Your function looks fairly simple to differentiate by hand, but if you have access to the gradient (or you estimate it numerically using scipy...), this function might do the job:

def hessian ( x, the_func, epsilon=1e-8):
    """Numerical approximation to the Hessian
    Parameters
    ------------
    x: array-like
        The evaluation point
    the_func: function
        The function. We assume that the function returns the function value and
        the associated gradient as the second return element
    epsilon: float
        The size of the step
    """

    N = x.size
    h = np.zeros((N,N))
    df_0 = the_func ( x )[1]
    for i in xrange(N):
        xx0 = 1.*x[i]
        x[i] = xx0 + epsilon
        df_1 = the_func ( x )[1]
        h[i,:] = (df_1 - df_0)/epsilon
        x[i] = xx0
    return h

Jose


On 8 August 2014 08:31, Kiko <kikocorreoso@gmail.com> wrote:
Hi all,

I am trying to calculate a Hessian. I am using numdifftools for this (https://pypi.python.org/pypi/Numdifftools).

My question is, is it possible to make it using pure numpy?.

The actual code is like this:

import numdifftools as nd
import numpy as np

def log_likelihood(params):
    sum1 = 0; sum2 = 0
    mu = params[0]; sigma = params[1]; xi = params[2]
    for z in data:
        x = 1 + xi * ((z-mu)/sigma)
        sum1 += np.log(x)
        sum2 += x**(-1.0/xi)
    return -((-len(data) * np.log(sigma)) - (1 + 1/xi)*sum1 - sum2) # negated so we can use 'minimum'

kk = nd.Hessian(log_likelihood)

Thanks in advance.

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