Numpy N-D Matrix to a 3D Mesh Graph

Question

I tried looking this up a lot and there are lot of information on specific examples but they are too specific to understand.

How do I put data in a Numpy N-D Matrix to a 3D graph. please refer below example

 import numpy as np
 X =20

 Y =  20
 Z = 2
 sample = np.zeros(((X,Y,Z)))
 sample[1][2][2]=45
 sample[1][3][0]=52
 sample[1][8][1]=42
 sample[1][15][1]=30
 sample[1][19][2]=15

I Want to use values on X,Y,Z positions to be on a 3D graph (plot).

Thanks in advance

 import numpy as np
 import matplotlib.pyplot as plt
 from mpl_toolkits.mplot3d import axes3d

 # Define size of data
 P= 25
 X = 70
 Y = 25
 Z = 3

 # Create meshgrid
 x,y = np.meshgrid(np.arange(X),np.arange(Y))

 # Create some random data (your example didn't work)
 sample = np.random.randn((((P,X,Y,Z))))

 # Create figure
 fig=plt.figure()
 ax=fig.add_subplot(111,projection='3d')
 fig.show()

 # Define colors
 colors=['b','r','g']

 # Plot for each entry of in Z
 for i in range(Z):
    ax.plot_wireframe(x, y, sample[:,:,:,i],color=colors[i])
    plt.draw()
 plt.show()

But I only want to draw X,Y,Z only. when I used above code python throws me lots of errors like ValueError: too many values to unpack

Explaining what you mean by a "mesh diagram" would be a start. Perhaps a set of test input and expected output. — Daniel F, Feb 16 '17 at 12:15
Are you looking for a graphical output? Then you probably want to flag `matplotlib` so those folks can help you. — Daniel F, Feb 16 '17 at 12:28
One of the erros in your code is resulting from `` sample = np.random.randn((((P.X,Y,Z))))``. ``P.X`` should be ``P,X``. Like I said in my answer below, the only way to visualize 4D data is through color-encoding the 4th dimension. Check out (this)[http://stackoverflow.com/questions/14995610/how-to-make-a-4d-plot-with-matplotlib-using-arbitrary-data] answer — alexblae, Feb 17 '17 at 14:19
Thanks @alexblae , that was a mistake that happend when I tried to copy and paste the code. but now I got it right and there is still the error of too many values to unpack — D_Wills, Feb 18 '17 at 00:19

score 1 · Answer 1 · answered Feb 16 '17 at 13:20

Are you looking for something like this?

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import axes3d

# Define size of data
X = 20
Y = 20
Z = 3

# Create meshgrid
x,y = np.meshgrid(np.arange(X),np.arange(Y))

# Create some random data (your example didn't work)
sample = np.random.randn(X,Y,Z)

# Create figure
fig=plt.figure()
ax=fig.add_subplot(111,projection='3d')
fig.show()

# Define colors
colors=['b','r','g']

# Plot for each entry of in Z
for i in range(Z):
    ax.plot_wireframe(x, y, sample[:,:,i],color=colors[i])
    plt.draw()
plt.show()

which would you give

There are plenty of other ways to display 3D data in matplotlib, see also here. However, you are always limited to 3 dimensions (or 4, if you do a 3D scatter plot where color encodes the 4th dimension). So you need to make a decision which dimensions you want to show or if you can summarize them somehow.

This code is exactly what I was looking for but I need it for a 4D matrix. when I edited your code to suit for a 4D matrix (pls look at the edited code in the post) it gives me lots of errors — D_Wills, Feb 17 '17 at 12:31

score 0 · Answer 2 · answered May 11 '22 at 13:28

I have got something it may work for you. To understand it I explain the process I go briefly. I have connected 4x4x4 = 64 point masses to each other and created a cube with dampers and springs and inner friction. I solved the kinematic and mechanical behaviour using numpy and then I need to visualise the cube all I have is X,Y,Z points for each time step of each mass.

What I have is 4x4x4 XYZ points of a cube for each time tn:

Here how it goes :

import matplotlib.pyplot as plt 

zeroPoint=points[50] # at time step 50 elastic cube in space 
surf0x=zeroPoint[0,:,:,0]
surf0y=zeroPoint[0,:,:,1]
surf0z=zeroPoint[0,:,:,2]

surf1x=zeroPoint[:,0,:,0]
surf1y=zeroPoint[:,0,:,1]
surf1z=zeroPoint[:,0,:,2]

surf2x=zeroPoint[:,:,0,0]
surf2y=zeroPoint[:,:,0,1]
surf2z=zeroPoint[:,:,0,2]

surf3x=zeroPoint[nmx-1,:,:,0]
surf3y=zeroPoint[nmx-1,:,:,1]
surf3z=zeroPoint[nmx-1,:,:,2]

surf4x=zeroPoint[:,nmy-1,:,0]
surf4y=zeroPoint[:,nmy-1,:,1]
surf4z=zeroPoint[:,nmy-1,:,2]

surf5x=zeroPoint[:,:,nmz-1,0]
surf5y=zeroPoint[:,:,nmz-1,1]
surf5z=zeroPoint[:,:,nmz-1,2]

fig = plt.figure(figsize=(10,10))
wf = plt.axes(projection ='3d')
wf.set_xlim(-0.5,2)
wf.set_ylim(-0.5,2)
wf.set_zlim(-0.5,2)
wf.plot_wireframe(surf0x, surf0y, surf0z, color ='green')
wf.plot_wireframe(surf1x, surf1y, surf1z, color ='red')
wf.plot_wireframe(surf2x, surf2y, surf2z, color ='blue')
wf.plot_wireframe(surf3x, surf3y, surf3z, color ='black')
wf.plot_wireframe(surf4x, surf4y, surf4z, color ='purple')
wf.plot_wireframe(surf5x, surf5y, surf5z, color ='orange')
 
 
# displaying the visualization
wf.set_title('Its a Cube :) ')
pyplot.show()

at time step 190 same cube (animation is 60 FPS) :

The trick is as you see you need to create surfaces from points before you go. You dont even need np.meshgrid to do that. People does it for parametric z values calculation. If you have all points you dont need it.

Numpy N-D Matrix to a 3D Mesh Graph

2 Answers2