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In this case, there are 3 ODE's that describe a SIR model. The issue comes in I want to calculate which beta and gamma values are the best to fit onto the datapoints from the x_axis and y_axisvalues. The method I'm currently using is to integrate the ODE's using odeintfrom the scipy library and the curve_fit method also from the same library. In this case, how would you calculate the values for beta and gamma to fit the datapoints?
P.S. the current error is this: ValueError: operands could not be broadcast together with shapes (3,) (14,)
#initial values
S_I_R = (0.762/763, 1/763, 0)
x_axis = [m for m in range(1,15)]
y_axis = [3,8,28,75,221,291,255,235,190,125,70,28,12,5]
# ODE's that describe the system
def equation(SIR_Values,t,beta,gamma):
Array = np.zeros((3))
SIR = SIR_Values
Array[0] = -beta * SIR[0] * SIR[1]
Array[1] = beta * SIR[0] * SIR[1] - gamma * SIR[1]
Array[2] = gamma * SIR[1]
return Array
# Results = spi.odeint(equation,S_I_R,time)
#fitting the values
beta_values,gamma_values = curve_fit(equation, x_axis,y_axis)
# Starting values
S0 = 762/763
I0 = 1/763
R0 = 0
x_axis = np.array([m for m in range(0,15)])
y_axis = np.array([1,3,8,28,75,221,291,255,235,190,125,70,28,12,5])
y_axis = np.divide(y_axis,763)
def sir_model(y, x, beta, gamma):
S = -beta * y[0] * y[1]
R = gamma * y[1]
I = beta * y[0] * y[1] - gamma * y[1]
return S, I, R
def fit_odeint(x, beta, gamma):
return spi.odeint(sir_model, (S0, I0, R0), x, args=(beta, gamma))[:,1]
popt, pcov = curve_fit(fit_odeint, x_axis, y_axis)
beta,gamma = popt
fitted = fit_odeint(x_axis,*popt)
plt.plot(x_axis,y_axis, 'o', label = "infected per day")
plt.plot(x_axis, fitted, label = "fitted graph")
plt.xlabel("Time (in days)")
plt.ylabel("Fraction of infected")
plt.title("Fitted beta and gamma values")
plt.legend()
plt.show()
As this example from scipy documentation, the function must output an array with the same size as x_axis and y_axis.
When I use this code for Single variable Linear regression, the theta is being evaluated correctly but when on the multi variable it is giving weird output for theta.
I am trying to convert my octave code, that I wrote when I took Andrew Ng's course.
This is the main calling file:
m = data.shape[0]
a = np.array(data[0])
a.shape = (m,1)
b = np.array(data[1])
b.shape = (m, 1)
x = np.append(a, b, axis=1)
y = np.array(data[2])
lr = LR.LinearRegression()
[X, mu, sigma] = lr.featureNormalize(x)
z = np.ones((m, 1), dtype=float)
X = np.append(z, X, axis=1)
alpha = 0.01
num_iters = 400
theta = np.zeros(shape=(3,1))
[theta, J_history] = lr.gradientDescent(X, y, theta, alpha, num_iters)
print(theta)
And here are the contents of class :
class LinearRegression:
def featureNormalize(self, data):#this normalizes the features
data = np.array(data)
x_norm = data
mu = np.zeros(shape=(1, data.shape[1]))#creates mu vector filled with zeros
sigma = np.zeros(shape=(1, data.shape[1]))
for i in range(0, data.shape[1]):
mu[0, i] = np.mean(data[:, i])
sigma[0, i] = np.std(data[:, i])
for i in range(0, data.shape[1]):
x_norm[:, i] = np.subtract(x_norm[:, i], mu[0, i])
x_norm[:, i] = np.divide(x_norm[:, i], sigma[0, i])
return [x_norm, mu, sigma]
def gradientDescent(self, X, y, theta, alpha, num_iters):
m = y.shape[0]
J_history = np.zeros(shape=(num_iters, 1))
for i in range(0, num_iters):
predictions = X.dot(theta) # X is 47*3 theta is 3*1 predictions is 47*1
theta = np.subtract(theta , (alpha / m) * np.transpose((np.transpose(np.subtract(predictions ,y))).dot(X))) #1*97 into 97*3
J_history[i] = self.computeCost(X, y, theta)
return [theta, J_history]
def computeCost(self, X, y, theta):
warnings.filterwarnings('ignore')
m = X.shape[0]
J = 0
predictions = X.dot(theta)
sqrErrors = np.power(predictions - y, 2)
J = 1 / (2 * m) * np.sum(sqrErrors)
return J
I expected a theta that'll be a 3*1 matrix. According to Andrew's course my octave implementation was producing a theta
334302.063993
100087.116006
3673.548451
But in python implementation I am getting very weird output:
[[384596.12996714 317274.97693463 354878.64955708 223121.53576488
519238.43603216 288423.05420641 302849.01557052 191383.45903309
203886.92061274 233219.70871976 230814.42009498 333720.57288972
317370.18827964 673115.35724932 249953.82390212 432682.6678475
288423.05420641 192249.97844569 480863.45534211 576076.72380674
243221.70859887 245241.34318985 233604.4010228 249953.82390212
551937.2817908 240336.51632605 446723.93690857 451051.7253178
456822.10986344 288423.05420641 336509.59208678 163398.05571747
302849.01557052 557707.6...................... this goes on for long
The same code is working absolutely fine in Single Variable dataset. It is also working fine in the octave but seems like I am missing some point for 2+ hours now. Happy to get your help.
Try in gradientDescent the following second line of the for loop:
theta=theta-(alpha/m)*X.T.dot(X.dot(theta)-y)
Also, if you want to add a column of ones, it is easier to do like so:
np.c_[np.ones((m,1)),data]
For the past few days, I have been trying to code this application of Gradient Descent for my final-year project in Mechanical Engineering. https://drive.google.com/open?id=1tIGqZ2Lb0sN4GEpgYEZLFvtmhigXnot0 The HTML file is attached above. Just download the file, and if you see the results. There are only 3 values in theta, whereas x has 3 independent variables. So it should have 4 values in theta.
The code is as follows. For the result, it is theta [-0.03312393 0.94409351 0.99853041]
import numpy as np
import random
import pandas as pd
def gradientDescent(x, y, theta, alpha, m, numIterations):
xTrans = x.transpose()
for i in range(0, numIterations):
hypothesis = np.dot(x, theta)
loss = hypothesis - y
# avg cost per example (the 2 in 2*m doesn't really matter here.
# But to be consistent with the gradient, I include it)
cost = np.sum(loss ** 2) / (2 * m)
print("Iteration %d | Cost: %f" % (i, cost))
# avg gradient per example
gradient = np.dot(xTrans, loss) / m
# update
theta = theta - alpha * gradient
return theta
df = pd.read_csv(r'C:\Users\WELCOME\Desktop\FinalYearPaper\ConferencePaper\NewTrain.csv', 'rU', delimiter=",",header=None)
x = df.loc[:,'0':'2']
y = df[3]
print (x)
m, n = np.shape(x)
numIterations= 200
alpha = 0.000001
theta = np.ones(n)
theta = gradientDescent(x, y, theta, alpha, m, numIterations)
print(theta)
I'm trying to implement in Python the first exercise of Andrew NG's Coursera Machine Learning course. In the course the exercise is with Matlab/Octave, but I wanted to implement it in Python as well.
The problem is that the line that updates theta values, does not seem to be working right, is returning values [[0.72088159] [0.72088159]] but should return [[-3.630291] [1.166362]]
I'm using a learning rate of 0.01 and the gradient loop was set to 1500 (the same values from the original exercise in Octave).
And obviously, with these wrong values for theta, the predictions are not correct as shown in the last chart.
In the rows in which I tesyo the cost function with theta values defined as [0; 0] and [-1; 2], the results are correct (the same as the exercise in Octave), so the error can only be in the function of the gradient, but I do not know what went wrong.
I wanted someone to help me figure out what I'm doing wrong. I'm grateful already.
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
def load_data():
X = np.genfromtxt('data.txt', usecols=(0), delimiter=',', dtype=None)
y = np.genfromtxt('data.txt', usecols=(1), delimiter=',', dtype=None)
X = X.reshape(1, X.shape[0])
y = y.reshape(1, y.shape[0])
ones = np.ones(X.shape)
X = np.append(ones, X, axis=0)
theta = np.zeros((2, 1))
return (X, y, theta)
alpha = 0.01
iter_num = 1500
debug_at_loop = 10
def plot(x, y, y_hat=None):
x = x.reshape(x.shape[0], 1)
plt.xlabel('x')
plt.ylabel('hΘ(x)')
plt.ylim(ymax = 25, ymin = -5)
plt.xlim(xmax = 25, xmin = 5)
plt.scatter(x, y)
if type(y_hat) is np.ndarray:
plt.plot(x, y_hat, '-')
plt.show()
plot(X[1], y)
def hip(X, theta):
return np.dot(theta.T, X)
def cost(X, y, theta):
m = y.shape[1]
return np.sum(np.square(hip(X, theta) - y)) / (2 * m)
print('With theta = [0 ; 0]')
print('Cost computed =', cost(X, y, np.array([0, 0])))
print()
print('With theta = [-1 ; 2]')
print('Cost computed =', cost(X, y, np.array([-1, 2])))
def grad(X, y, alpha, theta, iter_num=1500, debug_cost_at_each=10):
J = []
m = y.shape[1]
for i in range(iter_num):
theta -= ((alpha * 1) / m) * np.sum(np.dot(hip(X, theta) - y, X.T))
if i % debug_cost_at_each == 0:
J.append(round(cost(X, y, theta), 6))
return J, theta
X, y, theta = load_data()
J, fit_theta = grad(X, y, alpha, theta)
print('Theta found by Gradient Descent:', fit_theta)
# Predict values for population sizes of 35,000 and 70,000
predict1 = np.dot(np.array([[1], [3.5]]).T, fit_theta);
print('For population = 35,000, we predict a profit of \n', predict1 * 10000);
predict2 = np.dot(np.array([[1], [7]]).T, fit_theta);
print('For population = 70,000, we predict a profit of \n', predict2 * 10000);
pred_y = hip(X, fit_theta)
plot(X[1], y, pred_y.T)
The data I'm using is the following txt:
6.1101,17.592
5.5277,9.1302
8.5186,13.662
7.0032,11.854
5.8598,6.8233
8.3829,11.886
7.4764,4.3483
8.5781,12
6.4862,6.5987
5.0546,3.8166
5.7107,3.2522
14.164,15.505
5.734,3.1551
8.4084,7.2258
5.6407,0.71618
5.3794,3.5129
6.3654,5.3048
5.1301,0.56077
6.4296,3.6518
7.0708,5.3893
6.1891,3.1386
20.27,21.767
5.4901,4.263
6.3261,5.1875
5.5649,3.0825
18.945,22.638
12.828,13.501
10.957,7.0467
13.176,14.692
22.203,24.147
5.2524,-1.22
6.5894,5.9966
9.2482,12.134
5.8918,1.8495
8.2111,6.5426
7.9334,4.5623
8.0959,4.1164
5.6063,3.3928
12.836,10.117
6.3534,5.4974
5.4069,0.55657
6.8825,3.9115
11.708,5.3854
5.7737,2.4406
7.8247,6.7318
7.0931,1.0463
5.0702,5.1337
5.8014,1.844
11.7,8.0043
5.5416,1.0179
7.5402,6.7504
5.3077,1.8396
7.4239,4.2885
7.6031,4.9981
6.3328,1.4233
6.3589,-1.4211
6.2742,2.4756
5.6397,4.6042
9.3102,3.9624
9.4536,5.4141
8.8254,5.1694
5.1793,-0.74279
21.279,17.929
14.908,12.054
18.959,17.054
7.2182,4.8852
8.2951,5.7442
10.236,7.7754
5.4994,1.0173
20.341,20.992
10.136,6.6799
7.3345,4.0259
6.0062,1.2784
7.2259,3.3411
5.0269,-2.6807
6.5479,0.29678
7.5386,3.8845
5.0365,5.7014
10.274,6.7526
5.1077,2.0576
5.7292,0.47953
5.1884,0.20421
6.3557,0.67861
9.7687,7.5435
6.5159,5.3436
8.5172,4.2415
9.1802,6.7981
6.002,0.92695
5.5204,0.152
5.0594,2.8214
5.7077,1.8451
7.6366,4.2959
5.8707,7.2029
5.3054,1.9869
8.2934,0.14454
13.394,9.0551
5.4369,0.61705
Well, I got it after losing several strands of hair (the programming will still leave me bald).
It was on the gradient line, and the solution was this:
theta -= ((alpha * 1) / m) * np.dot(X, (hip(X, theta) - y).T)
I changed the place of X and transposed the error vector.
I am trying to mimic the gradient descent algorithm for linear regression from Andrew NG's Machine learning course to Python, but for some reason my implementation is not working correctly.
Here's my implementation in Octave, it works correctly:
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
J_history = zeros(num_iters, 1);
for iter = 1:num_iters
prediction = X*theta;
margin_error = prediction - y;
gradient = 1/m * (alpha * (X' * margin_error));
theta = theta - gradient;
J_history(iter) = computeCost(X, y, theta);
end
end
However, when I translate this to Python for some reason it is not giving me accurate results. The cost seems to be going up rather than descending.
Here's my implementation in Python:
def gradientDescent(x, y, theta, alpha, iters):
m = len(y)
J_history = np.matrix(np.zeros((iters,1)))
for i in range(iters):
prediction = x*theta.T
margin_error = prediction - y
gradient = 1/m * (alpha * (x.T * margin_error))
theta = theta - gradient
J_history[i] = computeCost(x,y,theta)
return theta,J_history
My code is compiling and there isn't anything wrong. Please note this is theta:
theta = np.matrix(np.array([0,0]))
Alpha and iters is set to this:
alpha = 0.01
iters = 1000
When I run it, opt_theta, cost = gradientDescent(x, y, theta, alpha, iters) and print out opt_theta, I get this:
matrix([[ 2.36890383e+16, -1.40798902e+16],
[ 2.47503758e+17, -2.36890383e+16]])
when I should get this:
matrix([[-3.24140214, 1.1272942 ]])
What am I doing wrong?
Edit:
Cost function
def computeCost(x, y, theta):
# Get length of data set
m = len(y)
# We get theta transpose because we are working with a numpy array [0,0] for example
prediction = x * theta.T
J = 1/(2*m) * np.sum(np.power((prediction - y), 2))
return J
Look there:
>>> A = np.matrix([3,3,3])
>>> B = np.matrix([[1,1,1], [2,2,2]])
>>> A-B
matrix([[2, 2, 2],
[1, 1, 1]])
Matrices are broadcasted together.
"it's because np.matrix inherits from np.array. np.matrix overrides multiplication, but not addition and subtraction"
In yours situation theta(1x2) subtract gradient(2x1) and in result you have got 2x2. Try to transpose gradient before subtracting.
theta = theta - gradient.T