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How to plot normal vectors in each point of the curve with a given length?
import matplotlib.pyplot as plt
import numpy as np
fig, ax = plt.subplots()
plt.rcParams["figure.figsize"] = [8, 8]
x = np.linspace(-1, 1, 100)
y = x**2
ax.set_ylim(-0.3, 1.06)
ax.plot(x, y)
plt.show()
To plot the normals, you need to calculate the slope at each point; from there, you get the tangent vector that you can rotate by pi/2.
here is one approach using python i/o np, which makes it probably easier to understand at first.
Changing the length will adjust the size of the normals to properly scale with your plot.
import matplotlib.pyplot as plt
import numpy as np
import math
def get_normals(length=.1):
for idx in range(len(x)-1):
x0, y0, xa, ya = x[idx], y[idx], x[idx+1], y[idx+1]
dx, dy = xa-x0, ya-y0
norm = math.hypot(dx, dy) * 1/length
dx /= norm
dy /= norm
ax.plot((x0, x0-dy), (y0, y0+dx)) # plot the normals
fig, ax = plt.subplots()
plt.rcParams["figure.figsize"] = [8, 8]
x = np.linspace(-1, 1, 100)
y = x**2
ax.set_ylim(-0.3, 1.06)
ax.plot(x, y)
get_normals()
plt.show()
or longer normals, directed downwards: get_normals(length=-.3)
(use ax.set_aspect('equal') to maintain angles)
import matplotlib.pyplot as plt
import numpy as np
fig, ax = plt.subplots()
plt.rcParams["figure.figsize"] = [8, 8]
x = np.linspace(-1, 1, 100)
y = x**2
# Calculating the gradient
L=.1 # gradient length
grad = np.ones(shape = (2, x.shape[0]))
grad[0, :] = -2*x
grad /= np.linalg.norm(grad, axis=0) # normalizing to unit vector
nx = np.vstack((x - L/2 * grad[0], x + L/2 * grad[0]))
ny = np.vstack((y - L/2 * grad[1], y + L/2 * grad[1]))
# ax.set_ylim(-0.3, 1.06)
ax.plot(x, y)
ax.plot(nx, ny, 'r')
ax.axis('equal')
plt.show()
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'm looking for help to draw a 3D cone using matplotlib.
My goal is to draw a HSL cone, then base on the vertex coordinats i will select the color.
from matplotlib import cm
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
theta1 = np.linspace(0, 2*np.pi, 100)
r1 = np.linspace(-2, 0, 100)
t1, R1 = np.meshgrid(theta1, r1)
X1 = R1*np.cos(t1)
Y1 = R1*np.sin(t1)
Z1 = 5+R1*2.5
theta2 = np.linspace(0, 2*np.pi, 100)
r2 = np.linspace(0, 2, 100)
t2, R2 = np.meshgrid(theta2, r2)
X2 = R2*np.cos(t2)
Y2 = R2*np.sin(t2)
Z2 = -5+R2*2.5
ax.set_xlabel('x axis')
ax.set_ylabel('y axis')
ax.set_zlabel('z axis')
# ax.set_xlim(-2.5, 2.5)
# ax.set_ylim(-2.5, 2.5)
# ax.set_zlim(0, 5)
ax.set_aspect('equal')
ax.plot_surface(X1, Y1, Z1, alpha=0.8, color="blue")
ax.plot_surface(X2, Y2, Z2, alpha=0.8, color="blue")
# ax.plot_surface(X, Y, Z, alpha=0.8)
#fig. savefig ("Cone.png", dpi=100, transparent = False)
plt.show()
HSL CONE
My cone
So my question now is how to define color of each element.
i have found a solution, maybe it will be usefull for others.
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
import matplotlib.pyplot as plt
import numpy as np
import colorsys
from matplotlib.tri import Triangulation
from mpl_toolkits.mplot3d.art3d import Poly3DCollection
n_angles = 80
n_radii = 20
# An array of radii
# Does not include radius r=0, this is to eliminate duplicate points
radii = np.linspace(0.0, 0.5, n_radii)
# An array of angles
angles = np.linspace(0, 2*np.pi, n_angles, endpoint=False)
# Repeat all angles for each radius
angles = np.repeat(angles[..., np.newaxis], n_radii, axis=1)
# Convert polar (radii, angles) coords to cartesian (x, y) coords
# (0, 0) is added here. There are no duplicate points in the (x, y) plane
x = np.append(0, (radii*np.cos(angles)).flatten())
y = np.append(0, (radii*np.sin(angles)).flatten())
# Pringle surface
z = 1+-np.sqrt(x**2+y**2)*2
print(x.shape, y.shape, angles.shape, radii.shape, z.shape)
# NOTE: This assumes that there is a nice projection of the surface into the x/y-plane!
tri = Triangulation(x, y)
triangle_vertices = np.array([np.array([[x[T[0]], y[T[0]], z[T[0]]],
[x[T[1]], y[T[1]], z[T[1]]],
[x[T[2]], y[T[2]], z[T[2]]]]) for T in tri.triangles])
x2 = np.append(0, (radii*np.cos(angles)).flatten())
y2 = np.append(0, (radii*np.sin(angles)).flatten())
# Pringle surface
z2 = -1+np.sqrt(x**2+y**2)*2
# NOTE: This assumes that there is a nice projection of the surface into the x/y-plane!
tri2 = Triangulation(x2, y2)
triangle_vertices2 = np.array([np.array([[x2[T[0]], y2[T[0]], z2[T[0]]],
[x2[T[1]], y2[T[1]], z2[T[1]]],
[x2[T[2]], y2[T[2]], z2[T[2]]]]) for T in tri2.triangles])
triangle_vertices = np.concatenate([triangle_vertices, triangle_vertices2])
midpoints = np.average(triangle_vertices, axis=1)
def find_color_for_point(pt):
c_x, c_y, c_z = pt
angle = np.arctan2(c_x, c_y)*180/np.pi
if (angle < 0):
angle = angle + 360
if c_z < 0:
l = 0.5 - abs(c_z)/2
#l=0
if c_z == 0:
l = 0.5
if c_z > 0:
l = (1 - (1-c_z)/2)
if c_z > 0.97:
l = (1 - (1-c_z)/2)
col = colorsys.hls_to_rgb(angle/360, l, 1)
return col
facecolors = [find_color_for_point(pt) for pt in midpoints] # smooth gradient
# facecolors = [np.random.random(3) for pt in midpoints] # random colors
coll = Poly3DCollection(
triangle_vertices, facecolors=facecolors, edgecolors=None)
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.add_collection(coll)
ax.set_xlim(-1, 1)
ax.set_ylim(-1, 1)
ax.set_zlim(-1, 1)
ax.elev = 50
plt.show()
Inspired from Jake Vanderplas with Python Data Science Handbook, when you are drawing some 3-D plot whose base is a circle, it is likely that you would try:
# Actually not sure about the math here though:
u, v = np.mgrid[0:2*np.pi:100j, 0:np.pi:20j]
x = np.cos(u)*np.sin(v)
y = np.sin(u)*np.sin(v)
and then think about the z-axis. Since viewing from the z-axis the cone is just a circle, so the relationships between z and x and y is clear, which is simply: z = np.sqrt(x ** 2 + y ** 2). Then you can draw the cone based on the codes below:
from mpl_toolkits import mplot3d
import numpy as np
import matplotlib.pyplot as plt
from matplotlib import cm
def f(x, y):
return np.sqrt(x ** 2 + y ** 2)
fig = plt.figure()
ax = plt.axes(projection='3d')
# Can manipulate with 100j and 80j values to make your cone looks different
u, v = np.mgrid[0:2*np.pi:100j, 0:np.pi:80j]
x = np.cos(u)*np.sin(v)
y = np.sin(u)*np.sin(v)
z = f(x, y)
ax.plot_surface(x, y, z, cmap=cm.coolwarm)
# Some other effects you may want to try based on your needs:
# ax.plot_surface(x, y, -z, cmap=cm.coolwarm)
# ax.scatter3D(x, y, z, color="b")
# ax.plot_wireframe(x, y, z, color="b")
# ax.plot_wireframe(x, y, -z, color="r")
# Can set your view from different angles.
ax.view_init(azim=15, elev=15)
ax.set_xlabel("x")
ax.set_ylabel("y")
ax.set_zlabel("z")
plt.show()
ax.set_ylabel("y")
ax.set_zlabel("z")
plt.show()
And from my side, the cone looks like:
and hope it helps.
Say I have a grid
xGrid = np.linspace(0.1, 1, 10)
yGrid = np.linspace(5, 10, 5)
and some data on that grid:
X, Y = np.meshgrid(xGrid, yGrid, indexing='ij')
Z = X*Y + 1
I could easily now plot Z(x, y). Now, there is a transformation t(x, y):
T = X+1+Y/2
and I would like to plot Z(t(x, y), y) instead. To do that, I need to project my Z data onto the t(x,y)-y plane. What'd be the best way of doing that?
Since I ultimately want to plot the data and not do any further work with it, direct methods of doing this in matplotlib (but actually drawing onto the correct new coordinates, not just relabeling the ticks) are also accepted.
You can use interpolation to compute the values in the projection, for example with scipy.interpolate.RectBivariateSpline:
import numpy as np
import scipy.interpolate
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
xGrid = np.linspace(0.1, 1, 10)
yGrid = np.linspace(5, 10, 5)
X, Y = np.meshgrid(xGrid, yGrid, indexing='ij')
Z = X * Y + 1
T = X + 1 + Y / 2
# Interpolate values
interp = scipy.interpolate.RectBivariateSpline(xGrid, yGrid, Z)
Zt = interp.ev(T.ravel(), Y.ravel()).reshape(Z.shape)
# Plot
fig = plt.figure(figsize=(8, 10))
ax1 = fig.add_subplot(211, projection='3d')
ax1.set_title('Original')
ax1.plot_surface(X, Y, Z)
ax2 = fig.add_subplot(212, projection='3d')
ax2.set_title('Projected')
ax2.plot_surface(T, Y, Zt)
fig.tight_layout()
Output:
If I understand your problem you could use pcolormesh that can be used for non regular meshes
In [8]: import numpy as np
...: import matplotlib.pyplot as plt
...: from matplotlib.collections import PatchCollection, QuadMesh
...: from matplotlib.patches import Rectangle
...:
...: np.random.seed(2018)
...: xGrid = np.linspace(0.1, 1, 10)
...: yGrid = np.linspace(5, 10, 6)
...: X, Y = np.meshgrid(xGrid, yGrid, indexing='ij')
...: Z = X*Y + 1
...: T = X+1+Y/2
...: Zt = T*Y + 1
...: plt.pcolormesh(T, Y, Zt)
...: plt.colorbar()
Out[8]: <matplotlib.colorbar.Colorbar at 0x7fda83cd4ef0>
that produces
If the bands are too ugly use plt.pcolormesh(T, Y, Zt, shading='gouraud')
Provided we have a contour on the xy plane, how can we plot "a curtain" raised from the contour to the limiting surface?
An example:
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
def figure():
fig = plt.figure(figsize=(8,6))
axes = fig.gca(projection='3d')
x = np.linspace(-2, 2, 100)
y = np.linspace(-2, 2, 100)
x, y = np.meshgrid(x, y)
t1 = np.linspace(0, 8/9, 100)
x1 = t1
y1 = (2*t1)**0.5
f1 = lambda x, y: y
plt.plot(x1, y1)
axes.plot_surface(x, y, f1(x, y),color ='red', alpha=0.1)
axes.set_xlim(-2,2)
axes.set_ylim(-2,2)
figure()
How to plot a surface from the given line to the limiting surface?
Somebody wanted help plotting an intersection here cylinder "cuts" a sphere in python you could use the vertical cylinder part. It uses u, v parameters to generate x, y, z values