numpy second derivative of a ndimensional array

Question

I have a set of simulation data where I would like to find the lowest slope in n dimensions. The spacing of the data is constant along each dimension, but not all the same (I co

Answer 1

The second derivatives are given by the Hessian matrix. Here is a Python implementation for ND arrays, that consists in applying the np.gradient twice and storing the output appropriately,

import numpy as np

def hessian(x):
    """
    Calculate the hessian matrix with finite differences
    Parameters:
       - x : ndarray
    Returns:
       an array of shape (x.dim, x.ndim) + x.shape
       where the array[i, j, ...] corresponds to the second derivative x_ij
    """
    x_grad = np.gradient(x) 
    hessian = np.empty((x.ndim, x.ndim) + x.shape, dtype=x.dtype) 
    for k, grad_k in enumerate(x_grad):
        # iterate over dimensions
        # apply gradient again to every component of the first derivative.
        tmp_grad = np.gradient(grad_k) 
        for l, grad_kl in enumerate(tmp_grad):
            hessian[k, l, :, :] = grad_kl
    return hessian

x = np.random.randn(100, 100, 100)
hessian(x)

Note that if you are only interested in the magnitude of the second derivatives, you could use the Laplace operator implemented by scipy.ndimage.filters.laplace, which is the trace (sum of diagonal elements) of the Hessian matrix.

Taking the smallest element of the the Hessian matrix could be used to estimate the lowest slope in any spatial direction.

Answer 2

Slopes, Hessians and Laplacians are related, but are 3 different things.
Start with 2d: a function( x, y ) of 2 variables, e.g. a height map of a range of hills,

• slopes aka gradients are direction vectors, a direction and length at each point x y.
  This can be given by 2 numbers dx dy in cartesian coordinates, or an angle θ and length sqrt( dx^2 + dy^2 ) in polar coordinates. Over a whole range of hills, we get a vector field.

• Hessians describe curvature near x y, e.g. a paraboloid or a saddle, with 4 numbers: dxx dxy dyx dyy.

• a Laplacian is 1 number, dxx + dyy, at each point x y. Over a range of hills, we get a scalar field. (Functions or hills with Laplacian = 0 are particularly smooth.)

Slopes are linear fits and Hessians quadratic fits, for tiny steps h near a point xy:

f(xy + h)  ~  f(xy)
        +  slope . h    -- dot product, linear in both slope and h
        +  h' H h / 2   -- quadratic in h

Here xy, slope and h are vectors of 2 numbers, and H is a matrix of 4 numbers dxx dxy dyx dyy.

N-d is similar: slopes are direction vectors of N numbers, Hessians are matrices of N^2 numbers, and Laplacians 1 number, at each point.

(You might find better answers over on math.stackexchange .)

Answer 3

You can see the Hessian Matrix as a gradient of gradient, where you apply gradient a second time for each component of the first gradient calculated here is a wikipedia link definig Hessian matrix and you can see clearly that is a gradient of gradient, here is a python implementation defining gradient then hessian :

import numpy as np
#Gradient Function
def gradient_f(x, f):
  assert (x.shape[0] >= x.shape[1]), "the vector should be a column vector"
  x = x.astype(float)
  N = x.shape[0]
  gradient = []
  for i in range(N):
    eps = abs(x[i]) *  np.finfo(np.float32).eps 
    xx0 = 1. * x[i]
    f0 = f(x)
    x[i] = x[i] + eps
    f1 = f(x)
    gradient.append(np.asscalar(np.array([f1 - f0]))/eps)
    x[i] = xx0
  return np.array(gradient).reshape(x.shape)

#Hessian Matrix
def hessian (x, the_func):
  N = x.shape[0]
  hessian = np.zeros((N,N)) 
  gd_0 = gradient_f( x, the_func)
  eps = np.linalg.norm(gd_0) * np.finfo(np.float32).eps 
  for i in range(N):
    xx0 = 1.*x[i]
    x[i] = xx0 + eps
    gd_1 =  gradient_f(x, the_func)
    hessian[:,i] = ((gd_1 - gd_0)/eps).reshape(x.shape[0])
    x[i] =xx0
  return hessian

As a test, the Hessian matrix of (x^2 + y^2) is 2 * I_2 where I_2 is the identity matrix of dimension 2