General aim
I am trying to write some plotting functionality that (at its core)
plots arbitrary paths with a constant width given in data coordinates
(i.e. unlike lines in matplotlib which have widths given in display coordinates).
Previous solutions
This answer achieves
the basic goal. However, this answer converts between display and data
coordinates and then uses a matplotlib line with adjusted
coordinates. The existing functionality in my code that I would like
to replace / extend inherits from matplotlib.patches.Polygon. Since
the rest of the code base makes extensive use of
matplotlib.patches.Polygon attributes and methods, I would like to
continue to inherit from that class.
Problem
My current implementation (code below) seems to come close. However,
the patch created by simple_test seems to be subtly thicker towards
the centre than it is at the start and end point, and I have no
explanation why that may be the case.
I suspect that the problem lies in the computation of the orthogonal vector.
As supporting evidence, I would like to point to the start and end points of the patch in the figure created by complicated_test, which do not seem exactly orthogonal to the path. However, the dot product of the orthonormal vector and the tangent vector is always zero, so I am not sure that what is going on here.
Output of simple_test:
Output of complicated_test:
Code
#!/usr/bin/env python
import numpy as np
import matplotlib.patches
import matplotlib.pyplot as plt
class CurvedPatch(matplotlib.patches.Polygon):
def __init__(self, path, width, *args, **kwargs):
vertices = self.get_vertices(path, width)
matplotlib.patches.Polygon.__init__(self, list(map(tuple, vertices)),
closed=True,
*args, **kwargs)
def get_vertices(self, path, width):
left = _get_parallel_path(path, -width/2)
right = _get_parallel_path(path, width/2)
full = np.concatenate([left, right[::-1]])
return full
def _get_parallel_path(path, delta):
# initialise output
offset = np.zeros_like(path)
# use the previous and the following point to
# determine the tangent at each point in the path;
for ii in range(1, len(path)-1):
offset[ii] += _get_shift(path[ii-1], path[ii+1], delta)
# handle start and end points
offset[0] = _get_shift(path[0], path[1], delta)
offset[-1] = _get_shift(path[-2], path[-1], delta)
return path + offset
def _get_shift(p1, p2, delta):
# unpack coordinates
x1, y1 = p1
x2, y2 = p2
# get orthogonal unit vector;
# adapted from https://stackoverflow.com/a/16890776/2912349
v = np.r_[x2-x1, y2-y1] # vector between points
v = v / np.linalg.norm(v) # unit vector
w = np.r_[-v[1], v[0]] # orthogonal vector
w = w / np.linalg.norm(w) # orthogonal unit vector
# check that vectors are indeed orthogonal
assert np.isclose(np.dot(v, w), 0.)
# rescale unit vector
dx, dy = delta * w
return dx, dy
def simple_test():
x = np.linspace(-1, 1, 1000)
y = np.sqrt(1. - x**2)
path = np.c_[x, y]
curve = CurvedPatch(path, 0.1, facecolor='red', alpha=0.5)
fig, ax = plt.subplots(1,1)
ax.add_artist(curve)
ax.plot(x, y) # plot path for reference
plt.show()
def complicated_test():
random_points = np.random.rand(10, 2)
# Adapted from https://stackoverflow.com/a/35007804/2912349
import scipy.interpolate as si
def scipy_bspline(cv, n=100, degree=3, periodic=False):
""" Calculate n samples on a bspline
cv : Array ov control vertices
n : Number of samples to return
degree: Curve degree
periodic: True - Curve is closed
"""
cv = np.asarray(cv)
count = cv.shape[0]
# Closed curve
if periodic:
kv = np.arange(-degree,count+degree+1)
factor, fraction = divmod(count+degree+1, count)
cv = np.roll(np.concatenate((cv,) * factor + (cv[:fraction],)),-1,axis=0)
degree = np.clip(degree,1,degree)
# Opened curve
else:
degree = np.clip(degree,1,count-1)
kv = np.clip(np.arange(count+degree+1)-degree,0,count-degree)
# Return samples
max_param = count - (degree * (1-periodic))
spl = si.BSpline(kv, cv, degree)
return spl(np.linspace(0,max_param,n))
x, y = scipy_bspline(random_points, n=1000).T
path = np.c_[x, y]
curve = CurvedPatch(path, 0.1, facecolor='red', alpha=0.5)
fig, ax = plt.subplots(1,1)
ax.add_artist(curve)
ax.plot(x, y) # plot path for reference
plt.show()
if __name__ == '__main__':
plt.ion()
simple_test()
complicated_test()
Related
I'm plotting the curve of a function, and it's tangent at point p. I would like to manage xlim for the curve and the tangent independently. In this code the tangent half-length should be 1:
from sympy import init_printing, symbols, N, plot
from sympy import diff
from sympy import log, cos, atan
init_printing()
x = symbols('x')
# Plot a tangent at point (p_x, p_y), of length l
def plot_line(p_x, p_y, x, a, l):
# Compute b, build tangent expression
b = p_y - a*p_x
t = a*x + b
# Limit line length
r = atan(a) # angle in rad
dx = N(l*cos(r)) # half range for x
lims = {'xlim': (p_x-dx, p_x+dx)}
# Build plot
t_plot = plot(t, show=False, **lims)
return t_plot
# Function
y = 2.1*log(x)
# Point
px = 7
py = y.subs(x, px)
# Plot curve and point
marker = {'args': [px, py, 'bo']}
lims = {'xlim': (0,10), 'ylim': (0,5)}
plots = plot(y, markers=[marker], show=False, **lims)
# Find derivative, plot tangent
y_d = diff(y)
a = y_d.subs(x, px)
plots.extend(plot_line(px, py, x, a, 1))
# Finalize and show plots
plots.aspect_ratio=(1,1)
plots.show()
However this is not the case...
SymPy's plot() function signature is something similar to this:
plot(expr, range, **kwargs)
where range is a 3-elements tuple: (symbol, min_val, max_val). The plot function will evaluate expr starting from min_val up to max_val.
One of the **kwargs is xlim, which is a 2-element tuple: xlim=(x_min, x_max). It is used to restrict the visualization along the x-axis from x_min to x_max. Nonetheless, the numerical values computed by the plot function go from min_val to max_val.
With that said, you need to remove xlim from inside plot_line and provide the range argument instead:
from sympy import init_printing, symbols, N, plot
from sympy import diff
from sympy import log, cos, atan
init_printing()
x = symbols('x')
# Plot a tangent at point (p_x, p_y), of length l
def plot_line(p_x, p_y, x, a, l):
# Compute b, build tangent expression
b = p_y - a*p_x
t = a*x + b
# Limit line length
r = atan(a) # angle in rad
dx = N(l*cos(r)) # half range for x
# Build plot
# Need to provide the range to limit the line length
t_plot = plot(t, (x, p_x-dx, p_x+dx), show=False)
return t_plot
# Function
y = 2.1*log(x)
# Point
px = 7
py = y.subs(x, px)
# Plot curve and point
marker = {'args': [px, py, 'bo']}
lims = {'xlim': (0,10), 'ylim': (0,5)}
plots = plot(y, markers=[marker], show=False, **lims)
# Find derivative, plot tangent
y_d = diff(y)
a = y_d.subs(x, px)
plots.extend(plot_line(px, py, x, a, 1))
# Finalize and show plots
plots.aspect_ratio=(1,1)
plots.show()
I wrote some code that creates randomised patches from graphs in matplotlib. Basically how it works is that you create a graph from nodes taken from a circle using the parametric equation for a circle and then you randomly displace the nodes along the vector of (0,0) to the node point on the circumference of the circle. That way you can be certain to avoid lines from crossing each other once the circle is drawn. In the end you just append the first (x,y) coordinate to the list of coordinates to close the circle.
What I want to do next is to find a way to fill that circular graph with a solid colour so that I can create a "stamp" that can be used to make randomised patches on a canvas that hopefully will not create crossing edges. I want to use this to make procedural risk maps in svg format, because a lot of those are uploaded with terrible edges using raster image formats using jpeg.
I am pretty sure that my information of the nodes should be sufficient to make that happen but I have no idea how to implement that. Can anyone help?
import numpy as np
import matplotlib.pyplot as plt
def node_circle(r=0.5,res=100):
# Create arrays (x and y coordinates) for the nodes on the circumference of a circle. Use parametric equation.
# x = r cos(t) y = r sin(t)
t = np.linspace(0,2*np.pi,res)
x = r*np.cos(t)
y = r*np.sin(t)
return t,x,y
def sgn(x,x_shift=-0.5,y_shift=1):
# A shifted sign function to use as a switching function
# in order to avoid shifts lower than -0.5 which is
# the radius of the circle.
return -0.5*(np.abs(x -x_shift)/(x -x_shift)) +y_shift
def displacer(x,y,low=-0.5,high=0.5,maxrad=0.5):
# Displaces the node points of the circle
shift = 0
shift_increment = 0
for i in range(len(x)):
shift_increment = np.random.uniform(low,high)
shift += shift_increment*sgn(maxrad)
x[i] += x[i]*shift
y[i] += y[i]*shift
x = np.append(x,x[0])
y = np.append(y,y[0])
return x,y
def plot():
# Actually visualises everything
fig, ax = plt.subplots(figsize=(4,4))
# np.random.seed(1)
ax.axis('off')
t,x,y = node_circle(res=100)
a = 0
x,y = displacer(x,y,low=-0.15,high=0.15)
ax.plot(x,y,'r-')
# ax.scatter(x,y,)
plt.show()
plot()
got it: the answer is to use matplotlib.Patches.Polygon
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.patches import Polygon
def node_circle(r=0.5,res=100):
# Create arrays (x and y coordinates) for the nodes on the circumference of a circle. Use parametric equation.
# x = r cos(t) y = r sin(t)
t = np.linspace(0,2*np.pi,res)
x = r*np.cos(t)
y = r*np.sin(t)
return x,y
def sgn(x,x_shift=-0.5,y_shift=1):
# A shifted sign function to use as a switching function
# in order to avoid shifts lower than -0.5 which is
# the radius of the circle.
return -0.5*(np.abs(x -x_shift)/(x -x_shift)) +y_shift
def displacer(x,y,low=-0.5,high=0.5,maxrad=0.5):
# Displaces the node points of the circle
shift = 0
shift_increment = 0
for i in range(len(x)):
shift_increment = np.random.uniform(low,high)
shift += shift_increment*sgn(maxrad)
x[i] += x[i]*shift
y[i] += y[i]*shift
x = np.append(x,x[0])
y = np.append(y,y[0])
return x,y
def patch_distributor(M,N,res,grid='square'):
# Distribute Patches based on a specified pattern/grid.
if grid == 'square':
data = np.zeros(shape=(M,N,2,res+1))
for i in range(M):
for j in range(N):
x,y = displacer(*node_circle(res=res),low=-0.2,high=0.2)
data[i,j,0,:] = x
data[i,j,1,:] = y
return data
def plot(res):
# Actually visualises everything
fig, ax = plt.subplots(figsize=(4,4))
# np.random.seed(1)
ax.axis('off')
# x,y = node_circle(res=res)
# x,y = displacer(x,y,low=-0.15,high=0.15)
# xy = np.zeros((len(x),2))
# xy[:,0] = x
# xy[:,1] = y
patch_data = patch_distributor(10,10,res)
for i in range(patch_data.shape[0]):
for j in range(patch_data.shape[1]):
x,y = patch_data[i,j]
x += i*0.5
y += j*0.5
xy = np.zeros((len(x),2))
xy[:,0] = x
xy[:,1] = y
patch = Polygon(xy,fc='w',ec='k',lw=2,zorder=np.random.randint(2),antialiased=False)
ax.add_patch(patch)
ax.autoscale_view()
# ax.plot(x,y,'r-')
# ax.scatter(x,y,)
plt.savefig('lol.png')
plot(res=40)
# Displace circle along the line of (0,0) -> (cos(t),sin(t))
# Make the previous step influence the next to avoid jaggedness
# limit displacement level to an acceptable amount
# Random displaced cubic grid as placing points for stamps.
say we have a 2D grid that is projected on a 3D surface, resulting in a 3D numpy array, like the below image. What is the most efficient way to calculate a surface normal for each point of this grid?
I can give you an example with simulated data:
I showed your way, with three points. With three points you can always calculate the cross product to get the perpendicular vector based on the two vectors created from three points. Order does not matter.
I took the liberty to also add the PCA approach using predefined sklearn functions. You can create your own PCA, good exercise to understand what happens under the hood but this works fine. The benefit of the approach is that it is easy to increase the number of neighbors and you are still able to calculate the normal vector. It is also possible to select the neighbors within a range instead of N nearest neighbors.
If you need more explanation about the working of the code please let me know.
from functools import partial
import numpy as np
from sklearn.neighbors import KDTree
from mpl_toolkits.mplot3d import axes3d
import matplotlib.pyplot as plt
from sklearn.decomposition import PCA
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
# Grab some test data.
X, Y, Z = axes3d.get_test_data(0.25)
X, Y, Z = map(lambda x: x.flatten(), [X, Y, Z])
plt.plot(X, Y, Z, '.')
plt.show(block=False)
data = np.array([X, Y, Z]).T
tree = KDTree(data, metric='minkowski') # minkowki is p2 (euclidean)
# Get indices and distances:
dist, ind = tree.query(data, k=3) #k=3 points including itself
def calc_cross(p1, p2, p3):
v1 = p2 - p1
v2 = p3 - p1
v3 = np.cross(v1, v2)
return v3 / np.linalg.norm(v3)
def PCA_unit_vector(array, pca=PCA(n_components=3)):
pca.fit(array)
eigenvalues = pca.explained_variance_
return pca.components_[ np.argmin(eigenvalues) ]
combinations = data[ind]
normals = list(map(lambda x: calc_cross(*x), combinations))
# lazy with map
normals2 = list(map(PCA_unit_vector, combinations))
## NEW ##
def calc_angle_with_xy(vectors):
'''
Assuming unit vectors!
'''
l = np.sum(vectors[:,:2]**2, axis=1) ** 0.5
return np.arctan2(vectors[:, 2], l)
dist, ind = tree.query(data, k=5) #k=3 points including itself
combinations = data[ind]
# map with functools
pca = PCA(n_components=3)
normals3 = list(map(partial(PCA_unit_vector, pca=pca), combinations))
print( combinations[10] )
print(normals3[10])
n = np.array(normals3)
n[calc_angle_with_xy(n) < 0] *= -1
def set_axes_equal(ax):
'''Make axes of 3D plot have equal scale so that spheres appear as spheres,
cubes as cubes, etc.. This is one possible solution to Matplotlib's
ax.set_aspect('equal') and ax.axis('equal') not working for 3D.
Input
ax: a matplotlib axis, e.g., as output from plt.gca().
FROM: https://stackoverflow.com/questions/13685386/matplotlib-equal-unit-length-with-equal-aspect-ratio-z-axis-is-not-equal-to
'''
x_limits = ax.get_xlim3d()
y_limits = ax.get_ylim3d()
z_limits = ax.get_zlim3d()
x_range = abs(x_limits[1] - x_limits[0])
x_middle = np.mean(x_limits)
y_range = abs(y_limits[1] - y_limits[0])
y_middle = np.mean(y_limits)
z_range = abs(z_limits[1] - z_limits[0])
z_middle = np.mean(z_limits)
# The plot bounding box is a sphere in the sense of the infinity
# norm, hence I call half the max range the plot radius.
plot_radius = 0.5*max([x_range, y_range, z_range])
ax.set_xlim3d([x_middle - plot_radius, x_middle + plot_radius])
ax.set_ylim3d([y_middle - plot_radius, y_middle + plot_radius])
ax.set_zlim3d([z_middle - plot_radius, z_middle + plot_radius])
u, v, w = n.T
fig = plt.figure()
ax = fig.add_subplot(projection='3d')
# ax.set_aspect('equal')
# Make the grid
ax.quiver(X, Y, Z, u, v, w, length=10, normalize=True)
set_axes_equal(ax)
plt.show()
The surface normal for a point cloud is not well defined. One way to define them is from the surface normal of a reconstructed mesh using triangulation (which can introduce artefacts regarding you specific input). A relatively simple and fast solution is to use VTK to do that, and more specifically, vtkSurfaceReconstructionFilter and vtkPolyDataNormals . Regarding your needs, it might be useful to apply other filters.
I'd like to plot two profiles through the highest intensity point in a 2D numpy array, which is an image of a blob (i.e. a line through the semi-major axis, and another line through the semi-minor axis). The blob is rotated at an angle theta counterclockwise from the standard x-axis and is asymmetric.
It is a 600x600 array with a max intensity of 1 (at only one pixel) that is located right at the center at (300, 300). The angle rotation from the x-axis (which then gives the location of the semi-major axis when rotated by that angle) is theta = 89.54 degrees. I do not want to use scipy.ndimage.rotate because it uses spline interpolation, and I do not want to change any of my pixel values. But I suppose a nearest-neighbor interpolation method would be okay.
I tried generating lines corresponding to the major and minor axes across the image, but the result was not right at all (the peak was far less than 1), so maybe I did something wrong. The code for this is below:
import numpy as np
import matplotlib.pyplot as plt
from scipy import ndimage
def profiles_at_angle(image, axis, theta):
theta = np.deg2rad(theta)
if axis == 'major':
x_0, y_0 = 0, 300-300*np.tan(theta)
x_1, y_1 = 599, 300+300*np.tan(theta)
elif axis=='minor':
x_0, y_0 = 300-300*np.tan(theta), 599
x_1, y_1 = 300+300*np.tan(theta), -599
num = 600
x, y = np.linspace(x_0, x_1, num), np.linspace(y_0, y_1, num)
z = ndimage.map_coordinates(image, np.vstack((x,y)))
fig, axes = plt.subplots(nrows=2)
axes[0].imshow(image, cmap='gray')
axes[0].axis('image')
axes[1].plot(z)
plt.xlim(250,350)
plt.show()
profiles_at_angle(image, 'major', theta)
Did I do something obviously wrong in my code above? Or how else can I accomplish this? Thank you.
Edit: Here are some example images. Sorry for the bad quality; my browser crashed every time I tried uploading them anywhere so I had to take photos of the screen.
Figure 1: This is the result of my code above, which is clearly wrong since the peak should be at 1. I'm not sure what I did wrong though.
Figure 2: I made this plot below by just taking the profiles through the standard x and y axes, ignoring any rotation (this only looks good coincidentally because the real angle of rotation is so close to 90 degrees, so I was able to just switch the labels and get this). I want my result to look something like this, but taking the correction rotation angle into account.
Edit: It could be useful to run tests on this method using data very much like my own (it's a 2D Gaussian with nearly the same parameters):
image = np.random.random((600,600))
def generate(data_set):
xvec = np.arange(0, np.shape(data_set)[1], 1)
yvec = np.arange(0, np.shape(data_set)[0], 1)
X, Y = np.meshgrid(xvec, yvec)
return X, Y
def gaussian_func(xy, x0, y0, sigma_x, sigma_y, amp, theta, offset):
x, y = xy
a = (np.cos(theta))**2/(2*sigma_x**2) + (np.sin(theta))**2/(2*sigma_y**2)
b = -np.sin(2*theta)/(4*sigma_x**2) + np.sin(2*theta)/(4*sigma_y**2)
c = (np.sin(theta))**2/(2*sigma_x**2) + (np.cos(theta))**2/(2*sigma_y**2)
inner = a * (x-x0)**2
inner += 2*b*(x-x0)*(y-y0)
inner += c * (y-y0)**2
return (offset + amp * np.exp(-inner)).ravel()
xx, yy = generate(image)
image = gaussian_func((xx.ravel(), yy.ravel()), 300, 300, 5, 4, 1, 1.56, 0)
image = np.reshape(image, (600, 600))
This should do it for you. You just did not properly compute your lines.
theta = 65
peak = np.argwhere(image==1)[0]
x = np.linspace(peak[0]-100,peak[0]+100,1000)
y = lambda x: (x-peak[1])*np.tan(np.deg2rad(theta))+peak[0]
y_maj = np.linspace(y(peak[1]-100),y(peak[1]+100),1000)
y = lambda x: -(x-peak[1])/np.tan(np.deg2rad(theta))+peak[0]
y_min = np.linspace(y(peak[1]-100),y(peak[1]+100),1000)
del y
z_min = scipy.ndimage.map_coordinates(image, np.vstack((x,y_min)))
z_maj = scipy.ndimage.map_coordinates(image, np.vstack((x,y_maj)))
fig, axes = plt.subplots(nrows=2)
axes[0].imshow(image)
axes[0].plot(x,y_maj)
axes[0].plot(x,y_min)
axes[0].axis('image')
axes[1].plot(z_min)
axes[1].plot(z_maj)
plt.show()
I solve a differential equation with vector inputs
y' = f(t,y), y(t_0) = y_0
where y0 = y(x)
using the explicit Euler method, which says that
y_(i+1) = y_i + h*f(t_i, y_i)
where t is a time vector, h is the step size, and f is the right-hand side of the differential equation.
The python code for the method looks like this:
for i in np.arange(0,n-1):
y[i+1,...] = y[i,...] + dt*myode(t[i],y[i,...])
The result is a k,m matrix y, where k is the size of the t dimension, and m is the size of y.
The vectors y and t are returned.
t, x, and y are passed to scipy.interpolate.RectBivariateSpline(t, x, y, kx=1, ky=1):
g = scipy.interpolate.RectBivariateSpline(t, x, y, kx=1, ky=1)
The resulting object g takes new vectors ti,xi ( g(p,q) ) to give y_int, which is y interpolated at the points defined by ti and xi.
Here is my problem:
The documentation for RectBivariateSpline describes the __call__ method in terms of x and y:
__call__(x, y[, mth]) Evaluate spline at the grid points defined by the coordinate arrays
The matplotlib documentation for plot_surface uses similar notation:
Axes3D.plot_surface(X, Y, Z, *args, **kwargs)
with the important difference that X and Y are 2D arrays which are generated by numpy.meshgrid().
When I compute simple examples, the input order is the same in both and the result is exactly what I would expect. In my explicit Euler example, however, the initial order is ti,xi, yet the surface plot of the interpolant output only makes sense if I reverse the order of the inputs, like so:
ax2.plot_surface(xi, ti, u, cmap=cm.coolwarm)
While I am glad that it works, I'm not satisfied because I cannot explain why, nor why (apart from the array geometry) it is necessary to swap the inputs. Ideally, I would like to restructure the code so that the input order is consistent.
Here is a working code example to illustrate what I mean:
# Heat equation example with explicit Euler method
import numpy as np
import matplotlib.pyplot as mplot
import matplotlib.cm as cm
import scipy.sparse as sp
import scipy.interpolate as interp
from mpl_toolkits.mplot3d import Axes3D
import pdb
# explicit Euler method
def eev(myode,tspan,y0,dt):
# Preprocessing
# Time steps
tspan[1] = tspan[1] + dt
t = np.arange(tspan[0],tspan[1],dt,dtype=float)
n = t.size
m = y0.shape[0]
y = np.zeros((n,m),dtype=float)
y[0,:] = y0
# explicit Euler recurrence relation
for i in np.arange(0,n-1):
y[i+1,...] = y[i,...] + dt*myode(t[i],y[i,...])
return y,t
# generate matrix A
# u'(t) = A*u(t) + g*u(t)
def a_matrix(n):
aa = sp.diags([1, -2, 1],[-1,0,1],(n,n))
return aa
# System of ODEs with finite differences
def f(t,u):
dydt = np.divide(1,h**2)*A.dot(u)
return dydt
# homogenous Dirichlet boundary conditions
def rbd(t):
ul = np.zeros((t,1))
return ul
# Initial value problem -----------
def main():
# Metal rod
# spatial discretization
# number of inner nodes
m = 20
x0 = 0
xn = 1
x = np.linspace(x0,xn,m+2)
# Step size
global h
h = x[1]-x[0]
# Initial values
u0 = np.sin(np.pi*x)
# A matrix
global A
A = a_matrix(m)
# Time
t0 = 0
tend = 0.2
# Time step width
dt = 0.0001
tspan = [t0,tend]
# Test r for stability
r = np.divide(dt,h**2)
if r <= 0.5:
u,t = eev(f,tspan,u0[1:-1],dt)
else:
print('r = ',r)
print('r > 0.5. Explicit Euler method will not be stable.')
# Add boundary values back
rb = rbd(t.size)
u = np.hstack((rb,u,rb))
# Interpolate heat values
# Create interpolant. Note the parameter order
fi = interp.RectBivariateSpline(t, x, u, kx=1, ky=1)
# Create vectors for interpolant
xi = np.linspace(x[0],x[-1],100)
ti = np.linspace(t0,tend,100)
# Compute function values from interpolant
u_int = fi(ti,xi)
# Change xi, ti in to 2D arrays
xi,ti = np.meshgrid(xi,ti)
# Create figure and axes objects
fig3 = mplot.figure(1)
ax3 = fig3.gca(projection='3d')
print('xi.shape =',xi.shape,'ti.shape =',ti.shape,'u_int.shape =',u_int.shape)
# Plot surface. Note the parameter order, compare with interpolant!
ax3.plot_surface(xi, ti, u_int, cmap=cm.coolwarm)
ax3.set_xlabel('xi')
ax3.set_ylabel('ti')
main()
mplot.show()
As I can see you define :
# Change xi, ti in to 2D arrays
xi,ti = np.meshgrid(xi,ti)
Change this to :
ti,xi = np.meshgrid(ti,xi)
and
ax3.plot_surface(xi, ti, u_int, cmap=cm.coolwarm)
to
ax3.plot_surface(ti, xi, u_int, cmap=cm.coolwarm)
and it works fine (if I understood well ).