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finding the closest value

Asked by FIR

FIR (view profile)

on 29 Nov 2012
i have valued as val(:,:,1) 8x8 matrix,,,,,,,,,,,,val(:,:,192)8x8 matrix
now i want to check which all values are nearby ,at a difference of 5 and for that we have to find mean value
suppose val(:,:,1) has mean value of 145.23 and val(:,:,56) has mean of 149.10and val(:,:,103) has mean of 143.05 for all these three we have to find the mean,and so on

please help

3 Comments

Azzi Abdelmalek

Azzi Abdelmalek (view profile)

on 29 Nov 2012

What about values like: 145 148 150 153; the difference between each 2 numbers is<5 but the total difference is 8

Jan Simon

Jan Simon (view profile)

on 29 Nov 2012

So actually you are talking about the 3D array val(:,:,n) but you have calculated the mean values already, such that you operate on a vector? Does this mean, that the thread http://www.mathworks.com/matlabcentral/answers/55028-finding-mean-for-values is solved already? If so, please mark mark this thread correspondingly.

I still cannot guess, what "val(:,:,1) 8x8 matrix,,,,,,,," exactly means. The notation [8 x 8 x N] is common in Matlab, and a Matlab command as "rand(8, 8, n)" would be clear also.

Image Analyst

Image Analyst (view profile)

on 29 Nov 2012

I know FIR deals with images so maybe val is a color image that is a moving average window, like he's scanned it with a 8x8 window to get the average. But usually windows are an odd number so you don't have a shift of the image in the filtered image vs. the original image like you would for an even window size. But anyway, that's the only thing I can figure out when he says "val(:,:,1) has mean value of 145.23" because an array element by itself can't have an average - it has only its value - so it must be the average of a group of elements. But then I don't know what he means by the mean of all three. Maybe he means (145.23+149.10+143.05)/3 or maybe he wants a weighted average like rgb2gray calculates. But he already knows how to do those things. This question is not clearly explained.

FIR

FIR (view profile)

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1 Answer

Answer by Jan Simon

Jan Simon (view profile)

on 29 Nov 2012

Are you searching for a k-means algorithm? If so, ask your favorite search engine, search in this forum and the FileExchange also.

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Jan Simon

Jan Simon (view profile)

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