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How to implement Kernel density estimation in multivariate/3D

I have dataset like the following fromat and im trying to find out the Kernel density estimation with optimal bandwidth.

data = np.array([[1, 4, 3], [2, .6, 1.2], [2, 1, 1.2],
         [2, 0.5, 1.4], [5, .5, 0], [0, 0, 0],
         [1, 4, 3], [5, .5, 0], [2, .5, 1.2]])

but I couldn't figure out how to approach it. also how to find the Σ matrix.

UPDATE

I tried KDE function from scikit-learn toolkits to find out univariate(1D) kde,

# kde function
def kde_sklearn(x, x_grid, bandwidth):
    kde = KernelDensity(kernel='gaussian', bandwidth=bandwidth).fit(x)
    log_pdf = kde.score_samples(x_grid[:, np.newaxis])
    return np.exp(log_pdf)

# optimal bandwidth selection
from sklearn.grid_search import GridSearchCV
grid = GridSearchCV(KernelDensity(), {'bandwidth': np.linspace(.1, 1.0, 30)}, cv=20)
grid.fit(x)
bw = grid.best_params_

# pdf using kde
pdf = kde_sklearn(x, x_grid, bw)
ax.plot(x_grid, pdf, label='bw={}'.format(bw))
ax.legend(loc='best')
plt.show()

Can any one help me to extend this to multivariate / in this case 3D data?

like image 554
jquery404 Avatar asked Sep 27 '22 16:09

jquery404


1 Answers

Interesting problem. You have a few options:

  1. Continue with scikit-learn
  2. Use a different library. For instance, if the kernel you are interested in is the gaussian - then you could use scipy.gaussian_kde which is arguably easier to understand / apply. There is a very good example of this technique in this question.
  3. roll your own from first principles. This is very difficult and I don't recommend it

This blog post goes into detail about the relative merits of various library implementations of Kernel Density Estimation (KDE).


I'm going to show you what (in my opinion - yes this is a bit opinion based) is the simplest way, which I think is option 2 in your case.

NOTE This method uses a rule of thumb as described in the linked docs to determine bandwidth. The exact rule used is Scott's rule. Your mention of the Σ matrix makes me think rule of thumb bandwidth selection is OK for you, but you also talk about optimal bandwidth and the example you present uses cross-validation to determine the best bandwidth. Therefore, if this method doesn't suit your purposes - let me know in comments.

import numpy as np
from scipy import stats
data = np.array([[1, 4, 3], [2, .6, 1.2], [2, 1, 1.2],
         [2, 0.5, 1.4], [5, .5, 0], [0, 0, 0],
         [1, 4, 3], [5, .5, 0], [2, .5, 1.2]])

data = data.T #The KDE takes N vectors of length K for K data points
              #rather than K vectors of length N

kde = stats.gaussian_kde(data)

# You now have your kde!!  Interpreting it / visualising it can be difficult with 3D data
# You might like to try 2D data first - then you can plot the resulting estimated pdf
# as the height in the third dimension, making visualisation easier.

# Here is the basic way to evaluate the estimated pdf on a regular n-dimensional mesh
# Create a regular N-dimensional grid with (arbitrary) 20 points in each dimension
minima = data.T.min(axis=0)
maxima = data.T.max(axis=0)
space = [np.linspace(mini,maxi,20) for mini, maxi in zip(minima,maxima)]
grid = np.meshgrid(*space)

#Turn the grid into N-dimensional coordinates for each point
#Note - coords will get very large as N increases...
coords = np.vstack(map(np.ravel, grid))

#Evaluate the KD estimated pdf at each coordinate
density = kde(coords)

#Do what you like with the density values here..
#plot them, output them, use them elsewhere...

Caveat

this may give terrible results, depending on your particular problem. Things to bear in mind are obviously:

  1. as your number of dimensions goes up, the number of observed data points you want will have to go up exponentially - your sample data of 9 points in 3 dimensions is pretty sparse - although I assume the dots indicate that in fact you have many more.

  2. As mentioned in the main body - the bandwidth is selected in a particular way - this may result in over- (or conceivably but unlikely under-) smoothing of the estimated pdf.

like image 188
J Richard Snape Avatar answered Oct 06 '22 23:10

J Richard Snape