Make contour of scatter
Question:
In python, If I have a set of data
x, y, z
I can make a scatter with
import matplotlib.pyplot as plt
plt.scatter(x,y,c=z)
How I can get a plt.contourf(x,y,z) of the scatter ?
Answers:
contour expects regularly gridded data. You thus need to interpolate your data first:
import numpy as np
from scipy.interpolate import griddata
import matplotlib.pyplot as plt
import numpy.ma as ma
from numpy.random import uniform, seed
# make up some randomly distributed data
seed(1234)
npts = 200
x = uniform(-2,2,npts)
y = uniform(-2,2,npts)
z = x*np.exp(-x**2-y**2)
# define grid.
xi = np.linspace(-2.1,2.1,100)
yi = np.linspace(-2.1,2.1,100)
# grid the data.
zi = griddata((x, y), z, (xi[None,:], yi[:,None]), method='cubic')
# contour the gridded data, plotting dots at the randomly spaced data points.
CS = plt.contour(xi,yi,zi,15,linewidths=0.5,colors='k')
CS = plt.contourf(xi,yi,zi,15,cmap=plt.cm.jet)
plt.colorbar() # draw colorbar
# plot data points.
plt.scatter(x,y,marker='o',c='b',s=5)
plt.xlim(-2,2)
plt.ylim(-2,2)
plt.title('griddata test (%d points)' % npts)
plt.show()
Note that I shamelessly stole this code from the excellent matplotlib cookbook
You can use tricontourf as suggested in case b. of this other answer:
import matplotlib.tri as tri
import matplotlib.pyplot as plt
plt.tricontour(x, y, z, 15, linewidths=0.5, colors='k')
plt.tricontourf(x, y, z, 15)
Old reply:
Use the following function to convert to the format required by contourf:
from numpy import linspace, meshgrid
from matplotlib.mlab import griddata
def grid(x, y, z, resX=100, resY=100):
"Convert 3 column data to matplotlib grid"
xi = linspace(min(x), max(x), resX)
yi = linspace(min(y), max(y), resY)
Z = griddata(x, y, z, xi, yi)
X, Y = meshgrid(xi, yi)
return X, Y, Z
Now you can do:
X, Y, Z = grid(x, y, z)
plt.contourf(X, Y, Z)
The solution will depend on how the data is organized.
Data on regular grid
If the x and y data already define a grid, they can be easily reshaped to a quadrilateral grid. E.g.
#x y z
4 1 3
6 1 8
8 1 -9
4 2 10
6 2 -1
8 2 -8
4 3 8
6 3 -9
8 3 0
4 4 -1
6 4 -8
8 4 8
can plotted as a contour using
import matplotlib.pyplot as plt
import numpy as np
x,y,z = np.loadtxt("data.txt", unpack=True)
plt.contour(x.reshape(4,3), y.reshape(4,3), z.reshape(4,3))
Arbitrary data
a. Interpolation
In case the data is not living on a quadrilateral grid, one can interpolate the data on a grid. One way to do so is scipy.interpolate.griddata
import numpy as np
from scipy.interpolate import griddata
xi = np.linspace(4, 8, 10)
yi = np.linspace(1, 4, 10)
zi = griddata((x, y), z, (xi[None,:], yi[:,None]), method='linear')
plt.contour(xi, yi, zi)
b. Non-gridded contour
Finally, one can plot a contour completely without the use of a quadrilateral grid. This can be done using tricontour.
plt.tricontour(x,y,z)
An example comparing the latter two methods is found on the matplotlib page.
In python, If I have a set of data
x, y, z
I can make a scatter with
import matplotlib.pyplot as plt
plt.scatter(x,y,c=z)
How I can get a plt.contourf(x,y,z) of the scatter ?
contour expects regularly gridded data. You thus need to interpolate your data first:
import numpy as np
from scipy.interpolate import griddata
import matplotlib.pyplot as plt
import numpy.ma as ma
from numpy.random import uniform, seed
# make up some randomly distributed data
seed(1234)
npts = 200
x = uniform(-2,2,npts)
y = uniform(-2,2,npts)
z = x*np.exp(-x**2-y**2)
# define grid.
xi = np.linspace(-2.1,2.1,100)
yi = np.linspace(-2.1,2.1,100)
# grid the data.
zi = griddata((x, y), z, (xi[None,:], yi[:,None]), method='cubic')
# contour the gridded data, plotting dots at the randomly spaced data points.
CS = plt.contour(xi,yi,zi,15,linewidths=0.5,colors='k')
CS = plt.contourf(xi,yi,zi,15,cmap=plt.cm.jet)
plt.colorbar() # draw colorbar
# plot data points.
plt.scatter(x,y,marker='o',c='b',s=5)
plt.xlim(-2,2)
plt.ylim(-2,2)
plt.title('griddata test (%d points)' % npts)
plt.show()
Note that I shamelessly stole this code from the excellent matplotlib cookbook
You can use tricontourf as suggested in case b. of this other answer:
import matplotlib.tri as tri
import matplotlib.pyplot as plt
plt.tricontour(x, y, z, 15, linewidths=0.5, colors='k')
plt.tricontourf(x, y, z, 15)
Old reply:
Use the following function to convert to the format required by contourf:
from numpy import linspace, meshgrid
from matplotlib.mlab import griddata
def grid(x, y, z, resX=100, resY=100):
"Convert 3 column data to matplotlib grid"
xi = linspace(min(x), max(x), resX)
yi = linspace(min(y), max(y), resY)
Z = griddata(x, y, z, xi, yi)
X, Y = meshgrid(xi, yi)
return X, Y, Z
Now you can do:
X, Y, Z = grid(x, y, z)
plt.contourf(X, Y, Z)
The solution will depend on how the data is organized.
Data on regular grid
If the x and y data already define a grid, they can be easily reshaped to a quadrilateral grid. E.g.
#x y z
4 1 3
6 1 8
8 1 -9
4 2 10
6 2 -1
8 2 -8
4 3 8
6 3 -9
8 3 0
4 4 -1
6 4 -8
8 4 8
can plotted as a contour using
import matplotlib.pyplot as plt
import numpy as np
x,y,z = np.loadtxt("data.txt", unpack=True)
plt.contour(x.reshape(4,3), y.reshape(4,3), z.reshape(4,3))
Arbitrary data
a. Interpolation
In case the data is not living on a quadrilateral grid, one can interpolate the data on a grid. One way to do so is scipy.interpolate.griddata
import numpy as np
from scipy.interpolate import griddata
xi = np.linspace(4, 8, 10)
yi = np.linspace(1, 4, 10)
zi = griddata((x, y), z, (xi[None,:], yi[:,None]), method='linear')
plt.contour(xi, yi, zi)
b. Non-gridded contour
Finally, one can plot a contour completely without the use of a quadrilateral grid. This can be done using tricontour.
plt.tricontour(x,y,z)
An example comparing the latter two methods is found on the matplotlib page.