# Gradient descent formula implementation in python

So I recently started with Andrew Ng's ML Course and this is the formula that Andrew lays out for calculating gradient descent on a linear model.

$$\theta_j = \theta_j - \alpha \frac{1}{m} \sum_{i=1}^m \left( h_\theta(x^{(i)}) - y^{(i)}\right)x_j^{(i)} \qquad \text{simultaneously update } \theta_j \text{ for all } j$$

As we see, the formula asks us to the sum over all the rows in data.

However, the below code doesn't work if I apply np.sum()

def gradientDescent(X, y, theta, alpha, num_iters):

# Initialize some useful values
m = y.shape[0]  # number of training examples

# make a copy of theta, to avoid changing the original array, since numpy arrays
# are passed by reference to functions
theta = theta.copy()

J_history = [] # Use a python list to save cost in every iteration

for i in range(num_iters):
temp = np.dot(X, theta) - y
temp = np.dot(X.T, temp)
theta = theta - ((alpha / m) * np.sum(temp))
# save the cost J in every iteration
J_history.append(computeCost(X, y, theta))

return theta, J_history


On the other hand, if I get rid of the np.sum(), the formula works perfectly.

def gradientDescent(X, y, theta, alpha, num_iters):

# Initialize some useful values
m = y.shape[0]  # number of training examples

# make a copy of theta, to avoid changing the original array, since numpy arrays
# are passed by reference to functions
theta = theta.copy()

J_history = [] # Use a python list to save cost in every iteration

for i in range(num_iters):
temp = np.dot(X, theta) - y
temp = np.dot(X.T, temp)
theta = theta - ((alpha / m) * temp)
# save the cost J in every iteration
J_history.append(computeCost(X, y, theta))

return theta, J_history


Your goal if to compute the gradients for the whole theta vector of size p (number of variables). Your temp is a vector also of size $$p$$, which contains the values of gradients of the cost function relative to each of your theta values.
Therefore, you want to substract point-wise the two vectors (with learning rate $$\alpha$$) to make an update, so no reason to sum the vector.
• @ManasTripathi if you refer to the $i$ in the formula, it’s just the sum over the training examples. The dot product already handles this (that’s why we say your code is “vectorized”) – Elliot Sep 9 '19 at 8:03