Related
I would like to contour data that are quite sparse and where a maximum is going diagonally through the picture; the matplotlib contour function invents minima between the sampled maxima.
Starting with the densely sampled case where everything looks as expected:
import matplotlib.pyplot as plt
import matplotlib.tri as tri
import numpy as np
x_1D = np.linspace(0., 10., 100)
y_1D = np.linspace(0., 10., 100)
x, y = np.meshgrid(x_1D, y_1D)
z = np.empty_like(x)
def peak(y, y0):
return np.exp(-(y-y0)**2)
for i in range(x_1D.size):
z[:,i] = peak(y_1D, i/x_1D.size*y_1D.max())
fig, ax = plt.subplots(ncols=3)
ax[0].set_title('measured data')
ax[0].scatter(x, y, marker='s', c=z, cmap=plt.cm.jet, s=25)
ax[1].set_title('contour')
ax[1].contourf(x, y, z, levels=14, cmap=plt.cm.jet)
# define grid
xi = np.linspace(x_1D.min()-0.1, x_1D.max()+0.1, 1000)
yi = np.linspace(y_1D.min()-0.1, y_1D.max()+0.1, 1000)
# grid the data
triang = tri.Triangulation(x.flatten(), y.flatten())
interpolator = tri.LinearTriInterpolator(triang, z.flatten())
Xi, Yi = np.meshgrid(xi, yi)
zi = interpolator(Xi, Yi)
ax[2].set_title('interpolated')
ax[2].contourf(xi, yi, zi, levels=14, cmap=plt.cm.jet)
plt.show()
yields
When x is sampled less by a factor 10, i.e. x_1D = np.linspace(0., 10., 10), minima appear between the sampled maxima in the contour plot.
Is there a way how to avoid this artefact and make the contour of the sparsely sampled data look like the one of the densely sampled data?
EDIT: Thanks for the answer that works very nicely on the example I provided. Unfortunately, I have simplified the problem too far. Rather than talking about one diagonal line, I should have enquired about an arbitrary number of peaks moving in arbitrary directions through the picture; e.g. replace the peak-generation by
z = np.zeros_like(x)
def peak(y, y0):
return np.exp(-(y-y0)**2)
for i in range(x_1D.size):
z[:,i] += peak(y_1D, np.cos(i/x_1D.size*np.pi)*y_1D.max()*0.05+y_1D.max()*0.8)
for i in range(x_1D.size):
z[:,i] += peak(y_1D, np.sin(i/x_1D.size*np.pi/2.)*y_1D.max()*0.5)
resulting in
The main issue with your approach is that the triangulation algorithm doesn't know that the peaks should be connecting to eachother between the "x-slices" (your line of dense data points for a constant x).
Simplifying a bit, the triangulation algorithm will look at the neighbours in the x and y direction and connect to those. Then, when trying to interpolate using this triangulation, the points between the peaks will be roughly an average of the nearest points in the x direction and hence the minima will appear. The best solution is to make your own triangulation, with the peaks connected directly.
Fortunately, we can actually hack the triangulation to make it connect to the peaks by shifting the coordinates in the y direction such that the peaks are all aligned horizontally. This works because the triangulation algorithm uses the coordinates that you pass it. In your example this is easy to accomplish because we can just apply the simple shift y_s = y - x. In general you would have to get the equation for your peak (call it y_p(x)) and then subtract that from y to get the y_s.
Now that you have a shifted triangulation, you can make a new denser grid (like you did) and apply the same shift. Then, you interpolate in the shifted mesh with the shifted dense grid to get the z values correctly interpolated. Finally, you un-shift the dense grid to get the correct y values and plot it.
Below is the code of applying this concept to your code and the final result. As you can see. It works quite well for this case.
import matplotlib.pyplot as plt
import matplotlib.tri as tri
import numpy as np
def peak(y, y0):
return np.exp(-(y-y0)**2)
x_1D = np.linspace(0., 10., 10)
y_1D = np.linspace(0., 10., 100)
x, y = np.meshgrid(x_1D, y_1D)
z = np.empty_like(x)
for i in range(x_1D.size):
z[:,i] = peak(y_1D, i/x_1D.size*y_1D.max())
fig, ax = plt.subplots(ncols=4)
ax[0].set_title('measured data')
ax[0].scatter(x, y, marker='s', c=z, cmap=plt.cm.jet, s=25)
ax[1].set_title('contour')
ax[1].contourf(x, y, z, levels=14, cmap=plt.cm.jet)
# define output grid
xi_1D = np.linspace(x_1D.min()-0.1, x_1D.max()+0.1, 1000)
yi_1D = np.linspace(y_1D.min()-0.1, y_1D.max()+0.1, 1000)
xi, yi = np.meshgrid(xi_1D, yi_1D)
# Old Linear Interpolation
triang = tri.Triangulation(x.flatten(), y.flatten())
interpolator = tri.LinearTriInterpolator(triang, z.flatten())
zi = interpolator(xi, yi)
ax[2].set_title('interpolated')
ax[2].contourf(xi, yi, zi, levels=14, cmap=plt.cm.jet)
# === SHIFTED LINEAR INTERPOLATION ===
# make shifted interpolating mesh for the data
y_s=y-x
triang_s = tri.Triangulation(x.flatten(), y_s.flatten())
interpolator_s = tri.LinearTriInterpolator(triang_s, z.flatten())
# interpolate in the shifted state
yi_s = yi-xi
zi_s = interpolator_s(xi, yi_s)
# unshift the fine mesh
yi_us = yi_s+xi
ax[3].set_title('interpolated (shifted)')
ax[3].contourf(xi, yi_us, zi_s, levels=14, cmap=plt.cm.jet)
plt.show()
I have a collection of 3D points. These points are sampled at constant levels (z=0,1,...,7). An image should make it clear:
These points are in a numpy ndarray of shape (N, 3) called X. The above plot is created using:
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
X = load('points.npy')
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.plot_wireframe(X[:,0], X[:,1], X[:,2])
ax.scatter(X[:,0], X[:,1], X[:,2])
plt.draw()
I'd like to instead triangulate only the surface of this object, and plot the surface. I do not want the convex hull of this object, however, because this loses subtle shape information I'd like to be able to inspect.
I have tried ax.plot_trisurf(X[:,0], X[:,1], X[:,2]), but this results in the following mess:
Any help?
Example data
Here's a snippet to generate 3D data that is representative of the problem:
import numpy as np
X = []
for i in range(8):
t = np.linspace(0,2*np.pi,np.random.randint(30,50))
for j in range(t.shape[0]):
# random circular objects...
X.append([
(-0.05*(i-3.5)**2+1)*np.cos(t[j])+0.1*np.random.rand()-0.05,
(-0.05*(i-3.5)**2+1)*np.sin(t[j])+0.1*np.random.rand()-0.05,
i
])
X = np.array(X)
Example data from original image
Here's a pastebin to the original data:
http://pastebin.com/YBZhJcsV
Here are the slices along constant z:
update 3
Here's a concrete example of what I describe in update 2. If you don't have mayavi for visualization, I suggest installing it via edm using edm install mayavi pyqt matplotlib.
Toy 2D contours stacked in 3D
Contours -> 3D surface
Code to generate the figures
from matplotlib import path as mpath
from mayavi import mlab
import numpy as np
def make_star(amplitude=1.0, rotation=0.0):
""" Make a star shape
"""
t = np.linspace(0, 2*np.pi, 6) + rotation
star = np.zeros((12, 2))
star[::2] = np.c_[np.cos(t), np.sin(t)]
star[1::2] = 0.5*np.c_[np.cos(t + np.pi / 5), np.sin(t + np.pi / 5)]
return amplitude * star
def make_stars(n_stars=51, z_diff=0.05):
""" Make `2*n_stars-1` stars stacked in 3D
"""
amps = np.linspace(0.25, 1, n_stars)
amps = np.r_[amps, amps[:-1][::-1]]
rots = np.linspace(0, 2*np.pi, len(amps))
zamps = np.linspace
stars = []
for i, (amp, rot) in enumerate(zip(amps, rots)):
star = make_star(amplitude=amp, rotation=rot)
height = i*z_diff
z = np.full(len(star), height)
star3d = np.c_[star, z]
stars.append(star3d)
return stars
def polygon_to_boolean(points, xvals, yvals):
""" Convert `points` to a boolean indicator mask
over the specified domain
"""
x, y = np.meshgrid(xvals, yvals)
xy = np.c_[x.flatten(), y.flatten()]
mask = mpath.Path(points).contains_points(xy).reshape(x.shape)
return x, y, mask
def plot_contours(stars):
""" Plot a list of stars in 3D
"""
n = len(stars)
for i, star in enumerate(stars):
x, y, z = star.T
mlab.plot3d(*star.T)
#ax.plot3D(x, y, z, '-o', c=(0, 1-i/n, i/n))
#ax.set_xlim(-1, 1)
#ax.set_ylim(-1, 1)
mlab.show()
if __name__ == '__main__':
# Make and plot the 2D contours
stars3d = make_stars()
plot_contours(stars3d)
xvals = np.linspace(-1, 1, 101)
yvals = np.linspace(-1, 1, 101)
volume = np.dstack([
polygon_to_boolean(star[:,:2], xvals, yvals)[-1]
for star in stars3d
]).astype(float)
mlab.contour3d(volume, contours=[0.5])
mlab.show()
update 2
I now do this as follows:
I use the fact that the paths in each z-slice are closed and simple and use matplotlib.path to determine points inside and outside of the contour. Using this idea, I convert the contours in each slice to a boolean-valued image, which is combined into a boolean-valued volume.
Next, I use skimage's marching_cubes method to obtain a triangulation of the surface for visualization.
Here's an example of the method. I think the data is slightly different, but you can definitely see that the results are much cleaner, and can handle surfaces that are disconnected or have holes.
Original answer
Ok, here's the solution I came up with. It depends heavily on my data being roughly spherical and sampled at uniformly in z I think. Some of the other comments provide more information about more robust solutions. Since my data is roughly spherical I triangulate the azimuth and zenith angles from the spherical coordinate transform of my data points.
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.tri as mtri
X = np.load('./mydatars.npy')
# My data points are strictly positive. This doesn't work if I don't center about the origin.
X -= X.mean(axis=0)
rad = np.linalg.norm(X, axis=1)
zen = np.arccos(X[:,-1] / rad)
azi = np.arctan2(X[:,1], X[:,0])
tris = mtri.Triangulation(zen, azi)
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_trisurf(X[:,0], X[:,1], X[:,2], triangles=tris.triangles, cmap=plt.cm.bone)
plt.show()
Using the sample data from the pastebin above, this yields:
I realise that you mentioned in your question that you didn't want to use the convex hull because you might lose some shape information. I have a simple solution that works pretty well for your 'jittered spherical' example data, although it does use scipy.spatial.ConvexHull. I thought I would share it here anyway, just in case it's useful for others:
from matplotlib.tri import triangulation
from scipy.spatial import ConvexHull
# compute the convex hull of the points
cvx = ConvexHull(X)
x, y, z = X.T
# cvx.simplices contains an (nfacets, 3) array specifying the indices of
# the vertices for each simplical facet
tri = Triangulation(x, y, triangles=cvx.simplices)
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.hold(True)
ax.plot_trisurf(tri, z)
ax.plot_wireframe(x, y, z, color='r')
ax.scatter(x, y, z, color='r')
plt.draw()
It does pretty well in this case, since your example data ends up lying on a more-or-less convex surface. Perhaps you could make some more challenging example data? A toroidal surface would be a good test case which the convex hull method would obviously fail.
Mapping an arbitrary 3D surface from a point cloud is a really tough problem. Here's a related question containing some links that might be helpful.
Suppose you have a 2D curve, given by e.g.:
from matplotlib import pylab
t = numpy.linspace(-1, 1, 21)
z = -t**2
pylab.plot(t, z)
which produces
I would like to perform a revolution to achieve a 3d plot (see http://reference.wolfram.com/mathematica/ref/RevolutionPlot3D.html). Plotting a 3d surface is not the problem, but it does not produce the result I'm expecting:
How can I perform a rotation of this blue curve in the 3d plot ?
Your plot on your figure seems to use cartesian grid. There is some examples on the matplotlib website of 3D cylindrical functions like Z = f(R) (here: http://matplotlib.org/examples/mplot3d/surface3d_radial_demo.html).
Is that what you looking for ?
Below is what I get with your function Z = -R**2 :
And to add cut off to your function, use the following example:
(matplotlib 1.2.0 required)
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
import matplotlib.pyplot as plt
import numpy as np
fig = plt.figure()
ax = fig.gca(projection='3d')
X = np.arange(-5, 5, 0.25)
Y = np.arange(-5, 5, 0.25)
X, Y = np.meshgrid(X, Y)
Z = -(abs(X) + abs(Y))
## 1) Initial surface
# Flatten mesh arrays, necessary for plot_trisurf function
X = X.flatten()
Y = Y.flatten()
Z = Z.flatten()
# Plot initial 3D surface with triangles (more flexible than quad)
#surfi = ax.plot_trisurf(X, Y, Z, cmap=cm.jet, linewidth=0.2)
## 2) Cut off
# Get desired values indexes
cut_idx = np.where(Z > -5)
# Apply the "cut off"
Xc = X[cut_idx]
Yc = Y[cut_idx]
Zc = Z[cut_idx]
# Plot the new surface (it would be impossible with quad grid)
surfc = ax.plot_trisurf(Xc, Yc, Zc, cmap=cm.jet, linewidth=0.2)
# You can force limit if you want to compare both graphs...
ax.set_xlim(-5,5)
ax.set_ylim(-5,5)
ax.set_zlim(-10,0)
plt.show()
Result for surfi:
and surfc:
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.
I have a set of X,Y data points (about 10k) that are easy to plot as a scatter plot but that I would like to represent as a heatmap.
I looked through the examples in Matplotlib and they all seem to already start with heatmap cell values to generate the image.
Is there a method that converts a bunch of x, y, all different, to a heatmap (where zones with higher frequency of x, y would be "warmer")?
If you don't want hexagons, you can use numpy's histogram2d function:
import numpy as np
import numpy.random
import matplotlib.pyplot as plt
# Generate some test data
x = np.random.randn(8873)
y = np.random.randn(8873)
heatmap, xedges, yedges = np.histogram2d(x, y, bins=50)
extent = [xedges[0], xedges[-1], yedges[0], yedges[-1]]
plt.clf()
plt.imshow(heatmap.T, extent=extent, origin='lower')
plt.show()
This makes a 50x50 heatmap. If you want, say, 512x384, you can put bins=(512, 384) in the call to histogram2d.
Example:
In Matplotlib lexicon, i think you want a hexbin plot.
If you're not familiar with this type of plot, it's just a bivariate histogram in which the xy-plane is tessellated by a regular grid of hexagons.
So from a histogram, you can just count the number of points falling in each hexagon, discretiize the plotting region as a set of windows, assign each point to one of these windows; finally, map the windows onto a color array, and you've got a hexbin diagram.
Though less commonly used than e.g., circles, or squares, that hexagons are a better choice for the geometry of the binning container is intuitive:
hexagons have nearest-neighbor symmetry (e.g., square bins don't,
e.g., the distance from a point on a square's border to a point
inside that square is not everywhere equal) and
hexagon is the highest n-polygon that gives regular plane
tessellation (i.e., you can safely re-model your kitchen floor with hexagonal-shaped tiles because you won't have any void space between the tiles when you are finished--not true for all other higher-n, n >= 7, polygons).
(Matplotlib uses the term hexbin plot; so do (AFAIK) all of the plotting libraries for R; still i don't know if this is the generally accepted term for plots of this type, though i suspect it's likely given that hexbin is short for hexagonal binning, which is describes the essential step in preparing the data for display.)
from matplotlib import pyplot as PLT
from matplotlib import cm as CM
from matplotlib import mlab as ML
import numpy as NP
n = 1e5
x = y = NP.linspace(-5, 5, 100)
X, Y = NP.meshgrid(x, y)
Z1 = ML.bivariate_normal(X, Y, 2, 2, 0, 0)
Z2 = ML.bivariate_normal(X, Y, 4, 1, 1, 1)
ZD = Z2 - Z1
x = X.ravel()
y = Y.ravel()
z = ZD.ravel()
gridsize=30
PLT.subplot(111)
# if 'bins=None', then color of each hexagon corresponds directly to its count
# 'C' is optional--it maps values to x-y coordinates; if 'C' is None (default) then
# the result is a pure 2D histogram
PLT.hexbin(x, y, C=z, gridsize=gridsize, cmap=CM.jet, bins=None)
PLT.axis([x.min(), x.max(), y.min(), y.max()])
cb = PLT.colorbar()
cb.set_label('mean value')
PLT.show()
Edit: For a better approximation of Alejandro's answer, see below.
I know this is an old question, but wanted to add something to Alejandro's anwser: If you want a nice smoothed image without using py-sphviewer you can instead use np.histogram2d and apply a gaussian filter (from scipy.ndimage.filters) to the heatmap:
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.cm as cm
from scipy.ndimage.filters import gaussian_filter
def myplot(x, y, s, bins=1000):
heatmap, xedges, yedges = np.histogram2d(x, y, bins=bins)
heatmap = gaussian_filter(heatmap, sigma=s)
extent = [xedges[0], xedges[-1], yedges[0], yedges[-1]]
return heatmap.T, extent
fig, axs = plt.subplots(2, 2)
# Generate some test data
x = np.random.randn(1000)
y = np.random.randn(1000)
sigmas = [0, 16, 32, 64]
for ax, s in zip(axs.flatten(), sigmas):
if s == 0:
ax.plot(x, y, 'k.', markersize=5)
ax.set_title("Scatter plot")
else:
img, extent = myplot(x, y, s)
ax.imshow(img, extent=extent, origin='lower', cmap=cm.jet)
ax.set_title("Smoothing with $\sigma$ = %d" % s)
plt.show()
Produces:
The scatter plot and s=16 plotted on top of eachother for Agape Gal'lo (click for better view):
One difference I noticed with my gaussian filter approach and Alejandro's approach was that his method shows local structures much better than mine. Therefore I implemented a simple nearest neighbour method at pixel level. This method calculates for each pixel the inverse sum of the distances of the n closest points in the data. This method is at a high resolution pretty computationally expensive and I think there's a quicker way, so let me know if you have any improvements.
Update: As I suspected, there's a much faster method using Scipy's scipy.cKDTree. See Gabriel's answer for the implementation.
Anyway, here's my code:
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.cm as cm
def data_coord2view_coord(p, vlen, pmin, pmax):
dp = pmax - pmin
dv = (p - pmin) / dp * vlen
return dv
def nearest_neighbours(xs, ys, reso, n_neighbours):
im = np.zeros([reso, reso])
extent = [np.min(xs), np.max(xs), np.min(ys), np.max(ys)]
xv = data_coord2view_coord(xs, reso, extent[0], extent[1])
yv = data_coord2view_coord(ys, reso, extent[2], extent[3])
for x in range(reso):
for y in range(reso):
xp = (xv - x)
yp = (yv - y)
d = np.sqrt(xp**2 + yp**2)
im[y][x] = 1 / np.sum(d[np.argpartition(d.ravel(), n_neighbours)[:n_neighbours]])
return im, extent
n = 1000
xs = np.random.randn(n)
ys = np.random.randn(n)
resolution = 250
fig, axes = plt.subplots(2, 2)
for ax, neighbours in zip(axes.flatten(), [0, 16, 32, 64]):
if neighbours == 0:
ax.plot(xs, ys, 'k.', markersize=2)
ax.set_aspect('equal')
ax.set_title("Scatter Plot")
else:
im, extent = nearest_neighbours(xs, ys, resolution, neighbours)
ax.imshow(im, origin='lower', extent=extent, cmap=cm.jet)
ax.set_title("Smoothing over %d neighbours" % neighbours)
ax.set_xlim(extent[0], extent[1])
ax.set_ylim(extent[2], extent[3])
plt.show()
Result:
Instead of using np.hist2d, which in general produces quite ugly histograms, I would like to recycle py-sphviewer, a python package for rendering particle simulations using an adaptive smoothing kernel and that can be easily installed from pip (see webpage documentation). Consider the following code, which is based on the example:
import numpy as np
import numpy.random
import matplotlib.pyplot as plt
import sphviewer as sph
def myplot(x, y, nb=32, xsize=500, ysize=500):
xmin = np.min(x)
xmax = np.max(x)
ymin = np.min(y)
ymax = np.max(y)
x0 = (xmin+xmax)/2.
y0 = (ymin+ymax)/2.
pos = np.zeros([len(x),3])
pos[:,0] = x
pos[:,1] = y
w = np.ones(len(x))
P = sph.Particles(pos, w, nb=nb)
S = sph.Scene(P)
S.update_camera(r='infinity', x=x0, y=y0, z=0,
xsize=xsize, ysize=ysize)
R = sph.Render(S)
R.set_logscale()
img = R.get_image()
extent = R.get_extent()
for i, j in zip(xrange(4), [x0,x0,y0,y0]):
extent[i] += j
print extent
return img, extent
fig = plt.figure(1, figsize=(10,10))
ax1 = fig.add_subplot(221)
ax2 = fig.add_subplot(222)
ax3 = fig.add_subplot(223)
ax4 = fig.add_subplot(224)
# Generate some test data
x = np.random.randn(1000)
y = np.random.randn(1000)
#Plotting a regular scatter plot
ax1.plot(x,y,'k.', markersize=5)
ax1.set_xlim(-3,3)
ax1.set_ylim(-3,3)
heatmap_16, extent_16 = myplot(x,y, nb=16)
heatmap_32, extent_32 = myplot(x,y, nb=32)
heatmap_64, extent_64 = myplot(x,y, nb=64)
ax2.imshow(heatmap_16, extent=extent_16, origin='lower', aspect='auto')
ax2.set_title("Smoothing over 16 neighbors")
ax3.imshow(heatmap_32, extent=extent_32, origin='lower', aspect='auto')
ax3.set_title("Smoothing over 32 neighbors")
#Make the heatmap using a smoothing over 64 neighbors
ax4.imshow(heatmap_64, extent=extent_64, origin='lower', aspect='auto')
ax4.set_title("Smoothing over 64 neighbors")
plt.show()
which produces the following image:
As you see, the images look pretty nice, and we are able to identify different substructures on it. These images are constructed spreading a given weight for every point within a certain domain, defined by the smoothing length, which in turns is given by the distance to the closer nb neighbor (I've chosen 16, 32 and 64 for the examples). So, higher density regions typically are spread over smaller regions compared to lower density regions.
The function myplot is just a very simple function that I've written in order to give the x,y data to py-sphviewer to do the magic.
If you are using 1.2.x
import numpy as np
import matplotlib.pyplot as plt
x = np.random.randn(100000)
y = np.random.randn(100000)
plt.hist2d(x,y,bins=100)
plt.show()
Seaborn now has the jointplot function which should work nicely here:
import numpy as np
import seaborn as sns
import matplotlib.pyplot as plt
# Generate some test data
x = np.random.randn(8873)
y = np.random.randn(8873)
sns.jointplot(x=x, y=y, kind='hex')
plt.show()
Here's Jurgy's great nearest neighbour approach but implemented using scipy.cKDTree. In my tests it's about 100x faster.
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.cm as cm
from scipy.spatial import cKDTree
def data_coord2view_coord(p, resolution, pmin, pmax):
dp = pmax - pmin
dv = (p - pmin) / dp * resolution
return dv
n = 1000
xs = np.random.randn(n)
ys = np.random.randn(n)
resolution = 250
extent = [np.min(xs), np.max(xs), np.min(ys), np.max(ys)]
xv = data_coord2view_coord(xs, resolution, extent[0], extent[1])
yv = data_coord2view_coord(ys, resolution, extent[2], extent[3])
def kNN2DDens(xv, yv, resolution, neighbours, dim=2):
"""
"""
# Create the tree
tree = cKDTree(np.array([xv, yv]).T)
# Find the closest nnmax-1 neighbors (first entry is the point itself)
grid = np.mgrid[0:resolution, 0:resolution].T.reshape(resolution**2, dim)
dists = tree.query(grid, neighbours)
# Inverse of the sum of distances to each grid point.
inv_sum_dists = 1. / dists[0].sum(1)
# Reshape
im = inv_sum_dists.reshape(resolution, resolution)
return im
fig, axes = plt.subplots(2, 2, figsize=(15, 15))
for ax, neighbours in zip(axes.flatten(), [0, 16, 32, 63]):
if neighbours == 0:
ax.plot(xs, ys, 'k.', markersize=5)
ax.set_aspect('equal')
ax.set_title("Scatter Plot")
else:
im = kNN2DDens(xv, yv, resolution, neighbours)
ax.imshow(im, origin='lower', extent=extent, cmap=cm.Blues)
ax.set_title("Smoothing over %d neighbours" % neighbours)
ax.set_xlim(extent[0], extent[1])
ax.set_ylim(extent[2], extent[3])
plt.savefig('new.png', dpi=150, bbox_inches='tight')
and the initial question was... how to convert scatter values to grid values, right?
histogram2d does count the frequency per cell, however, if you have other data per cell than just the frequency, you'd need some additional work to do.
x = data_x # between -10 and 4, log-gamma of an svc
y = data_y # between -4 and 11, log-C of an svc
z = data_z #between 0 and 0.78, f1-values from a difficult dataset
So, I have a dataset with Z-results for X and Y coordinates. However, I was calculating few points outside the area of interest (large gaps), and heaps of points in a small area of interest.
Yes here it becomes more difficult but also more fun. Some libraries (sorry):
from matplotlib import pyplot as plt
from matplotlib import cm
import numpy as np
from scipy.interpolate import griddata
pyplot is my graphic engine today,
cm is a range of color maps with some initeresting choice.
numpy for the calculations,
and griddata for attaching values to a fixed grid.
The last one is important especially because the frequency of xy points is not equally distributed in my data. First, let's start with some boundaries fitting to my data and an arbitrary grid size. The original data has datapoints also outside those x and y boundaries.
#determine grid boundaries
gridsize = 500
x_min = -8
x_max = 2.5
y_min = -2
y_max = 7
So we have defined a grid with 500 pixels between the min and max values of x and y.
In my data, there are lots more than the 500 values available in the area of high interest; whereas in the low-interest-area, there are not even 200 values in the total grid; between the graphic boundaries of x_min and x_max there are even less.
So for getting a nice picture, the task is to get an average for the high interest values and to fill the gaps elsewhere.
I define my grid now. For each xx-yy pair, i want to have a color.
xx = np.linspace(x_min, x_max, gridsize) # array of x values
yy = np.linspace(y_min, y_max, gridsize) # array of y values
grid = np.array(np.meshgrid(xx, yy.T))
grid = grid.reshape(2, grid.shape[1]*grid.shape[2]).T
Why the strange shape? scipy.griddata wants a shape of (n, D).
Griddata calculates one value per point in the grid, by a predefined method.
I choose "nearest" - empty grid points will be filled with values from the nearest neighbor. This looks as if the areas with less information have bigger cells (even if it is not the case). One could choose to interpolate "linear", then areas with less information look less sharp. Matter of taste, really.
points = np.array([x, y]).T # because griddata wants it that way
z_grid2 = griddata(points, z, grid, method='nearest')
# you get a 1D vector as result. Reshape to picture format!
z_grid2 = z_grid2.reshape(xx.shape[0], yy.shape[0])
And hop, we hand over to matplotlib to display the plot
fig = plt.figure(1, figsize=(10, 10))
ax1 = fig.add_subplot(111)
ax1.imshow(z_grid2, extent=[x_min, x_max,y_min, y_max, ],
origin='lower', cmap=cm.magma)
ax1.set_title("SVC: empty spots filled by nearest neighbours")
ax1.set_xlabel('log gamma')
ax1.set_ylabel('log C')
plt.show()
Around the pointy part of the V-Shape, you see I did a lot of calculations during my search for the sweet spot, whereas the less interesting parts almost everywhere else have a lower resolution.
Make a 2-dimensional array that corresponds to the cells in your final image, called say heatmap_cells and instantiate it as all zeroes.
Choose two scaling factors that define the difference between each array element in real units, for each dimension, say x_scale and y_scale. Choose these such that all your datapoints will fall within the bounds of the heatmap array.
For each raw datapoint with x_value and y_value:
heatmap_cells[floor(x_value/x_scale),floor(y_value/y_scale)]+=1
Very similar to #Piti's answer, but using 1 call instead of 2 to generate the points:
import numpy as np
import matplotlib.pyplot as plt
pts = 1000000
mean = [0.0, 0.0]
cov = [[1.0,0.0],[0.0,1.0]]
x,y = np.random.multivariate_normal(mean, cov, pts).T
plt.hist2d(x, y, bins=50, cmap=plt.cm.jet)
plt.show()
Output:
Here's one I made on a 1 Million point set with 3 categories (colored Red, Green, and Blue). Here's a link to the repository if you'd like to try the function. Github Repo
histplot(
X,
Y,
labels,
bins=2000,
range=((-3,3),(-3,3)),
normalize_each_label=True,
colors = [
[1,0,0],
[0,1,0],
[0,0,1]],
gain=50)
I'm afraid I'm a little late to the party but I had a similar question a while ago. The accepted answer (by #ptomato) helped me out but I'd also want to post this in case it's of use to someone.
''' I wanted to create a heatmap resembling a football pitch which would show the different actions performed '''
import numpy as np
import matplotlib.pyplot as plt
import random
#fixing random state for reproducibility
np.random.seed(1234324)
fig = plt.figure(12)
ax1 = fig.add_subplot(121)
ax2 = fig.add_subplot(122)
#Ratio of the pitch with respect to UEFA standards
hmap= np.full((6, 10), 0)
#print(hmap)
xlist = np.random.uniform(low=0.0, high=100.0, size=(20))
ylist = np.random.uniform(low=0.0, high =100.0, size =(20))
#UEFA Pitch Standards are 105m x 68m
xlist = (xlist/100)*10.5
ylist = (ylist/100)*6.5
ax1.scatter(xlist,ylist)
#int of the co-ordinates to populate the array
xlist_int = xlist.astype (int)
ylist_int = ylist.astype (int)
#print(xlist_int, ylist_int)
for i, j in zip(xlist_int, ylist_int):
#this populates the array according to the x,y co-ordinate values it encounters
hmap[j][i]= hmap[j][i] + 1
#Reversing the rows is necessary
hmap = hmap[::-1]
#print(hmap)
im = ax2.imshow(hmap)
Here's the result
None of these solutions worked for my application, so this is what I came up with. Essentially I am placing a 2D Gaussian at every single point:
import cv2
import numpy as np
import matplotlib.pyplot as plt
def getGaussian2D(ksize, sigma, norm=True):
oneD = cv2.getGaussianKernel(ksize=ksize, sigma=sigma)
twoD = np.outer(oneD.T, oneD)
return twoD / np.sum(twoD) if norm else twoD
def pt2heat(pts, shape, kernel=16, sigma=5):
heat = np.zeros(shape)
k = getGaussian2D(kernel, sigma)
for y,x in pts:
x, y = int(x), int(y)
for i in range(-kernel//2, kernel//2):
for j in range(-kernel//2, kernel//2):
if 0 <= x+i < shape[0] and 0 <= y+j < shape[1]:
heat[x+i, y+j] = heat[x+i, y+j] + k[i+kernel//2, j+kernel//2]
return heat
heat = pts2heat(pts, img.shape[:2])
plt.imshow(heat, cmap='heat')
Here are the points overlayed ontop of it's associated image, along with the resulting heat map: