I would like to plot the surface of my data which is given by 3D vectors in cartesian coordinates x,y,z. The data can not be represented by a smooth function.
So first we generate some dummy data with the function eq_points(N_count, r) which returns an array points with the x,y,z coordinates of each point on the surface of our object. The quantity omega is the solid angle, and not of interest right now.
#credit to Markus Deserno from MPI
#https://www.cmu.edu/biolphys/deserno/pdf/sphere_equi.pdf
def eq_points(N_count, r):
points = []
a = 4*np.pi*r**2/N_count
d = np.sqrt(a)
M_theta = int(np.pi/d)
d_theta = np.pi/M_theta
d_phi = a/d_theta
for m in range(M_theta):
theta = np.pi*(m+0.5)/M_theta
M_phi = int(2*np.pi*np.sin(theta)/d_phi)
for n in range(M_phi):
phi = 2*np.pi*n/M_phi
points.append(np.array([r*np.sin(theta)*np.cos(phi),
r*np.sin(theta)*np.sin(phi),
r*np.cos(theta)]))
omega = 4*np.pi/N_count
return np.array(points), omega
#starting plotting sequence
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
points, omega = eq_points(400, 1.)
ax.scatter(points[:,0], points[:,1], points[:,2])
ax.scatter(0., 0., 0., c="r")
ax.set_xlabel(r'$x$ axis')
ax.set_ylabel(r'$y$ axis')
ax.set_zlabel(r'$Z$ axis')
plt.savefig("./sphere.png", format="png", dpi=300)
plt.clf()
The result is a sphere shown in the following figure. The blue points mark the data from the points array, while the red point is the origin.
I would like to get something like this
taken from here. However the data in the mplot3d tutorial is always a result of a smooth function. Except to the ax.scatter() function which I used for my sphere plot.
So in the end my goal would be to plot some data showing only its surface. This data is produced by changing the radial distance to the origin of each blue point. Further more it would be necessary to make sure each point is in contact with the surface. How are the surfaces which are plotted here e.g. in plot_surface() constructed in detail? Some actual live data looks like this:
I would suggest finding the hull, and then plotting the simplices (i.e. the triangles forming the hull). Make sure to update the x,y,z-limits appropriately.
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from mpl_toolkits.mplot3d.art3d import Poly3DCollection
from scipy.spatial import ConvexHull
N = 1000
pts = np.random.randn(N, 3)
# exclude outliers
# obviously, this is data dependent
cutoff = 3.
is_outlier = np.any(np.abs(pts) > cutoff, axis=1)
pts = pts[~is_outlier]
# plot points
fig = plt.figure()
ax = Axes3D(fig)
ax.scatter(pts[:,0], pts[:,1], pts[:,2])
ax.set_xlim(-(cutoff +1), cutoff+1)
ax.set_ylim(-(cutoff +1), cutoff+1)
ax.set_zlim(-(cutoff +1), cutoff+1)
# get and plot hull
hull = ConvexHull(pts)
fig = plt.figure()
ax = Axes3D(fig)
vertices = [pts[s] for s in hull.simplices]
triangles = Poly3DCollection(vertices, edgecolor='k')
ax.add_collection3d(triangles)
ax.set_xlim(-(cutoff +1), cutoff+1)
ax.set_ylim(-(cutoff +1), cutoff+1)
ax.set_zlim(-(cutoff +1), cutoff+1)
plt.show()
Solution to the question with the new specification that all points are touching the surface. Assuming that the angles are set by the user as shown in the example, it is easy to precompute the indices of the points forming the simplices making up the surface by computing the simplices of the hull formed by points on the unit sphere with the same angles as in the data set of interest. We can then use these indices to get the surface of interest.
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from mpl_toolkits.mplot3d.art3d import Poly3DCollection
from scipy.spatial import ConvexHull
def eq_points(N_count, r):
points = []
a = 4*np.pi*r**2/N_count
d = np.sqrt(a)
M_theta = int(np.pi/d)
d_theta = np.pi/M_theta
d_phi = a/d_theta
for m in range(M_theta):
theta = np.pi*(m+0.5)/M_theta
M_phi = int(2*np.pi*np.sin(theta)/d_phi)
for n in range(M_phi):
phi = 2*np.pi*n/M_phi
points.append(np.array([r*np.sin(theta)*np.cos(phi),
r*np.sin(theta)*np.sin(phi),
r*np.cos(theta)]))
omega = 4*np.pi/N_count
return np.array(points), omega
def eq_points_with_random_radius(N_count, r):
points = []
a = 4*np.pi*r**2/N_count
d = np.sqrt(a)
M_theta = int(np.pi/d)
d_theta = np.pi/M_theta
d_phi = a/d_theta
for m in range(M_theta):
theta = np.pi*(m+0.5)/M_theta
M_phi = int(2*np.pi*np.sin(theta)/d_phi)
for n in range(M_phi):
phi = 2*np.pi*n/M_phi
rr = r * np.random.rand()
points.append(np.array([rr*np.sin(theta)*np.cos(phi),
rr*np.sin(theta)*np.sin(phi),
rr*np.cos(theta)]))
omega = 4*np.pi/N_count
return np.array(points), omega
N = 400
pts, _ = eq_points(N, 1.)
pts_rescaled, _ = eq_points_with_random_radius(N, 1.)
extremum = 2.
# plot points
fig = plt.figure()
ax = Axes3D(fig)
ax.scatter(pts_rescaled[:,0], pts_rescaled[:,1], pts_rescaled[:,2])
ax.set_xlim(-extremum, extremum)
ax.set_ylim(-extremum, extremum)
ax.set_zlim(-extremum, extremum)
# get indices of simplices making up the surface using points on unit sphere;
# index into rescaled points
hull = ConvexHull(pts)
vertices = [pts_rescaled[s] for s in hull.simplices]
fig = plt.figure()
ax = Axes3D(fig)
triangles = Poly3DCollection(vertices, edgecolor='k')
ax.add_collection3d(triangles)
ax.set_xlim(-extremum, extremum)
ax.set_ylim(-extremum, extremum)
ax.set_zlim(-extremum, extremum)
plt.show()
I need to plot contour and quiver plots of scalar and vector fields defined on an uneven grid in (r,theta) coordinates.
As a minimal example of the problem I have, consider the contour plot of a Stream function for a magnetic dipole, contours of such a function are streamlines of the corresponeding vector field (in this case, the magnetic field).
The code below takes an uneven grid in (r,theta) coordinates, maps it to the cartesian plane and plots a contour plot of the stream function.
import numpy as np
import matplotlib.pyplot as plt
r = np.logspace(0,1,200)
theta = np.linspace(0,np.pi/2,100)
N_r = len(r)
N_theta = len(theta)
# Polar to cartesian coordinates
theta_matrix, r_matrix = np.meshgrid(theta, r)
x = r_matrix * np.cos(theta_matrix)
y = r_matrix * np.sin(theta_matrix)
m = 5
psi = np.zeros((N_r, N_theta))
# Stream function for a magnetic dipole
psi = m * np.sin(theta_matrix)**2 / r_matrix
contour_levels = m * np.sin(np.linspace(0, np.pi/2,40))**2.
fig, ax = plt.subplots()
# ax.plot(x,y,'b.') # plot grid points
ax.set_aspect('equal')
ax.contour(x, y, psi, 100, colors='black',levels=contour_levels)
plt.show()
For some reason though, the plot I get doesn't look right:
If I interchange x and y in the contour function call, I get the desired result:
Same thing happens when I try to make a quiver plot of a vector field defined on the same grid and mapped to the x-y plane, except that interchanging x and y in the function call no longer works.
It seems like I made a stupid mistake somewhere but I can't figure out what it is.
If psi = m * np.sin(theta_matrix)**2 / r_matrix
then psi increases as theta goes from 0 to pi/2 and psi decreases as r increases.
So a contour line for psi should increase in r as theta increases. That results
in a curve that goes counterclockwise as it radiates out from the center. This is
consistent with the first plot you posted, and the result returned by the first version of your code with
ax.contour(x, y, psi, 100, colors='black',levels=contour_levels)
An alternative way to confirm the plausibility of the result is to look at a surface plot of psi:
import numpy as np
import matplotlib.pyplot as plt
import mpl_toolkits.mplot3d.axes3d as axes3d
r = np.logspace(0,1,200)
theta = np.linspace(0,np.pi/2,100)
N_r = len(r)
N_theta = len(theta)
# Polar to cartesian coordinates
theta_matrix, r_matrix = np.meshgrid(theta, r)
x = r_matrix * np.cos(theta_matrix)
y = r_matrix * np.sin(theta_matrix)
m = 5
# Stream function for a magnetic dipole
psi = m * np.sin(theta_matrix)**2 / r_matrix
contour_levels = m * np.sin(np.linspace(0, np.pi/2,40))**2.
fig = plt.figure()
ax = fig.add_subplot(1, 1, 1, projection='3d')
ax.set_aspect('equal')
ax.plot_surface(x, y, psi, rstride=8, cstride=8, alpha=0.3)
ax.contour(x, y, psi, colors='black',levels=contour_levels)
plt.show()
Using Matplotlib, I want to plot a 2D heat map. My data is an n-by-n Numpy array, each with a value between 0 and 1. So for the (i, j) element of this array, I want to plot a square at the (i, j) coordinate in my heat map, whose color is proportional to the element's value in the array.
How can I do this?
The imshow() function with parameters interpolation='nearest' and cmap='hot' should do what you want.
Please review the interpolation parameter details, and see Interpolations for imshow and Image antialiasing.
import matplotlib.pyplot as plt
import numpy as np
a = np.random.random((16, 16))
plt.imshow(a, cmap='hot', interpolation='nearest')
plt.show()
Seaborn is a high-level API for matplotlib, which takes care of a lot of the manual work.
seaborn.heatmap automatically plots a gradient at the side of the chart etc.
import numpy as np
import seaborn as sns
import matplotlib.pylab as plt
uniform_data = np.random.rand(10, 12)
ax = sns.heatmap(uniform_data, linewidth=0.5)
plt.show()
You can even plot upper / lower left / right triangles of square matrices. For example, a correlation matrix, which is square and is symmetric, so plotting all values would be redundant.
corr = np.corrcoef(np.random.randn(10, 200))
mask = np.zeros_like(corr)
mask[np.triu_indices_from(mask)] = True
with sns.axes_style("white"):
ax = sns.heatmap(corr, mask=mask, vmax=.3, square=True, cmap="YlGnBu")
plt.show()
I would use matplotlib's pcolor/pcolormesh function since it allows nonuniform spacing of the data.
Example taken from matplotlib:
import matplotlib.pyplot as plt
import numpy as np
# generate 2 2d grids for the x & y bounds
y, x = np.meshgrid(np.linspace(-3, 3, 100), np.linspace(-3, 3, 100))
z = (1 - x / 2. + x ** 5 + y ** 3) * np.exp(-x ** 2 - y ** 2)
# x and y are bounds, so z should be the value *inside* those bounds.
# Therefore, remove the last value from the z array.
z = z[:-1, :-1]
z_min, z_max = -np.abs(z).max(), np.abs(z).max()
fig, ax = plt.subplots()
c = ax.pcolormesh(x, y, z, cmap='RdBu', vmin=z_min, vmax=z_max)
ax.set_title('pcolormesh')
# set the limits of the plot to the limits of the data
ax.axis([x.min(), x.max(), y.min(), y.max()])
fig.colorbar(c, ax=ax)
plt.show()
For a 2d numpy array, simply use imshow() may help you:
import matplotlib.pyplot as plt
import numpy as np
def heatmap2d(arr: np.ndarray):
plt.imshow(arr, cmap='viridis')
plt.colorbar()
plt.show()
test_array = np.arange(100 * 100).reshape(100, 100)
heatmap2d(test_array)
This code produces a continuous heatmap.
You can choose another built-in colormap from here.
Here's how to do it from a csv:
import numpy as np
import matplotlib.pyplot as plt
from scipy.interpolate import griddata
# Load data from CSV
dat = np.genfromtxt('dat.xyz', delimiter=' ',skip_header=0)
X_dat = dat[:,0]
Y_dat = dat[:,1]
Z_dat = dat[:,2]
# Convert from pandas dataframes to numpy arrays
X, Y, Z, = np.array([]), np.array([]), np.array([])
for i in range(len(X_dat)):
X = np.append(X, X_dat[i])
Y = np.append(Y, Y_dat[i])
Z = np.append(Z, Z_dat[i])
# create x-y points to be used in heatmap
xi = np.linspace(X.min(), X.max(), 1000)
yi = np.linspace(Y.min(), Y.max(), 1000)
# Interpolate for plotting
zi = griddata((X, Y), Z, (xi[None,:], yi[:,None]), method='cubic')
# I control the range of my colorbar by removing data
# outside of my range of interest
zmin = 3
zmax = 12
zi[(zi<zmin) | (zi>zmax)] = None
# Create the contour plot
CS = plt.contourf(xi, yi, zi, 15, cmap=plt.cm.rainbow,
vmax=zmax, vmin=zmin)
plt.colorbar()
plt.show()
where dat.xyz is in the form
x1 y1 z1
x2 y2 z2
...
Use matshow() which is a wrapper around imshow to set useful defaults for displaying a matrix.
a = np.diag(range(15))
plt.matshow(a)
https://matplotlib.org/stable/api/_as_gen/matplotlib.axes.Axes.matshow.html
This is just a convenience function wrapping imshow to set useful defaults for displaying a matrix. In particular:
Set origin='upper'.
Set interpolation='nearest'.
Set aspect='equal'.
Ticks are placed to the left and above.
Ticks are formatted to show integer indices.
Here is a new python package to plot complex heatmaps with different kinds of row/columns annotations in Python: https://github.com/DingWB/PyComplexHeatmap
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: