Compute the Jacobian matrix in Python

Question

import numpy as np


a = np.array([[1,2,3],
              [4,5,6],
              [7,8,9]])


b = np.array([[1,2,3]]).T

c = a.dot(b) #function

jacobian = a # as partial         


        

Answer 1

You can use the Harvard autograd library (link), where grad and jacobian take a function as their argument:

import autograd.numpy as np
from autograd import grad, jacobian

x = np.array([5,3], dtype=float)

def cost(x):
    return x[0]**2 / x[1] - np.log(x[1])

gradient_cost = grad(cost)
jacobian_cost = jacobian(cost)

gradient_cost(x)
jacobian_cost(np.array([x,x,x]))

Otherwise, you could use the jacobian method available for matrices in sympy:

from sympy import sin, cos, Matrix
from sympy.abc import rho, phi

X = Matrix([rho*cos(phi), rho*sin(phi), rho**2])
Y = Matrix([rho, phi])

X.jacobian(Y)

Also, you may also be interested to see this low-level variant (link). MATLAB provides nice documentation on its jacobian function here.

Answer 2

Here is a Python implementation of the mathematical Jacobian of a vector function f(x), which is assumed to return a 1-D numpy array.

import numpy as np

def J(f, x, dx=1e-8):
    n = len(x)
    func = f(x)
    jac = np.zeros((n, n))
    for j in range(n):  # through columns to allow for vector addition
        Dxj = (abs(x[j])*dx if x[j] != 0 else dx)
        x_plus = [(xi if k != j else xi + Dxj) for k, xi in enumerate(x)]
        jac[:, j] = (f(x_plus) - func)/Dxj
    return jac

It is recommended to make dx ~ 10^-8.

Answer 3

In python 3, you can try sympy package：

import sympy as sym

def Jacobian(v_str, f_list):
    vars = sym.symbols(v_str)
    f = sym.sympify(f_list)
    J = sym.zeros(len(f),len(vars))
    for i, fi in enumerate(f):
        for j, s in enumerate(vars):
            J[i,j] = sym.diff(fi, s)
    return J

Jacobian('u1 u2', ['2*u1 + 3*u2','2*u1 - 3*u2'])

which gives out:

Matrix([[2,  3],[2, -3]])

Answer 4

If you want to find the Jacobian numerically for many points at once (for example, if your function accepts shape (n, x) and outputs (n, y)), here is a function. This is essentially the answer from James Carter but for many points. The dx may need to be adjusted based on absolute value as in his answer.

def numerical_jacobian(f, xs, dx=1e-6):
  """
      f is a function that accepts input of shape (n_points, input_dim)
      and outputs (n_points, output_dim)

      return the jacobian as (n_points, output_dim, input_dim)
  """
  if len(xs.shape) == 1:
    xs = xs[np.newaxis, :]
    
  assert len(xs.shape) == 2

  ys = f(xs)
  
  x_dim = xs.shape[1]
  y_dim = ys.shape[1]
  
  jac = np.empty((xs.shape[0], y_dim, x_dim))
  
  for i in range(x_dim):
    x_try = xs + dx * e(x_dim, i + 1)
    jac[:, :, i] = (f(x_try) - ys) / dx
  
  return jac

def e(n, i):
  ret = np.zeros(n)
  ret[i - 1] = 1.0
  return ret

Answer 5

The Jacobian is only defined for vector-valued functions. You cannot work with arrays filled with constants to calculate the Jacobian; you must know the underlying function and its partial derivatives, or the numerical approximation of these. This is obvious when you consider that the (partial) derivative of a constant (with respect to something) is 0.

In Python, you can work with symbolic math modules such as SymPy or SymEngine to calculate Jacobians of functions. Here's a simple demonstration of an example from Wikipedia:

Using the SymEngine module:

Python 2.7.11 (v2.7.11:6d1b6a68f775, Dec  5 2015, 20:40:30) [MSC v.1500 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>>
>>> import symengine
>>>
>>>
>>> vars = symengine.symbols('x y') # Define x and y variables
>>> f = symengine.sympify(['y*x**2', '5*x + sin(y)']) # Define function
>>> J = symengine.zeros(len(f),len(vars)) # Initialise Jacobian matrix
>>>
>>> # Fill Jacobian matrix with entries
... for i, fi in enumerate(f):
...     for j, s in enumerate(vars):
...         J[i,j] = symengine.diff(fi, s)
...
>>> print J
[2*x*y, x**2]
[5, cos(y)]
>>>
>>> print symengine.Matrix.det(J)
2*x*y*cos(y) - 5*x**2

Answer 6

While autograd is a good library, make sure to check out its upgraded version JAX which is very well documented (compared to autograd).

A simple example:

import jax.numpy as jnp
from jax import jacfwd

# Define some simple function.
def sigmoid(x):
    return 0.5 * (jnp.tanh(x / 2) + 1)
# Note that here, I want a derivative of a "vector" output function (inputs*a + b is a vector) wrt a input 
# "vector" a at a0: Derivative of vector wrt another vector is a matrix: The Jacobian
def simpleJ(a, b, inputs): #inputs is a matrix, a & b are vectors
    return sigmoid(jnp.dot(inputs, a) + b)

inputs = jnp.array([[0.52, 1.12,  0.77],
                   [0.88, -1.08, 0.15],
                   [0.52, 0.06, -1.30],
                   [0.74, -2.49, 1.39]])

b = jnp.array([0.2, 0.1, 0.3, 0.2])
a0 = jnp.array([0.1,0.7,0.7])

# Isolate the function: variables to be differentiated from the constant parameters
f = lambda a: simpleJ(a, b, inputs) # Now f is just a function of variable to be differentiated

J = jacfwd(f)
# Till now I have only calculated the derivative, it still needs to be evaluated at a0.
J(a0)