Resources

Stars

113 stars

Watchers

3 watching

Forks

Releases

Packages

Contributors

Languages

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Python-B-spline-examples

Examples in Python about evaluating and interpolating a B-spline curve and their comparaison using Numpy, Scipy and Matplotlib.

Requirements

  • Python 2.7.x or 3.x.x,
  • Numpy
  • Scipy
  • Matplotlib

Getting hands dirty

Assuming you already know a little about B-spline curves, otherwise google it or check this out.

Let consider a 9 points list named plist :

plist= [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),]

We will do two things :

  • A. Drawing a cubic B-Spline curve where plist is the control polygon.

  • B. Find(interpolate) and draw the B-spline curve that go trough plist points and or in other words a curve fitting using a cubic B-spline curve.

As we will work with Numpy , let's create a numpy array named ctr from plist ,and then split it into x and y arrays.

ctr=np.array(plist)
x=ctr[:,0]
y=ctr[:,1]

A. Drawing the B-spline Curve

We will draw a cubic B-spline which degree k=3 as it's the most used one. We need now to define the knots vector, with k ending equal knots:

l=len(x)
t=np.linspace(0,1,l-2,endpoint=True)
t=np.append([0,0,0],t)
t=np.append(t,[1,1,1])

Now we have defined all requirement to draw it. we can construct a sequence of length 3 containing the knots, coefficients, and degree of the spline to pass it as the tck argument to scipy.interpolate.splev , the function that will evaluate our curve:

tck=[t,[x,y],3]

And as we will evaluate the B-spline curve, we need a second parameter__u__ , an array of points :

u3=np.linspace(0,1,(max(l*2,70)),endpoint=True)

And we evaluate it :

out=interpolate.splev(u3,tck) 

Drawing it using matplotlib :

Cubic B-spline curve drawing

With this code :

plt.plot(x,y,'k--',label='Control polygon',marker='o',markerfacecolor='red')
plt.plot(out[0],out[1],'b',linewidth=2.0,label='B-spline curve')
plt.legend(loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('Cubic B-spline curve evaluation')
plt.show()

B. interpolate the B-spline Curve

We will interpolate now the B-spline cruve that should go through n points. As we done before, a numpy array must be created using plist and split into x and y arrays :

plist= [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),]
ctr=np.array(plist)
x=ctr[:,0]
y=ctr[:,1]

To interpolate the B-spline that go through this points. To do so, we will be using scipy.interpolate.splprep, to get the interpolated S-pline parameters :

tck,_=interpolate.splprep([x,y],k=3,s=0)

And we use them to evaluate it using scipy.interpolate.splev, we still have a to redefine u variable to get a clear plot :

u=np.linspace(0,1,num=50,endpoint=True)
out=interpolate.splev(u,tck)

and we finaly draw it : Cubic B-spline curve interplation

With this code :

plt.figure()
plt.plot(x, y, 'ro', out[0], out[1], 'b')
plt.legend(['Points', 'Interpolated B-spline', 'True'],loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('B-Spline interpolation')
plt.show()

Full code :

B-Spline evalutation

importnumpyasnpfromscipyimportinterpolateimportmatplotlib.pyplotaspltctr=np.array( [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),
(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),])
x=ctr[:,0]
y=ctr[:,1]
# uncomment both lines for a closed curve#x=np.append(x,[x[0]]) #y=np.append(y,[y[0]])l=len(x) t=np.linspace(0,1,l-2,endpoint=True)
t=np.append([0,0,0],t)
t=np.append(t,[1,1,1])
tck=[t,[x,y],3]
u3=np.linspace(0,1,(max(l*2,70)),endpoint=True)
out=interpolate.splev(u3,tck)
plt.plot(x,y,'k--',label='Control polygon',marker='o',markerfacecolor='red')
#plt.plot(x,y,'ro',label='Control points only')plt.plot(out[0],out[1],'b',linewidth=2.0,label='B-spline curve')
plt.legend(loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('Cubic B-spline curve evaluation')
plt.show()

B-Spline interpolation

importnumpyasnpfromscipyimportinterpolateimportmatplotlib.pyplotasplt#x = np.arange(0, 2*np.pi+np.pi/4, 2*np.pi/8)#y = np.sin(x)ctr=np.array( [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),
(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1)])
x=ctr[:,0]
y=ctr[:,1]
#x=np.append(x,x[0])#y=np.append(y,y[0])tck,u=interpolate.splprep([x,y],k=3,s=0)
u=np.linspace(0,1,num=50,endpoint=True)
out=interpolate.splev(u,tck)
plt.figure()
plt.plot(x, y, 'ro', out[0], out[1], 'b')
plt.legend(['Points', 'Interpolated B-spline', 'True'],loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('B-Spline interpolation')
plt.show()

You can comment about it here

Please help improving this code.

About

Examples in Python about plotting and interpolating a B-spline curve and their comparison using Numpy, Scipy and Matplotlib.

Resources

Stars

113 stars

Watchers

3 watching

Forks

Releases

Packages

Contributors

Languages

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Skip to content

Repository files navigation

Python-B-spline-examples

Examples in Python about evaluating and interpolating a B-spline curve and their comparaison using Numpy, Scipy and Matplotlib.

Requirements

  • Python 2.7.x or 3.x.x,
  • Numpy
  • Scipy
  • Matplotlib

Getting hands dirty

Assuming you already know a little about B-spline curves, otherwise google it or check this out.

Let consider a 9 points list named plist :

plist= [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),]

We will do two things :

  • A. Drawing a cubic B-Spline curve where plist is the control polygon.

  • B. Find(interpolate) and draw the B-spline curve that go trough plist points and or in other words a curve fitting using a cubic B-spline curve.

As we will work with Numpy , let's create a numpy array named ctr from plist ,and then split it into x and y arrays.

ctr=np.array(plist)
x=ctr[:,0]
y=ctr[:,1]

A. Drawing the B-spline Curve

We will draw a cubic B-spline which degree k=3 as it's the most used one. We need now to define the knots vector, with k ending equal knots:

l=len(x)
t=np.linspace(0,1,l-2,endpoint=True)
t=np.append([0,0,0],t)
t=np.append(t,[1,1,1])

Now we have defined all requirement to draw it. we can construct a sequence of length 3 containing the knots, coefficients, and degree of the spline to pass it as the tck argument to scipy.interpolate.splev , the function that will evaluate our curve:

tck=[t,[x,y],3]

And as we will evaluate the B-spline curve, we need a second parameter__u__ , an array of points :

u3=np.linspace(0,1,(max(l*2,70)),endpoint=True)

And we evaluate it :

out=interpolate.splev(u3,tck) 

Drawing it using matplotlib :

Cubic B-spline curve drawing

With this code :

plt.plot(x,y,'k--',label='Control polygon',marker='o',markerfacecolor='red')
plt.plot(out[0],out[1],'b',linewidth=2.0,label='B-spline curve')
plt.legend(loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('Cubic B-spline curve evaluation')
plt.show()

B. interpolate the B-spline Curve

We will interpolate now the B-spline cruve that should go through n points. As we done before, a numpy array must be created using plist and split into x and y arrays :

plist= [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),]
ctr=np.array(plist)
x=ctr[:,0]
y=ctr[:,1]

To interpolate the B-spline that go through this points. To do so, we will be using scipy.interpolate.splprep, to get the interpolated S-pline parameters :

tck,_=interpolate.splprep([x,y],k=3,s=0)

And we use them to evaluate it using scipy.interpolate.splev, we still have a to redefine u variable to get a clear plot :

u=np.linspace(0,1,num=50,endpoint=True)
out=interpolate.splev(u,tck)

and we finaly draw it : Cubic B-spline curve interplation

With this code :

plt.figure()
plt.plot(x, y, 'ro', out[0], out[1], 'b')
plt.legend(['Points', 'Interpolated B-spline', 'True'],loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('B-Spline interpolation')
plt.show()

Full code :

B-Spline evalutation

importnumpyasnpfromscipyimportinterpolateimportmatplotlib.pyplotaspltctr=np.array( [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),
(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),])
x=ctr[:,0]
y=ctr[:,1]
# uncomment both lines for a closed curve#x=np.append(x,[x[0]]) #y=np.append(y,[y[0]])l=len(x) t=np.linspace(0,1,l-2,endpoint=True)
t=np.append([0,0,0],t)
t=np.append(t,[1,1,1])
tck=[t,[x,y],3]
u3=np.linspace(0,1,(max(l*2,70)),endpoint=True)
out=interpolate.splev(u3,tck)
plt.plot(x,y,'k--',label='Control polygon',marker='o',markerfacecolor='red')
#plt.plot(x,y,'ro',label='Control points only')plt.plot(out[0],out[1],'b',linewidth=2.0,label='B-spline curve')
plt.legend(loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('Cubic B-spline curve evaluation')
plt.show()

B-Spline interpolation

importnumpyasnpfromscipyimportinterpolateimportmatplotlib.pyplotasplt#x = np.arange(0, 2*np.pi+np.pi/4, 2*np.pi/8)#y = np.sin(x)ctr=np.array( [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),
(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1)])
x=ctr[:,0]
y=ctr[:,1]
#x=np.append(x,x[0])#y=np.append(y,y[0])tck,u=interpolate.splprep([x,y],k=3,s=0)
u=np.linspace(0,1,num=50,endpoint=True)
out=interpolate.splev(u,tck)
plt.figure()
plt.plot(x, y, 'ro', out[0], out[1], 'b')
plt.legend(['Points', 'Interpolated B-spline', 'True'],loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('B-Spline interpolation')
plt.show()

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Python-B-spline-examples

Examples in Python about evaluating and interpolating a B-spline curve and their comparaison using Numpy, Scipy and Matplotlib.

Requirements

  • Python 2.7.x or 3.x.x,
  • Numpy
  • Scipy
  • Matplotlib

Getting hands dirty

Assuming you already know a little about B-spline curves, otherwise google it or check this out.

Let consider a 9 points list named plist :

plist= [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),]

We will do two things :

  • A. Drawing a cubic B-Spline curve where plist is the control polygon.

  • B. Find(interpolate) and draw the B-spline curve that go trough plist points and or in other words a curve fitting using a cubic B-spline curve.

As we will work with Numpy , let's create a numpy array named ctr from plist ,and then split it into x and y arrays.

ctr=np.array(plist)
x=ctr[:,0]
y=ctr[:,1]

A. Drawing the B-spline Curve

We will draw a cubic B-spline which degree k=3 as it's the most used one. We need now to define the knots vector, with k ending equal knots:

l=len(x)
t=np.linspace(0,1,l-2,endpoint=True)
t=np.append([0,0,0],t)
t=np.append(t,[1,1,1])

Now we have defined all requirement to draw it. we can construct a sequence of length 3 containing the knots, coefficients, and degree of the spline to pass it as the tck argument to scipy.interpolate.splev , the function that will evaluate our curve:

tck=[t,[x,y],3]

And as we will evaluate the B-spline curve, we need a second parameter__u__ , an array of points :

u3=np.linspace(0,1,(max(l*2,70)),endpoint=True)

And we evaluate it :

out=interpolate.splev(u3,tck) 

Drawing it using matplotlib :

Cubic B-spline curve drawing

With this code :

plt.plot(x,y,'k--',label='Control polygon',marker='o',markerfacecolor='red')
plt.plot(out[0],out[1],'b',linewidth=2.0,label='B-spline curve')
plt.legend(loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('Cubic B-spline curve evaluation')
plt.show()

B. interpolate the B-spline Curve

We will interpolate now the B-spline cruve that should go through n points. As we done before, a numpy array must be created using plist and split into x and y arrays :

plist= [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),]
ctr=np.array(plist)
x=ctr[:,0]
y=ctr[:,1]

To interpolate the B-spline that go through this points. To do so, we will be using scipy.interpolate.splprep, to get the interpolated S-pline parameters :

tck,_=interpolate.splprep([x,y],k=3,s=0)

And we use them to evaluate it using scipy.interpolate.splev, we still have a to redefine u variable to get a clear plot :

u=np.linspace(0,1,num=50,endpoint=True)
out=interpolate.splev(u,tck)

and we finaly draw it : Cubic B-spline curve interplation

With this code :

plt.figure()
plt.plot(x, y, 'ro', out[0], out[1], 'b')
plt.legend(['Points', 'Interpolated B-spline', 'True'],loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('B-Spline interpolation')
plt.show()

Full code :

B-Spline evalutation

importnumpyasnpfromscipyimportinterpolateimportmatplotlib.pyplotaspltctr=np.array( [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),
(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),])
x=ctr[:,0]
y=ctr[:,1]
# uncomment both lines for a closed curve#x=np.append(x,[x[0]]) #y=np.append(y,[y[0]])l=len(x) t=np.linspace(0,1,l-2,endpoint=True)
t=np.append([0,0,0],t)
t=np.append(t,[1,1,1])
tck=[t,[x,y],3]
u3=np.linspace(0,1,(max(l*2,70)),endpoint=True)
out=interpolate.splev(u3,tck)
plt.plot(x,y,'k--',label='Control polygon',marker='o',markerfacecolor='red')
#plt.plot(x,y,'ro',label='Control points only')plt.plot(out[0],out[1],'b',linewidth=2.0,label='B-spline curve')
plt.legend(loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('Cubic B-spline curve evaluation')
plt.show()

B-Spline interpolation

importnumpyasnpfromscipyimportinterpolateimportmatplotlib.pyplotasplt#x = np.arange(0, 2*np.pi+np.pi/4, 2*np.pi/8)#y = np.sin(x)ctr=np.array( [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),
(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1)])
x=ctr[:,0]
y=ctr[:,1]
#x=np.append(x,x[0])#y=np.append(y,y[0])tck,u=interpolate.splprep([x,y],k=3,s=0)
u=np.linspace(0,1,num=50,endpoint=True)
out=interpolate.splev(u,tck)
plt.figure()
plt.plot(x, y, 'ro', out[0], out[1], 'b')
plt.legend(['Points', 'Interpolated B-spline', 'True'],loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('B-Spline interpolation')
plt.show()

You can comment about it here

Please help improving this code.

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Examples in Python about plotting and interpolating a B-spline curve and their comparison using Numpy, Scipy and Matplotlib.

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3 watching

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Python-B-spline-examples

Examples in Python about evaluating and interpolating a B-spline curve and their comparaison using Numpy, Scipy and Matplotlib.

Requirements

  • Python 2.7.x or 3.x.x,
  • Numpy
  • Scipy
  • Matplotlib

Getting hands dirty

Assuming you already know a little about B-spline curves, otherwise google it or check this out.

Let consider a 9 points list named plist :

plist= [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),]

We will do two things :

  • A. Drawing a cubic B-Spline curve where plist is the control polygon.

  • B. Find(interpolate) and draw the B-spline curve that go trough plist points and or in other words a curve fitting using a cubic B-spline curve.

As we will work with Numpy , let's create a numpy array named ctr from plist ,and then split it into x and y arrays.

ctr=np.array(plist)
x=ctr[:,0]
y=ctr[:,1]

A. Drawing the B-spline Curve

We will draw a cubic B-spline which degree k=3 as it's the most used one. We need now to define the knots vector, with k ending equal knots:

l=len(x)
t=np.linspace(0,1,l-2,endpoint=True)
t=np.append([0,0,0],t)
t=np.append(t,[1,1,1])

Now we have defined all requirement to draw it. we can construct a sequence of length 3 containing the knots, coefficients, and degree of the spline to pass it as the tck argument to scipy.interpolate.splev , the function that will evaluate our curve:

tck=[t,[x,y],3]

And as we will evaluate the B-spline curve, we need a second parameter__u__ , an array of points :

u3=np.linspace(0,1,(max(l*2,70)),endpoint=True)

And we evaluate it :

out=interpolate.splev(u3,tck) 

Drawing it using matplotlib :

Cubic B-spline curve drawing

With this code :

plt.plot(x,y,'k--',label='Control polygon',marker='o',markerfacecolor='red')
plt.plot(out[0],out[1],'b',linewidth=2.0,label='B-spline curve')
plt.legend(loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('Cubic B-spline curve evaluation')
plt.show()

B. interpolate the B-spline Curve

We will interpolate now the B-spline cruve that should go through n points. As we done before, a numpy array must be created using plist and split into x and y arrays :

plist= [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),]
ctr=np.array(plist)
x=ctr[:,0]
y=ctr[:,1]

To interpolate the B-spline that go through this points. To do so, we will be using scipy.interpolate.splprep, to get the interpolated S-pline parameters :

tck,_=interpolate.splprep([x,y],k=3,s=0)

And we use them to evaluate it using scipy.interpolate.splev, we still have a to redefine u variable to get a clear plot :

u=np.linspace(0,1,num=50,endpoint=True)
out=interpolate.splev(u,tck)

and we finaly draw it : Cubic B-spline curve interplation

With this code :

plt.figure()
plt.plot(x, y, 'ro', out[0], out[1], 'b')
plt.legend(['Points', 'Interpolated B-spline', 'True'],loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('B-Spline interpolation')
plt.show()

Full code :

B-Spline evalutation

importnumpyasnpfromscipyimportinterpolateimportmatplotlib.pyplotaspltctr=np.array( [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),
(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),])
x=ctr[:,0]
y=ctr[:,1]
# uncomment both lines for a closed curve#x=np.append(x,[x[0]]) #y=np.append(y,[y[0]])l=len(x) t=np.linspace(0,1,l-2,endpoint=True)
t=np.append([0,0,0],t)
t=np.append(t,[1,1,1])
tck=[t,[x,y],3]
u3=np.linspace(0,1,(max(l*2,70)),endpoint=True)
out=interpolate.splev(u3,tck)
plt.plot(x,y,'k--',label='Control polygon',marker='o',markerfacecolor='red')
#plt.plot(x,y,'ro',label='Control points only')plt.plot(out[0],out[1],'b',linewidth=2.0,label='B-spline curve')
plt.legend(loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('Cubic B-spline curve evaluation')
plt.show()

B-Spline interpolation

importnumpyasnpfromscipyimportinterpolateimportmatplotlib.pyplotasplt#x = np.arange(0, 2*np.pi+np.pi/4, 2*np.pi/8)#y = np.sin(x)ctr=np.array( [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),
(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1)])
x=ctr[:,0]
y=ctr[:,1]
#x=np.append(x,x[0])#y=np.append(y,y[0])tck,u=interpolate.splprep([x,y],k=3,s=0)
u=np.linspace(0,1,num=50,endpoint=True)
out=interpolate.splev(u,tck)
plt.figure()
plt.plot(x, y, 'ro', out[0], out[1], 'b')
plt.legend(['Points', 'Interpolated B-spline', 'True'],loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('B-Spline interpolation')
plt.show()

You can comment about it here

Please help improving this code.

About

Examples in Python about plotting and interpolating a B-spline curve and their comparison using Numpy, Scipy and Matplotlib.

Resources

Stars

113 stars

Watchers

3 watching

Forks

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Packages

Contributors

Languages

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Repository files navigation

Python-B-spline-examples

Examples in Python about evaluating and interpolating a B-spline curve and their comparaison using Numpy, Scipy and Matplotlib.

Requirements

  • Python 2.7.x or 3.x.x,
  • Numpy
  • Scipy
  • Matplotlib

Getting hands dirty

Assuming you already know a little about B-spline curves, otherwise google it or check this out.

Let consider a 9 points list named plist :

plist= [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),]

We will do two things :

  • A. Drawing a cubic B-Spline curve where plist is the control polygon.

  • B. Find(interpolate) and draw the B-spline curve that go trough plist points and or in other words a curve fitting using a cubic B-spline curve.

As we will work with Numpy , let's create a numpy array named ctr from plist ,and then split it into x and y arrays.

ctr=np.array(plist)
x=ctr[:,0]
y=ctr[:,1]

A. Drawing the B-spline Curve

We will draw a cubic B-spline which degree k=3 as it's the most used one. We need now to define the knots vector, with k ending equal knots:

l=len(x)
t=np.linspace(0,1,l-2,endpoint=True)
t=np.append([0,0,0],t)
t=np.append(t,[1,1,1])

Now we have defined all requirement to draw it. we can construct a sequence of length 3 containing the knots, coefficients, and degree of the spline to pass it as the tck argument to scipy.interpolate.splev , the function that will evaluate our curve:

tck=[t,[x,y],3]

And as we will evaluate the B-spline curve, we need a second parameter__u__ , an array of points :

u3=np.linspace(0,1,(max(l*2,70)),endpoint=True)

And we evaluate it :

out=interpolate.splev(u3,tck) 

Drawing it using matplotlib :

Cubic B-spline curve drawing

With this code :

plt.plot(x,y,'k--',label='Control polygon',marker='o',markerfacecolor='red')
plt.plot(out[0],out[1],'b',linewidth=2.0,label='B-spline curve')
plt.legend(loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('Cubic B-spline curve evaluation')
plt.show()

B. interpolate the B-spline Curve

We will interpolate now the B-spline cruve that should go through n points. As we done before, a numpy array must be created using plist and split into x and y arrays :

plist= [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),]
ctr=np.array(plist)
x=ctr[:,0]
y=ctr[:,1]

To interpolate the B-spline that go through this points. To do so, we will be using scipy.interpolate.splprep, to get the interpolated S-pline parameters :

tck,_=interpolate.splprep([x,y],k=3,s=0)

And we use them to evaluate it using scipy.interpolate.splev, we still have a to redefine u variable to get a clear plot :

u=np.linspace(0,1,num=50,endpoint=True)
out=interpolate.splev(u,tck)

and we finaly draw it : Cubic B-spline curve interplation

With this code :

plt.figure()
plt.plot(x, y, 'ro', out[0], out[1], 'b')
plt.legend(['Points', 'Interpolated B-spline', 'True'],loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('B-Spline interpolation')
plt.show()

Full code :

B-Spline evalutation

importnumpyasnpfromscipyimportinterpolateimportmatplotlib.pyplotaspltctr=np.array( [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),
(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),])
x=ctr[:,0]
y=ctr[:,1]
# uncomment both lines for a closed curve#x=np.append(x,[x[0]]) #y=np.append(y,[y[0]])l=len(x) t=np.linspace(0,1,l-2,endpoint=True)
t=np.append([0,0,0],t)
t=np.append(t,[1,1,1])
tck=[t,[x,y],3]
u3=np.linspace(0,1,(max(l*2,70)),endpoint=True)
out=interpolate.splev(u3,tck)
plt.plot(x,y,'k--',label='Control polygon',marker='o',markerfacecolor='red')
#plt.plot(x,y,'ro',label='Control points only')plt.plot(out[0],out[1],'b',linewidth=2.0,label='B-spline curve')
plt.legend(loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('Cubic B-spline curve evaluation')
plt.show()

B-Spline interpolation

importnumpyasnpfromscipyimportinterpolateimportmatplotlib.pyplotasplt#x = np.arange(0, 2*np.pi+np.pi/4, 2*np.pi/8)#y = np.sin(x)ctr=np.array( [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),
(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1)])
x=ctr[:,0]
y=ctr[:,1]
#x=np.append(x,x[0])#y=np.append(y,y[0])tck,u=interpolate.splprep([x,y],k=3,s=0)
u=np.linspace(0,1,num=50,endpoint=True)
out=interpolate.splev(u,tck)
plt.figure()
plt.plot(x, y, 'ro', out[0], out[1], 'b')
plt.legend(['Points', 'Interpolated B-spline', 'True'],loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('B-Spline interpolation')
plt.show()

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Python-B-spline-examples

Examples in Python about evaluating and interpolating a B-spline curve and their comparaison using Numpy, Scipy and Matplotlib.

Requirements

  • Python 2.7.x or 3.x.x,
  • Numpy
  • Scipy
  • Matplotlib

Getting hands dirty

Assuming you already know a little about B-spline curves, otherwise google it or check this out.

Let consider a 9 points list named plist :

plist= [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),]

We will do two things :

  • A. Drawing a cubic B-Spline curve where plist is the control polygon.

  • B. Find(interpolate) and draw the B-spline curve that go trough plist points and or in other words a curve fitting using a cubic B-spline curve.

As we will work with Numpy , let's create a numpy array named ctr from plist ,and then split it into x and y arrays.

ctr=np.array(plist)
x=ctr[:,0]
y=ctr[:,1]

A. Drawing the B-spline Curve

We will draw a cubic B-spline which degree k=3 as it's the most used one. We need now to define the knots vector, with k ending equal knots:

l=len(x)
t=np.linspace(0,1,l-2,endpoint=True)
t=np.append([0,0,0],t)
t=np.append(t,[1,1,1])

Now we have defined all requirement to draw it. we can construct a sequence of length 3 containing the knots, coefficients, and degree of the spline to pass it as the tck argument to scipy.interpolate.splev , the function that will evaluate our curve:

tck=[t,[x,y],3]

And as we will evaluate the B-spline curve, we need a second parameter__u__ , an array of points :

u3=np.linspace(0,1,(max(l*2,70)),endpoint=True)

And we evaluate it :

out=interpolate.splev(u3,tck) 

Drawing it using matplotlib :

Cubic B-spline curve drawing

With this code :

plt.plot(x,y,'k--',label='Control polygon',marker='o',markerfacecolor='red')
plt.plot(out[0],out[1],'b',linewidth=2.0,label='B-spline curve')
plt.legend(loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('Cubic B-spline curve evaluation')
plt.show()

B. interpolate the B-spline Curve

We will interpolate now the B-spline cruve that should go through n points. As we done before, a numpy array must be created using plist and split into x and y arrays :

plist= [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),]
ctr=np.array(plist)
x=ctr[:,0]
y=ctr[:,1]

To interpolate the B-spline that go through this points. To do so, we will be using scipy.interpolate.splprep, to get the interpolated S-pline parameters :

tck,_=interpolate.splprep([x,y],k=3,s=0)

And we use them to evaluate it using scipy.interpolate.splev, we still have a to redefine u variable to get a clear plot :

u=np.linspace(0,1,num=50,endpoint=True)
out=interpolate.splev(u,tck)

and we finaly draw it : Cubic B-spline curve interplation

With this code :

plt.figure()
plt.plot(x, y, 'ro', out[0], out[1], 'b')
plt.legend(['Points', 'Interpolated B-spline', 'True'],loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('B-Spline interpolation')
plt.show()

Full code :

B-Spline evalutation

importnumpyasnpfromscipyimportinterpolateimportmatplotlib.pyplotaspltctr=np.array( [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),
(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1),])
x=ctr[:,0]
y=ctr[:,1]
# uncomment both lines for a closed curve#x=np.append(x,[x[0]]) #y=np.append(y,[y[0]])l=len(x) t=np.linspace(0,1,l-2,endpoint=True)
t=np.append([0,0,0],t)
t=np.append(t,[1,1,1])
tck=[t,[x,y],3]
u3=np.linspace(0,1,(max(l*2,70)),endpoint=True)
out=interpolate.splev(u3,tck)
plt.plot(x,y,'k--',label='Control polygon',marker='o',markerfacecolor='red')
#plt.plot(x,y,'ro',label='Control points only')plt.plot(out[0],out[1],'b',linewidth=2.0,label='B-spline curve')
plt.legend(loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('Cubic B-spline curve evaluation')
plt.show()

B-Spline interpolation

importnumpyasnpfromscipyimportinterpolateimportmatplotlib.pyplotasplt#x = np.arange(0, 2*np.pi+np.pi/4, 2*np.pi/8)#y = np.sin(x)ctr=np.array( [(3 , 1), (2.5, 4), (0, 1), (-2.5, 4),
(-3, 0), (-2.5, -4), (0, -1), (2.5, -4), (3, -1)])
x=ctr[:,0]
y=ctr[:,1]
#x=np.append(x,x[0])#y=np.append(y,y[0])tck,u=interpolate.splprep([x,y],k=3,s=0)
u=np.linspace(0,1,num=50,endpoint=True)
out=interpolate.splev(u,tck)
plt.figure()
plt.plot(x, y, 'ro', out[0], out[1], 'b')
plt.legend(['Points', 'Interpolated B-spline', 'True'],loc='best')
plt.axis([min(x)-1, max(x)+1, min(y)-1, max(y)+1])
plt.title('B-Spline interpolation')
plt.show()

You can comment about it here

Please help improving this code.

About

Examples in Python about plotting and interpolating a B-spline curve and their comparison using Numpy, Scipy and Matplotlib.

Resources

Stars

113 stars

Watchers

3 watching

Forks

Releases

Packages

Contributors

Languages