Speeding up newton-raphson in pandas/python

Question

I'm currently iterating through a very large set of data ~85GB (~600M lines) and simply using newton-raphson to compute a new parameter. As of right now my code is extremely slow, any tips on how to speed it up? The methods from BSCallClass & BSPutClass are closed-form, so there's nothing really to speed up there. Thanks.

class NewtonRaphson:

    def __init__(self, theObject):
        self.theObject = theObject

    def solve(self, Target, Start, Tolerance, maxiter=500):
        y = self.theObject.Price(Start)
        x = Start
        i = 0
        while (abs(y - Target) > Tolerance):
            i += 1
            d = self.theObject.Vega(x)
            x += (Target - y) / d
            y = self.theObject.Price(x)
            if i > maxiter:
                x = nan
                break
        return x

    def main():
        for row in a.iterrows():
            print row[1]["X.1"]
            T = (row[1]["X.7"] - row[1]["X.8"]).days
            Spot = row[1]["X.2"]
            Strike = row[1]["X.9"]
            MktPrice = abs(row[1]["X.10"]-row[1]["X.11"])/2
            CPflag = row[1]["X.6"]

            if CPflag == 'call':
                option = BSCallClass(0, 0, T, Spot, Strike)
            elif CPflag == 'put':
                option = BSPutClass(0, 0, T, Spot, Strike)

            a["X.15"][row[0]] = NewtonRaphson(option).solve(MktPrice, .05, .0001)

EDIT:

For those curious, I ended up speeding this entire process significantly by using the scipy suggestion, as well as using the multiprocessing module.

Answer 1

Don't code your own Newton-Raphson method in Python. You'll get better performance using one of the root finders in scipy.optimize such as brentq or newton. (Presumably, if you have pandas, you'd also install scipy.)

---

Back of the envelope calculation:

Making 600M calls to brentq should be manageable on standard hardware:

import scipy.optimize as optimize
def f(x):
    return x**2 - 2

In [28]: %timeit optimize.brentq(f, 0, 10)
100000 loops, best of 3: 4.86 us per loop

So if each call to optimize.brentq takes 4.86 microseconds, 600M calls will take about 4.86 * 600 ~ 3000 seconds ~ 1 hour.

---

newton may be slower, but still manageable:

def f(x):
    return x**2 - 2
def fprime(x):
    return 2*x

In [40]: %timeit optimize.newton(f, 10, fprime)
100000 loops, best of 3: 8.22 us per loop