fast_kde.py

import numpy as np
import scipy.sparse
import scipy.ndimage
import scipy.stats
import scipy.signal

import matplotlib.pyplot as plt

def main():
    x, y = generate_data(int(1e7))
    grid, extents, density = fast_kde(x, y, sample=True)

    image_example(grid, extents)
    scatter_example(x, y, density)

    plt.show()

def generate_data(num):
    x = 10 * np.random.random(num)
    y = x**2 + np.random.normal(0, 5, num)**2
    return x, y

def image_example(grid, extents):
    fig, ax = plt.subplots()
    im = ax.imshow(grid, origin='lower', extent=extents, aspect='auto',
                   cmap='gist_earth_r')
    fig.colorbar(im)

def scatter_example(x, y, density, num_points=10000):
    # Randomly draw a subset based on the _inverse_ of the estimated density
    prob = 1.0 / density
    prob /= prob.sum()
    subset = np.random.choice(np.arange(x.size), num_points, False, prob)
    x, y, density = x[subset], y[subset], density[subset]

    fig, ax = plt.subplots()
    ax.scatter(x, y, c=density, cmap='gist_earth_r')
    ax.axis('tight')

def fast_kde(x, y, gridsize=(400, 400), extents=None, weights=None,
             sample=False):
    """
    Performs a gaussian kernel density estimate over a regular grid using a
    convolution of the gaussian kernel with a 2D histogram of the data.

    This function is typically several orders of magnitude faster than
    scipy.stats.kde.gaussian_kde for large (>1e7) numbers of points and
    produces an essentially identical result.

    Input:
        x: array-like
            The x-coords of the input data points
        y: array-like
            The y-coords of the input data points
        gridsize: tuple, optional
            An (nx,ny) tuple of the size of the output
            grid. Defaults to (400, 400).
        extents: tuple, optional
            A (xmin, xmax, ymin, ymax) tuple of the extents of output grid.
            Defaults to min/max of x & y input.
        weights: array-like or None, optional
            An array of the same shape as x & y that weighs each sample (x_i,
            y_i) by each value in weights (w_i).  Defaults to an array of ones
            the same size as x & y.
        sample: boolean
            Whether or not to return the estimated density at each location.
            Defaults to False
    Output:
        density : 2D array of shape *gridsize*
            The estimated probability distribution function on a regular grid
        extents : tuple
            xmin, xmax, ymin, ymax
        sampled_density : 1D array of len(*x*)
            Only returned if *sample* is True.  The estimated density at each
            point.
    """
    #---- Setup --------------------------------------------------------------
    x, y = np.atleast_1d([x, y])
    x, y = x.reshape(-1), y.reshape(-1)

    if x.size != y.size:
        raise ValueError('Input x & y arrays must be the same size!')

    nx, ny = gridsize
    n = x.size

    if weights is None:
        # Default: Weight all points equally
        weights = np.ones(n)
    else:
        weights = np.squeeze(np.asarray(weights))
        if weights.size != x.size:
            raise ValueError('Input weights must be an array of the same size'
                    ' as input x & y arrays!')

    # Default extents are the extent of the data
    if extents is None:
        xmin, xmax = x.min(), x.max()
        ymin, ymax = y.min(), y.max()
    else:
        xmin, xmax, ymin, ymax = map(float, extents)
    extents = xmin, xmax, ymin, ymax
    dx = (xmax - xmin) / (nx - 1)
    dy = (ymax - ymin) / (ny - 1)

    #---- Preliminary Calculations -------------------------------------------

    # Most of this is a hack to re-implment np.histogram2d using `coo_matrix`
    # for better memory/speed performance with huge numbers of points.

    # First convert x & y over to pixel coordinates
    # (Avoiding np.digitize due to excessive memory usage!)
    ij = np.column_stack((y, x))
    ij -= [ymin, xmin]
    ij /= [dy, dx]
    ij = np.floor(ij, ij).T

    # Next, make a 2D histogram of x & y
    # Avoiding np.histogram2d due to excessive memory usage with many points
    grid = scipy.sparse.coo_matrix((weights, ij), shape=(ny, nx)).toarray()

    # Calculate the covariance matrix (in pixel coords)
    cov = image_cov(grid)

    # Scaling factor for bandwidth
    scotts_factor = np.power(n, -1.0 / 6) # For 2D

    #---- Make the gaussian kernel -------------------------------------------

    # First, determine how big the kernel needs to be
    std_devs = np.diag(np.sqrt(cov))
    kern_nx, kern_ny = np.round(scotts_factor * 2 * np.pi * std_devs)
    kern_nx, kern_ny = int(kern_nx), int(kern_ny)

    # Determine the bandwidth to use for the gaussian kernel
    inv_cov = np.linalg.inv(cov * scotts_factor**2)

    # x & y (pixel) coords of the kernel grid, with <x,y> = <0,0> in center
    xx = np.arange(kern_nx, dtype=np.float) - kern_nx / 2.0
    yy = np.arange(kern_ny, dtype=np.float) - kern_ny / 2.0
    xx, yy = np.meshgrid(xx, yy)

    # Then evaluate the gaussian function on the kernel grid
    kernel = np.vstack((xx.flatten(), yy.flatten()))
    kernel = np.dot(inv_cov, kernel) * kernel
    kernel = np.sum(kernel, axis=0) / 2.0
    kernel = np.exp(-kernel)
    kernel = kernel.reshape((kern_ny, kern_nx))

    #---- Produce the kernel density estimate --------------------------------

    # Convolve the gaussian kernel with the 2D histogram, producing a gaussian
    # kernel density estimate on a regular grid

    # Big kernel, use fft...
    if kern_nx * kern_ny > np.product(gridsize) / 4.0:
        grid = scipy.signal.fftconvolve(grid, kernel, mode='same')
    # Small kernel, use ndimage
    else:
        grid = scipy.ndimage.convolve(grid, kernel, mode='constant', cval=0)

    # Normalization factor to divide result by so that units are in the same
    # units as scipy.stats.kde.gaussian_kde's output.
    norm_factor = 2 * np.pi * cov * scotts_factor**2
    norm_factor = np.linalg.det(norm_factor)
    norm_factor = n * dx * dy * np.sqrt(norm_factor)

    # Normalize the result
    grid /= norm_factor

    if sample:
        i, j = ij.astype(int)
        return grid, extents, grid[i, j]
    else:
        return grid, extents

def image_cov(data):
    """Efficiently calculate the cov matrix of an image."""
    def raw_moment(data, ix, iy, iord, jord):
        data = data * ix**iord * iy**jord
        return data.sum()

    ni, nj = data.shape
    iy, ix = np.mgrid[:ni, :nj]
    data_sum = data.sum()

    m10 = raw_moment(data, ix, iy, 1, 0)
    m01 = raw_moment(data, ix, iy, 0, 1)
    x_bar = m10 / data_sum
    y_bar = m01 / data_sum

    u11 = (raw_moment(data, ix, iy, 1, 1) - x_bar * m01) / data_sum
    u20 = (raw_moment(data, ix, iy, 2, 0) - x_bar * m10) / data_sum
    u02 = (raw_moment(data, ix, iy, 0, 2) - y_bar * m01) / data_sum

    cov = np.array([[u20, u11], [u11, u02]])
    return cov

if __name__ == '__main__':
    main()

@MikeDacre - Sorry I didn't notice your comment earlier!

In that case, there's the correct way and then there's a shortcut.

The shortcut is to just plot the log of the values. If you're not displaying a colorbar, that's the route you'll want to take. For example, plot ax.scatter(x, y, c=np.log(density), cmap='gist_earth_r') and you should get something that looks similar to the R plot.

If you want to display a colorbar alongside it and still have the correct values, you'd want to change the Normalize type used for the color plotting. Something like ax.scatter(x, y, c=density, cmap='gist_earth_r', norm=LogNorm(density.min(), density.max())).

Hi @joferkington , My query is not related to the above case, apologies. I have determined the kde_values on grid points. Now, I want to plot the 1 sigma and 2 sigma contours based on the kde_values, the distribution of which is not necessarily have to be coming from a bivariate normal distribution. Basically, the contours may not be a perfect ellipse. How to go about that?