How to plot 3 arrays as a surface plot in Python - python

So, I've got three arrays of data, X, Y, and Z, each 225 numbers long. What I would like to do is to plot all three values at the same time on a surface plot. When I try to use
ax.plot_surface(X,Y,Z)
it tells me that Z needs to be a 2D array. I've looked it up and I've seen that it's possible to plot Z if it was a function of X and Y, but I need the first Z point to be associated with the first X and Y point etc. Is this possible in Python?

If your arrays are all 1-D, then I think what you want is
ax.plot_trisurf(X,Y,Z, triangles=tri.triangles, cmap=plt.cm.Spectral)
See more info at https://matplotlib.org/examples/mplot3d/trisurf3d_demo2.html

So, I've got three arrays of data, X, Y, and Z, each 225 numbers long. What I would like to do is to plot all three values at the same time on a surface plot.
So, from what i understood you want to plot a 3d surface plot.
But it seems you are only providing 3 1xn arrays. (in this case n == 255)
When plotting a surface plot, what you are doing in practice is getting each and every possible combination of a base (XY plane) and telling how high is a point Z on that given XY coordinates, hence Z is depicted as a function Z(i,j)
but I need the first Z point to be associated with the first X and Y point etc. Is this possible in Python?
Yes, but if you associate each Z point to the first X,Y and so on, you would only have the Z values for X==Y, which would be incomplete information for your surfaceplot!
A good (great) example of surface plot comes from matplotlib official docs
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
from matplotlib import cm
from matplotlib.ticker import LinearLocator, FormatStrFormatter
import numpy as np
fig = plt.figure()
ax = fig.gca(projection='3d')
# Make data.
X = np.arange(-5, 5, 0.25)
Y = np.arange(-5, 5, 0.25)
X, Y = np.meshgrid(X, Y)
R = np.sqrt(X**2 + Y**2)
Z = np.sin(R)
# Plot the surface.
surf = ax.plot_surface(X, Y, Z, cmap=cm.coolwarm,
linewidth=0, antialiased=False)
which results in:
What the code is actually doing:
Defining the input vectors X and Y (both range and interval)
Making a meshgrid out of those vectors (if unclear as to what a meshgrid is, print the output!)
Defining a function over the X,Y domain
Applying it to get Z
If you check, X,Y and Z are 2 dimensional arrays!
Hope it helps!

Related

How to change the axis dimension from pixel to length in matplotlib? is there any code in general?

Since the complete simulation is to big to post it right here only the code to plot the spectrum is given (I think this is enough)
d = i.sum(axis=2)
pylab.figure(figsize=(15,15))
pylab = imshow(d)
plt.axis('tight')
pylab.show()
This spectrum is given in pixel. But I would like to have this in the units of length. I will hope you may give me some advices.
Do you mean that you want axis ticks to show your custom dimensions instead of the number of pixels in d? If yes, use the extent keyword of imshow:
import numpy
import matplotlib
matplotlib.use('Agg')
import matplotlib.pyplot as plt
d = numpy.random.normal(size=(20, 40))
fig = plt.figure()
s = fig.add_subplot(1, 1, 1)
s.imshow(d, extent=(0, 1, 0, 0.5), interpolation='none')
fig.tight_layout()
fig.savefig('tt.png')
I'm guess a bit at what your problem is, so let's start by stating my interpretation/ You have some 2D data d that you plot using imshow and the units on the x and y axes are in the number of pixels. For example in the following we see the x axis labelled from 0 -> 10 for the number of data points:
import numpy as np
import matplotlib.pyplot as plt
# Generate a fake d
x = np.linspace(-1, 1, 10)
y = np.linspace(-1, 1, 10)
X, Y = np.meshgrid(x, y)
d = np.sin(X**2 + Y**2)
plt.imshow(d)
If this correctly describes your issue, then the solution is to avoid using imshow, which is designed to plot images. Firstly this will help as imshow attemps to interpolate to give a smoother image (which may hide features in the spectrum) and second because it is an image, there is no meaningful x and y data so it doesn't plot it.
The best alternative would be to use plt.pcolormesh which generate a psuedocolor plot of a 2D array and takes as arguments X and Y, which are both 2D arrays of points to which the values of d correspond.
For example:
# Generate a fake d
x = np.linspace(-1, 1, 10)
y = np.linspace(-1, 1, 10)
X, Y = np.meshgrid(x, y)
d = np.sin(X**2 + Y**2)
plt.pcolormesh(X, Y, d)
Now the x and y values correspond to the values of X and Y.

Plot of 3D matrix with colour scale - Python

I would like to plot a 3D matrix - essentially a box of numbers, each labelled by an x, y, z triad of coordinates- by assigning a different colour to each of the x, y, z point, according to its magnitude (for example, bigger numbers in red and smaller numbers in blue).
I cannot plot sections of the matrix, I rather need to plot the whole matrix together.
If we call matrix3D my matrix, its elements are built this way:
matrix3D[x][y][z] = np.exp(-(x**2+y**2+z**2))
How can I obtain the desired plot?
EDIT: Using Mayavi2 Contour3D(), I have tried to write the following:
from mayavi import mlab
X = np.arange(0, n_x, 1)
Y = np.arange(0, n_z, 1)
Z = np.arange(0, n_z, 1)
X, Y, Z = np.meshgrid(X, Y, Z)
obj = mlab.contour3d(X, Y, Z, matrix3D, contours=4, transparent=True)
where n_x, n_y, n_z are the dimension of the 3 axes. How can I actually see and/or save the image now?
If you need to plot the whole thing I think you're best taking a look at mayavi. This will let you plot a volume and you should be able to get the results you need.
I know you said you need to plot the whole thing at once, but this might still be of some use. You can use countourf to plot like this:
import numpy as np
import matplotlib.pyplot as plt
matrix3D = np.empty((10, 10, 10))
x = np.arange(10)
y = np.arange(10)
z = np.arange(10)
matrix3D[x][y][z] = np.exp(-(x**2+y**2+z**2))
fig = plt.figure()
ax = fig.add_subplot(plt.subplot(1, 1, 1))
ax.contourf(x, y, matrix3D[:, :, 3])
plt.show()
This gives you a slice of the 3D matrix (in this example the 4th slice).

How to plot a smooth 2D color plot for z = f(x, y)

I am trying to plot 2D field data using matplotlib. So basically I want something similar to this:
In my actual case I have data stored in a file on my harddrive. However for simplicity consider the function z = f(x, y). I want a smooth 2D plot where z is visualised using color. I managed the plotting with the following lines of code:
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-1, 1, 21)
y = np.linspace(-1, 1, 21)
z = np.array([i*i+j*j for j in y for i in x])
X, Y = np.meshgrid(x, y)
Z = z.reshape(21, 21)
plt.pcolor(X, Y, Z)
plt.show()
However, the plot I obtain is very coarse. Is there a very simple way to smooth the plot? I know something similar is possible with surface plots, however, those are 3D. I could change the camera angle to obtain a 2D representation, but I am convinced there is an easier way. I also tried imshow but then I have to think in graphic coordinates where the origin is in the upper left corner.
Problem solved
I managed to solve my problem using:
plt.imshow(Z,origin='lower',interpolation='bilinear')
If you can't change your mesh granularity, then try to go with imshow, which will essentially plot any 2D matrix as an image, where the values of each matrix cell represent the color to make that pixel. Using your example values:
In [3]: x = y = np.linspace(-1, 1, 21)
In [4]: z = np.array([i*i+j*j for j in y for i in x])
In [5]: Z = z.reshape(21, 21)
In [7]: plt.imshow(Z, interpolation='bilinear')
Out[7]: <matplotlib.image.AxesImage at 0x7f4864277650>
In [8]: plt.show()
you can use contourf
plt.contourf(X, Y, Z)
EDIT:
For more levels (smoother colour transitions), you can use more levels (contours)
For example:
plt.contourf(X, Y, Z, 100)

creating surface data for axes3d

Okay, apologies for this question but I'm pulling my hair out here.
I have a data structure loaded in python in the form:
[(1,0,#),(1,1,#),(1,2,#),(1,3,#),(2,0,#),(2,1,#) ... (26,3,#)]
with # being a different number each time that I wish to represent on the z-axis. You can see that x and y are always integers.
Plotting a scatter graph is simple:
x,y,z = zip(*data)
fig = plt.figure()
ax = fig.gca(projection = '3d')
surface = ax.scatter(x, y, z)
plt.show()
But when it comes to surfaces, I can see two methods:
1) Call ax.plot_trisurf(), which should work with 1D arrays similar to ax.scatter() and apparently works here, but for me gives me an error:
"AttributeError: Axes3D subplot object has not attribute 'plot_trisurf'"
This error also appears if I use the example source code at:
http://matplotlib.org/mpl_toolkits/mplot3d/tutorial.html#tri-surface-plots, suggesting it's something wrong with my installation - my Matplotlib version is 1.1.1rc,. This error does not appear if, for example, ax.plot_surface() is called, nor ax.scatter().
2) Use meshgrid() or griddata() in combination with ax.plot_surface() - in either case, after two days' of pouring over the documentation and examples, I still don't understand how to correctly use these in my case, particularly when it comes to generating the values for Z.
Any help would be much appreciated.
To address your first question (1) I believe you need to import Axes3D from the mplot3d library, even if you're not directly calling it. Maybe try adding
from mpl_toolkits.mplot3d import Axes3D
before your main code (this line triggered a memory while reading the tutorial).
As for (2), X, Y and Z need to be matrix (2d array) type objects. This can get confusing, but you may consider an example:
# two arrays - one for each axis
x = np.arange(-5, 5, 0.25)
y = np.arange(-5, 5, 0.25)
# create a mesh / matrix like object from the arrays
X, Y = np.meshgrid(x, y)
# create Z values - also in a mesh like shape
Z = np.sin(np.sqrt(X**2 + Y**2))
# plot!
surface = ax.plot_surface(X, Y, Z)
Here is an example of how could you extract your z-values from data
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
import matplotlib.pyplot as plt
import numpy as np
data = [(j,i,i**2 + j) for j in range(1,27) for i in range(4)]
print data
fig = plt.figure()
ax = fig.gca(projection='3d')
X = np.arange(0, 4, 1)
Y = np.arange(1, 27, 1)
X, Y = np.meshgrid(X, Y)
print X.shape
print Y.shape
Z = np.array([z for _,_,z in data]).reshape(26,4)
surf = ax.plot_surface(X, Y, Z, rstride=1, cstride=1, cmap=cm.coolwarm,
linewidth=0, antialiased=True)
fig.colorbar(surf, shrink=0.5, aspect=5)
plt.xlabel('X')
plt.ylabel('Y')
plt.show()

Plotting surface of implicitly defined volume

Having a volume implicitly defined by
x*y*z <= 1
for
-5 <= x <= 5
-5 <= y <= 5
-5 <= z <= 5
how would I go about plotting its outer surface using available Python modules, preferably mayavi?
I am aware of the function mlab.mesh, but I don't understand its input. It requires three 2D arrays, that I don't understand how to create having the above information.
EDIT:
Maybe my problem lies with an unsufficient understanding of the meshgrid()-function or the mgrid-class of numpy. I see that I have to use them in some way, but I do not completely grasp their purpose or what such a grid represents.
EDIT:
I arrived at this:
import numpy as np
from mayavi import mlab
x, y, z = np.ogrid[-5:5:200j, -5:5:200j, -5:5:200j]
s = x*y*z
src = mlab.pipeline.scalar_field(s)
mlab.pipeline.iso_surface(src, contours=[1., ],)
mlab.show()
This results in an isosurface (for x*y*z=1) of a volume though, which is not quite what I was looking for. What I am looking for is basically a method to draw an arbitrary surface, like a "polygon in 3d" if there is such a thing.
I created the following code, which plots a surface (works with mayavi, too). I would need to modify this code to my particular problem, but to do that I need to understand why and how a 3d surface is defined by three 2d-arrays? What do these arrays (x, y and z) represent?
import numpy as np
from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import axes3d, Axes3D
phi, theta = np.mgrid[0:np.pi:11j, 0:2*np.pi:11j]
x = np.sin(phi) * np.cos(theta)
y = np.sin(phi) * np.sin(theta)
z = np.cos(phi)
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_wireframe(x,y,z)
fig.show()
The outer surface, implicitly defined by
x*y*z = 1,
cannot be defined explicitly globally. To see this, consider x and y given, then:
z = 1/(x*y),
which is not defined for x = 0 or y = 0. Therefore, you can only define your surface locally for domains that do not include the singularity, e.g. for the domain
0 < x <= 5
0 < y <= 5
z is indeed defined (a hyperbolic surface). Similarly, you need to plot the surfaces for the other domains, until you have patched together
-5 <= x <= 5
-5 <= y <= 5
Note that your surface is not defined for x = 0 and y = 0, i.e. the axis of your coordinate system, so you cannot patch your surfaces together to get a globally defined surface.
Using numpy and matplotlib, you can plot one of these surfaces as follows (adopted from http://matplotlib.org/mpl_toolkits/mplot3d/tutorial.html#surface-plots):
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
import matplotlib.pyplot as plt
import numpy as np
fig = plt.figure()
ax = fig.gca(projection='3d')
X = np.arange(0.25, 5, 0.25)
Y = np.arange(0.25, 5, 0.25)
X, Y = np.meshgrid(X, Y)
Z = 1/(X*Y)
surf = ax.plot_surface(X, Y, Z, rstride=1, cstride=1, cmap=cm.coolwarm,
linewidth=0, antialiased=False)
ax.set_zlim(0, 10)
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z')
fig.colorbar(surf, shrink=0.5, aspect=5)
plt.show()
I'm not familiar with mayavi, but I would assume that creating the meshes with numpy would work the same.
The test case in the Mayavi docs where the function test_mesh() is defined is capable of producing a sphere. This is done by replacing
r = sin(m0*phi)**m1 + cos(m2*phi)**m3 + sin(m4*theta)**m5 + cos(m6*theta)**m7
with r = 1.0 say.
However, your problem is you need to understand that the equations you are writing define a volume when you want to draw a sphere. You need to reformulate them to give a parametric equation of a sphere. This is essentially what is done in the above example, but it may be worth your while to try it yourself. As a hint consider the equation of a circle and extend it.

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