Find the average between values in a matrix

Suppose im given a 1 X n matrix called a and i want to find matrix b which will give the average value of every element of a and its immediate neighbors

for example. a=[0 2 1 1 0].

b=[1 1 1.3 0.667];

Whats a smart way to do this?

Answer 1

[Star Strider]

I suggest:

a = [0 2 1 1 0];
for k1 = 1:length(a)-1
    ix1 = max(1,(k1-1));
    ix2 = min(length(a),k1+1);
    m(k1) = mean(a(ix1:ix2));
end

produces:

m =

       1.0000     1.0000     1.3333   666.6667

[Alexander]

Thank you! All though I am curious as to why the max and min in the for loop are written that way.

Could you explain please?

[Alexander]

Also, the majority of your code is correct, however, the k value in the for loop should be

for k1= 1:length(a) other than that it works like a charm.

[Star Strider]

My pleasure!

The (k1-1) originated in my initial desire to not get an index-out-of-bounds error. Then I thought of creating the ix1 and ix2 variables and forgot to reset the k1 limit in the for statement. My error.

An -1 index, and an index >length(a) would both throw errors, since neither exists. In order to include [a(1) a(2)] as the first and [a(end-1) a(end)] as the last sub-arrays to average (the others being [a(k1-1) a(k1) a(k1+1)]), the ix1 and ix2 calculations look for the minimum or maximum of the size of the arrays and only use the indices that are within the index range of the arrays. In other words, they prevent the occurrence of indices of (-1) and (end+1). (Since you specified ‘the average value of every element of a and its immediate neighbors’ and illustrated it in your example, that eliminated a(1) alone and a(end) alone being included in the algorithm.)

Answer 2

[Image Analyst]

You can try conv():

b = conv(a, [1,1,1]/3, 'full')
b = conv(a, [1,1,1]/3, 'valid')
b = conv(a, [1,1,1]/3, 'same')

Each of the shapes handles the "edge conditions" differently. It depends on what you want to do.

[Alexander]

is this for spherical coordinates?

[Image Analyst]

No. Who said anything about that? convolution is just a way of sliding an array past another array multiplying the values in the array together and adding them up. Works for any array. There are versions for 2D and up: conv2() and convn().

[Alexander]

I was just taking a guess, I was under the impression that conv meant conversion and there 3 values inside the parentheses were points on a plane.

Anyway, this is interesting, I will have to play around with the convolution function. Do you know of any real time application this would be used for? Systems of equations maybe?

[Image Analyst]

Oh my gosh. Convolution is used everywhere! It's the basis of linear systems theory and linear filtering. As just one example, if you have a sharp image and you blur it by the point spread function of the imaging system you will get an image that is the convolution of the image and the point spread function. And the whole basis of the duality of filtering in Fourier (spectral domain) and spatial (or time) domain is built upon convolution. Basically multiplication in the spatial (or time) domain is convolution in the Fourier domain, and multiplication in the spatial (or time) domain is convolution in the Fourier domain. As soon as you have your first course in linear systems, linear filtering, or optics you should get a heavy intensive education in convolution. Here's some background for you:

http://www.cv.nrao.edu/course/astr534/FourierTransforms.html

http://blogs.mathworks.com/steve/category/fourier-transforms/

http://en.wikipedia.org/wiki/Fourier_transform#Functional_relationships

http://www.mathworks.com/matlabcentral/fileexchange/28810-fourier-transform-demonstration/content/fourier_demo/html/fourier_demo.html

http://cns-alumni.bu.edu/~slehar/fourier/fourier.html