Re: [Numpy-discussion] Calculation of a hessian

2014-08-08 11:51 GMT+02:00 Jose Gomez-Dans jgomezdans@gmail.com:

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

Hi José,

Thanks for the answer.

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 already had something similar to that provided by NumDiffTools.

Thanks again.

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.

---

NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion

---

NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion