KDE in python with different mu, sigma / mapping a function to an array

Question

I have a 2-dimensional array of values that I would like to perform a Gaussian KDE on, with a catch: the points are assumed to have different variances. For that, I have a second 2-dimensional array (with the same shape) that is the variance of the Gaussian to be used for each point. In the simple example,

import numpy as np
data = np.array([[0.4,0.2],[0.1,0.5]])
sigma = np.array([[0.05,0.1],[0.02,0.3]])

there would be four gaussians, the first of which is centered at x=0.4 with σ=0.05. Note: Actual data is much larger than 2x2

I am looking for one of two things:

1. A Gaussian KDE solver that will allow for bandwidth to change for each point

or

2. A way to map the results of each Gaussian into a 3-dimensional array, with each Gaussian evaluated across a range of points (say, evaluate each center/σ pair along np.linspace(0,1,101)). In this case, I could e.g. have the KDE value at x=0.5 by taking outarray[:,:,51].

Answer 1

The best way I found to handle this is through array multiplication of a sigma array and a data array. Then, I stack the arrays for each value I want to solve the KDE for.

import numpy as np

def solve_gaussian(val,data_array,sigma_array):
    return (1. / sigma_array) * np.exp(- (val - data_array) * (val - data_array) / (2 * sigma_array * sigma_array))

def solve_kde(xlist,data_array,sigma_array):
    kde_array = np.array([])
    for xx in xlist:
        single_kde = solve_gaussian(xx,data_array,sigma_array)
        if np.ndim(kde_array) == 3:
            kde_array = np.concatenate((kde_array,single_kde[np.newaxis,:,:]),axis=0)
        else:
            kde_array = np.dstack(single_kde)
    return kde_array

xlist = np.linspace(0,1,101) #Adjust as needed
kde_array = solve_kde(xlist,data_array,sigma_array)
kde_vector = np.sum(np.sum(kde_array,axis=2),axis=1)
mode_guess = xlist[np.argmax(kde_vector)]

Caveat, for anyone attempting to use this code: the value of the Gaussian is along axis 0, not axis 2 as specified in the original question.