Finding average at each point

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Mayowa
Mayowa on 8 Oct 2014
Commented: dpb on 9 Oct 2014
I have a data say: 1 2 3 4 5 6 7 8
(8x1 vector). I want to find the average of each number using the 2 other numbers(one below and one above) and divide by 3. And for the first and the last numbers the average will be done by taking the average of the number with next below and the one above respectively and divide by 2. I want to write a script that can do this for this data and can also be used for large data (say 1000x1 vector). Thanks in advance

Accepted Answer

Mohammad Abouali
Mohammad Abouali on 8 Oct 2014
vector=rand(1000,1);
halfWindow=1; % 1 since you said one below and one above. You can change this
s=ones(2*halfWindow+1,1)/(2*halfWindow+1);
vector_average=conv(vector,s,'valid');
vector_average=[sum(vector(1:1+halfWindow,1))./(halfWindow+1); ...
vector_average(:); ...
sum(vector(end-halfWindow:end,1))./(halfWindow+1)];
  7 Comments
Mohammad Abouali
Mohammad Abouali on 9 Oct 2014
Edited: Mohammad Abouali on 9 Oct 2014
if we use the 'same' in convolution depending on the window size couple items in the beginning and in the end are going to be averaged with zero.
Let's say you have temperature
300, 301, 302, ...
If the halfWindow is 1 then the first entry would be calculated like this
[0 300 301].*[1/3 1/3 1/3]=200.333
So the average goes down artificially.
One approach is to switch the window size at the end zones (like what we did in the above code) another approach is to use padarray function to kinda replicate the boundary nodes. The second one still can be questionable. (Depends on the problem and boundary conditions)
dpb
dpb on 9 Oct 2014
Yabbut... :) I just suggested to OP he investigate 'cuz likelihood is it won't make a lick of difference on what those 2-5 or so BC points are in a series that is several K in length. And, I suggested he just throw perhaps half of those away to minimize the amount of actual zero padding that does occur.
Of course, he can just use the 'valid' option and not have any end effects.... :)

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More Answers (3)

Stephen23
Stephen23 on 8 Oct 2014
Edited: Stephen23 on 8 Oct 2014
You could try convolution :
>> A = 1:8;
>> [mean(A(1:2)),conv(A,[1,1,1],'valid')/3,mean(A(end-1:end))]
The end conditions require some special handling (and is of dubious mathematical value), so the simpler solution is to only have an output where three values are averaged:
>> conv(A,[1,1,1],'valid')/3
Another advantage of this is that the length of your moving window can be adjustable (not hardcoded):
>> N = 4;
>> conv(A,ones(1,N),'valid')/N
  2 Comments
Stephen23
Stephen23 on 8 Oct 2014
Edited: Stephen23 on 8 Oct 2014
Mayowa: a convolution is like a "moving window" multiply.
You can think of it like two vectors that are aligned together at one end, the corresponding elements are multiplied, and then the vectors shifted by one position relative to each other, and the process repeats until they are aligned at the other end. Dividing the resulting products by the length of the shorter "window" vector gives you the means of the original values of the longer vector.
The end values (which you defined as special cases) are calculated separately using the simple mean function.
Mohammad Abouali
Mohammad Abouali on 9 Oct 2014
except that you have to note that in convolution the second vector is flipped. for symmetric vectors (like in this case) it doesn't matter but if you are using non-symmetric kernel then that matters.
The one that does not flips the kernel is the cross correlation.

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Guillaume
Guillaume on 8 Oct 2014
What you're asking, except for the edges, is just a simple convolution by the vector [1/3 1/3 1/3]:
a = randi(100, 1, 20) %for example
b = conv(a, [1/3 1/3 1/3], 'same');
%now you can adjust for the edges:
b(1) = (a(1) + a(2)) / 2;
b(end) = (b(end) + b(end-1)) / 2
  1 Comment
Mayowa
Mayowa on 8 Oct 2014
Thanks for the answer. In a situation where instead of 3, I have 9, that implies I have to adjust the first and last 4 elements without repeating the last 2 lines 4 times?

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dpb
dpb on 8 Oct 2014
Edited: dpb on 8 Oct 2014
>> [mean(v(1:2)) conv(v,[1 1 1]/3,'valid') mean(v(end-1:end))]
ans =
1.5000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 7.5000
More general for N points is
a=[mean(v(1:N-1)) conv(v,ones(1,N)/N,'valid') mean(v(end-(N-1):end))];
  2 Comments
Stephen23
Stephen23 on 8 Oct 2014
Edited: Stephen23 on 8 Oct 2014
This generates an error "Error: Unexpected MATLAB expression.", pointing at that space character. An extra bracket behind after the mean is required.
Although this still misses the requirement "...for the first and the last numbers the average will be..."
dpb
dpb on 8 Oct 2014
Missed the closing parens on the mean() -- the extension for end effect ought to be self-evident altho didn't write it, granted.
Edited Answer to correct oversight and typo...

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