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I am trying to make a figure to visualize Lagranges multiplier method. This means I want to draw the graph of some function z = f(x,y), but also the constraint g(x,y) = c. Because I want to draw the graph of f, this must obviously be a 3D plot. But the constraint g(x,y) = c is a level curve of g, and should lie in the xy-plane.
I am using Python, and here is my current code:
import matplotlib.pyplot as plt
from matplotlib import cm
import numpy as np
fig, ax = plt.subplots(subplot_kw={"projection": "3d"})
X = np.arange(-5,5,0.5)
Y = X
X, Y = np.meshgrid(X, Y)
Z = 50 - X**2 - Y**2
surf = ax.plot_surface(X, Y, Z, cmap=cm.coolwarm)
ax.set_zlim(0, 50)
g = X**2+Y**2
CS = ax.contour(X,Y,g)
plt.show()
and this is the output:
Current plot
I only need one level curve of g in the xy-plane. Now, I have several, and none of them lies at z = 0. Ideally, I should also somehow mark out the points of z=f(x,y) that lies directly over g(x,y) = c.
I would really appreciate your feedback!
You need to add the optional argument "offset", so that the contour gets projected to a plane. To be in z=0:
CS = ax.contour(X,Y,g, offset = 0)
See here.
I'm trying to use quiver and I am struggling to understand how quiver is interpreting the input arrays U, V in its documentation https://matplotlib.org/3.5.0/api/_as_gen/matplotlib.pyplot.quiver.html
With the following code I expect the bottom-right arrow (element [2,0]) to point to the left. However, it's the top-left as in the attached figure.
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(0.3,.9,3)
y = x
nb_points = x.size
X,Y = np.meshgrid(x,y)
U = np.ones((nb_points, nb_points))
V = np.ones((nb_points, nb_points))
U[2,0] = -3
M = (np.hypot(U, V))
fig,ax = plt.subplots(1,1, figsize=(4,4))
ax.quiver(X, Y, U, V, M, units='width', pivot='middle')
ax.set_xlim(.2,1)
ax.set_ylim(.2,1)
Can anyone explain what's going on?
I completely misinterpreted the meaning of U,V's indeces
U[2,0] is supposed to be interpreted as the direction of X[2,0],Y[2,0] = (.3,.9), which is the top-left point
I have a 3d velocity vector field in a numpy array of shape (zlength, ylength, xlength, 3). The '3' contains the velocity components (u,v,w).
I can quite easily plot the vector field in the orthogonal x-y, x-z, and y-z planes using quiver, e.g.
X, Y = np.meshgrid(xvalues, yvalues)
xyfieldfig = plt.figure()
xyfieldax = xyfieldfig.add_subplot(111)
Q1 = xyfieldax.quiver(X, Y, velocity_field[zslice,:,:,0], velocity_field[zslice,:,:,1])
However, I'd like to be able to view the velocity field within an arbitrary plane.
I tried to project the velocity field onto a plane by doing:
projected_field = np.zeros(zlength,ylength,xlength,3)
normal = (nx,ny,nz) #normalised normal to the plane
for i in range(zlength):
for j in range(ylength):
for k in range(xlength):
projected_field[i,j,m] = velocity_field[i,j,m] - np.dot(velocity_field[i,j,m], normal)*normal
However, this (of course) still leaves me with a 3d numpy array with the same shape: (zlength, ylength, xlength, 3). The projected_field now contains velocity vectors at each (x,y,z) position that lie within planes at each local (x,y,z) position.
How do I project velocity_field onto a single plane? Or, how do I now plot my projected_field along one plane?
Thanks in advance!
You're close. Daniel F's suggestion was right, you just need to know how to do the interpolation. Here's a worked example
from mpl_toolkits.mplot3d import axes3d
import matplotlib.pyplot as plt
import numpy as np
import scipy.interpolate
def norm(v,axis=0):
return np.sqrt(np.sum(v**2,axis=axis))
#Original velocity field
xpoints = np.arange(-.2, .21, 0.05)
ypoints = np.arange(-.2, .21, 0.05)
zpoints = np.arange(-.2, .21, 0.05)
x, y, z = np.meshgrid(xpoints,ypoints,zpoints,indexing='ij')
#Simple example
#(u,v,w) are the components of your velocity field
u = x
v = y
w = z
#Setup a template for the projection plane. z-axis will be rotated to point
#along the plane normal
planex, planey, planez =
np.meshgrid(np.arange(-.2,.2001,.1),
np.arange(-.2,.2001,.1), [0.1],
indexing='ij')
planeNormal = np.array([0.1,0.4,.4])
planeNormal /= norm(planeNormal)
#pick an arbirtrary vector for projection x-axis
u0 = np.array([-(planeNormal[2] + planeNormal[1])/planeNormal[0], 1, 1])
u1 = -np.cross(planeNormal,u0)
u0 /= norm(u0)
u1 /= norm(u1)
#rotation matrix
rotation = np.array([u0,u1,planeNormal]).T
#Rotate plane to get projection vertices
rotatedVertices = rotation.dot( np.array( [planex.flatten(), planey.flatten(), planez.flatten()]) ).T
#Now you can interpolate gridded vector field to rotated vertices
uprime = scipy.interpolate.interpn( (xpoints,ypoints,zpoints), u, rotatedVertices, bounds_error=False )
vprime = scipy.interpolate.interpn( (xpoints,ypoints,zpoints), v, rotatedVertices, bounds_error=False )
wprime = scipy.interpolate.interpn( (xpoints,ypoints,zpoints), w, rotatedVertices, bounds_error=False )
#Projections
cosineMagnitudes = planeNormal.dot( np.array([uprime,vprime,wprime]) )
uProjected = uprime - planeNormal[0]*cosineMagnitudes
vProjected = vprime - planeNormal[1]*cosineMagnitudes
wProjected = wprime - planeNormal[2]*cosineMagnitudes
The number of lines could be reduced using some tensordot operations if you wanted to get fancy. Also this or some close variant it would work without indexing='ij' in meshgrid.
Original field:
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.quiver(x, y, z, u, v, w, length=0.1, normalize=True)
Projected field:
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.quiver(rotatedVertices[:,0], rotatedVertices[:,1], rotatedVertices[:,2],
uprime, vprime,wprime, length=0.5, color='blue', label='Interpolation only')
ax.quiver(rotatedVertices[:,0], rotatedVertices[:,1], rotatedVertices[:,2],
uProjected, vProjected, wProjected, length=0.5, color='red', label='Interpolation + Projection')
plt.legend()
I am taking a course on linear algebra and I want to visualize the vectors in action, such as vector addition, normal vector, so on.
For instance:
V = np.array([[1,1],[-2,2],[4,-7]])
In this case I want to plot 3 vectors V1 = (1,1), M2 = (-2,2), M3 = (4,-7).
Then I should be able to add V1,V2 to plot a new vector V12(all together in one figure).
when I use the following code, the plot is not as intended
import numpy as np
import matplotlib.pyplot as plt
M = np.array([[1,1],[-2,2],[4,-7]])
print("vector:1")
print(M[0,:])
# print("vector:2")
# print(M[1,:])
rows,cols = M.T.shape
print(cols)
for i,l in enumerate(range(0,cols)):
print("Iteration: {}-{}".format(i,l))
print("vector:{}".format(i))
print(M[i,:])
v1 = [0,0],[M[i,0],M[i,1]]
# v1 = [M[i,0]],[M[i,1]]
print(v1)
plt.figure(i)
plt.plot(v1)
plt.show()
How about something like
import numpy as np
import matplotlib.pyplot as plt
V = np.array([[1,1], [-2,2], [4,-7]])
origin = np.array([[0, 0, 0],[0, 0, 0]]) # origin point
plt.quiver(*origin, V[:,0], V[:,1], color=['r','b','g'], scale=21)
plt.show()
Then to add up any two vectors and plot them to the same figure, do so before you call plt.show(). Something like:
plt.quiver(*origin, V[:,0], V[:,1], color=['r','b','g'], scale=21)
v12 = V[0] + V[1] # adding up the 1st (red) and 2nd (blue) vectors
plt.quiver(*origin, v12[0], v12[1])
plt.show()
NOTE: in Python2 use origin[0], origin[1] instead of *origin
This may also be achieved using matplotlib.pyplot.quiver, as noted in the linked answer;
plt.quiver([0, 0, 0], [0, 0, 0], [1, -2, 4], [1, 2, -7], angles='xy', scale_units='xy', scale=1)
plt.xlim(-10, 10)
plt.ylim(-10, 10)
plt.show()
Your main problem is you create new figures in your loop, so each vector gets drawn on a different figure. Here's what I came up with, let me know if it's still not what you expect:
CODE:
import numpy as np
import matplotlib.pyplot as plt
M = np.array([[1,1],[-2,2],[4,-7]])
rows,cols = M.T.shape
#Get absolute maxes for axis ranges to center origin
#This is optional
maxes = 1.1*np.amax(abs(M), axis = 0)
for i,l in enumerate(range(0,cols)):
xs = [0,M[i,0]]
ys = [0,M[i,1]]
plt.plot(xs,ys)
plt.plot(0,0,'ok') #<-- plot a black point at the origin
plt.axis('equal') #<-- set the axes to the same scale
plt.xlim([-maxes[0],maxes[0]]) #<-- set the x axis limits
plt.ylim([-maxes[1],maxes[1]]) #<-- set the y axis limits
plt.legend(['V'+str(i+1) for i in range(cols)]) #<-- give a legend
plt.grid(b=True, which='major') #<-- plot grid lines
plt.show()
OUTPUT:
EDIT CODE:
import numpy as np
import matplotlib.pyplot as plt
M = np.array([[1,1],[-2,2],[4,-7]])
rows,cols = M.T.shape
#Get absolute maxes for axis ranges to center origin
#This is optional
maxes = 1.1*np.amax(abs(M), axis = 0)
colors = ['b','r','k']
for i,l in enumerate(range(0,cols)):
plt.axes().arrow(0,0,M[i,0],M[i,1],head_width=0.05,head_length=0.1,color = colors[i])
plt.plot(0,0,'ok') #<-- plot a black point at the origin
plt.axis('equal') #<-- set the axes to the same scale
plt.xlim([-maxes[0],maxes[0]]) #<-- set the x axis limits
plt.ylim([-maxes[1],maxes[1]]) #<-- set the y axis limits
plt.grid(b=True, which='major') #<-- plot grid lines
plt.show()
EDIT OUTPUT:
What did you expect the following to do?
v1 = [0,0],[M[i,0],M[i,1]]
v1 = [M[i,0]],[M[i,1]]
This is making two different tuples, and you overwrite what you did the first time... Anyway, matplotlib does not understand what a "vector" is in the sense you are using. You have to be explicit, and plot "arrows":
In [5]: ax = plt.axes()
In [6]: ax.arrow(0, 0, *v1, head_width=0.05, head_length=0.1)
Out[6]: <matplotlib.patches.FancyArrow at 0x114fc8358>
In [7]: ax.arrow(0, 0, *v2, head_width=0.05, head_length=0.1)
Out[7]: <matplotlib.patches.FancyArrow at 0x115bb1470>
In [8]: plt.ylim(-5,5)
Out[8]: (-5, 5)
In [9]: plt.xlim(-5,5)
Out[9]: (-5, 5)
In [10]: plt.show()
Result:
Thanks to everyone, each of your posts helped me a lot.
rbierman code was pretty straight for my question, I have modified a bit and created a function to plot vectors from given arrays. I'd love to see any suggestions to improve it further.
import numpy as np
import matplotlib.pyplot as plt
def plotv(M):
rows,cols = M.T.shape
print(rows,cols)
#Get absolute maxes for axis ranges to center origin
#This is optional
maxes = 1.1*np.amax(abs(M), axis = 0)
colors = ['b','r','k']
fig = plt.figure()
fig.suptitle('Vectors', fontsize=10, fontweight='bold')
ax = fig.add_subplot(111)
fig.subplots_adjust(top=0.85)
ax.set_title('Vector operations')
ax.set_xlabel('x')
ax.set_ylabel('y')
for i,l in enumerate(range(0,cols)):
# print(i)
plt.axes().arrow(0,0,M[i,0],M[i,1],head_width=0.2,head_length=0.1,zorder=3)
ax.text(M[i,0],M[i,1], str(M[i]), style='italic',
bbox={'facecolor':'red', 'alpha':0.5, 'pad':0.5})
plt.plot(0,0,'ok') #<-- plot a black point at the origin
# plt.axis('equal') #<-- set the axes to the same scale
plt.xlim([-maxes[0],maxes[0]]) #<-- set the x axis limits
plt.ylim([-maxes[1],maxes[1]]) #<-- set the y axis limits
plt.grid(b=True, which='major') #<-- plot grid lines
plt.show()
r = np.random.randint(4,size=[2,2])
print(r[0,:])
print(r[1,:])
r12 = np.add(r[0,:],r[1,:])
print(r12)
plotv(np.vstack((r,r12)))
Vector addition performed on random vectors
All nice solutions, borrowing and improvising for special case -> If you want to add a label near the arrowhead:
arr = [2,3]
txt = “Vector X”
ax.annotate(txt, arr)
ax.arrow(0, 0, *arr, head_width=0.05, head_length=0.1)
In order to match the vector lenght and angle with the x,y coordinates of the plot, you can use to following options to plt.quiver:
plt.figure(figsize=(5,2), dpi=100)
plt.quiver(0,0,250,100, angles='xy', scale_units='xy', scale=1)
plt.xlim(0,250)
plt.ylim(0,100)
Quiver is a good method once you figure out its annoying nuances, like not plotting vectors in their original scales. To do as far as I can tell you must pass these params to quiver call as many have pointed out: angles='xy', scale_units='xy', scale=1 AND you should set your plt.xlim and plt.ylim such that you get a square or near square grid. That is the only way I have gotten it to consistently plot the way I want. For instance passing a origin as *[0,0] and U, V as *[5,3] means the resulting plot should be a vector centered at 0,0 origin that goes over 5 units to the right on the x-axis and 3 units up on the y-axis.
I'd like to plot implicit equation F(x,y,z) = 0 in 3D. Is it possible in Matplotlib?
You can trick matplotlib into plotting implicit equations in 3D. Just make a one-level contour plot of the equation for each z value within the desired limits. You can repeat the process along the y and z axes as well for a more solid-looking shape.
from mpl_toolkits.mplot3d import axes3d
import matplotlib.pyplot as plt
import numpy as np
def plot_implicit(fn, bbox=(-2.5,2.5)):
''' create a plot of an implicit function
fn ...implicit function (plot where fn==0)
bbox ..the x,y,and z limits of plotted interval'''
xmin, xmax, ymin, ymax, zmin, zmax = bbox*3
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
A = np.linspace(xmin, xmax, 100) # resolution of the contour
B = np.linspace(xmin, xmax, 15) # number of slices
A1,A2 = np.meshgrid(A,A) # grid on which the contour is plotted
for z in B: # plot contours in the XY plane
X,Y = A1,A2
Z = fn(X,Y,z)
cset = ax.contour(X, Y, Z+z, [z], zdir='z')
# [z] defines the only level to plot for this contour for this value of z
for y in B: # plot contours in the XZ plane
X,Z = A1,A2
Y = fn(X,y,Z)
cset = ax.contour(X, Y+y, Z, [y], zdir='y')
for x in B: # plot contours in the YZ plane
Y,Z = A1,A2
X = fn(x,Y,Z)
cset = ax.contour(X+x, Y, Z, [x], zdir='x')
# must set plot limits because the contour will likely extend
# way beyond the displayed level. Otherwise matplotlib extends the plot limits
# to encompass all values in the contour.
ax.set_zlim3d(zmin,zmax)
ax.set_xlim3d(xmin,xmax)
ax.set_ylim3d(ymin,ymax)
plt.show()
Here's the plot of the Goursat Tangle:
def goursat_tangle(x,y,z):
a,b,c = 0.0,-5.0,11.8
return x**4+y**4+z**4+a*(x**2+y**2+z**2)**2+b*(x**2+y**2+z**2)+c
plot_implicit(goursat_tangle)
You can make it easier to visualize by adding depth cues with creative colormapping:
Here's how the OP's plot looks:
def hyp_part1(x,y,z):
return -(x**2) - (y**2) + (z**2) - 1
plot_implicit(hyp_part1, bbox=(-100.,100.))
Bonus: You can use python to functionally combine these implicit functions:
def sphere(x,y,z):
return x**2 + y**2 + z**2 - 2.0**2
def translate(fn,x,y,z):
return lambda a,b,c: fn(x-a,y-b,z-c)
def union(*fns):
return lambda x,y,z: np.min(
[fn(x,y,z) for fn in fns], 0)
def intersect(*fns):
return lambda x,y,z: np.max(
[fn(x,y,z) for fn in fns], 0)
def subtract(fn1, fn2):
return intersect(fn1, lambda *args:-fn2(*args))
plot_implicit(union(sphere,translate(sphere, 1.,1.,1.)), (-2.,3.))
Update: I finally have found an easy way to render 3D implicit surface with matplotlib and scikit-image, see my other answer. I left this one for whom is interested in plotting parametric 3D surfaces.
Motivation
Late answer, I just needed to do the same and I found another way to do it at some extent. So I am sharing this another perspective.
This post does not answer: (1) How to plot any implicit function F(x,y,z)=0? But does answer: (2) How to plot parametric surfaces (not all implicit functions, but some of them) using mesh with matplotlib?
#Paul's method has the advantage to be non parametric, therefore we can plot almost anything we want using contour method on each axe, it fully addresses (1). But matplotlib cannot easily build a mesh from this method, so we cannot directly get a surface from it, instead we get plane curves in all directions. This is what motivated my answer, I wanted to address (2).
Rendering mesh
If we are able to parametrize (this may be hard or impossible), with at most 2 parameters, the surface we want to plot then we can plot it with matplotlib.plot_trisurf method.
That is, from an implicit equation F(x,y,z)=0, if we are able to get a parametric system S={x=f(u,v), y=g(u,v), z=h(u,v)} then we can plot it easily with matplotlib without having to resort to contour.
Then, rendering such a 3D surface boils down to:
# Render:
ax = plt.axes(projection='3d')
ax.plot_trisurf(x, y, z, triangles=tri.triangles, cmap='jet', antialiased=True)
Where (x, y, z) are vectors (not meshgrid, see ravel) functionally computed from parameters (u, v) and triangles parameter is a Triangulation derived from (u,v) parameters to shoulder the mesh construction.
Imports
Required imports are:
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits import mplot3d
from matplotlib.tri import Triangulation
Some surfaces
Lets parametrize some surfaces...
Sphere
# Parameters:
theta = np.linspace(0, 2*np.pi, 20)
phi = np.linspace(0, np.pi, 20)
theta, phi = np.meshgrid(theta, phi)
rho = 1
# Parametrization:
x = np.ravel(rho*np.cos(theta)*np.sin(phi))
y = np.ravel(rho*np.sin(theta)*np.sin(phi))
z = np.ravel(rho*np.cos(phi))
# Triangulation:
tri = Triangulation(np.ravel(theta), np.ravel(phi))
Cone
theta = np.linspace(0, 2*np.pi, 20)
rho = np.linspace(-2, 2, 20)
theta, rho = np.meshgrid(theta, rho)
x = np.ravel(rho*np.cos(theta))
y = np.ravel(rho*np.sin(theta))
z = np.ravel(rho)
tri = Triangulation(np.ravel(theta), np.ravel(rho))
Torus
a, c = 1, 4
u = np.linspace(0, 2*np.pi, 20)
v = u.copy()
u, v = np.meshgrid(u, v)
x = np.ravel((c + a*np.cos(v))*np.cos(u))
y = np.ravel((c + a*np.cos(v))*np.sin(u))
z = np.ravel(a*np.sin(v))
tri = Triangulation(np.ravel(u), np.ravel(v))
Möbius Strip
u = np.linspace(0, 2*np.pi, 20)
v = np.linspace(-1, 1, 20)
u, v = np.meshgrid(u, v)
x = np.ravel((2 + (v/2)*np.cos(u/2))*np.cos(u))
y = np.ravel((2 + (v/2)*np.cos(u/2))*np.sin(u))
z = np.ravel(v/2*np.sin(u/2))
tri = Triangulation(np.ravel(u), np.ravel(v))
Limitation
Most of the time, Triangulation is required in order to coordinate mesh construction of plot_trisurf method, and this object only accepts two parameters, so we are limited to 2D parametric surfaces. It is unlikely we could represent the Goursat Tangle with this method.
Matplotlib expects a series of points; it will do the plotting if you can figure out how to render your equation.
Referring to Is it possible to plot implicit equations using Matplotlib? Mike Graham's answer suggests using scipy.optimize to numerically explore the implicit function.
There is an interesting gallery at http://xrt.wikidot.com/gallery:implicit showing a variety of raytraced implicit functions - if your equation matches one of these, it might give you a better idea what you are looking at.
Failing that, if you care to share the actual equation, maybe someone can suggest an easier approach.
As far as I know, it is not possible. You have to solve this equation numerically by yourself. Using scipy.optimize is a good idea. The simplest case is that you know the range of the surface that you want to plot, and just make a regular grid in x and y, and try to solve equation F(xi,yi,z)=0 for z, giving a starting point of z. Following is a very dirty code that might help you
from scipy import *
from scipy import optimize
xrange = (0,1)
yrange = (0,1)
density = 100
startz = 1
def F(x,y,z):
return x**2+y**2+z**2-10
x = linspace(xrange[0],xrange[1],density)
y = linspace(yrange[0],yrange[1],density)
points = []
for xi in x:
for yi in y:
g = lambda z:F(xi,yi,z)
res = optimize.fsolve(g, startz, full_output=1)
if res[2] == 1:
zi = res[0]
points.append([xi,yi,zi])
points = array(points)
Actually there is an easy way to plot implicit 3D surface with the scikit-image package. The key is the marching_cubes method.
import numpy as np
from skimage import measure
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import axes3d
Then we compute the function over a 3D meshgrid, in this example we use the goursat_tangle method #Paul defined in its answer:
xl = np.linspace(-3, 3, 50)
X, Y, Z = np.meshgrid(xl, xl, xl)
F = goursat_tangle(X, Y, Z)
The magic is happening here with marching_cubes:
verts, faces, normals, values = measure.marching_cubes(F, 0, spacing=[np.diff(xl)[0]]*3)
verts -= 3
We just need to correct vertices coordinates as they are expressed in Voxel coordinates (hence scaling using spacing switch and the subsequent origin shift).
Finally it is just about rendering the iso-surface using tri_surface:
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_trisurf(verts[:, 0], verts[:, 1], faces, verts[:, 2], cmap='jet', lw=0)
Which returns:
Have you looked at mplot3d on matplotlib?
Finally, I did it (I updated my matplotlib to 1.0.1).
Here is code:
import matplotlib.pyplot as plt
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
def hyp_part1(x,y,z):
return -(x**2) - (y**2) + (z**2) - 1
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
x_range = np.arange(-100,100,10)
y_range = np.arange(-100,100,10)
X,Y = np.meshgrid(x_range,y_range)
A = np.linspace(-100, 100, 15)
A1,A2 = np.meshgrid(A,A)
for z in A:
X,Y = A1, A2
Z = hyp_part1(X,Y,z)
ax.contour(X, Y, Z+z, [z], zdir='z')
for y in A:
X,Z= A1, A2
Y = hyp_part1(X,y,Z)
ax.contour(X, Y+y, Z, [y], zdir='y')
for x in A:
Y,Z = A1, A2
X = hyp_part1(x,Y,Z)
ax.contour(X+x, Y, Z, [x], zdir='x')
ax.set_zlim3d(-100,100)
ax.set_xlim3d(-100,100)
ax.set_ylim3d(-100,100)
Here is result:
Thank You, Paul!
MathGL (GPL plotting library) can plot it easily. Just create a data mesh with function values f[i,j,k] and use Surf3() function to make isosurface at value f[i,j,k]=0. See this sample.