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I have created this topic before: How to Add another subplot to show Solid of Revolution toward x-axis?
I want to plot the solid from a curve 3 + 2x - x^2 that is revolved about:
a. the x-axis
b. the y-axis
c. the line y = -1
d. the line x = 4
this is my MWE (the problem is to find the inverse of 3 + 2x - x^2 is not something easy thus I have no idea how to make this code works):
# Compare the plot at xy axis with the solid of revolution toward x and y axis
# For function x=(y)^(3/2)
import matplotlib.pyplot as plt
import numpy as np
n = 100
fig = plt.figure(figsize=(14, 7))
ax1 = fig.add_subplot(221)
ax2 = fig.add_subplot(222, projection='3d')
ax3 = fig.add_subplot(223)
ax4 = fig.add_subplot(224, projection='3d')
y = np.linspace(0, 9, n)
x = (y) ** (3 / 2)
t = np.linspace(0, np.pi * 2, n)
xn = np.outer(x, np.cos(t))
yn = np.outer(x, np.sin(t))
zn = np.zeros_like(xn)
for i in range(len(x)):
zn[i:i + 1, :] = np.full_like(zn[0, :], y[i])
ax1.plot(x, y)
ax1.set_title("$f(x)$")
ax2.plot_surface(xn, yn, zn)
ax2.set_title("$f(x)$: Revolution around $y$")
# find the inverse of the function
x_inverse = y
y_inverse = np.power(x_inverse, 3 / 2)
xn_inverse = np.outer(x_inverse, np.cos(t))
yn_inverse = np.outer(x_inverse, np.sin(t))
zn_inverse = np.zeros_like(xn_inverse)
for i in range(len(x_inverse)):
zn_inverse[i:i + 1, :] = np.full_like(zn_inverse[0, :], y_inverse[i])
ax3.plot(x_inverse, y_inverse)
ax3.set_title("Inverse of $f(x)$")
ax4.plot_surface(xn_inverse, yn_inverse, zn_inverse)
ax4.set_title("$f(x)$: Revolution around $x$")
plt.tight_layout()
plt.show()
I have this code modified from the topic here:
How to produce a revolution of a 2D plot with matplotlib in Python
The plot contains a subplot in the XY plane and another subplot of the solid of revolution toward the y-axis.
I want to add another subplot that is the solid of revolution toward the x-axis + how to add a legend to each subplot (above them), so there will be 3 subplots.
This is my MWE:
# Compare the plot at xy axis with the solid of revolution
# For function x=(y-2)^(1/3)
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np
n = 100
fig = plt.figure(figsize=(12,6))
ax1 = fig.add_subplot(121)
ax2 = fig.add_subplot(122,projection='3d')
y = np.linspace(np.pi/8, np.pi*40/5, n)
x = (y-2)**(1/3) # x = np.sin(y)
t = np.linspace(0, np.pi*2, n)
xn = np.outer(x, np.cos(t))
yn = np.outer(x, np.sin(t))
zn = np.zeros_like(xn)
for i in range(len(x)):
zn[i:i+1,:] = np.full_like(zn[0,:], y[i])
ax1.plot(x, y)
ax2.plot_surface(xn, yn, zn)
plt.show()
Option 1:
Simply reverse x and y to switch the axes of the function.
x = np.linspace(np.pi/8, np.pi*40/5, n)
y = (x-2)**(1/3)
Option 2:
It is a little complicated. You can also accomplish this by finding the inverse of the original function.
The inverse of f(x) = y = x^3 + 2 is f^{-1}(y) = (y - 2)^(1/3).
I modified the code you provided.
import matplotlib.pyplot as plt
import numpy as np
n = 100
fig = plt.figure(figsize=(14, 7))
ax1 = fig.add_subplot(221)
ax2 = fig.add_subplot(222, projection='3d')
ax3 = fig.add_subplot(223)
ax4 = fig.add_subplot(224, projection='3d')
y = np.linspace(np.pi / 8, np.pi * 40 / 5, n)
x = (y - 2) ** (1 / 3)
t = np.linspace(0, np.pi * 2, n)
xn = np.outer(x, np.cos(t))
yn = np.outer(x, np.sin(t))
zn = np.zeros_like(xn)
for i in range(len(x)):
zn[i:i + 1, :] = np.full_like(zn[0, :], y[i])
ax1.plot(x, y)
ax1.set_title("$f(x)$")
ax2.plot_surface(xn, yn, zn)
ax2.set_title("$f(x)$: Revolution around $y$")
# find the inverse of the function
x_inverse = y
y_inverse = np.power(x_inverse - 2, 1 / 3)
xn_inverse = np.outer(x_inverse, np.cos(t))
yn_inverse = np.outer(x_inverse, np.sin(t))
zn_inverse = np.zeros_like(xn_inverse)
for i in range(len(x_inverse)):
zn_inverse[i:i + 1, :] = np.full_like(zn_inverse[0, :], y_inverse[i])
ax3.plot(x_inverse, y_inverse)
ax3.set_title("Inverse of $f(x)$")
ax4.plot_surface(xn_inverse, yn_inverse, zn_inverse)
ax4.set_title("$f(x)$: Revolution around $x$")
plt.tight_layout()
plt.show()
Trying to plot two separate animations, i.e. in different windows as separate figures. Running this code for me rightly creates two windows, but animates the data on the second figure at the same time. Closing figure 1 results in only the intended data for figure 2 being animated, removing the overlap from the data intended for figure 1. Closing figure 2 results in only the intended data for figure 1 being animated, removing the overlap from the data intended for figure 2.
Minimum code below:
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.animation import FuncAnimation
dx, dv, N, Nb, decp = 2, 1.5, 100, 12, int(1)
Pd = np.zeros([N + 1, 2 * Nb])
Vd = np.zeros([N + 1, 2 * Nb])
Pd[:, 1] = 4
Vd[:, 3] = 2
t = np.zeros(N + 1)
t[0] = 0
for i in range(0, N):
t[i + 1] = (i + 1) * 0.1
Px = []
for i in range(0, (2 * Nb)):
PX = dx * (-Nb + i) / 4
Px.append(PX)
lblx = []
for i in range(0, int((Nb / 2) + 1)):
if i == (Nb / 4):
LBL = r"$\mu_x$"
lblx.append(LBL)
else:
LBL = r"${0}\sigma_x$".format(-(Nb / 4) + i)
lblx.append(LBL)
Pv = []
for i in range(0, (2 * Nb)):
PV = dv * (-Nb + i) / 4
Pv.append(PV)
lblv = []
for i in range(0, int((Nb / 2) + 1)):
if i == (Nb / 4):
LBL = r"$\mu_v$"
lblv.append(LBL)
else:
LBL = r"${0}\sigma_v$".format(-(Nb / 4) + i)
lblv.append(LBL)
fig1 = plt.figure(figsize=(8,6))
def animatex(i):
fig1.clear()
plt.bar(Px, Pd[i, :], width = dx / 4, align = 'edge', color = 'b', \
label = 't = {} seconds'.format(round(t[i], decp)))
s_ticks = np.arange(-3 * dx, (3 + 1) * dx, dx)
plt.xticks(s_ticks, lblx)
plt.ylim(0, np.max(Pd))
plt.xlim(-3 * dx, 3 * dx)
plt.legend()
plt.draw()
anix = FuncAnimation(fig1, animatex, repeat = True, interval = 200, frames = N + 1)
fig2 = plt.figure(figsize=(8,6))
def animatev(i):
fig2.clear()
plt.bar(Pv, Vd[i, :], width = dv / 4, align = 'edge', color = 'b', \
label = 't = {} seconds'.format(round(t[i], decp)))
s_ticks = np.arange(-3 * dv, (3 + 1) * dv, dv)
plt.xticks(s_ticks, lblv)
plt.ylim(0, np.max(Vd))
plt.xlim(-3 * dv, 3 * dv)
plt.legend()
plt.draw()
aniv = FuncAnimation(fig2, animatev, repeat = True, interval = 200, frames = N + 1)
plt.show()
As is probably clear, they are two bar plots, with different vertical and horizontal dimensions. I've seen some solutions for these kinds of problems where the data shares an axis through a shared variable, but here they are not (as can be seen).
For this minimum code, the solution involves having the two bars, one in Pd and the other in Vd, being on their respective intended figures, not both on the second figure.
Let me know if there are any issues with the information here i.e. minimal code requirements not met, more information etc. and I will update.
Ignore any wayward writing style, it is not relevant.
Simplifying your code:
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.animation import FuncAnimation
dx, dv, N, Nb, decp = 2, 1.5, 10, 12, int(1)
Px = np.arange(Nb)
Pd = np.random.randn(N, Nb)
Vd = np.random.randn(N, Nb)
fig1, ax1 = plt.subplots(figsize=(8, 6))
def animatex(i):
ax1.clear()
ax1.bar(Px, Pd[i, :], width=dx / 4, align='edge', color='b')
anix = FuncAnimation(fig1, animatex, repeat=True, interval=200, frames=N)
fig2, ax2 = plt.subplots(figsize=(8, 6))
def animatev(i):
ax2.clear()
ax2.bar(Px, Vd[i, :], width = dv / 4, align='edge', color='b')
aniv = FuncAnimation(fig2, animatev, repeat=True, interval=200, frames=N)
plt.show()
works fine for me. You can add the esthetic/data details back in...
I have the following code:
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-np.pi/2, np.pi/2, 30)
y = np.linspace(-np.pi/2, np.pi/2, 30)
x,y = np.meshgrid(x,y)
z = np.sin(x**2+y**2)[:-1,:-1]
fig,ax = plt.subplots()
ax.pcolormesh(x,y,z)
Which gives this image:
Now lets say I want to highlight the edge certain grid boxes:
highlight = (z > 0.9)
I could use the contour function, but this would result in a "smoothed" contour. I just want to highlight the edge of a region, following the edge of the grid boxes.
The closest I've come is adding something like this:
highlight = np.ma.masked_less(highlight, 1)
ax.pcolormesh(x, y, highlight, facecolor = 'None', edgecolors = 'w')
Which gives this plot:
Which is close, but what I really want is for only the outer and inner edges of that "donut" to be highlighted.
So essentially I am looking for some hybrid of the contour and pcolormesh functions - something that follows the contour of some value, but follows grid bins in "steps" rather than connecting point-to-point. Does that make sense?
Side note: In the pcolormesh arguments, I have edgecolors = 'w', but the edges still come out to be blue. Whats going on there?
EDIT:
JohanC's initial answer using add_iso_line() works for the question as posed. However, the actual data I'm using is a very irregular x,y grid, which cannot be converted to 1D (as is required for add_iso_line().
I am using data which has been converted from polar coordinates (rho, phi) to cartesian (x,y). The 2D solution posed by JohanC does not appear to work for the following case:
import numpy as np
import matplotlib.pyplot as plt
from scipy import ndimage
def pol2cart(rho, phi):
x = rho * np.cos(phi)
y = rho * np.sin(phi)
return(x, y)
phi = np.linspace(0,2*np.pi,30)
rho = np.linspace(0,2,30)
pp, rr = np.meshgrid(phi,rho)
xx,yy = pol2cart(rr, pp)
z = np.sin(xx**2 + yy**2)
scale = 5
zz = ndimage.zoom(z, scale, order=0)
fig,ax = plt.subplots()
ax.pcolormesh(xx,yy,z[:-1, :-1])
xlim = ax.get_xlim()
ylim = ax.get_ylim()
xmin, xmax = xx.min(), xx.max()
ymin, ymax = yy.min(), yy.max()
ax.contour(np.linspace(xmin,xmax, zz.shape[1]) + (xmax-xmin)/z.shape[1]/2,
np.linspace(ymin,ymax, zz.shape[0]) + (ymax-ymin)/z.shape[0]/2,
np.where(zz < 0.9, 0, 1), levels=[0.5], colors='red')
ax.set_xlim(*xlim)
ax.set_ylim(*ylim)
This post shows a way to draw such lines. As it is not straightforward to adapt to the current pcolormesh, the following code demonstrates a possible adaption.
Note that the 2d versions of x and y have been renamed, as the 1d versions are needed for the line segments.
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.collections import LineCollection
x = np.linspace(-np.pi / 2, np.pi / 2, 30)
y = np.linspace(-np.pi / 2, np.pi / 2, 30)
xx, yy = np.meshgrid(x, y)
z = np.sin(xx ** 2 + yy ** 2)[:-1, :-1]
fig, ax = plt.subplots()
ax.pcolormesh(x, y, z)
def add_iso_line(ax, value, color):
v = np.diff(z > value, axis=1)
h = np.diff(z > value, axis=0)
l = np.argwhere(v.T)
vlines = np.array(list(zip(np.stack((x[l[:, 0] + 1], y[l[:, 1]])).T,
np.stack((x[l[:, 0] + 1], y[l[:, 1] + 1])).T)))
l = np.argwhere(h.T)
hlines = np.array(list(zip(np.stack((x[l[:, 0]], y[l[:, 1] + 1])).T,
np.stack((x[l[:, 0] + 1], y[l[:, 1] + 1])).T)))
lines = np.vstack((vlines, hlines))
ax.add_collection(LineCollection(lines, lw=1, colors=color))
add_iso_line(ax, 0.9, 'r')
plt.show()
Here is an adaption of the second answer, which can work with only 2d arrays:
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.collections import LineCollection
from scipy import ndimage
x = np.linspace(-np.pi / 2, np.pi / 2, 30)
y = np.linspace(-np.pi / 2, np.pi / 2, 30)
x, y = np.meshgrid(x, y)
z = np.sin(x ** 2 + y ** 2)
scale = 5
zz = ndimage.zoom(z, scale, order=0)
fig, ax = plt.subplots()
ax.pcolormesh(x, y, z[:-1, :-1] )
xlim = ax.get_xlim()
ylim = ax.get_ylim()
xmin, xmax = x.min(), x.max()
ymin, ymax = y.min(), y.max()
ax.contour(np.linspace(xmin,xmax, zz.shape[1]) + (xmax-xmin)/z.shape[1]/2,
np.linspace(ymin,ymax, zz.shape[0]) + (ymax-ymin)/z.shape[0]/2,
np.where(zz < 0.9, 0, 1), levels=[0.5], colors='red')
ax.set_xlim(*xlim)
ax.set_ylim(*ylim)
plt.show()
I'll try to refactor add_iso_line method in order to make it more clear an open for optimisations. So, at first, there comes a must-do part:
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.collections import LineCollection
x = np.linspace(-np.pi/2, np.pi/2, 30)
y = np.linspace(-np.pi/2, np.pi/2, 30)
x, y = np.meshgrid(x,y)
z = np.sin(x**2+y**2)[:-1,:-1]
fig, ax = plt.subplots()
ax.pcolormesh(x,y,z)
xlim, ylim = ax.get_xlim(), ax.get_ylim()
highlight = (z > 0.9)
Now highlight is a binary array that looks like this:
After that we can extract indexes of True cells, look for False neighbourhoods and identify positions of 'red' lines. I'm not comfortable enough with doing it in a vectorised manner (like here in add_iso_line method) so just using simple loop:
lines = []
cells = zip(*np.where(highlight))
for x, y in cells:
if x == 0 or highlight[x - 1, y] == 0: lines.append(([x, y], [x, y + 1]))
if x == highlight.shape[0] or highlight[x + 1, y] == 0: lines.append(([x + 1, y], [x + 1, y + 1]))
if y == 0 or highlight[x, y - 1] == 0: lines.append(([x, y], [x + 1, y]))
if y == highlight.shape[1] or highlight[x, y + 1] == 0: lines.append(([x, y + 1], [x + 1, y + 1]))
And, finally, I resize and center coordinates of lines in order to fit with pcolormesh:
lines = (np.array(lines) / highlight.shape - [0.5, 0.5]) * [xlim[1] - xlim[0], ylim[1] - ylim[0]]
ax.add_collection(LineCollection(lines, colors='r'))
plt.show()
In conclusion, this is very similar to JohanC solution and, in general, slower. Fortunately, we can reduce amount of cells significantly, extracting contours only using python-opencv package:
import cv2
highlight = highlight.astype(np.uint8)
contours, hierarchy = cv2.findContours(highlight, cv2.RETR_TREE, cv2.CHAIN_APPROX_NONE)
cells = np.vstack(contours).squeeze()
This is an illustration of cells being checked:
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()