funtions with multiple arguments about findiff

[maroba]

The problem is that you tried to feed the function with two 1D-arrays of coordinate values, but actually you need a 2D grid in your example. You can create the 2D grid arrays from the coordinate arrays using the numpy.meshgrid function:

from findiff import FinDiff
import numpy as np
def f(x, y):
    return 5*x**2 - 5*x + 10*y**2 -10*y

diff = .01
x = np.linspace(-10,0,11)
y = np.linspace(0,10,11)

X, Y = np.meshgrid(x, y, indexing='ij')

d_dx = FinDiff(0, diff)

df_dx = d_dx(f(X, Y))

np.meshgrid works for any number of dimensions.

FinDiff determines the grid dimension by the arrays that you apply it to. If you pass a 1D-array to a FinDiff object, it thinks that your grid is only 1D. Incidentally, if you had calculated the partial derivative with respect to y instead of x in your original implementation, it would have thrown an IndexError exception, because there is no second index in the 1D array. All in all, if you use grid arrays, all should be fine. :-)

[RoyLarson]

This is the output I expect when I inputted what I put in.

  array([[-105.],
    [ -95.],
    [ -85.],
    [ -75.],
    [ -65.],
    [ -55.],
    [ -45.],
    [ -35.],
    [ -25.],
    [ -15.],
    [  -5.]])

This is the output that I get when I input what you say to

array([[-10500., -10500., -10500., -10500., -10500., -10500., -10500.,
    -10500., -10500., -10500., -10500.],
    [ -9500.,  -9500.,  -9500.,  -9500.,  -9500.,  -9500.,  -9500.,
    -9500.,  -9500.,  -9500.,  -9500.],
    [ -8500.,  -8500.,  -8500.,  -8500.,  -8500.,  -8500.,  -8500.,
    -8500.,  -8500.,  -8500.,  -8500.],
    [ -7500.,  -7500.,  -7500.,  -7500.,  -7500.,  -7500.,  -7500.,
    -7500.,  -7500.,  -7500.,  -7500.],
    [ -6500.,  -6500.,  -6500.,  -6500.,  -6500.,  -6500.,  -6500.,
    -6500.,  -6500.,  -6500.,  -6500.],
    [ -5500.,  -5500.,  -5500.,  -5500.,  -5500.,  -5500.,  -5500.,
    -5500.,  -5500.,  -5500.,  -5500.],
    [ -4500.,  -4500.,  -4500.,  -4500.,  -4500.,  -4500.,  -4500.,
    -4500.,  -4500.,  -4500.,  -4500.],
    [ -3500.,  -3500.,  -3500.,  -3500.,  -3500.,  -3500.,  -3500.,
    -3500.,  -3500.,  -3500.,  -3500.],
    [ -2500.,  -2500.,  -2500.,  -2500.,  -2500.,  -2500.,  -2500.,
    -2500.,  -2500.,  -2500.,  -2500.],
    [ -1500.,  -1500.,  -1500.,  -1500.,  -1500.,  -1500.,  -1500.,
    -1500.,  -1500.,  -1500.,  -1500.],
    [  -500.,   -500.,   -500.,   -500.,   -500.,   -500.,   -500.,
    -500.,   -500.,   -500.,   -500.]])

The numbers are still off, and there are more of them.

Also how do I get just one point

df_0_0_dx = d_dx(f((0,0)))

If you want to look at the jupyter notebook I am trying to figure this out on it is

https://github.com/RoyLarson/Simple_NG_Piping_Pressure_Drop/blob/initial_commit/tests/findiff_test.ipynb

[RoyLarson]

I figured this out.
I thought your program took at function evaluated the function along with the given dx and sent back the result at the point.
It instead evaluates the finite differences for already evaluated points from a function so the grid had to already be computed.