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Radially 'sweep out' a 2D contour plot to create a 3D plot (Python, Matplotlib)

I have three 1D arrays, which represent radius, height, and an intensity measured at that point. I have plotted these to create a 2D contour map. A simple example of the way in which the data is stored is below:
import numpy as np
import matplotlib.pyplot as plt
x = [1,1,1,2,2,2,3,3,3]
y = [1,2,3,1,2,3,1,2,3]
intensity = [5,6,8,9,9,11,15,5,2]
plt.xlabel('Radius')
plt.ylabel('Height')
plt.tricontourf(x,y,intensity)
plt.colorbar(label='Intensity')
plt.show()
(I have had to use plt.tricontourf rather than plt.contour, since the z data is not 2D)
I am looking to create a 3D plot by 'sweeping' the 2D plot through 360 degrees, creating a disk which is azimuthally symmetric. The image below illustrates what I am trying to do...
...with the data interpolated smoothly through the 360 degrees.
There are a couple of similar questions, notably this one, but this does not use three sets of data to create the contours.

Technically you cannot rotate a 2D plot and get a 3D surface. You can only rotate a 2D curve and get a 3D surface. If this is the case, you could do it as:
import numpy as np
from matplotlib import cm
import matplotlib.pyplot as plt
fig = plt.figure(figsize = (8, 6))
ax = fig.add_subplot(projection='3d')
N = 100
r = np.linspace(0, 1, N)
z = np.sqrt(1 - r**2)
intensity = np.linspace(0, 1, N).reshape(1, -1)
theta = np.linspace(0, 2*np.pi-1e-3, N)
X = np.outer(np.cos(theta), r)
Y = np.outer(np.sin(theta), r)
Z = np.repeat(z.reshape(1, -1), N, axis = 0)
surf = ax.plot_surface(X, Y, Z, facecolors=cm.jet(np.repeat(intensity, N, axis = 0)))
ax.axes.set_zlim3d(-1, 1)
plt.show()
In the code I rotated a curve to create half a unit sphere and color it according to intensity:
to
If you insist on plotting all the points, I would suggest a 3d scatter plot, I did some linear interpolation to show more points than the original 9:
from scipy.interpolate import interp2d
x = [1,1,1,2,2,2,3,3,3]
y = [1,2,3,1,2,3,1,2,3]
intensity = [5,6,8,9,9,11,15,5,2]
# number of points to interpolate in 3d space
N = 36
# number of points to interpolate in 2d space
N_2d = 10
f = interp2d(x, y, intensity)
# sample along the radius
r = np.linspace(1,3,N_2d)
# sample along z
z = np.linspace(1,3,N_2d)
intensity = f(r, z)
r,z = np.meshgrid(r, z)
theta = np.linspace(0, 2*np.pi, N)
X = np.outer(np.cos(theta), r)
Y = np.outer(np.sin(theta), r)
Z = np.repeat(z.reshape(1, -1), N, axis = 0)
fig = plt.figure(figsize = (10, 6))
ax = fig.add_subplot(projection='3d')
ax.scatter3D(X, Y, Z, c=np.tile(intensity.T, N).T, alpha = 0.5)
plt.show()

I want to know how to draw with python matplotlib with X Y Z values

I have an X Y Z measurement
I tried mplot3d example code but I don't know.
I would like to draw a chart that shows the difference in height with these values.
I want to know how to apply a value to an example.
have the image below
Image
x y z
```
X Y Z
0.0056 -149.00237 0.01375
-11.99014 -148.5314 0.01323
-23.90088 -147.08148 0.01294
-35.67262 -144.68161 0.01386
-47.19637 -141.34179 0.0139
-58.41713 -137.08002 0.01303
-69.24989 -131.93131 0.01319
-79.64266 -125.94063 0.01233
-89.52245 -119.11801 0.0124
-98.81825 -111.51743 0.01235
..........
```
Example code
```
# This import registers the 3D projection, but is otherwise unused.
from mpl_toolkits.mplot3d import Axes3D # noqa: F401 unused import
import matplotlib.pyplot as plt
import numpy as np
n_radii = 8
n_angles = 36
# Make radii and angles spaces (radius r=0 omitted to eliminate duplication).
radii = np.linspace(0.125, 1.0, n_radii)
angles = np.linspace(0, 2*np.pi, n_angles, endpoint=False)[..., np.newaxis]
# Convert polar (radii, angles) coords to cartesian (x, y) coords.
# (0, 0) is manually added at this stage, so there will be no duplicate
# points in the (x, y) plane.
x = np.append(0, (radii*np.cos(angles)).flatten())
y = np.append(0, (radii*np.sin(angles)).flatten())
# Compute z to make the pringle surface.
z = np.sin(-x*y)
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.plot_trisurf(x, y, z, linewidth=0.2, antialiased=True)
plt.show()
```

Python: How to revolve a surface around z axis and make a 3d plot?

I want to get 2d and 3d plots as shown below.
The equation of the curve is given.
How can we do so in python?
I know there may be duplicates but at the time of posting
I could not fine any useful posts.
My initial attempt is like this:
# Imports
import numpy as np
import matplotlib.pyplot as plt
# to plot the surface rho = b*cosh(z/b) with rho^2 = r^2 + b^2
z = np.arange(-3, 3, 0.01)
rho = np.cosh(z) # take constant b = 1
plt.plot(rho,z)
plt.show()
Some related links are following:
Rotate around z-axis only in plotly
The 3d-plot should look like this:

Ok so I think you are really asking to revolve a 2d curve around an axis to create a surface. I come from a CAD background so that is how i explain things.
and I am not the greatest at math so forgive any clunky terminology. Unfortunately you have to do the rest of the math to get all the points for the mesh.
Heres your code:
#import for 3d
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib.pyplot as plt
change arange to linspace which captures the endpoint otherwise arange will be missing the 3.0 at the end of the array:
z = np.linspace(-3, 3, 600)
rho = np.cosh(z) # take constant b = 1
since rho is your radius at every z height we need to calculate x,y points around that radius. and before that we have to figure out at what positions on that radius to get x,y co-ordinates:
#steps around circle from 0 to 2*pi(360degrees)
#reshape at the end is to be able to use np.dot properly
revolve_steps = np.linspace(0, np.pi*2, 600).reshape(1,600)
the Trig way of getting points around a circle is:
x = r*cos(theta)
y = r*sin(theta)
for you r is your rho, and theta is revolve_steps
by using np.dot to do matrix multiplication you get a 2d array back where the rows of x's and y's will correspond to the z's
theta = revolve_steps
#convert rho to a column vector
rho_column = rho.reshape(600,1)
x = rho_column.dot(np.cos(theta))
y = rho_column.dot(np.sin(theta))
# expand z into a 2d array that matches dimensions of x and y arrays..
# i used np.meshgrid
zs, rs = np.meshgrid(z, rho)
#plotting
fig, ax = plt.subplots(subplot_kw=dict(projection='3d'))
fig.tight_layout(pad = 0.0)
#transpose zs or you get a helix not a revolve.
# you could add rstride = int or cstride = int kwargs to control the mesh density
ax.plot_surface(x, y, zs.T, color = 'white', shade = False)
#view orientation
ax.elev = 30 #30 degrees for a typical isometric view
ax.azim = 30
#turn off the axes to closely mimic picture in original question
ax.set_axis_off()
plt.show()
#ps 600x600x600 pts takes a bit of time to render
I am not sure if it's been fixed in latest version of matplotlib but the setting the aspect ratio of 3d plots with:
ax.set_aspect('equal')
has not worked very well. you can find solutions at this stack overflow question

Only rotate the axis, in this case x
import numpy as np
import matplotlib.pyplot as plt
import mpl_toolkits.mplot3d.axes3d as axes3d
np.seterr(divide='ignore', invalid='ignore')
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
x = np.linspace(-3, 3, 60)
rho = np.cosh(x)
v = np.linspace(0, 2*np.pi, 60)
X, V = np.meshgrid(x, v)
Y = np.cosh(X) * np.cos(V)
Z = np.cosh(X) * np.sin(V)
ax.set_xlabel('eje X')
ax.set_ylabel('eje Y')
ax.set_zlabel('eje Z')
ax.plot_surface(X, Y, Z, cmap='YlGnBu_r')
plt.plot(x, rho, 'or') #Muestra la curva que se va a rotar
plt.show()
The result:

Ellipsoid creation in Python

I have ran into a problem relating to the drawing of the Ellipsoid.
The ellipsoid that I am drawing to draw is the following:
x**2/16 + y**2/16 + z**2/16 = 1.
So I saw a lot of references relating to calculating and plotting of an Ellipse void and in multiple questions a cartesian to spherical or vice versa calculation was mentioned.
Ran into a website that had a calculator for it, but I had no idea on how to successfully perform this calculation. Also I am not sure as to what the linspaces should be set to. Have seen the ones that I have there as defaults, but as I got no previous experience with these libraries, I really don't know what to expect from it.
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np
fig = plt.figure(figsize=plt.figaspect(1)) # Square figure
ax = fig.add_subplot(111, projection='3d')
multip = (1, 1, 1)
# Radii corresponding to the coefficients:
rx, ry, rz = 1/np.sqrt(multip)
# Spherical Angles
u = np.linspace(0, 2 * np.pi, 100)
v = np.linspace(0, np.pi, 100)
# Cartesian coordinates
#Lots of uncertainty.
#x =
#y =
#z =
# Plot:
ax.plot_surface(x, y, z, rstride=4, cstride=4, color='b')
# Axis modifications
max_radius = max(rx, ry, rz)
for axis in 'xyz':
getattr(ax, 'set_{}lim'.format(axis))((-max_radius, max_radius))
plt.show()

Your ellipsoid is not just an ellipsoid, it's a sphere.
Notice that if you use the substitution formulas written below for x, y and z, you'll get an identity. It is in general easier to plot such a surface of revolution in a different coordinate system (spherical in this case), rather than attempting to solve an implicit equation (which in most plotting programs ends up jagged, unless you take some countermeasures).
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np
phi = np.linspace(0,2*np.pi, 256).reshape(256, 1) # the angle of the projection in the xy-plane
theta = np.linspace(0, np.pi, 256).reshape(-1, 256) # the angle from the polar axis, ie the polar angle
radius = 4
# Transformation formulae for a spherical coordinate system.
x = radius*np.sin(theta)*np.cos(phi)
y = radius*np.sin(theta)*np.sin(phi)
z = radius*np.cos(theta)
fig = plt.figure(figsize=plt.figaspect(1)) # Square figure
ax = fig.add_subplot(111, projection='3d')
ax.plot_surface(x, y, z, color='b')

Plot stack of unit vectors in cylindrical coordinates - matplotlib

I have a python program that calculates angles for me and outputs them in a list.
What I would like to do is plot a stack of arrows that are unit vectors pointing in the direction of the angle. So I thought cylindrical coordinates would be best since they only have one angular coordinate.
I've tried pyplot.quiver but I don't think that can do anything in 3D, and a 3D line plot didn't work either.
Is there a way of doing this without laboriously converting each (length, height, angle) into a pair of vectors (a, b, c),(length*cos(angle), length*sin(angle), height)?

If you have a list of angles, you can easily calculate vectors associated with those angles using numpy.
import numpy as np
import matplotlib.pyplot as plt
angles = np.random.rand(100)
length = 1.
vectors_2d = np.vstack((length * np.cos(angles), length * np.sin(angles))).T
for x, y in vectors_2d:
plt.plot([0, x], [0, y])
plt.show()
If you really want it in cylindrical instead of polar coords, then
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
angles = np.random.rand(100)
length = 1.
heights = np.arange(len(angles))
vectors_3d = np.vstack((length * np.cos(angles),
length * np.sin(angles),
heights)).T
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
for x, y, z in vectors_3d:
ax.plot([0, x], [0, y], zs=[z, z])
plt.show()
Edit: I know how to put arrows on plots using pyplot.quiver. However, I don't think mplot3d plays nicely with quiver. Maybe someone like #tcaswell can help out with a work around. But in 2D, you can do
import numpy as np
import matplotlib.pyplot as plt
angles = np.random.rand(100)
# Define coords for arrow tails (the origin)
x0, y0 = np.zeros(100), np.zeros(100)
# Define coords for arrow tips (cos/sin)
x, y = np.cos(angles), np.sin(angles)
# in case you want colored arrows
colors = 'bgrcmyk'
colors *= colors * (len(x0) / len(colors) + 1)
plt.quiver(x0, y0, x, y, color=colors[:len(x0)], scale=1) #scale sets the length
plt.show()