#### [SOLVED] Make contour of scatter

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 ? #### @ImportanceOfBeingErnest 2017-04-14 07:54:56

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
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. #### @elyase 2013-09-12 12:56:34

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)
``````

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)
``````  #### @JuanPablo 2013-09-12 13:08:12

you import griddata from ? ... `from scipy.interpolate import griddata` or `from matplotlib.mlab import griddata` #### @elyase 2013-09-12 13:12:23

@JuanPablo, ups, you are right, fixed(`from matplotlib.mlab import griddata` is the right one). #### @Marti Nito 2018-11-26 12:17:31

`from matplotlib.mlab import griddata` is deprecated as of matplotlib 2.2 #### @endolith 2019-02-27 04:48:47

`Deprecated since version 2.2: The griddata function was deprecated in Matplotlib 2.2 and will be removed in 3.1. Use scipy.interpolate.griddata instead.` #### @Gustavo Lopes 2019-06-07 16:58:06

Here `plt.tricontourf(x, y, z, 15)`, what does the `15` means? #### @David Zwicker 2013-09-12 12:55:14

`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 #### @JuanPablo 2013-09-12 13:50:00

when I use gridata of `scipy.interpolate`, the program are running a long time, this never stop. #### @David Zwicker 2013-09-12 14:00:57

That of course depends on your data, which you didn't specify in your initial post. You should definitely try playing with the `method` argument of `griddata`. Try `method="nearest"`, which should give the fastest interpolation.

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