b/m gradient calculation

[gtrevi]

Hi,
the (2/N) factor could be pulled out of the for loop, since it's out of the sigma in the p-derivative equation, correct?
So something like this:

def step_gradient(b_current, m_current, points, learningRate): b_gradient = 0 m_gradient = 0 N = float(len(points)) for i in range(0, len(points)): x = points[i, 0] y = points[i, 1] b_gradient += -(y - ((m_current * x) + b_current)) # (2/N) outta here m_gradient += -x * (y - ((m_current * x) + b_current)) # (2/N) outta here new_b = b_current - (learningRate * ((2/N)*b_gradient)) # (2/N) to be used here new_m = m_current - (learningRate * ((2/N)*m_gradient)) # (2/N) to be used here return [new_b, new_m]

[kavinaravind]

Yes you're correct. The (2/N) should be pulled out of the for loop to match the equation. But since this value (2/N) is a constant, the end value will be the same. So the value of your code will be the same as Siraj's code.

[elmennani89]

yes, I think the only benifit of pulling out (2/N) is making the algo a little bit faster. making devision and multiplication 1 time instead of N times

[gtrevi]

yep, speed was my main point, but also, besides the 2/N in/out the "for" would give the same value, the code would be clearer in my opinion.