Plot a matrix equation in 2d or 3d using matplotlib - python

I have an equation as followed:
y = x^T * A * x + b^T * x + c
where x, b, c are vectors in n space and A is a nxn matrix.
I can plot a linear equation in matplotlib, but not sure how a matrix equation can be (if possible) shown also in a 3d plot.
I tried with following code, A is given matrix and w, c and b are column vectors. X and Y are mesh and Z is the solution.
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import axes3d
# if using a Jupyter notebook, include:
%matplotlib inline
fig = plt.figure(figsize=(10,6))
ax1 = fig.add_subplot(111, projection='3d')
n = 50
i = -5.0
j = 5.0
A = np.random.randint(i, j, size=(n, n))
w = np.random.randint(i, j, size=(n, 1))
c = b = np.random.randint(i, j, size=(n, 1))
X,Y = np.meshgrid(n,n)
Z = w.T*A*w + b.T*w + c
mycmap = plt.get_cmap('gist_earth')
surf1 = ax1.plot_surface(X, A, Z, cmap=mycmap)
fig.colorbar(surf1, ax=ax1, shrink=0.5, aspect=10)
plt.show()
The resulting plot does not seem to be a satisfied plot.

There are two problems in your code: 1) meshgrid was being used incorrectly (it needs two arrays, not two ints); 2) in the surface plot, you were using X, A, Z instead of X, Y, Z -- X, A, Z will work, and might make sense, but I'm guessing it wasn't your intention.
Here's a working solution:
fig = plt.figure(figsize=(10,6))
ax1 = fig.add_subplot(111, projection='3d')
n = 10
i = -5.0
j = 5.0
A = np.random.randint(i, j, size=(n, n))
w = np.random.randint(i, j, size=(n, 1))
c = b = np.random.randint(i, j, size=(n, 1))
X,Y = np.meshgrid(np.arange(n),np.arange(n))
Z = w.T*A*w + b.T*w + c
mycmap = plt.get_cmap('gist_earth')
surf1 = ax1.plot_surface(X, Y, Z, cmap=mycmap)
fig.colorbar(surf1, ax=ax1, shrink=0.5, aspect=10)

Related

Colormap a 3D curve in matplotlib

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()

How to create a numpy array filled with average value of a vector

I am not sure how to phrase my question in any way better. Basically, I have three lists of the same length x, y and z and I want to fill a 2D numpy array in the z/y plane with the average of the associated z values.
Here is how I can achieve what I wan to do:
import numpy as np
import matplotlib.pyplot as plt
x = [37.59390426045407, 38.00530354847739, 38.28412244348653, 38.74871247986305, 38.73175910429809, 38.869008864244016, 39.188234404976555, 39.92835838352555, 40.881394113153334, 41.686136269465884]
y = [0.1305391767832006, 0.13764519613447768, 0.14573326951792354, 0.15090729309032114, 0.16355823707239897, 0.17327106424274763, 0.17749746339532224, 0.17310384614773594, 0.16545780437882962, 0.1604752704890856]
z = [0.05738534353865021, 0.012572155256903583, -0.021709582561809437, -0.11191337750722108, -0.07931921785775153, -0.06241610118871843, 0.014216349927058225, 0.042002641153291886, -0.029354425271534645, 0.061894011359833856]
n = 5
image = np.zeros(shape=(n,n))
# Fill the 2D array
x0 = min(x)
y0 = min(y)
dx = (max(x) - min(x))/n
dy = (max(y) - min(y))/n
# Loop over each 2D cell
for index_x in range(n):
for index_y in range(n):
# find the limits of the cell
x1 = x0 + index_x * dx
x2 = x0 + (index_x+1) * dx
y1 = y0 + index_y * dy
y2 = y0 + (index_y+1) * dy
# find the points of z that lie within the range of the cell
vec_z = [z[idx] for idx in range(len(z)) if x[idx]>=x1 and x[idx]<x2 and y[idx]>=y1 and y[idx]<y2]
if vec_z:
image[index_x, index_y] = np.mean(vec_z)
# In the end, used to create a surface plot
fig, ax = plt.subplots()
ax.imshow(image, cmap=plt.cm.gray, interpolation='nearest')
plt.show()
Is there a more easy way to achieve this? I can imagine there is a numpy method for that.
If I understand correctly what you want to do, maybe a 2D interpolation from scipy.interpolate.interp2d is what you are looking for.
You define the interpolation function of your points:
f = interp2d(x = x, y = y, z = z)
Then you define the X and Y meshgrid:
N = 50
x_axis = np.linspace(np.min(x), np.max(x), N)
y_axis = np.linspace(np.min(y), np.max(y), N)
X, Y = np.meshgrid(x_axis, y_axis)
Finally you can compute Z interpolated values on the meshgrid:
Z = np.zeros((N, N))
for i in range(N):
for j in range(N):
Z[i, j] = f(X[i, j], Y[i, j])
If you plot in 3D the interpolated surface, you get:
fig = plt.figure()
ax = fig.add_subplot(projection = '3d')
ax.plot_surface(X, Y, Z, cmap = 'jet', shade = False)
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z')
plt.show()
Interpolated surface compared to interpolation data points:
ax.scatter(x, y, z, color = 'black', s = 100, alpha = 1)

How to add colors to each individual face of a cylinder using matplotlib

I am trying to color each individual face of a cylinder, however I am not sure how to go about it, I have tried the following:
for i in range(10):
col.append([])
for i in range(10):
for j in range(20):
col[i].append(plt.cm.Blues(0.4))
ax.plot_surface(X, Y, Z,facecolors = col,edgecolor = "red")
I want each face to be assigned its own color, so I would think I would supply an array of colors for each of the faces in a 2d array.
But this gives an error:
in plot_surface
colset.append(fcolors[rs][cs])
IndexError: list index out of range
Here is the full 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
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
origin = np.array([0, 0, 0])
#axis and radius
p0 = np.array([1, 3, 2])
p1 = np.array([8, 5, 9])
R = 5
#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]]
col = []
for i in range(10):
col.append([])
for i in range(10):
for j in range(20):
col[i].append(plt.cm.Blues(0.4))
ax.plot_surface(X, Y, Z,facecolors = col,edgecolor = "red")
#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()
Your Z array is of size 100x200, yet you are only specifying 10x20 colors. A quicker way to make col (with the right dimensions) might be something like:
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
col2 = plt.cm.Blues(np.linspace(0,1,100)) # linear gradient along the theta-axis
col2 = np.repeat(col2[:, np.newaxis, :], 200, axis=1) # expand over the t-axis
ax=plt.subplot(121, projection='3d')
ax.plot_surface(X, Y, Z, facecolors=col1)
ax=plt.subplot(122, projection='3d')
ax.plot_surface(X, Y, Z, facecolors=col2)
Which produces:

Find and draw regression plane to a set of points

I want to fit a plane to some data points and draw it. My current code is this:
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
points = [(1.1,2.1,8.1),
(3.2,4.2,8.0),
(5.3,1.3,8.2),
(3.4,2.4,8.3),
(1.5,4.5,8.0)]
xs, ys, zs = zip(*points)
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.scatter(xs, ys, zs)
point = np.array([0.0, 0.0, 8.1])
normal = np.array([0.0, 0.0, 1.0])
d = -point.dot(normal)
xx, yy = np.meshgrid([-5,10], [-5,10])
z = (-normal[0] * xx - normal[1] * yy - d) * 1. /normal[2]
ax.plot_surface(xx, yy, z, alpha=0.2, color=[0,1,0])
ax.set_xlim(-10,10)
ax.set_ylim(-10,10)
ax.set_zlim( 0,10)
plt.show()
which results in the following:
As you can see at the moment I create the plane manually. How can I calculate it? I guess it is possible with scipy.optimize.minimize somehow. The kind of error function is not that important to me at the moment. I think least squares (vertical point-plane-distance) would be fine. It would be cool if one of you could show me how to do it.
Oh, the idea just came to my mind. It's quite easy. :-)
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import scipy.optimize
import functools
def plane(x, y, params):
a = params[0]
b = params[1]
c = params[2]
z = a*x + b*y + c
return z
def error(params, points):
result = 0
for (x,y,z) in points:
plane_z = plane(x, y, params)
diff = abs(plane_z - z)
result += diff**2
return result
def cross(a, b):
return [a[1]*b[2] - a[2]*b[1],
a[2]*b[0] - a[0]*b[2],
a[0]*b[1] - a[1]*b[0]]
points = [(1.1,2.1,8.1),
(3.2,4.2,8.0),
(5.3,1.3,8.2),
(3.4,2.4,8.3),
(1.5,4.5,8.0)]
fun = functools.partial(error, points=points)
params0 = [0, 0, 0]
res = scipy.optimize.minimize(fun, params0)
a = res.x[0]
b = res.x[1]
c = res.x[2]
xs, ys, zs = zip(*points)
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.scatter(xs, ys, zs)
point = np.array([0.0, 0.0, c])
normal = np.array(cross([1,0,a], [0,1,b]))
d = -point.dot(normal)
xx, yy = np.meshgrid([-5,10], [-5,10])
z = (-normal[0] * xx - normal[1] * yy - d) * 1. /normal[2]
ax.plot_surface(xx, yy, z, alpha=0.2, color=[0,1,0])
ax.set_xlim(-10,10)
ax.set_ylim(-10,10)
ax.set_zlim( 0,10)
plt.show()
Sorry for asking unnecessarily.
Another way is with a straight forward least squares solution.
The equation for a plane is: ax + by + c = z. So set up matrices like this with all your data:
x_0 y_0 1
A = x_1 y_1 1
...
x_n y_n 1
And
a
x = b
c
And
z_0
B = z_1
...
z_n
In other words: Ax = B. Now solve for x which are your coefficients. But since (I assume) you have more than 3 points, the system is over-determined so you need to use the left pseudo inverse. So the answer is:
a
b = (A^T A)^-1 A^T B
c
And here is some simple Python code with an example:
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
N_POINTS = 10
TARGET_X_SLOPE = 2
TARGET_y_SLOPE = 3
TARGET_OFFSET = 5
EXTENTS = 5
NOISE = 5
# create random data
xs = [np.random.uniform(2*EXTENTS)-EXTENTS for i in range(N_POINTS)]
ys = [np.random.uniform(2*EXTENTS)-EXTENTS for i in range(N_POINTS)]
zs = []
for i in range(N_POINTS):
zs.append(xs[i]*TARGET_X_SLOPE + \
ys[i]*TARGET_y_SLOPE + \
TARGET_OFFSET + np.random.normal(scale=NOISE))
# plot raw data
plt.figure()
ax = plt.subplot(111, projection='3d')
ax.scatter(xs, ys, zs, color='b')
# do fit
tmp_A = []
tmp_b = []
for i in range(len(xs)):
tmp_A.append([xs[i], ys[i], 1])
tmp_b.append(zs[i])
b = np.matrix(tmp_b).T
A = np.matrix(tmp_A)
fit = (A.T * A).I * A.T * b
errors = b - A * fit
residual = np.linalg.norm(errors)
print "solution:"
print "%f x + %f y + %f = z" % (fit[0], fit[1], fit[2])
print "errors:"
print errors
print "residual:"
print residual
# plot plane
xlim = ax.get_xlim()
ylim = ax.get_ylim()
X,Y = np.meshgrid(np.arange(xlim[0], xlim[1]),
np.arange(ylim[0], ylim[1]))
Z = np.zeros(X.shape)
for r in range(X.shape[0]):
for c in range(X.shape[1]):
Z[r,c] = fit[0] * X[r,c] + fit[1] * Y[r,c] + fit[2]
ax.plot_wireframe(X,Y,Z, color='k')
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z')
plt.show()
Thanks #Ben for sharing! Since np.matrix is deprecated, I edited your code so it works with np arrays
import matplotlib.pyplot as plt
import numpy as np
from numpy.linalg import inv
# Pass the function array of points, shape (3, X)
def plane_from_points(points):
# Create this matrix correctly without transposing it later?
A = np.array([
points[0,:],
points[1,:],
np.ones(points.shape[1])
]).T
b = np.array([points[2, :]]).T
# fit = (A.T * A).I * A.T * b
fit = np.dot(np.dot(inv(np.dot(A.T, A)), A.T), b)
# errors = b - np.dot(A, fit)
# residual = np.linalg.norm(errors)
return fit
N_POINTS = 10
TARGET_X_SLOPE = 2
TARGET_y_SLOPE = 3
TARGET_OFFSET = 5
EXTENTS = 5
NOISE = 3
# create random data
xs = [np.random.uniform(2*EXTENTS)-EXTENTS for i in range(N_POINTS)]
ys = [np.random.uniform(2*EXTENTS)-EXTENTS for i in range(N_POINTS)]
zs = []
for i in range(N_POINTS):
zs.append(xs[i]*TARGET_X_SLOPE + \
ys[i]*TARGET_y_SLOPE + \
TARGET_OFFSET + np.random.normal(scale=NOISE))
points = np.array([xs, ys, zs])
# plot raw data
plt.figure()
ax = plt.subplot(111, projection='3d')
ax.scatter(xs, ys, zs, color='b')
fit = plane_from_points(points)
# plot plane
xlim = ax.get_xlim()
ylim = ax.get_ylim()
X,Y = np.meshgrid(np.arange(xlim[0], xlim[1]),
np.arange(ylim[0], ylim[1]))
Z = np.zeros(X.shape)
for r in range(X.shape[0]):
for c in range(X.shape[1]):
Z[r,c] = fit[0] * X[r,c] + fit[1] * Y[r,c] + fit[2]
ax.plot_wireframe(X,Y,Z, color='k')
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z')
plt.show()
I'm surprised nobody has mentioned lsq_linear. There you can more or less directly plug in the data points and get the plane coefficients out:
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
points = np.array([[1.1,2.1,8.1],
[3.2,4.2,8.0],
[5.3,1.3,8.2],
[3.4,2.4,8.3],
[1.5,4.5,8.0]])
A = np.hstack((points[:,:2], np.ones((len(xs),1))))
b = points[:,2]
res = scipy.optimize.lsq_linear(A, b)
assert res.success
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.scatter(xs, ys, zs)
XnY = np.linspace(-5,10,10)
X, Y = np.meshgrid(XnY, XnY)
Z = res.x[0] * X + res.x[1] * Y + res.x[2]
surf = ax.plot_surface(X, Y, Z, alpha=0.2, color=[0,1,0])
ax.set_xlim(-5,10)
ax.set_ylim(-5,10)
ax.set_zlim( 0,10)
plt.show()

imshow and histogram2d: can't get them to work

I'm learning Python and this is my first question here. I've read other topics related to the usage of imshow but didn't find anything useful. Sorry for my bad English.
I have plotted a set of points here, left graphic:
points (left) and image (right)
Now I'd like to see an image of the density of points, so I used imshow and histogram2d, and I got the image to the right in the previous link.
The image doesn't correspond to the distribution of points. How is this possible? I've followed the instructions in the help and even changed some parameters but nothing worked :(
The code is:
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.cm as cm
j, h, k = np.loadtxt("test.dat", usecols=(2, 4, 6), \
unpack=True)
# límites
xmin = -0.5
xmax = 3.0
ymin = -0.5
ymax = 4.0
# colores
j_h = j - h
h_k = h - k
# no todas las estrellas son graficadas
x1 = 0.5
y1 = 0.5
b = 2.2
c = y1 - b * x1
x = y = np.array([])
for xi, yi in zip(h_k, j_h):
if xi < (yi - c) / b:
x = np.append(x, xi)
y = np.append(y, yi)
# gráfico
fig = plt.figure(figsize=(8, 7))
ax = fig.add_subplot(111)
#ax.plot(x, y, "go")
ax.set_xlabel(r"X", fontsize=14)
ax.set_ylabel(r"Y", fontsize=14)
ax.axis([xmin, xmax, ymin, ymax])
# imagen
rango = [[xmin, xmax], [ymin, ymax]]
binsx = int((xmax - xmin) / 0.05)
binsy = int((ymax - ymin) / 0.05)
binsxy = [binsx, binsy]
H, xedges, yedges = np.histogram2d(x, y, range=rango, bins=binsxy)
extent = [yedges[0], yedges[-1], xedges[0], xedges[-1]]
cp = ax.imshow(H, interpolation='bilinear', extent=extent, cmap=cm.jet)
fig.colorbar(cp)
plt.show()
The links for the data used is here:
https://dl.dropbox.com/u/10411539/python/test.dat
Any help is appreciated!
Try different interpolation, and transpose the matrix to get it in the same axis:
cp = ax.imshow(H.transpose()[::-1], interpolation='nearest', extent=extent, cmap=cm.jet)
Is this what you want to get? You can use pcolor (and pcolormesh) if you want to pass the x and y coordinates.
import urllib
import numpy as np
import matplotlib.pyplot as plt
f = urllib.urlopen('https://dl.dropbox.com/u/10411539/python/test.dat')
j, h, k = np.loadtxt(f, usecols=(2, 4, 6), \
unpack=True)
j, h, k
j_h = j - h
h_k = h - k
H, xedges, yedges = np.histogram2d(j_h, h_k, bins=100)
plt.pcolor(xedges, yedges, H)
For imshow you have to reverse the first dimension, because imshow uses zero-based row, column indices to the x, y. Drawing from the top down.
plt.imshow(H[::-1,:], extent=(0,5, 0,2.5)) # typed in extent by hand.

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