Develop Reference

Cubic spline for non-monotonic data (not a 1d function)

I have a curve as shown below:
The x coordinates and the y coordinates for this plot are:
path_x= (4.0, 5.638304088577984, 6.785456961280076, 5.638304088577984, 4.0)
path_y =(0.0, 1.147152872702092, 2.7854569612800755, 4.423761049858059, 3.2766081771559668)
And I obtained the above picture by:
x_min =min(path_x)-1
x_max =max(path_x)+1
y_min =min(path_y)-1
y_max =max(path_y)+1
num_pts = len(path_x)
fig = plt.figure(figsize=(8,8))
#fig = plt.figure()
plt.suptitle("Curve and the boundary")
ax = fig.add_subplot(1,1,1)
ax.set_xlim([min(x_min,y_min),max(x_max,y_max)])
ax.set_ylim([min(x_min,y_min),max(x_max,y_max)])
ax.plot(path_x,path_y)
Now my intention is to draw a smooth curve using cubic splines. But looks like for cubic splines you need the x coordinates to be on ascending order. whereas in this case, neither x values nor y values are in the ascending order.
Also this is not a function. That is an x value is mapped with more than one element in the range.
I also went over this post. But I couldn't figure out a proper method to solve my problem.
I really appreciate your help in this regard

As suggested in the comments, you can always parameterize any curve/surface with an arbitrary (and linear!) parameter.
For example, define t as a parameter such that you get x=x(t) and y=y(t). Since t is arbitrary, you can define it such that at t=0, you get your first path_x[0],path_y[0], and at t=1, you get your last pair of coordinates, path_x[-1],path_y[-1].
Here is a code using scipy.interpolate
import numpy
import scipy.interpolate
import matplotlib.pyplot as plt
path_x = numpy.asarray((4.0, 5.638304088577984, 6.785456961280076, 5.638304088577984, 4.0),dtype=float)
path_y = numpy.asarray((0.0, 1.147152872702092, 2.7854569612800755, 4.423761049858059, 3.2766081771559668),dtype=float)
# defining arbitrary parameter to parameterize the curve
path_t = numpy.linspace(0,1,path_x.size)
# this is the position vector with
# x coord (1st row) given by path_x, and
# y coord (2nd row) given by path_y
r = numpy.vstack((path_x.reshape((1,path_x.size)),path_y.reshape((1,path_y.size))))
# creating the spline object
spline = scipy.interpolate.interp1d(path_t,r,kind='cubic')
# defining values of the arbitrary parameter over which
# you want to interpolate x and y
# it MUST be within 0 and 1, since you defined
# the spline between path_t=0 and path_t=1
t = numpy.linspace(numpy.min(path_t),numpy.max(path_t),100)
# interpolating along t
# r[0,:] -> interpolated x coordinates
# r[1,:] -> interpolated y coordinates
r = spline(t)
plt.plot(path_x,path_y,'or')
plt.plot(r[0,:],r[1,:],'-k')
plt.xlabel('x')
plt.ylabel('y')
plt.show()
With output

For non-ascending x splines can be easily computed if you make both x and y functions of another parameter t: x(t), y(t).
In your case you have 5 points so t should be just enumeration of these points, i.e. t = 0, 1, 2, 3, 4 for 5 points.
So if x = [5, 2, 7, 3, 6] then x(t) = x(0) = 5, x(1) = 2, x(2) = 7, x(3) = 3, x(4) = 6. Same for y.
Then compute spline function for both x(t) and y(t). Afterwards compute values of splines in all many intermediate t points. Lastly just use all calculated values x(t) and y(t) as a function y(x).
Once before I implemented cubic spline computation from scratch using Numpy, so I use this code in my example below if you don't mind (it could be useful for you to learn about spline math), replace with your library functions. Also in my code you can see numba lines commented out, if you want you can use these Numba annotations to speed up computation.
You have to look at main() function at the bottom of code, it shows how to compute and use x(t) and y(t).
Try it online!
import numpy as np, matplotlib.pyplot as plt
# Solves linear system given by Tridiagonal Matrix
# Helper for calculating cubic splines
##numba.njit(cache = True, fastmath = True, inline = 'always')
def tri_diag_solve(A, B, C, F):
n = B.size
assert A.ndim == B.ndim == C.ndim == F.ndim == 1 and (
A.size == B.size == C.size == F.size == n
) #, (A.shape, B.shape, C.shape, F.shape)
Bs, Fs = np.zeros_like(B), np.zeros_like(F)
Bs[0], Fs[0] = B[0], F[0]
for i in range(1, n):
Bs[i] = B[i] - A[i] / Bs[i - 1] * C[i - 1]
Fs[i] = F[i] - A[i] / Bs[i - 1] * Fs[i - 1]
x = np.zeros_like(B)
x[-1] = Fs[-1] / Bs[-1]
for i in range(n - 2, -1, -1):
x[i] = (Fs[i] - C[i] * x[i + 1]) / Bs[i]
return x
# Calculate cubic spline params
##numba.njit(cache = True, fastmath = True, inline = 'always')
def calc_spline_params(x, y):
a = y
h = np.diff(x)
c = np.concatenate((np.zeros((1,), dtype = y.dtype),
np.append(tri_diag_solve(h[:-1], (h[:-1] + h[1:]) * 2, h[1:],
((a[2:] - a[1:-1]) / h[1:] - (a[1:-1] - a[:-2]) / h[:-1]) * 3), 0)))
d = np.diff(c) / (3 * h)
b = (a[1:] - a[:-1]) / h + (2 * c[1:] + c[:-1]) / 3 * h
return a[1:], b, c[1:], d
# Spline value calculating function, given params and "x"
##numba.njit(cache = True, fastmath = True, inline = 'always')
def func_spline(x, ix, x0, a, b, c, d):
dx = x - x0[1:][ix]
return a[ix] + (b[ix] + (c[ix] + d[ix] * dx) * dx) * dx
# Compute piece-wise spline function for "x" out of sorted "x0" points
##numba.njit([f'f{ii}[:](f{ii}[:], f{ii}[:], f{ii}[:], f{ii}[:], f{ii}[:], f{ii}[:])' for ii in (4, 8)],
# cache = True, fastmath = True, inline = 'always')
def piece_wise_spline(x, x0, a, b, c, d):
xsh = x.shape
x = x.ravel()
ix = np.searchsorted(x0[1 : -1], x)
y = func_spline(x, ix, x0, a, b, c, d)
y = y.reshape(xsh)
return y
def main():
x0 = np.array([4.0, 5.638304088577984, 6.785456961280076, 5.638304088577984, 4.0])
y0 = np.array([0.0, 1.147152872702092, 2.7854569612800755, 4.423761049858059, 3.2766081771559668])
t0 = np.arange(len(x0)).astype(np.float64)
plt.plot(x0, y0)
vs = []
for e in (x0, y0):
a, b, c, d = calc_spline_params(t0, e)
x = np.linspace(0, t0[-1], 100)
vs.append(piece_wise_spline(x, t0, a, b, c, d))
plt.plot(vs[0], vs[1])
plt.show()
if __name__ == '__main__':
main()
Output:

Transfrom matrix from scipy.spatial.procrustes [duplicate]

Is there something like Matlab's procrustes function in NumPy/SciPy or related libraries?
For reference. Procrustes analysis aims to align 2 sets of points (in other words, 2 shapes) to minimize square distance between them by removing scale, translation and rotation warp components.
Example in Matlab:
X = [0 1; 2 3; 4 5; 6 7; 8 9]; % first shape
R = [1 2; 2 1]; % rotation matrix
t = [3 5]; % translation vector
Y = X * R + repmat(t, 5, 1); % warped shape, no scale and no distortion
[d Z] = procrustes(X, Y); % Z is Y aligned back to X
Z
Z =
0.0000 1.0000
2.0000 3.0000
4.0000 5.0000
6.0000 7.0000
8.0000 9.0000
Same task in NumPy:
X = arange(10).reshape((5, 2))
R = array([[1, 2], [2, 1]])
t = array([3, 5])
Y = dot(X, R) + t
Z = ???
Note: I'm only interested in aligned shape, since square error (variable d in Matlab code) is easily computed from 2 shapes.

I'm not aware of any pre-existing implementation in Python, but it's easy to take a look at the MATLAB code using edit procrustes.m and port it to Numpy:
def procrustes(X, Y, scaling=True, reflection='best'):
"""
A port of MATLAB's `procrustes` function to Numpy.
Procrustes analysis determines a linear transformation (translation,
reflection, orthogonal rotation and scaling) of the points in Y to best
conform them to the points in matrix X, using the sum of squared errors
as the goodness of fit criterion.
d, Z, [tform] = procrustes(X, Y)
Inputs:
------------
X, Y
matrices of target and input coordinates. they must have equal
numbers of points (rows), but Y may have fewer dimensions
(columns) than X.
scaling
if False, the scaling component of the transformation is forced
to 1
reflection
if 'best' (default), the transformation solution may or may not
include a reflection component, depending on which fits the data
best. setting reflection to True or False forces a solution with
reflection or no reflection respectively.
Outputs
------------
d
the residual sum of squared errors, normalized according to a
measure of the scale of X, ((X - X.mean(0))**2).sum()
Z
the matrix of transformed Y-values
tform
a dict specifying the rotation, translation and scaling that
maps X --> Y
"""
n,m = X.shape
ny,my = Y.shape
muX = X.mean(0)
muY = Y.mean(0)
X0 = X - muX
Y0 = Y - muY
ssX = (X0**2.).sum()
ssY = (Y0**2.).sum()
# centred Frobenius norm
normX = np.sqrt(ssX)
normY = np.sqrt(ssY)
# scale to equal (unit) norm
X0 /= normX
Y0 /= normY
if my < m:
Y0 = np.concatenate((Y0, np.zeros(n, m-my)),0)
# optimum rotation matrix of Y
A = np.dot(X0.T, Y0)
U,s,Vt = np.linalg.svd(A,full_matrices=False)
V = Vt.T
T = np.dot(V, U.T)
if reflection != 'best':
# does the current solution use a reflection?
have_reflection = np.linalg.det(T) < 0
# if that's not what was specified, force another reflection
if reflection != have_reflection:
V[:,-1] *= -1
s[-1] *= -1
T = np.dot(V, U.T)
traceTA = s.sum()
if scaling:
# optimum scaling of Y
b = traceTA * normX / normY
# standarised distance between X and b*Y*T + c
d = 1 - traceTA**2
# transformed coords
Z = normX*traceTA*np.dot(Y0, T) + muX
else:
b = 1
d = 1 + ssY/ssX - 2 * traceTA * normY / normX
Z = normY*np.dot(Y0, T) + muX
# transformation matrix
if my < m:
T = T[:my,:]
c = muX - b*np.dot(muY, T)
#transformation values
tform = {'rotation':T, 'scale':b, 'translation':c}
return d, Z, tform

There is a Scipy function for it: scipy.spatial.procrustes
I'm just posting its example here:
>>> import numpy as np
>>> from scipy.spatial import procrustes
>>> a = np.array([[1, 3], [1, 2], [1, 1], [2, 1]], 'd')
>>> b = np.array([[4, -2], [4, -4], [4, -6], [2, -6]], 'd')
>>> mtx1, mtx2, disparity = procrustes(a, b)
>>> round(disparity)
0.0

You can have both Ordinary Procrustes Analysis and Generalized Procrustes Analysis in python with something like this:
import numpy as np
def opa(a, b):
aT = a.mean(0)
bT = b.mean(0)
A = a - aT
B = b - bT
aS = np.sum(A * A)**.5
bS = np.sum(B * B)**.5
A /= aS
B /= bS
U, _, V = np.linalg.svd(np.dot(B.T, A))
aR = np.dot(U, V)
if np.linalg.det(aR) < 0:
V[1] *= -1
aR = np.dot(U, V)
aS = aS / bS
aT-= (bT.dot(aR) * aS)
aD = (np.sum((A - B.dot(aR))**2) / len(a))**.5
return aR, aS, aT, aD
def gpa(v, n=-1):
if n < 0:
p = avg(v)
else:
p = v[n]
l = len(v)
r, s, t, d = np.ndarray((4, l), object)
for i in range(l):
r[i], s[i], t[i], d[i] = opa(p, v[i])
return r, s, t, d
def avg(v):
v_= np.copy(v)
l = len(v_)
R, S, T = [list(np.zeros(l)) for _ in range(3)]
for i, j in np.ndindex(l, l):
r, s, t, _ = opa(v_[i], v_[j])
R[j] += np.arccos(min(1, max(-1, np.trace(r[:1])))) * np.sign(r[1][0])
S[j] += s
T[j] += t
for i in range(l):
a = R[i] / l
r = [np.cos(a), -np.sin(a)], [np.sin(a), np.cos(a)]
v_[i] = v_[i].dot(r) * (S[i] / l) + (T[i] / l)
return v_.mean(0)
For testing purposes, the output of each algorithm can be visualized as follows:
import matplotlib.pyplot as p; p.rcParams['toolbar'] = 'None';
def plt(o, e, b):
p.figure(figsize=(10, 10), dpi=72, facecolor='w').add_axes([0.05, 0.05, 0.9, 0.9], aspect='equal')
p.plot(0, 0, marker='x', mew=1, ms=10, c='g', zorder=2, clip_on=False)
p.gcf().canvas.set_window_title('%f' % e)
x = np.ravel(o[0].T[0])
y = np.ravel(o[0].T[1])
p.xlim(min(x), max(x))
p.ylim(min(y), max(y))
a = []
for i, j in np.ndindex(len(o), 2):
a.append(o[i].T[j])
O = p.plot(*a, marker='x', mew=1, ms=10, lw=.25, c='b', zorder=0, clip_on=False)
O[0].set(c='r', zorder=1)
if not b:
O[2].set_color('b')
O[2].set_alpha(0.4)
p.axis('off')
p.show()
# Fly wings example (Klingenberg, 2015 | https://en.wikipedia.org/wiki/Procrustes_analysis)
arr1 = np.array([[588.0, 443.0], [178.0, 443.0], [56.0, 436.0], [50.0, 376.0], [129.0, 360.0], [15.0, 342.0], [92.0, 293.0], [79.0, 269.0], [276.0, 295.0], [281.0, 331.0], [785.0, 260.0], [754.0, 174.0], [405.0, 233.0], [386.0, 167.0], [466.0, 59.0]])
arr2 = np.array([[477.0, 557.0], [130.129, 374.307], [52.0, 334.0], [67.662, 306.953], [111.916, 323.0], [55.119, 275.854], [107.935, 277.723], [101.899, 259.73], [175.0, 329.0], [171.0, 345.0], [589.0, 527.0], [591.0, 468.0], [299.0, 363.0], [306.0, 317.0], [406.0, 288.0]])
def opa_out(a):
r, s, t, d = opa(a[0], a[1])
a[1] = a[1].dot(r) * s + t
return a, d, False
plt(*opa_out([arr1, arr2, np.matrix.copy(arr2)]))
def gpa_out(a):
g = gpa(a, -1)
D = [avg(a)]
for i in range(len(a)):
D.append(a[i].dot(g[0][i]) * g[1][i] + g[2][i])
return D, sum(g[3])/len(a), True
plt(*gpa_out([arr1, arr2]))

Probably you want to try this package with various flavors of different Procrustes methods, https://github.com/theochem/procrustes.

root-finding algorithm for a complex polynomial equation in python

I am trying to solve for the following equation with a simple algorithm. I am not sure if the algorithm that I'm using is the best one or not but it is the only way that I could think of.
In this equation, everything other than P is known, and I am trying to solve for that. N is an array of counts, an i is the channel number. S(n) is the probability of having a certain n and C is binomial coefficient of (n, r). Pi is the probability in i channel and Pj is the probability in the previous channels with D distance to i. The code itself is not working but I believe that the main problem is in the way that I am trying to solve for it.
import numpy as np
import matplotlib.pyplot as plt
import math as ms
from scipy.misc import derivative
import scipy as sp
def next_guess(f, x):
slop = derivative(f, x, dx = 0.01)
return x - float(f(x))/slop
def my_newton(f, guess):
for i in range(30):
#print(guess)
guess = next_guess(f, guess)
return guess
def binomial(n, r):
dif = ms.factorial(n - r)
n = ms.factorial(n)
r = ms.factorial(r)
return (n/(r*dif))
def wrap_func(x, S = np.array([0.1, 0.5, 0.2, 0.1, 0.1]), D = 1, N = np.array([10, 15, 20, 1, 13])):
if type(x) == float:
z = np.zeros(1)
else:
z = np.zeros(x.shape[0])
N_tot = N.sum()
n_max = S.shape[0]
for i in range(z.shape[0]):
z[i] += my_newton(func(x, S = S, D = 1, N = N[i], n_max = n_max, N_tot = N_tot, i = i), i/100)
return z
def func(x, S = np.array([0.1, 0.5, 0.2, 0.1, 0.1]), D = 1, N = 0, n_max = 5, N_tot = 10, i = 0):
S_sum = 0
binom_sum = 0
y = 0
for n in range(n_max):
S_sum += S[n]
for r in range(n):
binom_sum += binomial(n, r)
y += S_sum * binom_sum * (x**r) * (1 - x - summ_x(x, D, i, S, N, n_max, N_tot))**(n-r)
return N_tot * y - N
def summ_x(x, D, i, S, N, n_max, N_tot):
j_min = max(i - D - 1, 0)
j_max = i - 1
x_values = 0
if i == 0:
return x_values
else:
for j in range(j_min, j_max):
x_values += func(x, S, D, N, n_max, N_tot, i)
return x_values
x = np.linspace(0, 1, 1000)
S = np.array([0.1, 0.5, 0.2, 0.1, 0.1])
N = np.random.choice(50, size = 1000)
print(my_newton(wrap_func, 0.1))
plt.plot(x, wrap_func(x, S = S, D = 1, N = N ))
plt.axhline(0, lw = 0.5, color = 'grey')
#plt.plot(my_newton(wrap_func, 1), wrap_func(my_newton(wrap_func, 1), S = S, D = 1, N = N), 'd')
plt.show()

How to calculate error in Polynomial Linear Regression?

I am trying to calculate the error rate of the training data I'm using.
I believe I'm calculating the error incorrectly. The formula is as shown:
y is calculated as shown:
I am calculating this in the function fitPoly(M) at line 49. I believe I am incorrectly calculating y(x(n)), but I don't know what else to do.
Below is the Minimal, Complete, and Verifiable example.
import numpy as np
import matplotlib.pyplot as plt
dataTrain = [[2.362761180904257019e-01, -4.108125266714775847e+00],
[4.324296163702689988e-01, -9.869308732049049127e+00],
[6.023323504115264404e-01, -6.684279243433971729e+00],
[3.305079685397107614e-01, -7.897042003779912278e+00],
[9.952423271981121200e-01, 3.710086310489402628e+00],
[8.308127402955634011e-02, 1.828266768673480147e+00],
[1.855495407116576345e-01, 1.039713135916495501e+00],
[7.088332047815845138e-01, -9.783208407540947560e-01],
[9.475723071629885697e-01, 1.137746192425550085e+01],
[2.343475721257285427e-01, 3.098019704040922750e+00],
[9.338350584099475160e-02, 2.316408265530458976e+00],
[2.107903139601833287e-01, -1.550451474833406396e+00],
[9.509966727520677843e-01, 9.295029459100994984e+00],
[7.164931165416982273e-01, 1.041025972594300075e+00],
[2.965557300301902011e-03, -1.060607693351102121e+01]]
def strip(L, xt):
ret = []
for i in L:
ret.append(i[xt])
return ret
x1 = strip(dataTrain, 0)
y1 = strip(dataTrain, 1)
# HELP HERE
def getY(m, w, D):
y = w[0]
y += np.sum(w[1:] * D[:m])
return y
# HELP ABOVE
def dataMatrix(X, M):
Z = []
for x in range(len(X)):
row = []
for m in range(M + 1):
row.append(X[x][0] ** m)
Z.append(row)
return Z
def fitPoly(M):
t = []
for i in dataTrain:
t.append(i[1])
w, _, _, _ = np.linalg.lstsq(dataMatrix(dataTrain, M), t)
w = w[::-1]
errTrain = np.sum(np.subtract(t, getY(M, w, x1)) ** 2)/len(x1)
print('errTrain: %s' % (errTrain))
return([w, errTrain])
#fitPoly(8)
def plotPoly(w):
plt.ylim(-15, 15)
x, y = zip(*dataTrain)
plt.plot(x, y, 'bo')
xw = np.arange(0, 1, .001)
yw = np.polyval(w, xw)
plt.plot(xw, yw, 'r')
#plotPoly(fitPoly(3)[0])
def bestPoly():
m = 0
plt.figure(1)
plt.xlim(0, 16)
plt.ylim(0, 250)
plt.xlabel('M')
plt.ylabel('Error')
plt.suptitle('Question 3: training and Test error')
while m < 16:
plt.figure(0)
plt.subplot(4, 4, m + 1)
plotPoly(fitPoly(m)[0])
plt.figure(1)
plt.plot(fitPoly(m)[1])
#plt.plot(fitPoly(m)[2])
m+= 1
plt.figure(3)
plt.xlabel('t')
plt.ylabel('x')
plt.suptitle('Question 3: best-fitting polynomial (degree = 8)')
plotPoly(fitPoly(8)[0])
print('Best M: %d\nBest w: %s\nTraining error: %s' % (8, fitPoly(8)[0], fitPoly(8)[1], ))
bestPoly()

Updated: This solution uses numpy's np.interp which will connect the points as a kind of "best fit". We then use your error function to find the difference between this interpolated line and the predicted y values for the degree of each polynomial.
import numpy as np
import matplotlib.pyplot as plt
import itertools
dataTrain = [
[2.362761180904257019e-01, -4.108125266714775847e+00],
[4.324296163702689988e-01, -9.869308732049049127e+00],
[6.023323504115264404e-01, -6.684279243433971729e+00],
[3.305079685397107614e-01, -7.897042003779912278e+00],
[9.952423271981121200e-01, 3.710086310489402628e+00],
[8.308127402955634011e-02, 1.828266768673480147e+00],
[1.855495407116576345e-01, 1.039713135916495501e+00],
[7.088332047815845138e-01, -9.783208407540947560e-01],
[9.475723071629885697e-01, 1.137746192425550085e+01],
[2.343475721257285427e-01, 3.098019704040922750e+00],
[9.338350584099475160e-02, 2.316408265530458976e+00],
[2.107903139601833287e-01, -1.550451474833406396e+00],
[9.509966727520677843e-01, 9.295029459100994984e+00],
[7.164931165416982273e-01, 1.041025972594300075e+00],
[2.965557300301902011e-03, -1.060607693351102121e+01]
]
data = np.array(dataTrain)
data = data[data[:, 0].argsort()]
X,y = data[:, 0], data[:, 1]
fig,ax = plt.subplots(4, 4)
indices = list(itertools.product([0,1,2,3], repeat=2))
for i,loc in enumerate(indices, start=1):
xx = np.linspace(X.min(), X.max(), 1000)
yy = np.interp(xx, X, y)
w = np.polyfit(X, y, i)
y_pred = np.polyval(w, xx)
ax[loc].scatter(X, y)
ax[loc].plot(xx, y_pred)
ax[loc].plot(xx, yy, 'r--')
error = np.square(yy - y_pred).sum() / X.shape[0]
print(error)
plt.show()
This prints out:
2092.19807848
1043.9400277
1166.94550318
252.238810889
225.798905379
155.785478366
125.662973726
143.787869281
6553.66570273
10805.6609259
15577.8686283
13536.1755299
108074.871771
213513916823.0
472673224393.0
1.01198058355e+12
Visually, it plots out this:
From here, it's just a matter of saving those errors to a list and finding the minimum.

I may contribute :
def pol_y(x, w):
y = 0; power = 0;
for i in w:
y += i*(x**power);
power += 1;
return y
The M is included implicitly because it is the final index of w. So if w = [0, 0, 1], then pol_y(x, w) is as same as f(x) = x^2.
If you want to map the 1st column of the dataTrain :
get_Y = [pol_y(i, w) for i in x1 ]
The error may be calculated by
vec_error = [(y1[i] - getY[i])**2 for i in range(0, len(y1)];
train_error = np.sum(vec_error)/len(y1);
Hope this helps.

Matplotlib contour plot of matrix type

I'm getting hung trying to create a contour plot of three [1, m] numpy matrices. When I try to plot:
plt.contour(reg.theta[0,:], reg.theta[1,:], reg.J[0,:])
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
I think this has something to do with the fact that
In[167]: shape(reg.theta[0,:])
Out[167]: (1, 15000)
and contour wants something along the lines of (15000,)
I've tried to create a meshgrid similar to how I would in matlab using
X , Y = meshgrid(reg.theta[0,:], reg.theta[1,:]
ValueError: total size of new array must be unchanged[1,:])
However, I have no idea how to interpret this.
Please advise stackoverflow!
Here is my complete code, tested on Python 2.7
from numpy import *
import matplotlib.pyplot as plt
class LinearRegression:
def __init__(self, data):
self.data_loc = data
self.max_it = 15000
self.theta = matrix(ones((2, self.max_it)))
self.alpha = .01
self.J = matrix(zeros((1, self.max_it)))
def importData(self):
Data = loadtxt(self.data_loc, delimiter = ',')
rawData = matrix(Data)
x = rawData[:,0]
y = rawData[:,1]
[m,n] = shape(x)
x_0 = matrix(ones((m,1)))
x = concatenate((x_0, x), axis = 1)
return x, y, m, n
def center(self, x):
x = x / mean(x)
return x
def scale(self, x):
x = x/ std(x)
return x
def funct(self, m, b):
return lambda x: m*x + b
def cost(self, x, y, m, mx, b):
f = self.funct(mx, b)
err = power( (f(x) - y), 2).sum()
err = err / (2*len(x))
return err
def gradientDesc(self, x, y, m):
for i in range(0, self.max_it -1):
error = (1.0/m) * transpose(x) * ((x * self.theta[:,i]) - y )
delta = self.theta[:,i] - self.alpha*error
self.J[0,i] = (self.cost( x, y, m, self.theta[1,i], self.theta[0,i]))
self.theta[:, i+1] = delta
print('Calculation Complete')
return self.theta, error
reg = LinearRegression('/users/michael/documents/machine_learning/ex1/ex1data1.txt')
x, y, m ,n = reg.importData()
theta, error = reg.gradientDesc(x,y,m)
print (reg.theta[:,reg.max_it -1])
XX = linspace(0,m,m)
J = reg.theta[0,reg.max_it-1] + reg.theta[1, reg.max_it-1] * XX
plt.plot(XX, J)
plt.scatter(x[:,1],y)`