# [Numpy-discussion] Calculation of a hessian

Kiko kikocorreoso at gmail.com
Mon Aug 11 02:02:48 EDT 2014

```2014-08-08 11:51 GMT+02:00 Jose Gomez-Dans <jgomezdans at gmail.com>:

> Your function looks fairly simple to differentiate by hand, but if you
> 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
>
>
Hi José,

My idea would be to generalise the calculation of the Hessian, not just to
differentiate the example I posted and I was wondering if Numpy/Scipy

Thanks again.

>
> On 8 August 2014 08:31, Kiko <kikocorreoso at 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)*
>>
>>
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>>
>>
>
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