Iterating over an array to produce an array of calculated variances based on the next 9 entries and previous 9 entries

Code

% Extract x coordinates from 1st column of h_coordinates
x_coordinates = h_coordinates(:,1);

    for i=1:length(x_coordinates);
        if(i <= (length(x_coordinates)-9))
       var_x(:,i) = x_coordinates(i) - mean(x_coordinates(i:i+9));
        else
       var_x2(:,i) = x_coordinates(i) - mean(x_coordinates(i:i-9)); 
        end
    end
    
    finalvar_x = horzcat(var_x,var_x2);

Description

I am examining variation in x_coordinates and want to calculate the variation of every entry compared to the 9 entries in front of it, until I get to the last 9 entries in the array then I want to calculate variation compared to the previous 9 entries.

The method I am trying above is not working for me, Var_x2 is only composed of 0s and Nans.

Thanks

[Walter Roberson]

Suppose that you have reached the 4th last entry in the array. Which values do you want to take the mean of in order to do the calculation for that output?

[James Morris]

Sorry I forgot to include that information. For the last 9 values in the array I would like to take the mean of the 9 previous entries.

[James Morris]

Just tried this but it didn't work. Var_x2 is only composed of 0s and Nans.

Code:

for i=1:length(x_coordinates);

if(i <= (length(x_coordinates)-9))

var_x(:,i) = x_coordinates(i) - mean(x_coordinates(i:i+9));

else

var_x2(:,i) = x_coordinates(i) - mean(x_coordinates(i:i-9));

end

end

finalvar_x = horzcat(var_x,var_x2);

[Walter Roberson]

The mean of the previous 10 values has already been calculated and can just be recalled: when you were at K you took the mean of K:K+9 and when you reach K+9 and need to calculate backwards, (K+9)-9:(K+9) is same location as before and therefore you can just pull out var_x(:, k-9)

[James Morris]

I get what you're saying about the 10 previous means being calculated, but var_x for the 10 previous entries would involve the 10 previous x_coordinate values in the calculation and not the last 10 that I want to iterate through wouldn't it?

[James Morris]

Plus I haven't created a variable to store the mean so all the calculated means for every iteration wouldn't be stored would they?

[James Morris]

Thanks for all your help. I found some user created moving deviation and moving variance functions that are perfect for me!

[Walter Roberson]

movmean is part of MATLAB these days.

Answer 1

[Image Analyst]

Replace

for i=1:length(x_coordinates);

with

for i = 1 : length(x_coordinates) - 9

[James Morris]

Thank you so much, but the nature of my question has changed. I was omitting some important information as to what I required.

[Image Analyst]

The last code I saw, up above in the comments section you had

for i=1:length(x_coordinates);

and within the loop you had

var_x(:,i) = x_coordinates(i) - mean(x_coordinates(i:i+9));

so let's examine what would happen if i was length(x_coordinates)

mean(mean(x_coordinates(i:i+9)) is equal to

mean(x_coordinates(length(x_coordinates) : length(x_coordinates)+9))

so you're going from the last element of x_coordinates to 9 elements PAST the end of x_coordinates. Why are you doing that?

[James Morris]

Sorry yes that was a mistake, for the last 9 elements of x_coordinates I want to calculate variance based on the mean of the previous 9 elements.

So while i = (length(x_coordinates) - 9) I want an array to be filled with variances (x_coordinates(i) - mean(x_coordinates(i + 9))).

but when i >= (length(x_coordinates) - 9) I want the array to be filled with variances (x_coordinates(i) - mean(x_coordinates(i - 9))).

I hope this makes sense.

[Image Analyst]

Why not simply use movstd() or movvar()?

[James Morris]

Just researched these and they seem perfect for what I want, but I am using MATLAB R2015b due to working on a project that was developed on the 2015 version and has dependencies. I'm seeing if there are alternative built in functions.

[James Morris]

Thanks for all your help. I found some user created moving deviation and moving variance functions that are perfect for me!