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 <kikocorre...@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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> NumPy-Discussion@scipy.org
> http://mail.scipy.org/mailman/listinfo/numpy-discussion
>
>
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