import numpy as np
import matplotlib.pyplot as plt
from scipy import interpolate

# Cubic-spline

x = np.arange(0, 2*np.pi+np.pi/4, 2*np.pi/8)
y = np.sin(x)
tck = interpolate.splrep(x, y, s=0)
xnew = np.arange(0, 2*np.pi, np.pi/50)
ynew = interpolate.splev(xnew, tck, der=0)

plt.figure()
plt.plot(x, y, 'x', xnew, ynew, xnew, np.sin(xnew), x, y, 'b')
plt.legend(['Linear', 'Cubic Spline', 'True'])
plt.axis([-0.05, 6.33, -1.05, 1.05])
plt.title('Cubic-spline interpolation')
plt.show()

# Derivative of spline

yder = interpolate.splev(xnew, tck, der=1)
plt.figure()
plt.plot(xnew, yder, xnew, np.cos(xnew),'--')
plt.legend(['Cubic Spline', 'True'])
plt.axis([-0.05, 6.33, -1.05, 1.05])
plt.title('Derivative estimation from spline')
plt.show()

# Integral of spline

def integ(x, tck, constant=-1):
    x = np.atleast_1d(x)
    out = np.zeros(x.shape, dtype=x.dtype)
    for n in xrange(len(out)):
        out[n] = interpolate.splint(0, x[n], tck)
    out += constant
    return out

yint = integ(xnew, tck)
plt.figure()
plt.plot(xnew, yint, xnew, -np.cos(xnew), '--')
plt.legend(['Cubic Spline', 'True'])
plt.axis([-0.05, 6.33, -1.05, 1.05])
plt.title('Integral estimation from spline')
plt.show()

# Roots of spline

interpolate.sproot(tck)
# array([3.1416])

# Notice that `sproot` failed to find an obvious solution at the edge of the
# approximation interval, :math:`x = 0`. If we define the spline on a slightly
# larger interval, we recover both roots :math:`x = 0` and :math:`x = 2\pi`:

x = np.linspace(-np.pi/4, 2.*np.pi + np.pi/4, 21)
y = np.sin(x)
tck = interpolate.splrep(x, y, s=0)
interpolate.sproot(tck)
# array([0., 3.1416])

# Parametric spline

t = np.arange(0, 1.1, .1)
x = np.sin(2*np.pi*t)
y = np.cos(2*np.pi*t)
tck, u = interpolate.splprep([x, y], s=0)
unew = np.arange(0, 1.01, 0.01)
out = interpolate.splev(unew, tck)
plt.figure()
plt.plot(x, y, 'x', out[0], out[1], np.sin(2*np.pi*unew), np.cos(2*np.pi*unew), x, y, 'b')
plt.legend(['Linear', 'Cubic Spline', 'True'])
plt.axis([-1.05, 1.05, -1.05, 1.05])
plt.title('Spline of parametrically-defined curve')
plt.show()