Develop Reference

matplotlib.plt adds an unwanted line when using fill_between

Question about using fill_between:
Using fill_between in the following snippet adds an unwanted vertical and horizontal line. Is it because of the parametric equation? How can I fix it?
import matplotlib.pyplot as plt
import numpy as np
t = np.linspace(0, 2 * np.pi, 100)
x = 16 * np.sin(t)**3
y = 13 * np.cos(t) - 5 * np.cos(2*t) - 2*np.cos(3*t) - np.cos(4*t)
plt.plot(x,y, 'red', linewidth=2)
plt.fill_between(x, y, color = 'red', alpha = 0.2)
Here is the outcome:

You can use plt.fill to fill an area bounded by xy values. plt.fill_between fills between zero and a curve defined as y being a function of x.
import matplotlib.pyplot as plt
import numpy as np
t = np.linspace(0, 2 * np.pi, 100)
x = 16 * np.sin(t) ** 3
y = 13 * np.cos(t) - 5 * np.cos(2 * t) - 2 * np.cos(3 * t) - np.cos(4 * t)
plt.plot(x, y, 'deeppink', linewidth=2)
plt.fill(x, y, color='deeppink', alpha=0.2)
plt.axis('off')
plt.show()
Or in a loop:
t = np.linspace(0, 2 * np.pi, 100)
x = 16 * np.sin(t) ** 3
y = 13 * np.cos(t) - 5 * np.cos(2 * t) - 2 * np.cos(3 * t) - np.cos(4 * t)
for r in np.linspace(1, 0.01, 50):
plt.fill(r * x, r * y, color=plt.cm.Reds(r))
plt.axis('off')
plt.show()

How to draw sphere with cylinder by formula with matplotlib

I have two formulas:
x^2 + y^2+z^2 = 6
(x-3)^2+y^2=4
So I have to draw this shapes by matplotlib on python.
My solution:
`
import math
import matplotlib.pyplot as plt
from pylab import *
import numpy as np
import matplotlib.patches as patches
from mpl_toolkits.mplot3d import Axes3D
from itertools import product, combinations
import mpl_toolkits.mplot3d.art3d as art3d
def task3():
fig = plt.figure()
ax = plt.axes(projection='3d')
coefs = (1, 1, 1) # Coefficients in a0/c x**2 + a1/c y**2 + a2/c z**2 = 1
# Radii corresponding to the coefficients:
rx, ry, rz = 1 / np.sqrt(coefs)
# Set of all spherical angles:
u = np.linspace(0, 2 * np.pi, 100)
v = np.linspace(0, np.pi, 100)
# Cartesian coordinates that correspond to the spherical angles:
# (this is the equation of an ellipsoid):
x = rx * np.outer(np.cos(u), np.sin(v))
y = ry * np.outer(np.sin(u), np.sin(v))
z = rz * np.outer(np.ones_like(u), np.cos(v))
ax.plot_surface(x, y, z, rstride=6, cstride=6,
cmap='viridis', edgecolor='none')
# ax.scatter(X, Y, Z, c=Z, cmap='viridis', label="xy")
radius = 0.3
height = 2.4
elevation = -1
resolution = 720
color = 'r'
x_center = 0.7
y_center = -0
a = np.linspace(x_center - radius, x_center + radius, resolution)
b = np.linspace(elevation, elevation + height, resolution)
X, Z = np.meshgrid(a, b)
Y = np.sqrt(radius ** 2 - (X - x_center) ** 2) + y_center # Pythagorean theorem
ax.plot_surface(X, Y, Z, rstride=6, cstride=6,
cmap='viridis', edgecolor='none')
ax.plot_surface(X, (2 * y_center - Y), Z, rstride=6, cstride=6,
cmap='viridis', edgecolor='none')
plt.show()
[![Result. I cant post images, but you can compile result and then you will see that my solution too approximate)
I want to do it correctly

Matplotlib - Forcing a 3D plot to use the available figsize

I'd like to create a plot with a specified figure size in which the 3D axes try to use the entire available space while maintaining an equal aspect ratio on all axis.
My current attempt shows a clipping rectangle that hides part of the mesh.
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np
fig = plt.figure(figsize=(5, 2.5))
ax = fig.add_subplot(projection='3d')
ax.get_proj = lambda: np.dot(Axes3D.get_proj(ax), np.diag([1, 1, 1, 0.5]))
u = v = np.linspace(0, 2 * np.pi, 50)
u, v = np.meshgrid(u, v)
X = np.cos(v) * (6 - (5/4 + np.sin(3 * u)) * np.sin(u - 3 * v))
Y = (6 - (5/4 + np.sin(3 * u)) * np.sin(u - 3 * v)) * np.sin(v)
Z = -np.cos(u - 3 * v) * (5/4 + np.sin(3 * u))
ax.plot_surface(X, Y, Z, rstride=1, cstride=1, color=[0.7] * 3, linewidth=0.25, edgecolor="k")
ax.set_box_aspect([ub - lb for lb, ub in (getattr(ax, f'get_{a}lim')() for a in 'xyz')])
plt.show()
What can I do?

I copied the code you provided and just add 4 lines at the end.
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np
plt.close()
fig = plt.figure(figsize=(5, 2.5))
ax = fig.add_subplot(projection='3d')
ax.get_proj = lambda: np.dot(Axes3D.get_proj(ax), np.diag([1, 1, 1, 0.5]))
u = v = np.linspace(0, 2 * np.pi, 50)
u, v = np.meshgrid(u, v)
X = np.cos(v) * (6 - (5/4 + np.sin(3 * u)) * np.sin(u - 3 * v))
Y = (6 - (5/4 + np.sin(3 * u)) * np.sin(u - 3 * v)) * np.sin(v)
Z = -np.cos(u - 3 * v) * (5/4 + np.sin(3 * u))
ax.plot_surface(X, Y, Z, rstride=1, cstride=1, color=[0.7] * 3, linewidth=0.25, edgecolor="k")
# ax.set_box_aspect([ub - lb for lb, ub in (getattr(ax, f'get_{a}lim')() for a in 'xyz')])
left, right = plt.xlim()
ax.set_zlim(left, right)
ax.set_ylim(left, right)
plt.tight_layout()
I comment out the line ax.set_box_aspect since it gives me an error. The output of the above is:
--- edit ---
I have an idea for a workaround to make it works in matplotlib v3.4.2:
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np
import os
plt.close()
# your code starts here, with a little modification
fig = plt.figure(figsize=(5,5))
ax = fig.add_subplot(projection='3d')
u = v = np.linspace(0, 2 * np.pi, 50)
u, v = np.meshgrid(u, v)
X = np.cos(v) * (6 - (5/4 + np.sin(3 * u)) * np.sin(u - 3 * v))
Y = (6 - (5/4 + np.sin(3 * u)) * np.sin(u - 3 * v)) * np.sin(v)
Z = -np.cos(u - 3 * v) * (5/4 + np.sin(3 * u))
ax.plot_surface(X, Y, Z, rstride=1, cstride=1, color=[0.7] * 3, linewidth=0.25, edgecolor="k")
# set the axes limits
left, right = plt.xlim()
ax.set_zlim(left, right)
ax.set_ylim(left, right)
# zoom in to the plot
ax.dist = 6
# make everything other than the plot itself transparent
fig.patch.set_alpha(0)
ax.patch.set_alpha(0)
ax.axis('off')
plt.tight_layout()
# save plot as image
plt.savefig('plotted.png')
# remove the axes where the image was plotted
ax.remove()
# resize figsize
fig = matplotlib.pyplot.gcf()
fig.set_size_inches(5, 2.5)
# add fake axes for gridlines as in 3d plot to make it look like a real plot
# skip this part if the gridlines are unnecessary
ax_bg = fig.add_subplot(111, projection='3d')
ax_bg.dist = 3
# add axes in cartesian coordinates (xy-plane) for the image
ax = fig.add_subplot(111)
fig.patch.set_alpha(1)
fig.subplots_adjust(left=0, bottom=0, right=1, top=1, wspace=0, hspace=0)
im = plt.imread('plotted.png')
h, w, dc = im.shape # (height=500, width=500, depth/color=4)
im_cropped = im[120:390, :, :] # this is manually adjusted
ax.axis('off')
ax.imshow(im_cropped)
# delete the saved image
os.remove('plotted.png')
Output is:
I don't know if it will work in your particular context, but it works for just fulfilling your question.
Let me know if something is unclear.

python matplotlib: drawing 3D sphere with circumferences

I'm trying to draw a sphere like this one using matplotlib:
but I can't find a way of having a dashed lines on the back and the vertical circumference looks a bit strange
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
fig = plt.figure(figsize=(12,12), dpi=300)
ax = fig.add_subplot(111, projection='3d')
ax.set_aspect('equal')
u = np.linspace(0, 2 * np.pi, 100)
v = np.linspace(0, np.pi, 100)
x = 1 * np.outer(np.cos(u), np.sin(v))
y = 1 * np.outer(np.sin(u), np.sin(v))
z = 1 * np.outer(np.ones(np.size(u)), np.cos(v))
#for i in range(2):
# ax.plot_surface(x+random.randint(-5,5), y+random.randint(-5,5), z+random.randint(-5,5), rstride=4, cstride=4, color='b', linewidth=0, alpha=0.5)
ax.plot_surface(x, y, z, rstride=4, cstride=4, color='b', linewidth=0, alpha=0.5)
ax.plot(np.sin(theta),np.cos(u),0,color='k')
ax.plot([0]*100,np.sin(theta),np.cos(u),color='k')

In the example you show, I don't think that the circles can be perpendicular to one another (i.e. one is the equator and one runs through the north pole and south pole). If the horizontal circle is the equator, then the north pole must be somewhere on a vertical line drawn through the center of the yellow circle that represents the sphere. Otherwise, the right side of the equator would look higher or lower than the left. However, the ellipse that represents the polar circle only crosses through that center line at the top and bottom of the yellow circle. Therefore, the north pole is at the top of the sphere, which means that we must be looking straight on the equator, which means it should look like a line, not an ellipse.
Here's some code to reproduce something similar to the figure you posted:
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.set_aspect('equal')
u = np.linspace(0, 2 * np.pi, 100)
v = np.linspace(0, np.pi, 100)
x = 1 * np.outer(np.cos(u), np.sin(v))
y = 1 * np.outer(np.sin(u), np.sin(v))
z = 1 * np.outer(np.ones(np.size(u)), np.cos(v))
#for i in range(2):
# ax.plot_surface(x+random.randint(-5,5), y+random.randint(-5,5), z+random.randint(-5,5), rstride=4, cstride=4, color='b', linewidth=0, alpha=0.5)
elev = 10.0
rot = 80.0 / 180 * np.pi
ax.plot_surface(x, y, z, rstride=4, cstride=4, color='b', linewidth=0, alpha=0.5)
#calculate vectors for "vertical" circle
a = np.array([-np.sin(elev / 180 * np.pi), 0, np.cos(elev / 180 * np.pi)])
b = np.array([0, 1, 0])
b = b * np.cos(rot) + np.cross(a, b) * np.sin(rot) + a * np.dot(a, b) * (1 - np.cos(rot))
ax.plot(np.sin(u),np.cos(u),0,color='k', linestyle = 'dashed')
horiz_front = np.linspace(0, np.pi, 100)
ax.plot(np.sin(horiz_front),np.cos(horiz_front),0,color='k')
vert_front = np.linspace(np.pi / 2, 3 * np.pi / 2, 100)
ax.plot(a[0] * np.sin(u) + b[0] * np.cos(u), b[1] * np.cos(u), a[2] * np.sin(u) + b[2] * np.cos(u),color='k', linestyle = 'dashed')
ax.plot(a[0] * np.sin(vert_front) + b[0] * np.cos(vert_front), b[1] * np.cos(vert_front), a[2] * np.sin(vert_front) + b[2] * np.cos(vert_front),color='k')
ax.view_init(elev = elev, azim = 0)
plt.show()

Matlab plot equivalent in Python [duplicate]

I'd like to plot pulse propagation in such a way at each step, it plots the pulse shape. In other words, I want a serie of x-z plots, for each values of y. Something like this (without color):
How can I do this using matplotlib (or Mayavi)? Here is what I did so far:
def drawPropagation(beta2, C, z):
""" beta2 in ps / km
C is chirp
z is an array of z positions """
T = numpy.linspace(-10, 10, 100)
sx = T.size
sy = z.size
T = numpy.tile(T, (sy, 1))
z = numpy.tile(z, (sx, 1)).T
U = 1 / numpy.sqrt(1 - 1j*beta2*z * (1 + 1j * C)) * numpy.exp(- 0.5 * (1 + 1j * C) * T * T / (1 - 1j*beta2*z*(1 + 1j*C)))
fig = pyplot.figure()
ax = fig.add_subplot(1,1,1, projection='3d')
surf = ax.plot_wireframe(T, z, abs(U))

Change to:
ax.plot_wireframe(T, z, abs(U), cstride=1000)
and call:
drawPropagation(1.0, 1.0, numpy.linspace(-2, 2, 10))
will create the following graph:
If you need the curve been filled with white color:
import numpy
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import pyplot
from matplotlib.collections import PolyCollection
def drawPropagation(beta2, C, z):
""" beta2 in ps / km
C is chirp
z is an array of z positions """
T = numpy.linspace(-10, 10, 100)
sx = T.size
sy = z.size
T = numpy.tile(T, (sy, 1))
z = numpy.tile(z, (sx, 1)).T
U = 1 / numpy.sqrt(1 - 1j*beta2*z * (1 + 1j * C)) * numpy.exp(- 0.5 * (1 + 1j * C) * T * T / (1 - 1j*beta2*z*(1 + 1j*C)))
fig = pyplot.figure()
ax = fig.add_subplot(1,1,1, projection='3d')
U = numpy.abs(U)
verts = []
for i in xrange(T.shape[0]):
verts.append(zip(T[i, :], U[i, :]))
poly = PolyCollection(verts, facecolors=(1,1,1,1), edgecolors=(0,0,1,1))
ax.add_collection3d(poly, zs=z[:, 0], zdir='y')
ax.set_xlim3d(numpy.min(T), numpy.max(T))
ax.set_ylim3d(numpy.min(z), numpy.max(z))
ax.set_zlim3d(numpy.min(U), numpy.max(U))
drawPropagation(1.0, 1.0, numpy.linspace(-2, 2, 10))
pyplot.show()