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Plotting inequalities in Python

I want to shade the area of the XY plane where Y<18 x^2.
This is my code:
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-.25, .25, 25)
y = np.linspace(0, 1, 20)
X,Y = np.meshgrid(x, y)
Z = masqarr(1,X, Y)
plt.imshow( (Y< 18*X**2).astype(float) ,
extent=(X.min(),X.max(),Y.min(),Y.max()),origin="lower", cmap="Greys", alpha = 0.3, aspect='auto', interpolation = 'hanning');
However, the output I get produces a plot with plateau as in:
How can I gat a proper smooth plot?

This is a resolution problem, you are defining your axes with too few poitns. I tried this:
import numpy as np
import matplotlib.pyplot as plt
resolution = 200
x = np.linspace(-.25, .25, resolution)
y = np.linspace(0, 1, resolution)
X,Y = np.meshgrid(x, y)
plt.imshow( (Y< 18*X**2).astype(float) ,
extent=(X.min(),X.max(),Y.min(),Y.max()),origin="lower", cmap="Greys", alpha = 0.3, aspect='auto', interpolation = 'hanning');
plt.show()
resolution = 25
resolution = 100
resolution = 200

Python: Plotting Evenly Spaced Spheres in Matplotlib

I'm trying to create a plot a bit like this:
Where there are spheres above all the minima.
The surface can be approximated with a sin(x)*sin(y) plot:
import numpy as np
import matplotlib.pyplot as plt
def func(x, y):
return np.sin(2*np.pi*x)*np.sin(2*np.pi*y) / 3
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
x = y = np.arange(-1.0, 1.0, 0.05)
X, Y = np.meshgrid(x, y)
zs = np.array([func(x,y) for x,y in zip(np.ravel(X), np.ravel(Y))])
Z = zs.reshape(X.shape)
ax.plot_surface(X, Y, Z, color="grey")
ax.set_zlim3d(-1,1)
plt.show()
However I'm unsure how to add evenly spaced spheres into this. Would anyone be able to help?

Using matplotlib one will inevitably run into problems of objects being hidden behind others. This is also stated in the matplotlib 3d FAQ and the recommendation is to use mayavi.
In mayavi the solution would look like this:
from mayavi import mlab
import numpy as np
### SURFACE '''
x,y = np.meshgrid(np.linspace(-2.5,2), np.linspace(-2,2))
f = lambda x,y: .4*np.sin(2*np.pi*x)*np.sin(2*np.pi*y)
z=f(x,y)
mlab.surf(x.T,y.T,z.T, colormap="copper")
### SPHERES '''
px,py = np.meshgrid(np.arange(-2,2)+.25, np.arange(-2,2)+.75)
px,py = px.flatten(),py.flatten()
pz = np.ones_like(px)*0.05
r = np.ones_like(px)*.4
mlab.points3d(px,py,pz,r, color=(0.9,0.05,.3), scale_factor=1)
mlab.show()

You need to determine the minima of the function, which are (with your parametrization) at (x = integer + 0.25, y=integer + 0.75) or the other way round. Then you can simply parametrize the spheres using spherical coordinates (for example as done here: python matplotlib: drawing 3D sphere with circumferences) and plot the spheres.
Now comes some good news and some bad news:
1.) The good news is that the minima are correctly determined and that the spheres are created. In the below plot you can see that they are right above the blue parts of the surface plot (where the blue parts show indeed the minima).
2.) The bad news is that you will have a hard time looking for another angle where the spheres are actually correctly rendered. I do not know a solution to this rather annoying behaviour, therefore you will probably have to play around until you have found the right angle. Have fun!
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
def func(x, y):
return np.sin(2*np.pi*x)*np.sin(2*np.pi*y) / 3
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
x = y = np.arange(-2.0, 2.0, 0.05)
# Get the minima of the function.
minsx1 = np.arange(int(np.amin(x)) + 0.25, int(np.amax(x)) + 0.25 + 1, 1)
minsy1 = np.arange(int(np.amin(y)) + 0.75, int(np.amax(y)) + 0.75 + 1, 1)
minsx2 = np.arange(int(np.amin(x)) + 0.75, int(np.amax(x)) + 0.75 + 1, 1)
minsy2 = np.arange(int(np.amin(y)) + 0.25, int(np.amax(y)) + 0.25 + 1, 1)
X, Y = np.meshgrid(x, y)
zs = np.array([func(x,y) for x,y in zip(np.ravel(X), np.ravel(Y))])
Z = zs.reshape(X.shape)
# Color map for better detection of minima (blue)
ax.plot_surface(X, Y, Z, cmap="viridis")
ax.set_zlim3d(-1,1)
# Spherical coordinates
r = 0.15
phi = np.linspace(0, 2 * np.pi, 30)
theta = np.linspace(0, np.pi, 30)
# Write spherical coordinates in cartesian coordinates.
x = r * np.outer(np.cos(phi), np.sin(theta))
y = r * np.outer(np.sin(phi), np.sin(theta))
z = r * np.outer(np.ones(np.size(phi)), np.cos(theta))
# Plot the spheres.
for xp in minsx1:
for yp in minsy1:
sphere = ax.plot_surface(x+xp, y+yp, z+0.35, color='r')
for xp in minsx2:
for yp in minsy2:
sphere = ax.plot_surface(x+xp, y+yp, z+0.35, color='r')
ax.view_init(elev=90, azim=0)
plt.savefig('test.png')
plt.show()

Plotting contours over pcolormesh data

I have some 2D data that I am displaying using pcolormesh that I would like to display a few contours on top of. I create the gridded data using
import numpy as np
import matplotlib.pyplot as plt
def bin(x, y, nbins, weights=None):
hist, X, Y = np.histogram2d(x, y, bins=nbins, weights=weights)
x_grid, y_grid = np.meshgrid(X,Y)
return hist, x_grid, y_grid
data = ... # read from binary file
h,x_grid,y_grid = bin(data.x,data.y,512)
# do some calculations with h
h = masked_log(h) # "safe" log that replaces <0 elements by 0 in output
pcm = plt.pcolormesh(x_grid,y_grid,h,cmap='jet')
# Just pretend that the data are lying on the center of the grid
# points, rather than on the edges
cont = plt.contour(x_grid[0:-1,0:-1],y_grid[0:-1,0:-1],h,4,colors='k',origin='lower')
When I plot only the output of pcolormesh, everything looks great. Adding the contours makes a giant mess.
I have read through the contour demo, the API examples, the pcolormesh levels example, and this closely-related SO post (my data is already gridded, so the solution doesn't help). But nothing I have tried thus far has created 4 simple contour lines atop my pcolormesh data.

I've put together minimal example with Gaussian filter (and scipy) which I think looks like it may do what you want. First, set up some dummy data (a Gaussian) and add noise,
import matplotlib
import numpy as np
import matplotlib.mlab as mlab
import matplotlib.pyplot as plt
delta = 0.025
x = np.arange(-3.0, 3.0, delta)
y = np.arange(-2.0, 2.0, delta)
X, Y = np.meshgrid(x, y)
Z = mlab.bivariate_normal(X, Y, 1.0, 1.0, 0.0, 0.0)
Z += 0.1*np.random.random(Z.shape)
and try to pcolormesh/contour,
plt.figure()
CS = plt.pcolormesh(X, Y, Z)
plt.contour(X, Y, Z, 4, colors='k')
plt.colorbar(CS)
plt.show()
which looks like this,
If we add filtering as follows,
import matplotlib
import numpy as np
import matplotlib.mlab as mlab
import matplotlib.pyplot as plt
from scipy.ndimage.filters import gaussian_filter
delta = 0.025
x = np.arange(-3.0, 3.0, delta)
y = np.arange(-2.0, 2.0, delta)
X, Y = np.meshgrid(x, y)
Z = mlab.bivariate_normal(X, Y, 1.0, 1.0, 0.0, 0.0)
Z += 0.1*np.random.random(Z.shape)
plt.figure()
plt.pcolormesh(X, Y, Z)
CS = plt.contour(X, Y, gaussian_filter(Z, 5.), 4, colors='k',interpolation='none')
plt.colorbar()
plt.show()
it looks much better,

Python Point density plots in polar stereographic projection

I have a point cloud of magnetization directions with azimut (declination between 0° and 360°) and inclination between 0° and 90°. I display these points in a polar azimuthal equidistant projection (using matplotlib basemap). That means 90° inclination will point directly in the center of the plot and the declination runs clockwise.
My problem is that I want to also plot isolines around these point clouds, which should represent where the highest density of point/directions is located. What is the easiest way to do this? Nice would be to mark the isoline which encircles 50% is my data. If Iam not mistaken - this would be the median.
So far I've fiddled around with gaussian_kde and the outlier detection of sklearn (1 and 2), but the results are not as expected.
Any ideas?
Edit #1:
First gaussian_kde
import numpy as np
import matplotlib.pyplot as plt
import scipy.stats as stats
from mpl_toolkits.basemap import Basemap
m = Basemap(projection='spaeqd',boundinglat=0,lon_0=180,resolution='l',round=True)
m.drawparallels(np.arange(-80.,1.,10.),labels=[False,True,True,False])
m.drawmeridians(np.arange(-180.,181.,30.),labels=[True,False,False,True])
#data
x, y = m(m1,-m2) #m2 is negative because I to plot in the southern hemisphere!
#set up the grid for evaluation of the KDE
yi = np.arange(0,360.1,1)
xi = np.arange(-90,1,1)
xx,yy = np.meshgrid(xi,yi)
X, Y = m(xx,yy) # to have it in my basemap projection
#setup the gaussian kde and evaluate it
#pretty much similiar to the scipy.stats docs
positions = np.vstack([X.ravel(), Y.ravel()])
values = np.vstack([x, y])
kernel = stats.gaussian_kde(values)
Z = np.reshape(kernel(positions).T, X.shape)
#plot orginal points and probaility density function
ax = plt.gca()
ax.scatter(x,y,c = 'Crimson')
TOT = ax.contour(X,Y,Z,cmap=plt.cm.Reds)
plt.show()
Then sklearn:
import numpy as np
import matplotlib.pyplot as plt
import scipy.stats as stats
from mpl_toolkits.basemap import Basemap
from sklearn import svm
from sklearn.covariance import EllipticEnvelope
m = Basemap(projection='spaeqd',boundinglat=0,lon_0=180,resolution='l',round=True)
m.drawparallels(np.arange(-80.,1.,10.),labels=[False,True,True,False])
m.drawmeridians(np.arange(-180.,181.,30.),labels=[True,False,False,True])
#data
x, y = m(m1,-m2) #m2 is negative because I to plot in the southern hemisphere!
#Similar to examples in sklearn docs
outliers_fraction = 0.5
oneclass_svm = svm.OneClassSVM(nu=0.95 * outliers_fraction + 0.05,\
kernel="rbf", gamma=0.1,verbose=True)
#seup grid
yi = np.arange(0,360.1,1)
xi = np.arange(-90,1,1)
R,T = np.meshgrid(xi,yi)
xx, yy = m(T,R)
x, y = m(m1,-m2)
#standardize data as suggested by docs
x_std = (x-x.mean())/x.std()
y_std = (y-y.mean())/y.std()
values = np.vstack([x_std, y_std])
#fit data and calculate threshold - this should mark my median - according to value of outliers_fraction
oneclass_svm.fit(values.T)
y_pred = oneclass_svm.decision_function(values.T).ravel()
threshold = stats.scoreatpercentile(y_pred, 100 * outliers_fraction)
y_pred = y_pred > threshold
#Target vector for evaluation
TV = np.c_[xx.ravel(), yy.ravel()]
TV = (TV-TV.mean(axis=0))/TV.std(axis=0) #must be standardized as well
# evaluation - This is now shifted in the plot ad does not fit my point cloud anymore - because of the standadrization
Z = oneclass_svm.decision_function(TV)
Z = Z.reshape(xx.shape)
#plotting - very similar to the example in the docs
ax = plt.gca()
ax.contourf(xx, yy, Z, levels=np.linspace(Z.min(), threshold, 7), \
cmap=plt.cm.Blues_r)
ax.contour(xx, yy, Z, levels=[threshold],
linewidths=2, colors='red')
ax.contourf(xx, yy, Z, levels=[threshold, Z.max()],
colors='orange')
ax.scatter(x, y,s=30, marker='s',c = 'RoyalBlue',label = 'Mr')
plt.show()
The EllipticEvelope works, but it is not that want I want.

Ok, I think I might found a solution. But it should not work in every case. It should fail in my opinion when the data is multimodal distributed.
Nevertheless, here is my though process:
So the Probalibity Density Function (PDF) is essentially the same as a continuous histogram. So I used np.percentile to calculate the upper and lower 25% percentile of both vectors. The I've searched for the value of the PDF at these perctiles and this should be the Isoline that i want.
Of course this should also work in the polar stereographic (or any other) projection.
Here is a litte example code of two gamma distributed data sets in a crossplot:
import numpy as np
import matplotlib.pyplot as plt
import scipy.stats as stats
from scipy.interpolate import LinearNDInterpolator, RegularGridInterpolator
#generate some data
x = np.random.gamma(10,0.8,1e4)
y = np.random.gamma(4,0.3,1e4)
#set up the data and grid for the 2D PDF
values = np.vstack([x,y])
pdf_x = np.linspace(x.min(),x.max(),1e2)
pdf_y = np.linspace(y.min(),y.max(),1e2)
X,Y = np.meshgrid(pdf_x,pdf_y)
kernel = stats.gaussian_kde(values)
#evaluate the PDF at every grid location
positions = np.vstack([X.ravel(), Y.ravel()])
Z = np.reshape(kernel(positions).T, X.shape)
#upper and lower quartiles of x and y data
xql = np.percentile(x,25)
xqu = np.percentile(x,75)
yql = np.percentile(y,25)
yqu = np.percentile(y,75)
#set up the interpolator - I could also use RegularGridInterpolator - should be faster
Interp = LinearNDInterpolator((X.flatten(),Y.flatten()),Z.flatten())
#1D example to illustrate what I mean
plt.figure()
kernel2 = stats.gaussian_kde(x)
plt.hist(x,30,normed=True)
plt.plot(pdf_x,kernel2(pdf_x),'r--',linewidth=2)
#plot vertical lines at the upper and lower quartiles
plt.vlines(np.percentile(x,25),0,0.2,color='red')
plt.vlines(np.percentile(x,75),0,0.2,color='red')
#Scatterplot / Crossplot with PDF and 25 and 75% isolines
plt.figure()
plt.scatter(x,y)
#search for the isolines defining the upper and lower quartiles
#the lower quartiles isoline should encircle 75% of the data
levels = [Interp(xql,yql),Interp(xqu,yqu)]
plt.contour(X,Y,Z,levels=levels,colors='orange')
plt.show()
To finish up I will give a quick example of what it looks in a polar stereographic projection:
import numpy as np
import matplotlib.pyplot as plt
import scipy.stats as stats
from scipy.interpolate import LinearNDInterpolator
from mpl_toolkits.basemap import Basemap
#set up the coordinate projection
m = Basemap(projection='spaeqd',boundinglat=0,lon_0=180,\
resolution='l',round=True,suppress_ticks=True)
parallelGrid = np.arange(-80.,1.,10.)
meridianGrid = np.arange(-180.0,180.1,30)
m.drawparallels(parallelGrid,labels=[False,False,False,False])
m.drawmeridians(meridianGrid,labels=[False,False,False,False],labelstyle='+/-',fmt='%i')
#Found this on stackoverflow - labels it exactly how I want it
ax = plt.gca()
ax.text(0.5,1.025,'N',transform=ax.transAxes,\
horizontalalignment='center',verticalalignment='bottom',size=25)
for para in np.arange(30,360,30):
x= (1.1*0.5*np.sin(np.deg2rad(para)))+0.5
y= (1.1*0.5*np.cos(np.deg2rad(para)))+0.5
ax.text(x,y,u'%i\N{DEGREE SIGN}'%para,transform=ax.transAxes,\
horizontalalignment='center',verticalalignment='center')
#generate some data
x = np.random.randint(180,225,size=15)
y = np.random.randint(30,40,size=15)
#into projection
x,y = m(x,-y)
values = np.vstack([x,y])
pdf_x = np.arange(0,361,1)
pdf_y = np.arange(0,91,1)
#into projection
X,Y = np.meshgrid(pdf_x,pdf_y)
X,Y = m(X,-Y)
kernel = stats.gaussian_kde(values)
positions = np.vstack([X.ravel(), Y.ravel()])
Z = np.reshape(kernel(positions).T, X.shape)
xql = np.percentile(x,25)
xqu = np.percentile(x,75)
yql = np.percentile(y,25)
yqu = np.percentile(y,75)
Interp = LinearNDInterpolator((X.flatten(),Y.flatten()),Z.flatten())
ax = plt.gca()
ax.scatter(x,y)
levels = [Interp(xql,yql),Interp(xqu,yqu)]
ax.contour(X,Y,Z,levels=levels,colors='red')
plt.show()

Scattered x,y,z via python's matplotlib.pyplot.contourf

Most pyplot examples out there use linear data, but what if data is scattered?
x = 3,7,9
y = 1,4,5
z = 20,3,7
better meshgrid for contourf
xi = np.linspace(min(x)-1, max(x)+1, 9)
yi = np.linspace(min(y)-1, max(y)+1, 9)
X, Y = np.meshgrid(xi, yi)
Now "z" data got to be interpolated onto the meshgrid.
numpy.interp does little help here, while both linear and nn interpolaton of
zi = matplotlib.mlab.griddata(x,y,z,xi,yi,interp="linear")
returns rather strange results
scipy.interpolate.griddata cubic from second answer below needs something else to return data rather than nils
With custom levels data expected be looking something like this

This is what happens:
Although contour requires grid data, we can caste scatter data to a grid and then using masked arrays mask out the blank regions. I simulate this in the code below, by creating a random array, then using this to mask a test dataset (shown at bottom). The bulk of the code is taken from this matplotlib demo page.
import matplotlib
import numpy as np
import matplotlib.mlab as mlab
import matplotlib.pyplot as plt
matplotlib.rcParams['xtick.direction'] = 'out'
matplotlib.rcParams['ytick.direction'] = 'out'
delta = 0.025
x = np.arange(-3.0, 3.0, delta)
y = np.arange(-2.0, 2.0, delta)
X, Y = np.meshgrid(x, y)
Z1 = mlab.bivariate_normal(X, Y, 1.0, 1.0, 0.0, 0.0)
Z2 = mlab.bivariate_normal(X, Y, 1.5, 0.5, 1, 1)
# difference of Gaussians
Z = 10.0 * (Z2 - Z1)
from numpy.random import *
import numpy.ma as ma
J = random_sample(X.shape)
mask = J > 0.7
X = ma.masked_array(X, mask=mask)
Y = ma.masked_array(Y, mask=mask)
Z = ma.masked_array(Z, mask=mask)
plt.figure()
CS = plt.contour(X, Y, Z, 20)
plt.clabel(CS, inline=1, fontsize=10)
plt.title('Simplest default with labels')
plt.savefig('cat.png')
plt.show()

countourf will only work with a grid of data. If you're data is scattered, then you'll need to create an interpolated grid matching your data, like this: (note you'll need scipy to perform the interpolation)
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
# your data
x = [3,7,9]
y = [1,4,5]
z = [20,3,7]
# define grid.
xi = np.linspace(0,10,300)
yi = np.linspace(0,6,300)
# 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(min(x),max(x))
plt.ylim(min(y),max(y))
plt.title('griddata test (%d points)' % len(x))
plt.show()
See here for the origin of that code.