approx_fprime — SciPy v1.15.2 Manual (original) (raw)
scipy.optimize.
scipy.optimize.approx_fprime(xk, f, epsilon=1.4901161193847656e-08, *args)[source]#
Finite difference approximation of the derivatives of a scalar or vector-valued function.
If a function maps from \(R^n\) to \(R^m\), its derivatives form an m-by-n matrix called the Jacobian, where an element \((i, j)\) is a partial derivative of f[i] with respect to xk[j]
.
Parameters:
xkarray_like
The coordinate vector at which to determine the gradient of f.
fcallable
Function of which to estimate the derivatives of. Has the signaturef(xk, *args)
where xk is the argument in the form of a 1-D array and args is a tuple of any additional fixed parameters needed to completely specify the function. The argument xk passed to this function is an ndarray of shape (n,) (never a scalar even if n=1). It must return a 1-D array_like of shape (m,) or a scalar.
Suppose the callable has signature f0(x, *my_args, **my_kwargs)
, wheremy_args
and my_kwargs
are required positional and keyword arguments. Rather than passing f0
as the callable, wrap it to accept only x
; e.g., pass fun=lambda x: f0(x, *my_args, **my_kwargs)
as the callable, where my_args
(tuple) and my_kwargs
(dict) have been gathered before invoking this function.
Changed in version 1.9.0: f is now able to return a 1-D array-like, with the \((m, n)\)Jacobian being estimated.
epsilon{float, array_like}, optional
Increment to xk to use for determining the function gradient. If a scalar, uses the same finite difference delta for all partial derivatives. If an array, should contain one value per element of_xk_. Defaults to sqrt(np.finfo(float).eps)
, which is approximately 1.49e-08.
*argsargs, optional
Any other arguments that are to be passed to f.
Returns:
jacndarray
The partial derivatives of f to xk.
See also
Check correctness of gradient function against approx_fprime.
Notes
The function gradient is determined by the forward finite difference formula:
f(xk[i] + epsilon[i]) - f(xk[i])
f'[i] = --------------------------------- epsilon[i]
Examples
import numpy as np from scipy import optimize def func(x, c0, c1): ... "Coordinate vector
x
should be an array of size two." ... return c0 * x[0]*2 + c1x[1]**2
x = np.ones(2) c0, c1 = (1, 200) eps = np.sqrt(np.finfo(float).eps) optimize.approx_fprime(x, func, [eps, np.sqrt(200) * eps], c0, c1) array([ 2. , 400.00004208])