Logistic Regression Gradient Descent [closed]

Question

I have to do Logistic regression using batch gradient descent.

import numpy as np

X = np.asarray([
[0.50],[0.75],[1.00],[1.25],[1.50],[1.75],[1.75],
[2.00],[2.25],[2.50],[2.75],[3.00],[3.25],[3.50],
[4.00],[4.25],[4.50],[4.75],[5.00],[5.50]])

y = np.asarray([0,0,0,0,0,0,1,0,1,0,1,0,1,0,1,1,1,1,1,1])

m = len(X)

def sigmoid(a):
    return 1.0 / (1 + np.exp(-a))

def gradient_Descent(theta, alpha, X , y):
    for i in range(0,m):
        cost = ((-y) * np.log(sigmoid(X[i]))) - ((1 - y) * np.log(1 - sigmoid(X[i])))
    grad = theta - alpha * (1.0/m) * (np.dot(cost,X[i]))
    theta = theta - alpha * grad

gradient_Descent(0.1,0.005,X,y)   

The way I have to do it is like this but I can't seem to understand how to make it work.

Answer 1

It looks like you have some stuff mixed up in here. It's critical when doing this that you keep track of the shape of your vectors and makes sure you're getting sensible results. For example, you are calculating cost with:

cost = ((-y) * np.log(sigmoid(X[i]))) - ((1 - y) * np.log(1 - sigmoid(X[i])))

In your case y is vector with 20 items and X[i] is a single value. This makes your cost calculation a 20 item vector which doesn't makes sense. Your cost should be a single value. (you're also calculating this cost a bunch of times for no reason in your gradient descent function).

Also, if you want this to be able to fit your data you need to add a bias terms to X. So let's start there.

X = np.asarray([
    [0.50],[0.75],[1.00],[1.25],[1.50],[1.75],[1.75],
    [2.00],[2.25],[2.50],[2.75],[3.00],[3.25],[3.50],
    [4.00],[4.25],[4.50],[4.75],[5.00],[5.50]])

ones = np.ones(X.shape)
X = np.hstack([ones, X])
# X.shape is now (20, 2)

Theta will now need 2 values for each X. So initialize that and Y:

Y = np.array([0,0,0,0,0,0,1,0,1,0,1,0,1,0,1,1,1,1,1,1]).reshape([-1, 1])
# reshape Y so it's column vector so matrix multiplication is easier
Theta = np.array([[0], [0]])

Your sigmoid function is good. Let's also make a vectorized cost function:

def sigmoid(a):
    return 1.0 / (1 + np.exp(-a))

def cost(x, y, theta):
    m = x.shape[0]
    h = sigmoid(np.matmul(x, theta))
    cost = (np.matmul(-y.T, np.log(h)) - np.matmul((1 -y.T), np.log(1 - h)))/m
    return cost

The cost function works because Theta has a shape of (2, 1) and X has a shape of (20, 2) so matmul(X, Theta) will be shaped (20, 1). The then matrix multiply the transpose of Y (y.T shape is (1, 20)), which result in a single value, our cost given a particular value of Theta.

We can then write a function that performs a single step of batch gradient descent:

def gradient_Descent(theta, alpha, x , y):
    m = x.shape[0]
    h = sigmoid(np.matmul(x, theta))
    grad = np.matmul(X.T, (h - y)) / m;
    theta = theta - alpha * grad
    return theta

Notice np.matmul(X.T, (h - y)) is multiplying shapes (2, 20) and (20, 1) which results in a shape of (2, 1) — the same shape as Theta, which is what you want from your gradient. This allows you to multiply is by your learning rate and subtract it from the initial Theta, which is what gradient descent is supposed to do.

So now you just write a loop for a number of iterations and update Theta until it looks like it converges:

n_iterations = 500
learning_rate = 0.5

for i in range(n_iterations):
    Theta = gradient_Descent(Theta, learning_rate, X, Y)
    if i % 50 == 0:
        print(cost(X, Y, Theta))

This will print the cost every 50 iterations resulting in a steadily decreasing cost, which is what you hope for:

[[ 0.6410409]]
[[ 0.44766253]]
[[ 0.41593581]]
[[ 0.40697167]]
[[ 0.40377785]]
[[ 0.4024982]]
[[ 0.40195]]
[[ 0.40170533]]
[[ 0.40159325]]
[[ 0.40154101]]

You can try different initial values of Theta and you will see it always converges to the same thing.

Now you can use your newly found values of Theta to make predictions:

h = sigmoid(np.matmul(X, Theta))
print((h > .5).astype(int) )

This prints what you would expect for a linear fit to your data:

[[0]
 [0]
 [0]
 [0]
 [0]
 [0]
 [0]
 [0]
 [0]
 [0]
 [1]
 [1]
 [1]
 [1]
 [1]
 [1]
 [1]
 [1]
 [1]
 [1]]