With other words I got a set of data-points (x,y,z) associated to a value b and I would like to interpolate this data as accurate as possible.
Scipy.interpolate.griddata only can do a linear interpolation, what are the other options?
How about interpolating x, y, z separatly? I modified this tutorial example and added interpolation to it:
import matplotlib as mpl
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib.pyplot as plt
from scipy.interpolate import InterpolatedUnivariateSpline
mpl.rcParams['legend.fontsize'] = 10
# let's take only 20 points for original data:
n = 20
fig = plt.figure()
ax = fig.gca(projection='3d')
theta = np.linspace(-4 * np.pi, 4 * np.pi, n)
z = np.linspace(-2, 2, n)
r = z**2 + 1
x = r * np.sin(theta)
y = r * np.cos(theta)
ax.plot(x, y, z, label='rough curve')
# this variable represents distance along the curve:
t = np.arange(n)
# now let's refine it to 100 points:
t2 = np.linspace(t.min(), t.max(), 100)
# interpolate vector components separately:
x2 = InterpolatedUnivariateSpline(t, x)(t2)
y2 = InterpolatedUnivariateSpline(t, y)(t2)
z2 = InterpolatedUnivariateSpline(t, z)(t2)
ax.plot(x2, y2, z2, label='interpolated curve')
ax.legend()
plt.show()
The result looks like this:
UPDATE
Didn't understand the question at the first time, sorry.
You are probably looking for tricubic interpolation. Try this.
Related
I would like to plot the function x^3 - 3xy + y^3 = 0 (Folium of Descartes) but with different colors in quadrant 2 and 4, and at the peak of the leaf I want the colors to change. I was thinking about converting to a piecewise parametric equation to make the plotting in different colors easier, but I don't know how to convert into a parametric form.
I was originally using a contour to plot the function only in one color, but I have no idea how to get the color changes.
Below is what I have for the single color, but I would like to have separate colors for each of the segments listed above.
import matplotlib.pyplot as plt
import numpy as np
xrange = np.arange(-5, 5, .025)
yrange = np.arange(-5, 5, .025)
X, Y = np.meshgrid(xrange, yrange)
plt.contour(X, Y, X**3 - 3*X*Y + Y**3, levels=[0], colors=['#000000'])
plt.show()
This website gives parametric Cartesian and polar equations of the curve.
The parametric equation isn't symmetric in x and y, which is annoying for drawing nice curves.
The polar equation gives a radius rho for a given angle theta. Some experimenting shows that the tails are formed by theta negative, or larger than pi/2. The cusp is at pi/4.
To visualize the polar curve, x=rho*cos(theta) and y=rho*sin(theta) can be used. Coloring can happen depending on theta.
import matplotlib.pyplot as plt
from matplotlib.colors import LinearSegmentedColormap, ListedColormap
import numpy as np
a = 1
tail = 0.5
theta = np.linspace(-tail, np.pi / 2 + tail, 500)
rho = 3 * a * np.sin(theta) * np.cos(theta) / (np.cos(theta) ** 3 + np.sin(theta) ** 3)
x = rho * np.cos(theta)
y = rho * np.sin(theta)
cmap = LinearSegmentedColormap.from_list('', ['red', 'gold', 'blue'])
# cmap = ListedColormap(['red', 'blue'])
# cmap = 'Spectral'
plt.scatter(x, y, c=theta, cmap=cmap, s=1)
plt.axis('equal')
plt.axis('off')
plt.show()
Copying the leave 4 times, and filling them in decreasing order creates the following plot.
import matplotlib.pyplot as plt
import numpy as np
for a in np.linspace(1, 0.001, 50):
theta = np.linspace(0, np.pi / 2, 500)
rho = 3 * a * np.sin(theta) * np.cos(theta) / (np.cos(theta) ** 3 + np.sin(theta) ** 3)
x = rho * np.cos(theta)
y = rho * np.sin(theta)
for i in [-1, 1]:
for j in [-1, 1]:
plt.fill(i * x, j * y, c=plt.cm.summer(1 - a))
plt.axis('equal')
plt.axis('off')
plt.show()
I have 4 arrays x, y, z and T of length n and I want to plot a 3D curve using matplotlib. The (x, y, z) are the points positions and T is the value of each point (which is plotted as color), like the temperature of each point. How can I do it?
Example code:
import numpy as np
from matplotlib import pyplot as plt
fig = plt.figure()
ax = fig.gca(projection='3d')
n = 100
cmap = plt.get_cmap("bwr")
theta = np.linspace(-4 * np.pi, 4 * np.pi, n)
z = np.linspace(-2, 2, n)
r = z**2 + 1
x = r * np.sin(theta)
y = r * np.cos(theta)
T = (2*np.random.rand(n) - 1) # All the values are in [-1, 1]
What I found over the internet:
It's possible to use cmap with scatter like shown in the docs and in this stackoverflow question
ax = plt.gca()
ax.scatter(x, y, z, cmap=cmap, c=T)
The problem is that scatter is a set of points, not a curve.
In this stackoverflow question the solution was divide in n-1 intervals and each interval we use a different color like
t = (T - np.min(T))/(np.max(T)-np.min(T)) # Normalize
for i in range(n-1):
plt.plot(x[i:i+2], y[i:i+2], z[i:i+2], c=cmap(t[i])
The problem is that each segment has only one color, but it should be an gradient. The last value is not even used.
Useful links:
Matplotlib - Colormaps
Matplotlib - Tutorial 3D
This is a case where you probably need to use Line3DCollection. This is the recipe:
create segments from your array of coordinates.
create a Line3DCollection object.
add that collection to the axis.
set the axis limits.
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d.art3d import Line3DCollection
from matplotlib.cm import ScalarMappable
from matplotlib.colors import Normalize
def get_segments(x, y, z):
"""Convert lists of coordinates to a list of segments to be used
with Matplotlib's Line3DCollection.
"""
points = np.ma.array((x, y, z)).T.reshape(-1, 1, 3)
return np.ma.concatenate([points[:-1], points[1:]], axis=1)
fig = plt.figure()
ax = fig.add_subplot(projection='3d')
n = 100
cmap = plt.get_cmap("bwr")
theta = np.linspace(-4 * np.pi, 4 * np.pi, n)
z = np.linspace(-2, 2, n)
r = z**2 + 1
x = r * np.sin(theta)
y = r * np.cos(theta)
T = np.cos(theta)
segments = get_segments(x, y, z)
c = Line3DCollection(segments, cmap=cmap, array=T)
ax.add_collection(c)
fig.colorbar(c)
ax.set_xlim(x.min(), x.max())
ax.set_ylim(y.min(), y.max())
ax.set_zlim(z.min(), z.max())
plt.show()
I am trying to plot the following function on a unit sphere, the points should be on the sphere and fill up the whole sphere however some of the points are falling off. Any suggestions why? I believe it is because the sphere is not spanning 1,1,1 3D grid but I am not sure how to edit my code to fix this.
from itertools import product, combinations
import matplotlib
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
def d(kx,ky):
M = 1
B = 1
vf = 1
kxx,kyy = np.meshgrid(kx,ky)
x = (vf*kxx)/(np.sqrt(((((vf**2)*(kxx**2)))+((vf**2)*(kyy**2))+(M-B*(kxx**2+(kyy**2)))**2)))
y = (vf*kxx)/(np.sqrt(((((vf**2)*(kxx**2)))+((vf**2)*(kyy**2))+(M-B*(kxx**2+(kyy**2)))**2)))
z = (M-B*(kxx**2+(kyy**2)))/(np.sqrt(((((vf**2)*(kxx**2)))+((vf**2)*(kyy**2))+(M-B*(kxx**2+(kyy**2)))**2)))
return x,y,z
kx = np.linspace(-2, 2, 10)
ky = np.linspace(-2, 2, 10)
xi, yi, zi = d(kx,ky)
phi = np.linspace(0, np.pi, 100)
theta = np.linspace(0, 2*np.pi, 100)
phi, theta = np.meshgrid(phi, theta)
x = np.sin(phi) * np.cos(theta)
y = np.sin(phi) * np.sin(theta)
z = np.cos(phi)
fig = plt.figure(figsize=plt.figaspect(1.))
ax = fig.add_subplot(111, projection='3d')
ax.plot_surface(x, y, z, color="w", rstride=1, cstride=1)
ax.scatter(xi,yi,zi,color="k",s=20)
plt.show()
Thank you kindly,
I can draw a circle by scatter, which has been shown in the image. But I want to draw them buy a line, because there are many circles in total, I need to link nodes together for a certain circle. Thanks.
I the order of the points is random, you can change X-Y to polar, and sort the data by angle:
create some random order points first:
import pylab as pl
import numpy as np
angle = np.arange(0, np.pi*2, 0.05)
r = 50 + np.random.normal(0, 2, angle.shape)
x = r * np.cos(angle)
y = r * np.sin(angle)
idx = np.random.permutation(angle.shape[0])
x = x[idx]
y = y[idx]
Then use arctan2() to calculate the angle, and sort the data by it:
angle = np.arctan2(x, y)
order = np.argsort(angle)
x = x[order]
y = y[order]
fig, ax = pl.subplots()
ax.set_aspect(1.0)
x2 = np.r_[x, x[0]]
y2 = np.r_[y, y[0]]
ax.plot(x, y, "o")
ax.plot(x2, y2, "r", lw=2)
here is the output:
Here is one way to do it. This answer uses different methods than the linked possible duplicate, so may be worth keeping.
import matplotlib.pyplot as plt
from matplotlib import patches
fig = plt.figure(figsize=plt.figaspect(1.0))
ax = fig.add_subplot(111)
cen = (2.0,1.0); r = 3.0
circle = patches.Circle(cen, r, facecolor='none')
ax.add_patch(circle)
ax.set_xlim(-6.0, 6.0)
ax.set_ylim(-6.0, 6.0)
If all you have are the x and y points, you can use PathPatch. Here's a tutorial
If your data points are already in order, the plot command should work fine. If you're looking to generate a circle from scratch, you can use a parametric equation.
>>> import numpy as np
>>> import matplotlib.pyplot as plt
>>> t = np.linspace(0,2*np.pi, 100)
>>> x = np.cos(t)
>>> y = np.sin(t)
>>> plt.plot(x,y)
Does anyone have sample code for plotting ellipsoids? There is one for sphere on matplotlib site, but nothing for ellipsoids. I am trying to plot
x**2 + 2*y**2 + 2*z**2 = c
where c is a constant (like 10) that defines an ellipsoid. I tried the meshgrid(x,y) route, reworked the equation so z is on one side, but the sqrt is a problem. The matplotlib sphere example works with angles, u,v, but I am not sure how to work that for ellipsoid.
Here is how you can do it via spherical coordinates:
# from mpl_toolkits.mplot3d import Axes3D # Not needed with Matplotlib 3.6.3
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')
coefs = (1, 2, 2) # 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))
# Plot:
ax.plot_surface(x, y, z, rstride=4, cstride=4, color='b')
# Adjustment of the axes, so that they all have the same span:
max_radius = max(rx, ry, rz)
for axis in 'xyz':
getattr(ax, 'set_{}lim'.format(axis))((-max_radius, max_radius))
plt.show()
The resulting plot is similar to
The program above actually produces a nicer looking "square" graphics.
This solution is strongly inspired from the example in Matplotlib's gallery.
Building on EOL's answer. Sometimes you have an ellipsoid in matrix format:
A and c Where A is the ellipsoid matrix and c is a vector representing the centre of the ellipsoid.
import numpy as np
import numpy.linalg as linalg
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
# your ellispsoid and center in matrix form
A = np.array([[1,0,0],[0,2,0],[0,0,2]])
center = [0,0,0]
# find the rotation matrix and radii of the axes
U, s, rotation = linalg.svd(A)
radii = 1.0/np.sqrt(s)
# now carry on with EOL's answer
u = np.linspace(0.0, 2.0 * np.pi, 100)
v = np.linspace(0.0, np.pi, 100)
x = radii[0] * np.outer(np.cos(u), np.sin(v))
y = radii[1] * np.outer(np.sin(u), np.sin(v))
z = radii[2] * np.outer(np.ones_like(u), np.cos(v))
for i in range(len(x)):
for j in range(len(x)):
[x[i,j],y[i,j],z[i,j]] = np.dot([x[i,j],y[i,j],z[i,j]], rotation) + center
# plot
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_wireframe(x, y, z, rstride=4, cstride=4, color='b', alpha=0.2)
plt.show()
plt.close(fig)
del fig
So, not too much new here, but helpful if you've got an ellipsoid in matrix form which is rotated and perhaps not centered at 0,0,0 and want to plot it.
If you have an ellipsoid specified by an arbitrary covariance matrix cov and offset bias, you do not need to figure out the intuitive parameters of the ellipsoid to get the shape. Specifically, you don't need the individual axes or rotations. The whole point of the matrix is that it transforms a unit sphere (represented by the identity matrix) into your ellipse.
Starting with
u = np.linspace(0, 2 * np.pi, 100)
v = np.linspace(0, np.pi, 100)
Make a unit sphere
x = np.outer(np.cos(u), np.sin(v))
y = np.outer(np.sin(u), np.sin(v))
z = np.outer(np.ones_like(u), np.cos(v))
Now transform the sphere:
ellipsoid = (cov # np.stack((x, y, z), 0).reshape(3, -1) + bias).reshape(3, *x.shape)
You can plot the result pretty much as before:
ax.plot_surface(*ellipsoid, rstride=4, cstride=4, color='b', alpha=0.75)