Field interpolation

[GenB31415]

Hello everyone!
In my pde-equation problem, I calculate some localized state, which I then want to distribute (copy) to different point(s) of the original space (grid). Then I will use this multi-state as a new initial state. In Matlab's case, I can capture such a localized state and then copy it to different places as some kind of interpolation task using the "pdeInterpolant" function. Can I use the "Field Interpolation" function from very promising py-pde for this? Thank you.

[david-zwicker]

Would the method interpolate_to_grid be helpful?

[GenB31415]

I tried to use the method "interpolate_to_grid", but something goes wrong and the exception occurs "AttributeError: 'CartesianGrid' object has no attribute 'interpolate to grid'". I appreciate for a help to indicate how to correct this. Thank you in advance.

My code
import numpy as np
from pde import CartesianGrid, ScalarField
L=10; W=2
gridOld = CartesianGrid([[0, 2 * L]] * 2, 64) #initial 2D grid
myFun = "exp(-((x-" + str(L) + ")**2+(y-" + str(L)+")*2)/(2" + str(W) + "**2))"
print("\nmyFun = "+myFun+"\n")
stateOld = ScalarField.from_expression(gridOld, myFun) # <--- old state
stateOld.plot()

Now try to create the interpolator to new grid

gridNew = CartesianGrid([[0, 2 * L]] * 2, 2*64) # <--- new grid - - -

and interpolate myFun to the new grid:

fieldCollNew = gridNew.interpolate_to_grid(gridNew, "scipy", "linear", "New state")

But Error: ---> AttributeError: 'CartesianGrid' object has no attribute 'interpolate_to_grid'

fieldCollNew.data.plot()

[david-zwicker]

Note that the method is defined for fields, not grids.

[GenB31415]

Thank you. On Tuesday, May 11, 2021, 10:30:24 AM CDT, David Zwicker ***@***.***> wrote: Note that the method is defined for fields, not grids. — You are receiving this because you authored the thread. Reply to this email directly, view it on GitHub, or unsubscribe.

[GenB31415]

Thanks for the advice. Now the code runs correctly. So the question is ### Could I indicate the insertion point(s) [x,y] of the interpolated data to new grid? ### Thank you,

My code:

from matplotlib.pyplot import fill
import numpy as np
from pde import CartesianGrid, ScalarField, FieldCollection

L, W = 10, 2
gridOld = CartesianGrid([[0, 2 * L]] * 2, 64) #initial 2D grid
myFun = "exp(-((x-" + str(L) + ")**2+(y-" + str(L)+")*2)/(2" + str(W) + "**2))"
print("\nmyFun = "+myFun+"\n")
stateOld = ScalarField.from_expression(gridOld, myFun) # old state
stateOld.plot()
fieldCollectionOld = FieldCollection(fields = [stateOld])
gridNew = CartesianGrid([[0, 1.5 * L],[0, 1.5 * L]] , 2*64) # create new 2D grid - - -
print("Create the interpolator (please wait)...")
fieldCollNew = fieldCollectionOld.interpolate_to_grid(gridNew)
print("Done\n")
print("Plotting new grid:")
fieldCollNew.plot()
print("\nFinish\n")

[GenB31415]

I found simple solution that is good for my problem

My code:
from pde import CartesianGrid, ScalarField
L, W = 10, 3 #parameters of Gaussian
myFun = "exp(-((x-" + str(L) + ")**2+(y-" + str(L)+")*2)/(2" + str(W) + "**2))"
print("\n"+"myFun = " + myFun )
gridOld = CartesianGrid([[0, 2 * L]] * 2, 64) #initial 2D grid
stateOld = ScalarField.from_expression(gridOld, myFun, label="Init state") # old state
stateOld.plot()
dataOld = stateOld.data
gridNew = CartesianGrid([[0, 1.8 * L],[0, 1.8 * L]] , 2*64) # create new 2D grid
print("Create new 2x2 state:")
fieldNew = ScalarField(gridNew, 1, label="Final 2x2 state")
for r in range(0,64+1,64):
for c in range(0,64+1,64):
fieldNew.data[r:r+dataOld.shape[0], c:c+dataOld.shape[1]] = dataOld
fieldNew.plot()
print("Finish\n")

[david-zwicker]

That's a good solution. I guess we cannot supply a method for each special requirement, but since you can simply access the field data as a numpy array, you can certainly create your own solutions. Good job!

[GenB31415]

Thank you! On Friday, May 14, 2021, 12:30:56 PM CDT, David Zwicker ***@***.***> wrote: That's a good solution. I guess we cannot supply a method for each special requirement, but since you can simply access the field data as a numpy array, you can certainly create your own solutions. Good job! — You are receiving this because you authored the thread. Reply to this email directly, view it on GitHub, or unsubscribe.