I am trying to remove the edge color in the plot of a cylinder where I have set an alpha and facecolors. However, if I also set the facecolors, I can still see the edge colors. If I remove the alpha = 0.5 statement then the problem is resolved, however I need the alpha to be <1 . Here is an example:
You can still see the blue edgecolors even tough I have set the edgecolor to None.
This is the code where I use plot_surface()
ax.plot_surface(X, Y,Z, edgecolor = "None", facecolors = col1, alpha = 0.5)
Yet the edge colors are still there? However, if I remove the facecolors statement inside plot_surface() then the edge colors are no longer there. Here is the complete code:
import numpy as np
from matplotlib import cm
from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from scipy.linalg import norm
from mpl_toolkits.mplot3d.art3d import Poly3DCollection
import random
import numpy as np
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
origin = np.array([0, 0, 0])
#axis and radius
p0 = np.array([0, 0, 0])
p1 = np.array([8, 8, 8])
R = 4
#vector in direction of axis
v = p1 - p0
#find magnitude of vector
mag = norm(v)
#unit vector in direction of axis
v = v / mag
#make some vector not in the same direction as v
not_v = np.array([1, 0, 0])
if (v == not_v).all():
not_v = np.array([0, 1, 0])
#make vector perpendicular to v
n1 = np.cross(v, not_v)
#normalize n1
n1 /= norm(n1)
#make unit vector perpendicular to v and n1
n2 = np.cross(v, n1)
#surface ranges over t from 0 to length of axis and 0 to 2*pi
t = np.linspace(0, mag, 200)
theta = np.linspace(0, 2 * np.pi, 100)
#use meshgrid to make 2d arrays
t, theta = np.meshgrid(t, theta)
#generate coordinates for surface
X, Y, Z = [p0[i] + v[i] * t + R * np.sin(theta) * n1[i] + R * np.cos(theta) * n2[i] for i in [0, 1, 2]]
col1 = plt.cm.Blues(np.linspace(0,1,200)) # linear gradient along the t-axis
col1 = np.repeat(col1[np.newaxis,:, :], 100, axis=0) # expand over the theta- axis
ax.plot_surface(X, Y,Z, edgecolor = None, facecolors = col1, alpha = 0.5)
#plot axis
ax.plot(*zip(p0, p1), color = 'red')
ax.set_xlim(0, 10)
ax.set_ylim(0, 10)
ax.set_zlim(0, 10)
plt.axis('off')
ax.axes.get_xaxis().set_visible(False)
ax.axes.get_yaxis().set_visible(False)
plt.show()
Setting linewidth=0 in plot_surface() solves this problem:
ax.plot_surface(X, Y, Z, edgecolor=None, facecolors=col1, alpha=0.5, linewidth=0)
p.s.: I didn't find this worth an answer, but per: Question with no answers, but issue solved in the comments (or extended in chat), I added it as a quick answer so the question can be marked as solved
Related
How to rotate a function by the desired angle (for instance, 30 degrees)?
import matplotlib.pyplot as plt
import numpy as np
from numpy import exp, sin
def g(y):
return exp(-y)*sin(4*y)
y = np.linspace(0, 1.8, 501)
values = g(y)
fig, ax = plt.subplots(figsize=(5,5))
plt.plot(y, values)
plt.show()
Using the cosine and sine of the angle, you can create a rotation matrix. Multiplying each point (y, g(y)) with that matrix create a rotation around 0,0.
Here is some Python/numpy code to illustrate how everything could work together:
import matplotlib.pyplot as plt
import numpy as np
def g(y):
return np.exp(-y) * np.sin(4 * y)
y = np.linspace(0, 1.8, 501)
values = g(y)
theta = np.radians(30)
c, s = np.cos(theta), np.sin(theta)
rot_matrix = np.array(((c, s), (-s, c)))
xy = np.array([y, values]).T # rot_matrix
fig, ax = plt.subplots(figsize=(5, 5))
plt.plot(y, values)
plt.plot(xy[:, 0], xy[:, 1])
plt.axis('equal') # so angles on the screen look like the real angles
plt.show()
PS: To rotate around another point, first subtract the rotation center, do the rotation and then add it again:
center = np.array([0.9, 0])
xy = (np.array([y, values]).T - center) # rot_matrix + center
I want to plot a quantity which is given on a parametric surface in 3d space (for example the temperature distribution on a sphere). I can plot a parametric 3D plot of the sphere (as a function of the two parameters phi and theta) but I don't know how to make the colors of the polygons making up the sphere depend on the parameters theta and phi (normally, the color of a polygon is simply determined by the z-Position of the polygon).
Here's a basic example which plots a torus with colormap:
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
angle = np.linspace(0, 2 * np.pi, 32)
theta, phi = np.meshgrid(angle, angle)
r, R = .25, 1.
X = (R + r * np.cos(phi)) * np.cos(theta)
Y = (R + r * np.cos(phi)) * np.sin(theta)
Z = r * np.sin(phi)
# Display the mesh
fig = plt.figure()
ax = fig.gca(projection = '3d')
ax.set_xlim3d(-1, 1)
ax.set_ylim3d(-1, 1)
ax.set_zlim3d(-1, 1)
ax.plot_surface(X, Y, Z, rstride = 1, cstride = 1,cmap="hot")
plt.show()
However, the colors of the files are given by the z position of the tile, I want the color to be given by a function f(x,y).
Does anyone know how I can achieve this dependency in Matplotlib?
Thanks very much!
Ok, if anyone else is looking for a solution to this problem here's a possible solution:
The colors of the individual faces making up the surface plot can be set using the keyword argument facecolors. The following code will use the function X**2+Y**2 for coloring the faces of the parametric surface:
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.colors as mcolors
from matplotlib import cm
import matplotlib.pyplot as plt
import numpy as np
# Generate torus mesh
angle = np.linspace(0, 2 * np.pi, 32)
theta, phi = np.meshgrid(angle, angle)
r, R = .25, 1.
X = (R + r * np.cos(phi)) * np.cos(theta)
Y = (R + r * np.cos(phi)) * np.sin(theta)
Z = r * np.sin(phi)
colorfunction=(X**2+Y**2)
norm=mcolors.Normalize(colorfunction.min(),colorfunction.max())
# Display the mesh
fig = plt.figure(figsize=(7, 7))
ax = fig.add_subplot(projection='3d')
ax.set_xlim3d(-1, 1)
ax.set_ylim3d(-1, 1)
ax.set_zlim3d(-1, 1)
ax.plot_surface(X, Y, Z, rstride = 1, cstride = 1, facecolors=cm.jet(norm(colorfunction)))
plt.show()
I want to plot a truncated cone by using exactly the same method used in
Plotting a solid cylinder centered on a plane in Matplotlib; which plots a cylinder when two points on the center of each base and the radius are known. On the other hand, I want to plot a truncated cone when the coordinates of the two points on the center of its bases and the radius of each base are known.
It seems that I just should change the second last line of the function in the following program which plots a cylinder, but I could not do this in all of my efforts.
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
from scipy.linalg import norm
import pylab as pllt
fig = pllt.figure()
ax = fig.add_subplot(1,1,1, projection='3d')
#ax = pllt.subplot2grid((2,2), (0,0), rowspan=2, projection='3d')
#axis and radius
def cylinder(p0,p1,R,ccc):
#vector in direction of axis
v = p1 - p0
#find magnitude of vector
mag = norm(v)
#unit vector in direction of axis
v = v / mag
#make some vector not in the same direction as v
not_v = np.array([1, 1, 0])
if (v == not_v).all():
not_v = np.array([0, 1, 0])
#make vector perpendicular to v
n1 = np.cross(v, not_v)
#print n1,'\t',norm(n1)
#normalize n1
n1 /= norm(n1)
#make unit vector perpendicular to v and n1
n2 = np.cross(v, n1)
#surface ranges over t from 0 to length of axis and 0 to 2*pi
t = np.linspace(0, mag, 80)
theta = np.linspace(0, 2 * np.pi, 80)
#use meshgrid to make 2d arrays
t, theta = np.meshgrid(t, theta)
#generate coordinates for surface
X, Y, Z = [p0[i] + v[i] * t + R * np.sin(theta) * n1[i] + R * np.cos(theta) * n2[i] for i in [0, 1, 2]]
ax.plot_surface(X, Y, Z,color=ccc,linewidth=0, antialiased=False)
A0 = np.array([1, 3, 2])
A1 = np.array([8, 5, 9])
ax.set_xlim(0,10)
ax.set_ylim(0,10)
ax.set_zlim(0,10)
cylinder(A0,A1,1,'blue')
pllt.show()
I think I should change the radius as a function of v=p1-p0 as mentioned in:
http://mathworld.wolfram.com/ConicalFrustum.html to be able to do this.
Please let me know if there is any way to do this.
Instead of a constant radius, R, make it change from R0 to R1:
R = np.linspace(R0, R1, n)
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
from scipy.linalg import norm
import pylab as plt
fig = plt.figure()
ax = fig.add_subplot(1, 1, 1, projection='3d')
def truncated_cone(p0, p1, R0, R1, color):
"""
Based on https://stackoverflow.com/a/39823124/190597 (astrokeat)
"""
# vector in direction of axis
v = p1 - p0
# find magnitude of vector
mag = norm(v)
# unit vector in direction of axis
v = v / mag
# make some vector not in the same direction as v
not_v = np.array([1, 1, 0])
if (v == not_v).all():
not_v = np.array([0, 1, 0])
# make vector perpendicular to v
n1 = np.cross(v, not_v)
# print n1,'\t',norm(n1)
# normalize n1
n1 /= norm(n1)
# make unit vector perpendicular to v and n1
n2 = np.cross(v, n1)
# surface ranges over t from 0 to length of axis and 0 to 2*pi
n = 80
t = np.linspace(0, mag, n)
theta = np.linspace(0, 2 * np.pi, n)
# use meshgrid to make 2d arrays
t, theta = np.meshgrid(t, theta)
R = np.linspace(R0, R1, n)
# generate coordinates for surface
X, Y, Z = [p0[i] + v[i] * t + R *
np.sin(theta) * n1[i] + R * np.cos(theta) * n2[i] for i in [0, 1, 2]]
ax.plot_surface(X, Y, Z, color=color, linewidth=0, antialiased=False)
A0 = np.array([1, 3, 2])
A1 = np.array([8, 5, 9])
ax.set_xlim(0, 10)
ax.set_ylim(0, 10)
ax.set_zlim(0, 10)
truncated_cone(A0, A1, 1, 5, 'blue')
plt.show()
I am trying to make a 'closed' cylinder in matplotlib but I am not sure how to go about doing this. So far I have a cylinder with the ends open, the code for this is as follows:
#make a cylinder without the ends closed
import numpy as np
from matplotlib import cm
from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from scipy.linalg import norm
from mpl_toolkits.mplot3d.art3d import Poly3DCollection
import numpy as np
import math
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
origin = [0,0,0]
#radius = R
p0 = np.array(origin)
p1 = np.array([8, 8, 8])
origin = np.array(origin)
R = 4
#vector in direction of axis
v = p1 - p0
#find magnitude of vector
mag = norm(v)
#unit vector in direction of axis
v = v / mag
#make some vector not in the same direction as v
not_v = np.array([1, 0, 0])
if (v == not_v).all():
not_v = np.array([0, 1, 0])
#make vector perpendicular to v
n1 = np.cross(v, not_v)
#normalize n1
n1 /= norm(n1)
#make unit vector perpendicular to v and n1
n2 = np.cross(v, n1)
#surface ranges over t from 0 to length of axis and 0 to 2*pi
t = np.linspace(0, mag, 600)
theta = np.linspace(0, 2 * np.pi, 100)
#use meshgrid to make 2d arrays
t, theta = np.meshgrid(t, theta)
#generate coordinates for surface
X, Y, Z = [p0[i] + v[i] * t + R * np.sin(theta) * n1[i] + R * np.cos(theta) * n2[i] for i in [0, 1, 2]]
#make the color for the faces
col1 = plt.cm.autumn(np.ones(600)) # linear gradient along the t-axis
col1 = np.repeat(col1[np.newaxis,:, :], 100, axis=0) # expand over the theta-axis
ax.plot_surface(X, Y,Z, facecolors = col1, shade = True,edgecolors = "None", alpha = 0.4, linewidth = 0)
plt.show()
Running this code produces the following image
How would I close the ends of the cylinder with a solid circle (i.e. disk)?
A quick and easy way that's similar to your other code is to generate a surface using strips from r=0 to r=R. Right before plt.show() add the following lines:
R = np.array([0,R])
# cap at t=0
X, Y, Z = [p0[i] + np.outer(R, np.sin(theta)) * n1[i] + np.outer(R, np.cos(theta))*n2[i] for i in [0, 1, 2]]
ax.plot_surface(X, Y, Z, edgecolors = "r", alpha=.4, linewidth = .1)
# cap at t=mag
X, Y, Z = [p0[i] + v[i]*mag + np.outer(R, np.sin(theta)) * n1[i] + np.outer(R, np.cos(theta))*n2[i] for i in [0, 1, 2]]
ax.plot_surface(X, Y, Z, edgecolors = "r", alpha=.4, linewidth = .1)
Here the colors are more for illustrative purposes, mostly so you can see the strips. The result looks like:
I have the following code which produces a cylinder-like object using matplotlib:
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d.art3d import Poly3DCollection
import numpy as np
fig = plt.figure()
ax = fig.gca(projection='3d')
nphi,nz=7,20
r=1 # radius of cylinder
phi = np.linspace(0,360, nphi)/180.0*np.pi
z= np.linspace(0,1.0,nz)
print z
cols=[]
verts2 = []
for i in range(len(phi)-1):
cp0= r*np.cos(phi[i])
cp1= r*np.cos(phi[i+1])
sp0= r*np.sin(phi[i])
sp1= r*np.sin(phi[i+1])
for j in range(len(z)-1):
z0=z[j]
z1=z[j+1]
verts=[]
verts.append((cp0, sp0, z0))
verts.append((cp1, sp1, z0))
verts.append((cp1, sp1, z1))
verts.append((cp0, sp0, z1))
verts2.append(verts)
value=np.random.rand()
#print value
col=plt.cm.rainbow(0.9)
#print col
cols.append(col)
poly3= Poly3DCollection(verts2, facecolor=cols,edgecolor = "none" )
poly3.set_alpha(0.8)
ax.add_collection3d(poly3)
ax.set_xlabel('X')
ax.set_xlim3d(-1, 1)
ax.set_ylabel('Y')
ax.set_ylim3d(-1, 1)
ax.set_zlabel('Z')
ax.set_zlim3d(0, 1)
plt.show()
This code produces the following image:
However as you can see the are sharp corners in the figure. Is there anyway to make these edges rounder so that the figure looks like a proper cylinder with a circular cross-section as opposed to a hexagonal cross-section?
The third argument to
np.linspace
controls how many values you want it to generate. Thus, nphi controls the
number of values in phi, and nz controls the number of values in z:
phi = np.linspace(0,360, nphi)/180.0*np.pi
z = np.linspace(0,1.0,nz)
So if you increase nphi, then you'll get more points along the circle:
cp0 = r*np.cos(phi[i])
sp0 = r*np.sin(phi[i])
For example, try changing nphi, nz = 7,20 to nphi, nz = 70, 2.
Note that there is no need for nz to be greater than 2 since the sides of the
cylinder are flat in the z direction.
By the way, the double for-loop can be replaced by:
PHI, Z = np.meshgrid(phi, z)
CP = r * np.cos(PHI)
SP = r * np.sin(PHI)
XYZ = np.dstack([CP, SP, Z])
verts = np.stack(
[XYZ[:-1, :-1], XYZ[:-1, 1:], XYZ[1:, 1:], XYZ[1:, :-1]], axis=-2).reshape(-1, 4, 3)
So, for example,
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d.art3d import Poly3DCollection
import numpy as np
fig = plt.figure()
ax = fig.gca(projection='3d')
nphi, nz = 70, 2
r = 1 # radius of cylinder
phi = np.linspace(0, 360, nphi) / 180.0 * np.pi
z = np.linspace(0, 1.0, nz)
PHI, Z = np.meshgrid(phi, z)
CP = r * np.cos(PHI)
SP = r * np.sin(PHI)
XYZ = np.dstack([CP, SP, Z])
verts = np.stack(
[XYZ[:-1, :-1], XYZ[:-1, 1:], XYZ[1:, 1:], XYZ[1:, :-1]], axis=-2).reshape(-1, 4, 3)
cmap = plt.cm.rainbow
cols = cmap(np.random.random())
poly3 = Poly3DCollection(verts, facecolor=cols, edgecolor="none")
poly3.set_alpha(0.8)
ax.add_collection3d(poly3)
ax.set_xlabel('X')
ax.set_xlim3d(-1, 1)
ax.set_ylabel('Y')
ax.set_ylim3d(-1, 1)
ax.set_zlabel('Z')
ax.set_zlim3d(0, 1)
plt.show()
yields