I am trying to plot the error of this algorithm against h and I have run into a problem, for this error calculation, it cant use the first value, as it divides 0/0. How do I go about ignoring the first value where x =0? I basically need to start the summation on i=2 on line 46 (the absolute error one). Any help is much appreciated
import numpy
import matplotlib.pyplot as pyplot
from scipy.optimize import fsolve
from matplotlib import rcParams
rcParams['font.family'] = 'serif'
rcParams['font.size'] = 16
rcParams['figure.figsize'] = (12,6)
printing = False
def rk3(A, bvector, y0, interval, N):
h = (interval[1] - interval[0]) / N
x = numpy.linspace(interval[0], interval[1], N+1)
y = numpy.zeros((len(y0), N+1))
y[:, 0] = y0
b = bvector
for i in range(N):
y_1 = y[:, i] + h *(numpy.dot(A, y[:, i]) + b(x[i]))
y_2= (3/4)*y[:, i] + 0.25*y_1+0.25* h* (numpy.dot(A,y_1)+b(x[i]+h))
y[:, i+1] = (1/3)*y[:, i] + (2/3)*y_2 + (2/3)*h*(numpy.dot(A,y_2)+b(x[i]+h))
return x, y
def exact( interval, N):
w = numpy.linspace(interval[0], interval[1], N+1)
z = numpy.array([numpy.exp(-1000*w),(1000/999)*(numpy.exp(-w)-numpy.exp(-1000*w))])
return w, z
A=numpy.array([[-1000,0],[1000,-1]])
def bvector(x):
return numpy.zeros(2)
y0=numpy.array([1,0])
interval=numpy.array([0,0.1])
N=numpy.arange(40,401,40)
h=numpy.zeros(len(N))
abs_err = numpy.zeros(len(N))
for i in range(len(N)):
interval=numpy.array([0,0.1])
h[i]=(interval[1] - interval[0]) / N[i]
x, y = rk3(A, bvector, y0, interval, N[i])
w,z=exact(interval,N[i])
abs_err[i] = h[i]*numpy.sum(numpy.abs((y[1,:]-z[1,:])/z[1,:]))
p = numpy.polyfit(numpy.log(h), numpy.log(abs_err),1)
fig = pyplot.figure(figsize = (12, 8), dpi = 50)
pyplot.loglog(h, abs_err, 'kx')
pyplot.loglog(h, numpy.exp(p[1]) * h**(p[0]), 'b-')
pyplot.xlabel('$h$', size = 16)
pyplot.ylabel('$|$Error$|$', size = 16)
pyplot.show()
Simply add an if for the value which is zero. so for example if the dividing variable is x.
if x>0:
#code here for the calculation
The above code will use all positive non-zero value. to only skip zero use this
if x!=0:
You can also us the three arguments of a for loop:
for a in range(start_value, end_value, increment):
so this means
for a in range(2,10,2):
print a
will give you the below result
2
4
6
8
Related
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 try to use Lagrange multiplier to optimize a function, and I am trying to loop through the function to get a list of number, however I got the error
ValueError: setting an array element with a sequence.
Here is my code, where do I go wrong? If the n is not an array I can get the result correctly though
import numpy as np
from scipy.optimize import fsolve
n = np.arange(10000,100000,10000)
def func(X):
x = X[0]
y = X[1]
L = X[2]
return (x + y + L * (x**2 + y**2 - n))
def dfunc(X):
dLambda = np.zeros(len(X))
h = 1e-3
for i in range(len(X)):
dX = np.zeros(len(X))
dX[i] = h
dLambda[i] = (func(X+dX)-func(X-dX))/(2*h);
return dLambda
X1 = fsolve(dfunc, [1, 1, 0])
print (X1)
Helps would be appreciated, thank you very much
First, check func = fsolve()
Second, print(func([1,1,0]))` - result in not number ([2 2 2 2 2 2 2 2 2]), beause "n" is list. if you want to iterate n try:
import numpy as np
from scipy.optimize import fsolve
n = np.arange(10000,100000,10000)
def func(X,n):
x = X[0]
y = X[1]
L = X[2]
return (x + y + L * (x**2 + y**2 - n))
def dfunc(X,n):
dLambda = np.zeros(len(X))
h = 1e-3
r = 0
for i in range(len(X)):
dX = np.zeros(len(X))
dX[i] = h
dLambda[i] = (func(X+dX,n)-func(X-dX,n))/(2*h)
return dLambda
for iter_n in n:
print("for n = {0} dfunc = {1}".format(iter_n,dfunc([0.8,0.4,0.3],iter_n)))
I implemented logistic regression and use it on a data set. (This is an exercise in Coursera's ML course Week #3 (which normally uses matlab and octave) using python (so this isn't cheating)).
I started with the implementation in sklearn to classify the data set used in week three of this course (http://pastie.org/10872959). Here is a small, reproducible example for anyone to try out what I used (it relies only on numpy and sklearn):
It takes the data set, splits it into the feature matrix and the output matrix, and then constructs 26 more features from the original 2 (i.e from
). I then use logistic regression in sklearn, but this does not give the contour plot desired (please see below).
from sklearn.linear_model import LogisticRegression as expit
import numpy as np
def thetaFunc(y, theta, x):
deg = 6
spot = 0
sum = 0
for i in range(1, deg + 1):
for j in range(i + 1):
sum += theta[spot] * x**(i - j) * y**(j)
spot += 1
return sum
def constructVariations(X, deg):
features = np.zeros((len(X), 27))
spot = 0
for i in range(1, deg + 1):
for j in range(i + 1):
features[:, spot] = X[:,0]**(i - j) * X[:,1]**(j)
spot += 1
return features
if __name__ == '__main__':
data = np.loadtxt("ex2points.txt", delimiter = ",")
X,Y = np.split(data, [len(data[0,:]) - 1], 1)
X = reg.constructVariations(X, 6)
oneArray = np.ones((len(X),1))
X = np.hstack((oneArray, X))
trial = expit(solver = 'sag')
trial = trial.fit(X = X,y = np.ravel(Y))
print(trial.coef_)
# everything below has been edited in
from matplotlib import pyplot as plt
txt = open("RegLogTheta", "r").read()
txt = txt.split()
theta = np.array(txt, float)
x = np.linspace(-1, 1.5, 100)
y = np.linspace(-1,1.5,100)
z = np.empty((100,100))
xx,yy = np.meshgrid(x,y)
for i in range(len(x)):
for j in range(len(y)):
z[i][j] = thetaFunc(yy[i][j], theta, xx[i][j])
plt.contour(xx,yy,z, levels = [0])
plt.show()
Here are the coefficients of the generic feature terms.
http://pastie.org/10872957 (i.e the coefficients to terms
and the contour it generates:
One potential source of error is that I'm misinterpreting the 7 X 4 matrix coefficient matrix stored in trial._coeff. I believe that these 28 values are the coefficients to the 28 "variations" above, and I've mapped the coefficients to the variations both column-wise and row-wise. By column-wise, I mean that [:][0] get mapped to the first 7 variations, [:][1] to the next 7 and so on, and my function constructVariations explains how the variations are systematically created. Now the API maintains than an array of shape (n_classes, n_features) is stored in trial._coeff, so should I infer that fit classified the data into four classes? Or have I run through this problem poorly in another way?
Update
My interpretation (and/or use) of the weights must be at fault:
Instead of relying on the prediction built into sklearn, I myself tried to calculate the values that set the following to 1/2
The values of theta are those found from printing trial._coeff and x and y are scalars. Those x,y are then plotted to give the contour.
The code I used (but did not originally add in) attempts to do this. What is wrong with the math behind it?
One potential source of error is that I'm misinterpreting the 7 X 4 matrix coefficient matrix stored in trial._coeff
This matrix is not 7x4, it is 1x28 (check print(trial.coef_.shape)). One coefficient for each of your 28 features (27 returned by constructVariations and 1 added manually).
so should I infer that fit classified the data into four classes?
No, you missinterpreted the array, it has a single row (for binary classificaation there is no point in having two).
Or have I run through this problem poorly in another way?
Code is fine, interpretation not. In particular, see actual decision boundary from your model (plotted by calling "predict" and plotting contour)
from sklearn.linear_model import LogisticRegression as expit
import numpy as np
def constructVariations(X, deg):
features = np.zeros((len(X), 27))
spot = 0
for i in range(1, deg + 1):
for j in range(i + 1):
features[:, spot] = X[:,0]**(i - j) * X[:,1]**(j)
spot += 1
return features
if __name__ == '__main__':
data = np.loadtxt("ex2points.txt", delimiter = ",")
X,Y = np.split(data, [len(data[0,:]) - 1], 1)
rawX = np.copy(X)
X = constructVariations(X, 6)
oneArray = np.ones((len(X),1))
X = np.hstack((oneArray, X))
trial = expit(solver = 'sag')
trial = trial.fit(X = X,y = np.ravel(Y))
print(trial.coef_)
from matplotlib import pyplot as plt
h = 0.01
x_min, x_max = rawX[:, 0].min() - 1, rawX[:, 0].max() + 1
y_min, y_max = rawX[:, 1].min() - 1, rawX[:, 1].max() + 1
xx, yy = np.meshgrid(np.arange(x_min, x_max, h),
np.arange(y_min, y_max, h))
data = constructVariations(np.c_[xx.ravel(), yy.ravel()], 6)
oneArray = np.ones((len(data),1))
data = np.hstack((oneArray, data))
Z = trial.predict(data)
Z = Z.reshape(xx.shape)
plt.figure()
plt.scatter(rawX[:, 0], rawX[:, 1], c=Y, linewidth=0, s=50)
plt.contourf(xx, yy, Z, cmap=plt.cm.Paired, alpha=0.8)
plt.show()
Update
In the code provided you forgot (in visualization) that you added column of "1"s to your data representation, thus your thetas are one "off", as theta[0] is a bias, theta1 is related to your 0'th variable etc.
def thetaFunc(y, theta, x):
deg = 6
spot = 0
sum = theta[spot]
spot += 1
for i in range(1, deg + 1):
for j in range(i + 1):
sum += theta[spot] * x**(i - j) * y**(j)
spot += 1
return sum
you also forgot about intercept term from logisticregression itself, thus
xx,yy = np.meshgrid(x,y)
for i in range(len(x)):
for j in range(len(y)):
z[i][j] = thetaFunc(yy[i][j], theta, xx[i][j])
z -= trial.intercept_
(image generated using fixed code of yours)
import numpy as np
from sklearn.linear_model import LogisticRegression as expit
def thetaFunc(y, theta, x):
deg = 6
spot = 0
sum = theta[spot]
spot += 1
for i in range(1, deg + 1):
for j in range(i + 1):
sum += theta[spot] * x**(i - j) * y**(j)
spot += 1
return np.exp(-sum)
def constructVariations(X, deg):
features = np.zeros((len(X), 27))
spot = 0
for i in range(1, deg + 1):
for j in range(i + 1):
features[:, spot] = X[:,0]**(i - j) * X[:,1]**(j)
spot += 1
return features
if __name__ == '__main__':
data = np.loadtxt("ex2points.txt", delimiter = ",")
X,Y = np.split(data, [len(data[0,:]) - 1], 1)
X = constructVariations(X, 6)
rawX = np.copy(X)
oneArray = np.ones((len(X),1))
X = np.hstack((oneArray, X))
trial = expit(solver = 'sag')
trial = trial.fit(X = X,y = np.ravel(Y))
from matplotlib import pyplot as plt
theta = trial.coef_.ravel()
x = np.linspace(-1, 1.5, 100)
y = np.linspace(-1,1.5,100)
z = np.empty((100,100))
xx,yy = np.meshgrid(x,y)
for i in range(len(x)):
for j in range(len(y)):
z[i][j] = thetaFunc(yy[i][j], theta, xx[i][j])
z -= trial.intercept_
plt.contour(xx,yy,z > 1,cmap=plt.cm.Paired, alpha=0.8)
plt.scatter(rawX[:, 0], rawX[:, 1], c=Y, linewidth=0, s=50)
plt.show()
I'm required to take a Python module for my course and I get this error for my script. It's plotting the trajectory of a projectile and calculating a few other variables. I've typed the script exactly as in the booklet we are given.
Because I am an absolute beginner I can't understand other answers to this error. I would appreciate it an awful lot if someone could give me a quick fix, I don't have time at the moment to learn enough to fix it myself.
Code:
import matplotlib.pyplot as plt
import numpy as np
import math # need math module for trigonometric functions
g = 9.81 #gravitational constant
dt = 1e-3 #integration time step (delta t)
v0 = 40 # initial speed at t = 0
angle = math.pi/4 #math.pi = 3.14, launch angle in radians
time = np.arange(0,10,dt) #time axis
vx0 = math.cos(angle)*v0 # starting velocity along x axis
vy0 = math.sin(angle)*v0 # starting velocity along y axis
xa = vx0*time # compute x coordinates
ya = -0.5*g*time**2 + vy0*time # compute y coordinates
fig1 = plt.figure()
plt.plot(xa, ya) # plot y versus x
plt.xlabel ("x")
plt.ylabel ("y")
plt.ylim(0, 50)
plt.show()
def traj(angle, v0): # function for trajectory
vx0 = math.cos(angle) * v0 # for some launch angle and starting velocity
vy0 = math.sin(angle) * v0 # compute x and y component of starting velocity
x = np.zeros(len(time)) #initialise x and y arrays
y = np.zeros(len(time))
x[0], y[0], 0 #projecitle starts at 0,0
x[1], y[1] = x[0] + vx0 * dt, y[0] + vy0 * dt # second elements of x and
# y are determined by initial
# velocity
i = 1
while y[i] >= 0: # conditional loop continuous until
# projectile hits ground
x[i+1] = (2 * x[i] - x[i - 1]) # numerical integration to find x[i + 1]
y[i+1] = (2 * y[i] - y[i - 1]) - g * dt ** 2 # and y[i + 1]
i = [i + 1] # increment i for next loop
x = x[0:i+1] # truncate x and y arrays
y = y[0:i+1]
return x, y, (dt*i), x[i] # return x, y, flight time, range of projectile
x, y, duration, distance = traj(angle, v0)
print "Distance:" ,distance
print "Duration:" ,duration
n = 5
angles = np.linspace(0, math.pi/2, n)
maxrange = np.zeros(n)
for i in range(n):
x,y, duration, maxrange [i] = traj(angles[i], v0)
angles = angles/2/math.pi*360 #convert rad to degress
print "Optimum angle:", angles[np.where(maxrange==np.max(maxrange))]
The error explicitly:
File "C:/Users/***** at *****", line 52, in traj
x = x[0:i+1] # truncate x and y arrays
TypeError: can only concatenate list (not "int") to list
As is pointed out in the comments, this is the offending line
i = [i + 1] # increment i for next loop
Here, i is not actually being incremented as the comment suggests. When i is 1, it's being set to [1 + 1], which evaluates to [2], the list containing only the number 2. Remove the brackets.