I try to draw a sphere to represent a qubit using matplotlib
import numpy as np
import matplotlib.pyplot as plt
theta = [0, np.pi]
phi = [0, 2* np.pi]
N=100
theta_array = np.linspace(theta[0], theta[1], N)
phi_array = np.linspace(phi[0], phi[1], N)
theta_grid, phi_grid = np.meshgrid(theta_array,phi_array)
x = np.sin(theta_grid) * np.cos(phi_grid)
y = np.sin(theta_grid) * np.sin(phi_grid)
z = np.cos(theta_grid)
fig = plt.figure(figsize=(6,6))
ax = fig.gca(projection='3d')
ax.plot_surface(x, y, z, rstride=1, cstride=1, shade=False,linewidth=0)
plt.show()
I want to add tube arrows on the sphere with directions parallel with xyz axis, like this:
I am not an expert in matplotlib, so it's seem pretty tough to me. Can anyone help me? thanks in advance!
You can use the quiver function to to what you want.
See code below:
import numpy as np
import matplotlib.pyplot as plt
theta = [0, np.pi]
phi = [0, 2* np.pi]
N=100
theta_array = np.linspace(theta[0], theta[1], N)
phi_array = np.linspace(phi[0], phi[1], N)
theta_grid, phi_grid = np.meshgrid(theta_array,phi_array)
x = np.sin(theta_grid) * np.cos(phi_grid)
y = np.sin(theta_grid) * np.sin(phi_grid)
z = np.cos(theta_grid)
fig = plt.figure(figsize=(6,6))
ax = fig.gca(projection='3d')
ax.view_init(azim=60)
ax.plot_surface(x, y, z, rstride=1, cstride=1, shade=False,linewidth=0)
#Set up arrows
ax.quiver(1,0,0,1,0,0,color = 'k', alpha = .8, lw = 3) #x arrow
ax.text(2.4,0,0,'Sx',fontsize=20)
ax.quiver(0,1,0,0,1,0,color = 'k', alpha = .8, lw = 3)#y arrow
ax.text(0,2.4,0,'Sy',fontsize=20)
ax.quiver(0,0,1,0,0,1,color = 'k', alpha = .8, lw = 3)#z arrow
ax.text(-0.3,0,1.8,'Sz',fontsize=20)
plt.show()
And the output gives:
Related
I want to draw a spin systems. I have the picture below. However, I want the quivers to be INSIDE the scatter circles and the "Main arrow" to be a bright one. Thus, how can I order three plots?
Thank You in advance!
Here is the code:
import matplotlib.pyplot as plt
from matplotlib import cm, colors
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
############################## A sphere #################################
r = 4
pi = np.pi
cos = np.cos
sin = np.sin
phi, theta = np.mgrid[0.0:pi:100j, 0.0:2.0*pi:100j]
x = r*sin(phi)*cos(theta)
y = r*sin(phi)*sin(theta)
z = r*cos(phi)
#Set colours and render
fig = plt.figure(figsize=(10,10))
ax = fig.add_subplot(111, projection='3d')
ax.plot_surface(x, y, z, rstride=1, cstride=1, color='red', alpha=0.6, linewidth=0)
############################ One arrow ########################
x1 = 0
y1 = 0
z1 = 0
dx1 = 0
dy1 = 0
dz1 = 2
ax.quiver(x1, y1, z1, dx1, dy1, dz1, pivot='middle', color="blue", linewidths=5.0,length=1.4)
ax.scatter(0, 0, 0,color="blue", s=800)
############################ Array of arrows ############################################
ax.set_xlim([-4,4])
ax.set_ylim([-4,4])
ax.set_zlim([-4,4])
ax.set_aspect("equal")
plt.tight_layout()
N = 5 # number of points - 5x5x5
scale = 0.4 # scale for the increment of an arrow
ax = fig.gca(projection='3d')
xx = np.linspace(-3.5, 3.5, N)
yy = np.linspace(-3.5, 3.5, N)
zz = np.linspace(-3.5, 3.5, N)
dxx = scale*np.random.rand(N)
dyy = scale*np.random.rand(N)
dzz = scale*np.random.rand(N)
x, y, z = np.meshgrid(yy, zz, xx)
dx, dy, dz = np.meshgrid(dxx, dyy, dzz)
ax.scatter(x, y, z,color="orange", s=200)
ax.quiver(x, y, z, dx, dy, dz, pivot='middle', length=1.6, color="blue")
plt.show()
here is the code im using and I've also attached the output. I'd like to plot a two dimensional lognorm function as a 3d surface, the above code is supposed to do this however the output results in the entire plane being skewed rather than just the z values. any help or suggestions would be greatly appreciated.
dx = 90 - (-90)
dy = 90 - (-90)
c = [dx + dx/2.0, dy+dy/2.0]
z = np.zeros((400, 400))
x = np.linspace(-90, 90, 400)
y = x.copy()
for i in range(len(x)):
for j in range(len(y)):
p =[x[i], y[j]]
d = math.sqrt((p[0]-c[0])**2 + (p[1]-c[1])**2)
t = d
z[i][j] = lognorm.pdf(t, 1.2)
fig = plt.figure()
ax = fig.add_subplot(111, projection = '3d')
ax.plot_surface(x,y, z, cmap = 'viridis')
plt.show()
output of the provided code
ideally I'd like for it to look something like this.
this is the image here
I think you wanted to plot a 3D surface and here is an example:
#!/usr/bin/python3
# 2018/10/25 14:44 (+0800)
# Plot a 3D surface
from scipy.stats import norm, lognorm
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
xy = np.linspace(-5, 5, 400)
xx, yy = np.meshgrid(xy)
t = np.sqrt(xx**2 + yy**2)
zz = lognorm.pdf(t, 1.2)
fig = plt.figure()
ax = fig.add_subplot(111, projection = '3d')
ax.plot_surface(xx,yy, zz, cmap = 'viridis')
plt.show()
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
I'm looking to plot a 2D probability distribution with one of its marginals in a single plot using Python and matplotlib. I'm almost there, but the line in the plot is always drawn in front of the surface, instead of being occluded properly. How do I fix this?
import numpy as np
import scipy.stats as stats
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import axes3d
fig = plt.figure()
ax = fig.gca(projection='3d')
delta = 0.05
f = 0.5
X, Y = np.meshgrid(np.arange(-3.0, 3.0, delta),
np.arange(-3.0, 3.0, delta))
xy = np.hstack((X.flatten()[:, None], Y.flatten()[:, None]))
p1 = stats.multivariate_normal.pdf(xy, mean=[1, -1], cov=(np.eye(2) * 0.28 * f))
p2 = stats.multivariate_normal.pdf(xy, mean=[-1, 1], cov=(np.eye(2) * 0.5 * f))
p = 0.3 * p1 + 0.7 * p2
Z = p.reshape(len(X), len(X))
plt.plot(X[0, :], np.zeros(len(X)) + 3, np.sum(Z, 0) * 0.05) # , color='red')
ax.plot_surface(X, Y, Z, alpha=1.0, cmap='jet', linewidth=0.1, rstride=2, cstride=2)
ax.set_xlabel('Object colour')
ax.set_ylabel('Illumination colour')
ax.set_zlabel('Probability density')
ax.set_zlim(min(cont_offset, np.min(Z)), max(np.max(Z), cont_offset))
plt.show()
The built-in contour function at least gets the z-order right; if you don't want the full thing, you could cheat with a calculated Z. To start, replacing your call to plt.plot with this:
from matplotlib import cm
cset = ax.contour(X, Y, Z, zdir='y', offset=3, cmap='binary')
cset = ax.contour(X, Y, Z, zdir='x', offset=-3, cmap='Blues')
Faking up Z's for the contours, one way:
from matplotlib import cm
Zys = np.zeros_like(Z)
Zys[60,:] = Z.max(0)
cset = ax.contour(X, Y, Zys, zdir='y', offset=3, cmap='binary')
Zys = np.zeros_like(Z)
Zys[:,60] = Z.max(1)
cset = ax.contour(X, Y, Zys, zdir='x', offset=-3, cmap='Blues')
More ambitiously, somewhere in the contour code they're calculating the z-order...
I can't find a way to draw errorbars in a 3D scatter plot in matplotlib.
Basically, for the following piece of code
from mpl_toolkits.mplot3d import axes3d
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
X, Y, Z = axes3d.get_test_data(1)
ax.scatter(X, Y, zs = Z, zdir = 'z')
I am looking for something like
ax.errorbar(X,Y, zs = Z, dY, dX, zserr = dZ)
Is there a way to do this in mplot3d? If not, are there other libraries with this function?
There is clearly example on forum http://mple.m-artwork.eu/home/posts/simple3dplotwith3derrorbars
Here is the code but is not built-in functionality:
import numpy as np
import matplotlib.pyplot as plt
import mpl_toolkits.mplot3d.axes3d as axes3d
fig = plt.figure(dpi=100)
ax = fig.add_subplot(111, projection='3d')
#data
fx = [0.673574075,0.727952994,0.6746285]
fy = [0.331657721,0.447817839,0.37733386]
fz = [18.13629648,8.620699842,9.807536512]
#error data
xerror = [0.041504064,0.02402152,0.059383144]
yerror = [0.015649804,0.12643117,0.068676131]
zerror = [3.677693713,1.345712547,0.724095592]
#plot points
ax.plot(fx, fy, fz, linestyle="None", marker="o")
#plot errorbars
for i in np.arange(0, len(fx)):
ax.plot([fx[i]+xerror[i], fx[i]-xerror[i]], [fy[i], fy[i]], [fz[i], fz[i]], marker="_")
ax.plot([fx[i], fx[i]], [fy[i]+yerror[i], fy[i]-yerror[i]], [fz[i], fz[i]], marker="_")
ax.plot([fx[i], fx[i]], [fy[i], fy[i]], [fz[i]+zerror[i], fz[i]-zerror[i]], marker="_")
#configure axes
ax.set_xlim3d(0.55, 0.8)
ax.set_ylim3d(0.2, 0.5)
ax.set_zlim3d(8, 19)
plt.show()
I ended up writing the method for matplotlib: official example for 3D errorbars:
import matplotlib.pyplot as plt
import numpy as np
ax = plt.figure().add_subplot(projection='3d')
# setting up a parametric curve
t = np.arange(0, 2*np.pi+.1, 0.01)
x, y, z = np.sin(t), np.cos(3*t), np.sin(5*t)
estep = 15
i = np.arange(t.size)
zuplims = (i % estep == 0) & (i // estep % 3 == 0)
zlolims = (i % estep == 0) & (i // estep % 3 == 2)
ax.errorbar(x, y, z, 0.2, zuplims=zuplims, zlolims=zlolims, errorevery=estep)
ax.set_xlabel("X label")
ax.set_ylabel("Y label")
ax.set_zlabel("Z label")
plt.show()