I have to do a second degree interpolation using an already existing code and changing the values for mine, but for some reason when i go ahead and gragth the interpolation, the fuction suddently stops (it is not continuous). Can someone help me figuring out whats wrong? I believe it has something to do with line 43 (evaluation on new data points), but I am not sure.
Source code:
import numpy as np
import matplotlib.pyplot as plt
plt.style.use('seaborn-poster')
%matplotlib inline
def divided_diff(x, y):
'''
function to calculate the divided
differences table
'''
n = len(y)
coef = np.zeros([n, n])
# the first column is y
coef[:,0] = y
for j in range(1,n):
for i in range(n-j):
coef[i][j] = \
(coef[i+1][j-1] - coef[i][j-1]) / (x[i+j]-x[i])
return coef
def newton_poly(coef, x_data, x):
'''
evaluate the newton polynomial
at x
'''
n = len(x_data) - 1
p = coef[n]
for k in range(1,n+1):
p = coef[n-k] + (x -x_data[n-k])*p
return p
x = np.array([-5, -1, 0, 2])
y = np.array([-2, 6, 1, 3])
# get the divided difference coef
a_s = divided_diff(x, y)[0, :]
# evaluate on new data points
x_new = np.arange(-5, 2.1, .1)
y_new = newton_poly(a_s, x, x_new)
plt.figure(figsize = (12, 8))
plt.plot(x, y, 'bo')
plt.plot(x_new, y_new)
My code (adjusted for data points (0,0);(6.4,1.9);(10.6,4.3)):
import numpy as np
import matplotlib.pyplot as plt
plt.style.use('seaborn-poster')
%matplotlib inline
def divided_diff(x, y):
'''
function to calculate the divided
differences table
'''
n = len(y)
coef = np.zeros([n, n])
# the first column is y
coef[:,0] = y
for j in range(1,n):
for i in range(n-j):
coef[i][j] = \
(coef[i+1][j-1] - coef[i][j-1]) / (x[i+j]-x[i])
return coef
def newton_poly(coef, x_data, x):
'''
evaluate the newton polynomial
at x
'''
n = len(x_data) - 1
p = coef[n]
for k in range(1,n+1):
p = coef[n-k] + (x -x_data[n-k])*p
return p
x = np.array([0, 6.4, 10.6])
y = np.array([0, 1.9, 4.3])
# get the divided difference coef
a_s = divided_diff(x, y)[0, :]
# evaluate on new data points
x_new = np.arange(-5, 2.1, .1)
y_new = newton_poly(a_s, x, x_new)
plt.figure(figsize = (12, 8))
plt.plot(x, y, 'bo')
plt.plot(x_new, y_new)
Related
How would i generate 1000 2D samples for a Gaussian with mu = (2,0), and sigma = [[1,0],[0,1]]? This is my code below so far:
import numpy as np
import matplotlib.pyplot as plt
from matplotlib import cm
from mpl_toolkits.mplot3d import Axes3D
# Our 2-dimensional distribution will be over variables X and Y
N = 40
X = np.linspace(-2, 2, N)
Y = np.linspace(-2, 2, N)
X, Y = np.meshgrid(X, Y)
# Mean vector and covariance matrix
mu = np.array([2, 0])
Sigma = np.array([[ 1 , 0], [0, 1]])
# Pack X and Y into a single 3-dimensional array
pos = np.empty(X.shape + (2,))
pos[:, :, 0] = X
pos[:, :, 1] = Y
def multivariate_gaussian(pos, mu, Sigma):
"""Return the multivariate Gaussian distribution on array pos."""
n = mu.shape[0]
Sigma_det = np.linalg.det(Sigma)
Sigma_inv = np.linalg.inv(Sigma)
N = np.sqrt((2*np.pi)**n * Sigma_det)
# This einsum call calculates (x-mu)T.Sigma-1.(x-mu) in a vectorized
# way across all the input variables.
fac = np.einsum('...k,kl,...l->...', pos-mu, Sigma_inv, pos-mu)
return np.exp(-fac / 2) / N
# The distribution on the variables X, Y packed into pos.
Z = multivariate_gaussian(pos, mu, Sigma)
Does this look correct? How can I better this code?
I have a set of data and I have fit a normal distribution to the data using numpy and scipy. But when I try to plot the pdf, I get the following error:
I have tried to change the dtype of Z, but that did not work. Any suggestions will help. Thanks.
import matplotlib.pyplot as plt
from mpl_toolkits import mplot3d
import numpy as np
import random
from sympy.matrices import Matrix
from sympy import symbols, pprint, N
from scipy.stats import multivariate_normal
from target import true_target_trajectory, target_posiion
def plot_gaussian(X, Y, Z):
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_surface(X, Y, Z)
plt.show()
def covariance(x, y):
sigma1 = np.std(x, dtype=np.float64)
sigma2 = np.std(y, dtype=np.float64)
cov = np.matrix([[sigma1, sigma1*sigma2], [sigma1*sigma2, sigma2]])
min_eig = np.min(np.real(np.linalg.eigvals(cov)))
if min_eig < 0:
cov -= 10*min_eig * np.eye(*cov.shape)
return cov
def gaussian(x, mu, cov):
rv = multivariate_normal(mu, cov)
return rv.pdf(x)
#plot_gaussian()
vin = 300
qin = 9
x = []
y = []
time = np.linspace(0, 2*np.pi, 100)
for t in (time):
cc = target_posiion(vin, qin, t)
x.append(cc.T[0])
y.append(cc.T[1])
mu = np.array([np.mean(x), np.mean(y)])
cov = covariance(x, y)
X, Y = np.meshgrid(x, y)
pos = np.dstack((X, Y))
Z = gaussian(pos, mu, cov)
plot_gaussian(X, Y, Z)
I tried to reproduce the issue with x = np.linspace(-1, 3, 100) and y = np.linspace(0, 4, 100). But that did not give any error and i got the bell curve as expected.
So i am attaching the code for target position.
The code for target_position:
import random
import numpy as np
from sympy.vector.coordsysrect import CoordSys3D
from sympy.physics.mechanics import dynamicsymbols
from sympy import symbols, sin, pprint, Derivative, Identity, N
from sympy.matrices import Matrix, BlockMatrix, block_collapse
C = CoordSys3D('C')
i, j, k = C.base_vectors()
def evaluate_matrix(m, v_in, q_in, tk):
w, t = symbols('w t')
v0, q = symbols('v0 q')
params = {v0:v_in, q:q_in, t:tk}
return Matrix([[N(m[0].subs(params)), N(m[1].subs(params))]]).T
def true_target_trajectory(v_in, q_in, tk):
w, t = symbols('w t')
v0, q, A = symbols('v0 q A')
r, v, a, x, y = dynamicsymbols('r v a x y')
A = (v0**2)/q
w = q/(2*v0)
x = A*sin(w*t)*i
y = A*sin(2*w*t)*j
r = x + y
r_m = Matrix(r.to_matrix(C)[:2])
v = Derivative(r, t).doit()
v_m = Matrix(v.to_matrix(C)[:2])
a = Derivative(v, t).doit()
a_m = Matrix(a.to_matrix(C)[:2])
x_k = BlockMatrix([[r_m.T, v_m.T, a_m.T]]).T
I = Identity(2)
H = BlockMatrix([[I, I, I]])
z = evaluate_matrix(block_collapse(H*x_k), v_in, q_in, tk)
return z
def target_posiion(v_in, q_in, tk):
sigma = 50
u_k = Matrix([[random.gauss(0,1), random.gauss(0,1)]]).T
z = true_target_trajectory(v_in, q_in, tk)
z_c_k = z + sigma*u_k
return z_c_k
The problem
The problem is that your x and y are lists of type sympy.core.numbers.Float, not regular Python float. Numpy doesn't know how to convert Sympy numeric types, so meshgrid ends up returning X and Y arrays of dtype=object. Down the line, this ends up screwing up the call to ax.plot_surface.
The fix
Just convert x and y to standard Numpy arrays of np.float64 before you pass them into meshgrid:
X, Y = np.meshgrid(np.array(x).astype(float), np.array(y).astype(float))
Once you do that, everything should be fine. Here's the output:
Currently, I solve the following ODE system of equations using odeint
dx/dt = (-x + u)/2.0
dy/dt = (-y + x)/5.0
initial conditions: x = 0, y = 0
However, I would like to use solve_ivp which seems to be the recommended option for this type of problems, but honestly I don't know how to adapt the code...
Here is the code I'm using with odeint:
import numpy as np
from scipy.integrate import odeint, solve_ivp
import matplotlib.pyplot as plt
def model(z, t, u):
x = z[0]
y = z[1]
dxdt = (-x + u)/2.0
dydt = (-y + x)/5.0
dzdt = [dxdt, dydt]
return dzdt
def main():
# initial condition
z0 = [0, 0]
# number of time points
n = 401
# time points
t = np.linspace(0, 40, n)
# step input
u = np.zeros(n)
# change to 2.0 at time = 5.0
u[51:] = 2.0
# store solution
x = np.empty_like(t)
y = np.empty_like(t)
# record initial conditions
x[0] = z0[0]
y[0] = z0[1]
# solve ODE
for i in range(1, n):
# span for next time step
tspan = [t[i-1], t[i]]
# solve for next step
z = odeint(model, z0, tspan, args=(u[i],))
# store solution for plotting
x[i] = z[1][0]
y[i] = z[1][1]
# next initial condition
z0 = z[1]
# plot results
plt.plot(t,u,'g:',label='u(t)')
plt.plot(t,x,'b-',label='x(t)')
plt.plot(t,y,'r--',label='y(t)')
plt.ylabel('values')
plt.xlabel('time')
plt.legend(loc='best')
plt.show()
main()
It's important that solve_ivp expects f(t, z) as right-hand side of the ODE. If you don't want to change your ode function and also want to pass your parameter u, I recommend to define a wrapper function:
def model(z, t, u):
x = z[0]
y = z[1]
dxdt = (-x + u)/2.0
dydt = (-y + x)/5.0
dzdt = [dxdt, dydt]
return dzdt
def odefun(t, z):
if t < 5:
return model(z, t, 0)
else:
return model(z, t, 2)
Now it's easy to call solve_ivp:
def main():
# initial condition
z0 = [0, 0]
# number of time points
n = 401
# time points
t = np.linspace(0, 40, n)
# step input
u = np.zeros(n)
# change to 2.0 at time = 5.0
u[51:] = 2.0
res = solve_ivp(fun=odefun, t_span=[0, 40], y0=z0, t_eval=t)
x = res.y[0, :]
y = res.y[1, :]
# plot results
plt.plot(t,u,'g:',label='u(t)')
plt.plot(t,x,'b-',label='x(t)')
plt.plot(t,y,'r--',label='y(t)')
plt.ylabel('values')
plt.xlabel('time')
plt.legend(loc='best')
plt.show()
main()
Note that without passing t_eval=t, the solver will automatically choose the time points inside tspan at which the solution will be stored.
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.
I am building a SVM in python. However, my implemtation is generating the wrong plane. I think it has something to do with my parameters(langrage multipliers) being so small but I am not sure. I think I am doing the convex optimization right. Maybe my data isn't in the right format. I based my code on theses tutorials:http://tullo.ch/articles/svm-py/ and http://www.mblondel.org/journal/2010/09/19/support-vector-machines-in-python/.
Here is my code and output:
import numpy
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import cvxopt
from matplotlib import cm
from cvxopt import matrix, solvers
from itertools import izip
#http://www.tristanfletcher.co.uk/SVM%20Explained.pdf
class SVM:
def __init__(self,X,y):
self.X = X
self.y = y
def findParameters(self,X,y):
# min 1/2 x^T P x + q^T x
#Ax = b
#y's are answer vectors
#put in cvxopt
#
"""P = cvxopt.matrix(np.outer(self.y,self.y)* self.gramMatrix())
q = cvxopt.matrix((numpy.ones(len(self.y))).T)
#G =
#h =
limits = np.asarray(self.y)
A = cvxopt.matrix(limits.T)
#genrates matrix of zzeros
b = cvxopt.matrix(numpy.zeros(len(self.y)))
# actually comp
param = cvxopt.solvers.qp(P,q,G,h,A,b);"""
n_samples, n_features = X.shape
K = self.gramMatrix(X)
P = cvxopt.matrix(np.outer(y, y) * K)
q = cvxopt.matrix(-1 * np.ones(n_samples))
Gtry = cvxopt.matrix(np.diag(np.ones(n_samples) * -1))
htry = cvxopt.matrix(np.zeros(n_samples))
A = cvxopt.matrix(y, (1, n_samples))
b = cvxopt.matrix(0.0)
param = cvxopt.solvers.qp(P, q, Gtry, htry, A, b)
array = param['x']
return array
def WB_calculator(self,X,y):
#calculates w vector
yi = self.y
X = np.asarray(X)
y = np.asarray(y)
important = self.findParameters(X,y)
print("these are parameters")
print(important)
firstsum = [0 for x in range(0,len(y))]
for point in range(0,len(important)):
liste = X[point]*important[point]*yi[point]
firstsum = [x + y for x, y in zip(firstsum,liste)]
#this part calculates bias
#this is a very naive implementation of bias
#xstuff is the x_coordinate vector we find this by transpose
b = 0
for i in range(0,len(important)):
b = b+ (yi[i]- np.dot(firstsum,X[i]))
avgB = b/len(important)
answer = (firstsum , avgB)
print("w vector")
print(firstsum)
return answer
def polynomialK(self,u,v,b):
return (np.dot(u,v)+b)**2
#Guassian Kernal Funciton
def gaussianK(self,v1, v2, sigma):
return np.exp(-norm(v1-v2, 2)**2/(2.*sigma**2))
#computes the gramMatrix given a set of all points included in the data
#this is basicly a matrix of dot prodducts
def gramMatrix(self,X):
gramMatrix = []
data = np.asarray(self.X)
dataTran = data
#print(dataTran)
for x in dataTran:
row = []
#print(row)
for y in dataTran:
row.append(np.dot(x,y))
gramMatrix.append(row)
#print(row)
return gramMatrix
def determineAcceptance(self,point,X,y):
# I'm not sure if this is the proper bounding lets checl
cutoff = self.WB_calculator(X,y)
if(np.dot(cutoff[0],point)+cutoff[1] >0):
print("You got in")
elif(np.dot(cutoff[0],point)+cutoff[1]<0):
print("Study")
# plots plane and points
def Graph(self,X,y):
important_stuff = self.WB_calculator(X,y)
weights = important_stuff[0]
c = important_stuff[1]
#here we actaually graph the functionb
graphable = X.T
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
xs = graphable[0]
ys = graphable[1]
zs = graphable[2]
colors = self.y
ax.scatter(xs,ys,zs,c=colors)
ax.set_xlabel("A")
ax.set_ylabel("B")
ax.set_zlabel("C")
#this changes orientation and look of surface
ax.view_init(azim = 180+40,elev = 22)
X = np.arange(-2, 2, 0.25)
Y = np.arange(-2, 2, 0.25)
X, Y = np.meshgrid(X, Y)
Z = ((-weights[0]*X + -weights[1]*Y - c)/(weights[2]))
surf = ax.plot_surface(X, Y, Z, rstride=1, cstride=1, cmap=cm.coolwarm,
linewidth=0, antialiased=True)
plt.show()
#list of points to test
a = [[-.1,-.1,-.1],[-.2,-.2,-.2],[.15,.15,.15],[.9,.9,.9],[.95,.95,.95]]
check = np.asarray(a)
b = [.01,.01,.01,1,1]
bigger =np.asarray(b)
d = SVM(a,b)
print(d.gramMatrix(check)[0])
print("parameters ya")
print(d.findParameters(check,bigger))
print(d.WB_calculator(check,bigger))
d.Graph(check,bigger)
d.determineAcceptance([.01,.01,.01],check,bigger)