Inverse of numpy.gradient function

问题

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Provide a function which is the opposite to to the numpy.gradient function.

Takes in a 2D array of derivative values (which were originally created by the numpy.gradient function) and returns the original scalar values.

I need to create a function which would be the inverse of the np.gradient function.

Where the Vx,Vy arrays (Velocity component vectors) are the input and the output would be an array of anti-derivatives (Arrival Time) at the datapoints x,y.

I have data on a (x,y) grid with scalar values (time) at each point.

I have used the numpy gradient function and linear interpolation to determine the gradient vector Velocity (Vx,Vy) at each point (See below).

I have achieved this by:

 #LinearTriInterpolator applied to a delaunay triangular mesh
 LTI= LinearTriInterpolator(masked_triang, time_array)

 #Gradient requested at the mesh nodes:
 (Vx, Vy) = LTI.gradient(triang.x, triang.y)

The first image below shows the velocity vectors at each point, and the point labels represent the time value which formed the derivatives (Vx,Vy)

The next image shows the resultant scalar value of the derivatives (Vx,Vy) plotted as a colored contour graph with associated node labels.

So my challenge is:

I need to reverse the process!

Using the gradient vectors (Vx,Vy) or the resultant scalar value to determine the original Time-Value at that point.

Is this possible?

Knowing that the numpy.gradient function is computed using second order accurate central differences in the interior points and either first or second order accurate one-sides (forward or backwards) differences at the boundaries, I am sure there is a function which would reverse this process.

I was thinking that taking a line derivative between the original point (t=0 at x1,y1) to any point (xi,yi) over the Vx,Vy plane would give me the sum of the velocity components. I could then divide this value by the distance between the two points to get the time taken..

Would this approach work? And if so, which numpy integrate function would be best applied?

An example of my data can be found here [http://www.filedropper.com/calculatearrivaltimefromgradientvalues060820]

Your help would be greatly appreciated

EDIT:

Maybe this simplified drawing might help understand where I'm trying to get to..

EDIT:

Thanks to @Aguy who has contibuted to this code.. I Have tried to get a more accurate representation using a meshgrid of spacing 0.5 x 0.5m and calculating the gradient at each meshpoint, however I am not able to integrate it properly. I also have some edge affects which are affecting the results that I don't know how to correct.

import numpy as np
from scipy import interpolate
from matplotlib import pyplot
from mpl_toolkits.mplot3d import Axes3D

#Createmesh grid with a spacing of 0.5 x 0.5
stepx = 0.5
stepy = 0.5
xx = np.arange(min(x), max(x), stepx)
yy = np.arange(min(y), max(y), stepy)
xgrid, ygrid = np.meshgrid(xx, yy)
grid_z1 = interpolate.griddata((x,y), Arrival_Time, (xgrid, ygrid), method='linear') #Interpolating the Time values

#Formatdata
X = np.ravel(xgrid)
Y= np.ravel(ygrid)
zs = np.ravel(grid_z1)
Z = zs.reshape(X.shape)

#Calculate Gradient
(dx,dy) = np.gradient(grid_z1) #Find gradient for points on meshgrid

Velocity_dx= dx/stepx #velocity ms/m
Velocity_dy= dy/stepx #velocity ms/m

Resultant = (Velocity_dx**2 + Velocity_dy**2)**0.5 #Resultant scalar value ms/m

Resultant = np.ravel(Resultant)

#Plot Original Data F(X,Y) on the meshgrid
fig = pyplot.figure()
ax = fig.add_subplot(projection='3d')
ax.scatter(x,y,Arrival_Time,color='r')
ax.plot_trisurf(X, Y, Z)
ax.set_xlabel('X-Coordinates')
ax.set_ylabel('Y-Coordinates')
ax.set_zlabel('Time (ms)')
pyplot.show()

#Plot the Derivative of f'(X,Y) on the meshgrid
fig = pyplot.figure()
ax = fig.add_subplot(projection='3d')
ax.scatter(X,Y,Resultant,color='r',s=0.2)
ax.plot_trisurf(X, Y, Resultant)
ax.set_xlabel('X-Coordinates')
ax.set_ylabel('Y-Coordinates')
ax.set_zlabel('Velocity (ms/m)')
pyplot.show()

#Integrate to compare the original data input
dxintegral = np.nancumsum(Velocity_dx, axis=1)*stepx
dyintegral = np.nancumsum(Velocity_dy, axis=0)*stepy

valintegral = np.ma.zeros(dxintegral.shape)
for i in range(len(yy)):
    for j in range(len(xx)):
        valintegral[i, j] = np.ma.sum([dxintegral[0, len(xx) // 2], 
    dyintegral[i, len(yy)  // 2], dxintegral[i, j], - dxintegral[i, len(xx) // 2]])
valintegral = valintegral * np.isfinite(dxintegral)

Now the np.gradient is applied at every meshnode (dx,dy) = np.gradient(grid_z1)

Now in my process I would analyse the gradient values above and make some adjustments (There is some unsual edge effects that are being create which I need to rectify) and would then integrate the values to get back to a surface which would be very similar to f(x,y) shown above.

I need some help adjusting the integration function:

#Integrate to compare the original data input
dxintegral = np.nancumsum(Velocity_dx, axis=1)*stepx
dyintegral = np.nancumsum(Velocity_dy, axis=0)*stepy

valintegral = np.ma.zeros(dxintegral.shape)
for i in range(len(yy)):
    for j in range(len(xx)):
        valintegral[i, j] = np.ma.sum([dxintegral[0, len(xx) // 2], 
    dyintegral[i, len(yy)  // 2], dxintegral[i, j], - dxintegral[i, len(xx) // 2]])
valintegral = valintegral * np.isfinite(dxintegral)

And now I need to calculate the new 'Time' values at the original (x,y) point locations.

UPDATE: I am getting some promising results using the help from @Aguy. The results can be seen below (with the red contours representing the original data, and the blue contours representing the integrated values). I am getting some inaccuracy in the areas of min(y) and max(y) - I am unsure how to overcome this..

Code here: https://pastebin.com/BRj61UKC

回答1:

Here is one approach:

First, in order to be able to do integration, it's good to be on a regular grid. Using here variable names x and y as short for your triang.x and triang.y we can first create a grid:

import numpy as np
n = 200 # Grid density
stepx = (max(x) - min(x)) / n
stepy = (max(y) - min(y)) / n
xspace = np.arange(min(x), max(x), stepx)
yspace = np.arange(min(y), max(y), stepy)
xgrid, ygrid = np.meshgrid(xspace, yspace)

Then we can interpolate dx and dy on the grid using the same LinearTriInterpolator function:

fdx = LinearTriInterpolator(masked_triang, dx)
fdy = LinearTriInterpolator(masked_triang, dy)

dxgrid = fdx(xgrid, ygrid)
dygrid = fdy(xgrid, ygrid)

Now comes the integration part. In principle, any path we choose should get us to the same value. In practice, since there are missing values and different densities, the choice of path is very important to get a reasonably accurate answer.

Below I choose to integrate over dxgrid in the x direction from 0 to the middle of the grid at n/2. Then integrate over dygrid in the y direction from 0 to the i point of interest. Then over dxgrid again from n/2 to the point j of interest. This is a simple way to make sure most of the path of integration is inside the bulk of available data by simply picking a path that goes mostly in the "middle" of the data range. Other alternative consideration would lead to different path selections.

So we do:

dxintegral = np.nancumsum(dxgrid, axis=1) * stepx
dyintegral = np.nancumsum(dygrid, axis=0) * stepy

and then (by somewhat brute force for clarity):

valintegral = np.ma.zeros(dxintegral.shape)
for i in range(n):
    for j in range(n):
        valintegral[i, j] = np.ma.sum([dxintegral[0, n // 2],  dyintegral[i, n // 2], dxintegral[i, j], - dxintegral[i, n // 2]])
valintegral = valintegral * np.isfinite(dxintegral)

valintegral would be the result up to an arbitrary constant which can help put the "zero" where you want.

With your data shown here:

ax.tricontourf(masked_triang, time_array)

This is what I'm getting reconstructed when using this method:

ax.contourf(xgrid, ygrid, valintegral)

Hopefully this is somewhat helpful.

If you want to revisit the values at the original triangulation points, you can use interp2d on the valintegral regular grid data.

EDIT:

In reply to your edit, your adaptation above has a few errors:

1. Change the line (dx,dy) = np.gradient(grid_z1) to (dy,dx) = np.gradient(grid_z1)

2. In the integration loop change the dyintegral[i, len(yy) // 2] term to dyintegral[i, len(xx) // 2]

3. Better to replace the line valintegral = valintegral * np.isfinite(dxintegral) with valintegral[np.isnan(dx)] = np.nan

来源：https://stackoverflow.com/questions/63220629/inverse-of-numpy-gradient-function