Gaussian filtering a image with Nan in Python

Question

From a list of 2D coordinates, and a third variable (velocity), I have created a 2D numpy array covering the whole sampled area. My intention is to create an image, in which each pixel contains the mean velocity of the points lying within it. After that filter that image with a gaussian filter.

The problem is that the area is not uniformly sampled. Therefore I have several pixels without information (Nan) in the middle of the image. When I try to filter the array through a gaussian filter, the Nan propagate ruining the whole image.

I need to filter this image, but rejecting all pixels without information. In other words, If a pixel does not contain information, then it should be not taken into account for the filtering.

Here is an example of my code for averaging:

Mean_V = np.zeros([len(x_bins), len(y_bins)])

for i, x_bin in enumerate(x_bins[:-1]):
    bin_x = (x > x_bins[i]) & (x <= x_bins[i+1])
    for j, y_bin in enumerate(y_bins[:-1]):
        bin_xy = (y[bin_x] > y_bins[j]) & (y[bin_x] <= y_bins[j+1])
        if (sum(x > 0 for x in bin_xy) > 0) :
            Mean_V[i,j]=np.mean(V[bin_x][bin_xy])
        else:
            Mean_V[i,j]=np.nan

EDIT:

Surfing the web I have ended into this question I made in 2013. The solution to this problem can be found in the astropy library:

http://docs.astropy.org/en/stable/convolution/

Astropy's convolution replaces the NaN pixels with a kernel-weighted interpolation from their neighbors.

Thanks folks!!

Answer 1

in words:

A Gaussian filter which ignores NaNs in a given array U can be easily obtained by applying a standard Gaussian filter to two auxiliary arrays V and W and by taking the ratio of the two to get the result Z.

Here, V is copy of the original U with NaNs replaced by zeros and W is an array of ones with zeros indicating the positions of NaNs in the original U.

The idea is that replacing the NaNs by zeros introduces an error in the filtered array which can, however, be compensated by applying the same Gaussian filter to another auxiliary array and combining the two.

in Python:

import numpy as np
import scipy as sp
import scipy.ndimage

sigma=2.0                  # standard deviation for Gaussian kernel
truncate=4.0               # truncate filter at this many sigmas

U=sp.randn(10,10)          # random array...
U[U>2]=np.nan              # ...with NaNs for testing

V=U.copy()
V[np.isnan(U)]=0
VV=sp.ndimage.gaussian_filter(V,sigma=sigma,truncate=truncate)

W=0*U.copy()+1
W[np.isnan(U)]=0
WW=sp.ndimage.gaussian_filter(W,sigma=sigma,truncate=truncate)

Z=VV/WW

in numbers:

Here coefficients of the Gaussian filter are set to [0.25,0.50,0.25] for demonstration purposes and they sum up to one 0.25+0.50+0.25=1, without loss of generality.

After replacing the NaNs by zeros and applying the Gaussian filter (cf. VV below) it is clear that the zeros introduce an error, i.e., due to the "missing" data the coefficients 0.25+0.50=0.75 do not sum up to one anymore and therefore underestimate the "true" value.

However, this can be compensated by using the second auxiliary array (cf. WW below) which, after filtering with the same Gaussian, just contains the sum of coefficients.

Therefore, dividing the two filtered auxiliary arrays rescales the coefficients such that they sum up to one while the NaN positions are ignored.

array U         1   2   NaN 1   2    
auxiliary V     1   2   0   1   2    
auxiliary W     1   1   0   1   1
position        a   b   c   d   e

filtered VV_b   = 0.25*V_a  + 0.50*V_b  + 0.25*V_c
                = 0.25*1    + 0.50*2    + 0
                = 1.25

filtered WW_b   = 0.25*W_a  + 0.50*W_b  + 0.25*W_c
                = 0.25*1    + 0.50*1    + 0
                = 0.75

ratio Z         = VV_b / WW_b  
                = (0.25*1 + 0.50*2) / (0.25*1    + 0.50*1)
                = 0.333*1 + 0.666*2
                = 1.666

update - division-by-zero:

The following incorporates useful questions and answers by @AndyL and @amain from the comments below, thanks!

Large areas of NaNs may lead to a zero denominator (WW=0) at some positions when there are only NaN entries within the support of the Gaussian kernel (in theory that support is infinite, but in practice the kernel is usually truncated, see 'truncate' parameter in code example above). In that situation, the nominator becomes zero as well (VV=0) so that numpy throws a 'RuntimeWarning: invalid value encountered in true_divide' and returns NaN at the corresponding positions.

This is probably the most consistent/meaningful result and if you can live with a numpy warning, no further adjustments are required.