Add margin in a 2D scatter plot - python

How would you add a margin between the axis lines and the actual origin? In 3D scatter, like this one, the origin of x = 6 and y = -10 are not on the same point. How to do the same thing but in a 2D scatter (something similar to the graphic of p. 122 (132 of pdf) in this matplotlib doc, where the origin x = 0 and y = 0 are not located at the same place).

A margin can be easily added to any plot by supplying the plt.set_xmargin(m) with a float between 0 and 1 indicating the relative size of the margin:
import matplotlib.pyplot as plt
import numpy as np
fig =plt.figure()
ax = plt.subplot(111)
x = np.linspace(0, 10, 100)
y = np.cos(x)
ax.set_xmargin(0.2)
ax.set_ymargin(0.2)
ax.plot(x, y)
Note in practise this needs to be called before ax.plot(.. as this is when the autoscaling is done. For more information see here

Related

Matplotlib automatically scale vertical height of subplots for shared x-axis figure

I want to automatically scale the vertical height of subplots for shared x-axis figures based on their data span! I want to compare the relative intensity of the displayed data. If i use the sharey=True kwarg for the subbplots the data is displayed in a way that the relative intensity is recognizable:
import matplotlib.pyplot as plt
from matplotlib import gridspec
import numpy as np
SIZE = (12, 8) #desired overall figure size
# Simple data to display in various forms
x = np.linspace(0, 2 * np.pi, 400)
y = np.sin(x ** 2)
y2 = 2*(np.sin(x ** 2))
y3 = 3*(np.sin(x ** 2))
fig, ax = plt.subplots(3,ncols=1, sharex=True, sharey=True)
fig.set_size_inches(SIZE[1], SIZE[0])
fig.subplots_adjust(hspace=0.001)
ax[0].plot(x, y)
ax[1].plot(x, y2)
ax[2].plot(x, y3)
plt.show()
All subplots have the same height now and the data span in the y-Axis is recognizable as the data is displayed with the correct relative proportion.
What i would like to achieve is that the scales of each plot end where the data ends. Essentially eliminating the not used white space. The size of the subplot would than represent the relative height ratios of the data. They should still have the same scaling on the Y axis in order for the viewer to estimate the relative data height ( which cold be a countrate for example).
I found the following links to similar problems but none really helped me to solve my issue:
Link1 Link2
Here an example that determines the ratio for you and creates the subplots accordingly:
import matplotlib.pyplot as plt
from matplotlib import gridspec
import numpy as np
SIZE = (12, 8) #desired overall figure size
# Simple data to display in various forms
x = np.linspace(0, 2 * np.pi, 400)
# the maximum multiplier for the function
N = 3
# the y-ranges:
ys = [i * np.sin(x**2) for i in range(1,N+1)]
# the maximum extent of the plot in y-direction (cast as int)
hs = [int(np.ceil(np.max(np.abs(y)))) for y in ys]
# determining the size of the GridSpec:
gs_size = np.sum(hs)
gs = gridspec.GridSpec(gs_size,1)
# the figure
fig = plt.figure(figsize = SIZE)
# creating the subplots
base = 0
ax = []
for y,h in zip(ys,hs):
ax.append(fig.add_subplot(gs[base:h+base,:]))
base += h
ax[-1].plot(x,y)
##fig, ax = plt.subplots(3,ncols=1, sharex=True, sharey=True)
##fig.set_size_inches(SIZE[1], SIZE[0])
fig.subplots_adjust(hspace=0.001)
##ax[0].plot(x, ys[0])
##ax[1].plot(x, ys[1])
##ax[2].plot(x, ys[2])
plt.show()
The code determines the maximum y-extend for each set of data, casts it into an integer and then divides the figure into subplots using the sum of these extends as scale for the GridSpec.
The resulting figure looks like this:
Tested on Python 3.5
EDIT:
If the maximum and minimum extents of your data are not comparable, it may be better to change the way hs is calculated into
hs = [int(np.ceil(np.max(y))) - int(np.floor(np.min(y))) for y in ys]

Matplotlib how to center a 2D image at origin on XY surface in 3D plot?

I have an image whose size is 330x330 in pixel and 5cm x 5cm in reality. When I tried to use the function plot_surface, the left top corner of the image was stuck to the origin. However, I want the center of the image to locate at the origin like this,
Is there any way I can do this? And how can I change the unit system into cm instead of pixels?
EDIT:
Here is my current code:
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
xx, yy = np.meshgrid(np.linspace(0, 1, 330), np.linspace(0, 1, 330))
X = xx
Y = yy
ax.plot_surface(X, Y, 0, rstride=10, cstride=10, facecolors=plt.imread('pattern.jpg') / 255., shade=False)
plt.show()
You can set the size of a figure in matplotlib using the figsize=(width, height) argument. The figure size in pixel is determined by the dpi = 80 argument. For example
plt.figure(figsize=(8,6))
...
plt.savefig("out.png", dpi=100)
will result in a 8 times 6 inch figure, which has 800 x 600 pixels.
You cannot change the units from inch to something else, but since 1 inch = 2.54 cm you can calculate your desired figure size if needed.
All of this has nothing to do with the plotting units that matplotlib uses and
plotting in matplotlib is in no way related to real length units, i.e. you can plot the size of an atom next to the map of the solar system, all in one plot of the size of a sheet of paper.
The position of the origin in a plot is determined by the axis limits. Roughly speeking if your axis goes from 0 to 2, the origin is on the left side of the figure, if your plot goes from -1 to 1, the origin is in the middle.
When ax is your axes object, you can set the origin using
ax.set_xlim([-1,1])
ax.set_ylim([-1,1])
in the 2D case or, for the 3D case via
ax.set_xlim3d([-1,1])
ax.set_ylim3d([-1,1])
ax.set_zlim3d([-1,1])
Now that we have some code to work with, we can set the image in the center by using appropriate coordinates. As in the case of the axis limits if the image is defines between 0 and 1 it will of course sit there and not be centered at the origin.
When we define the image to go from -1 to 1, it should work as expected.
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
xx, yy = np.meshgrid(np.linspace(-1, 1, 330), np.linspace(-1, 1, 330))
X = xx
Y = yy
c = np.random.rand(330,330,3)
ax.plot_surface(X, Y, 0, rstride=30, cstride=30,
facecolors=c, shade=False)
plt.show()

Plotting a 2D Array with Matplotlib

So I have a 2D array (named Data) that looks like:
Shape 0 Shape 1 ... Shape N
------- ------- -------
Scale 0 | Value00 , Value01 ... Value0N |
Scale 1 | Value10 , Value11 ... Value1N |
.
.
.
Scale N | ValueN0 , ValueN1 ... ValueNN |
And I want to create a 3D plot where the ValueXXs are the Z axis. I've tried two attempts, but each give me a surface that is rotated with respect to the other one, so I've gotten myself a bit confused. Here is my 1st attempt at a solution:
x,y = numpy.mgrid[0:50:50j,0:50:50j]
f = Data
fig = plt.figure()
ax = Axes3D(fig)
ax.plot_surface(x,y,f,rstride=1,cstride=1)
Here is my second attempt:
nx, ny = 50, 50
x = range(nx)
y = range(ny)
hf = plt.figure()
ha = hf.add_subplot(111, projection='3d')
X, Y = numpy.meshgrid(x, y)
ha.plot_surface(X,Y,Data,rstride=1,cstride=1)
Examining X and Y does no help really because its a square. I'm not sure when X represents my 'Scale' vs when it is representing my 'Shape'.
So, what is really going on with these two examples? Is there a better way to plot this array?
Thanks!
If I understand you right, the confusion is which axis is which, right? If this is the case, you can easily plot a known asymmetric shape and the plot will tell you everything. For example, adopting an example from the gallery:
# By Armin Moser
from mpl_toolkits.mplot3d import Axes3D
import matplotlib
import numpy as np
from matplotlib import cm
from matplotlib import pyplot as plt
step = 0.04
maxval = 1.0
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
# create supporting points in polar coordinates
r = np.linspace(0,1.25,50)
p = np.linspace(0,2*np.pi,50)
R,P = np.meshgrid(r,p)
# transform them to cartesian system
X,Y = R*np.cos(P),R*np.sin(P)
#Z = ((R**2 - 1)**2)
Z = (X**2 + 0.2*Y**2 -1)**2 # <------- edit
ax.plot_surface(X, Y, Z, rstride=1, cstride=1, cmap=cm.jet)
#ax.set_zlim3d(0, 1)
ax.set_xlabel(r'$\phi_\mathrm{real}$')
ax.set_ylabel(r'$\phi_\mathrm{im}$')
ax.set_zlabel(r'$V(\phi)$')
plt.show()

repeat y axis scale along grid line of graph ( matplotlib)

I am new to matplotlib and I am trying to figure out if I can repeat the y axis scale
values along the grid lines of the line graph.
The graph has 2 axis,
x-axis has hourly values and y-axis has temperature values.
I need to show the graph for 48 hours, so it results in a long horizontal graph. when user scrolls through the graph horizontally he has x-axis scale available for reference but
y axis scale is way towards left and is not visible.
I need a way to repeat the y-axis scale(temperature values) along all the graph. Is there any way to achieve this?
Is there any better solution to this problem, apart from repeating the values?
You might take a look at the colorbar from this example:
import matplotlib.pyplot as plt
import numpy as np
from matplotlib.collections import EllipseCollection
x = np.arange(10)
y = np.arange(15)
X, Y = np.meshgrid(x, y)
XY = np.hstack((X.ravel()[:,np.newaxis], Y.ravel()[:,np.newaxis]))
ww = X/10.0
hh = Y/15.0
aa = X*9
ax = plt.subplot(1,1,1)
ec = EllipseCollection(
ww,
hh,
aa,
units='x',
offsets=XY,
transOffset=ax.transData)
ec.set_array((X+Y).ravel())
ax.add_collection(ec)
ax.autoscale_view()
ax.set_xlabel('X')
ax.set_ylabel('y')
cbar = plt.colorbar(ec)
cbar.set_label('X+Y')
plt.show()
A quick experiment shows me that you can pan/zoom the main window and the colorbar will stay constant.

Generate a heatmap using a scatter data set

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:

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