Numpy gradient for a 2 dimensions array-CodePudding

I'm not sure of how to use numpy.gradient().

to compute the partial derivatives (2nd order) I was using for loops :

for j in range(1, nx-1): 
            d2px[:, j] = (p[:, j - 1] - 2 * p[:, j]   p[:, j   1]) / dx ** 2
for i in range(1, ny-1): 
            d2py[i, :] = (p[i - 1, :] - 2 * p[i, :]   p[i   1, :]) / dy ** 2

And I tried to replace it with numpy.gradient : (for x here)

dpx = np.gradient(p, [1, dx], axis = 0)
d2px = np.gradient(dpx, [1, dx])

But I always have the same error message :

"ValueError: when 1d, distances must match the length of the corresponding dimension"

with the following code :

x = np.linspace(0, nx, nx) # coordonnées selon x...
dpx = np.gradient(p, x, axis = 0)
d2px_test = np.gradient(dpx, x, axis = 0)

the input p is :

[[ 0.00000000e 00  0.00000000e 00  0.00000000e 00  0.00000000e 00
   0.00000000e 00]
 [ 0.00000000e 00  6.53832270e-23 -1.19328961e-22  6.53832270e-23
   0.00000000e 00]
 [ 0.00000000e 00 -1.19328961e-22  2.07190726e-22 -1.19328961e-22
   0.00000000e 00]
 [ 0.00000000e 00  6.53832270e-23 -1.19328961e-22  6.53832270e-23
   0.00000000e 00]
 [ 0.00000000e 00  0.00000000e 00  0.00000000e 00  0.00000000e 00
   0.00000000e 00]]

The expected output is :

[[ 0.00000000e 00  0.00000000e 00  0.00000000e 00  0.00000000e 00
   0.00000000e 00]
 [ 0.00000000e 00 -2.50095415e-20  3.69424377e-20 -2.50095415e-20
   0.00000000e 00]
 [ 0.00000000e 00  4.45848648e-20 -6.53039374e-20  4.45848648e-20
   0.00000000e 00]
 [ 0.00000000e 00 -2.50095415e-20  3.69424377e-20 -2.50095415e-20
   0.00000000e 00]
 [ 0.00000000e 00  0.00000000e 00  0.00000000e 00  0.00000000e 00
   0.00000000e 00]]

And the actual output is :

[[ 0.00000000e 00 -8.00305329e-23  1.42671568e-22 -8.00305329e-23
   0.00000000e 00]
 [ 0.00000000e 00 -2.09226327e-23  3.81852676e-23 -2.09226327e-23
   0.00000000e 00]
 [ 0.00000000e 00  3.81852676e-23 -6.63010323e-23  3.81852676e-23
   0.00000000e 00]
 [ 0.00000000e 00 -2.09226327e-23  3.81852676e-23 -2.09226327e-23
   0.00000000e 00]
 [ 0.00000000e 00 -8.00305329e-23  1.42671568e-22 -8.00305329e-23
   0.00000000e 00]]

In terms of visualisation : the expected output is :

And the actual output (with np.gradient) is :

Thanks for your help.

CodePudding user response：

You can simply vectorize the operation

d2px2 = (p[:, :-2] - 2 * p[:, 1:-1]   p[:, 2:]) / dx ** 2
d2py2 = (p[:-2, :] - 2 * p[1:-1, :]   p[2:, :]) / dy ** 2

np.allclose(d2px2, d2px[:, 1:])
# True
np.allclose(d2py2, d2py[1:, :])
# True