Iterating using a specific interval, to intermittently average data in a scatter plot.

Trying to iterate through two large sets of row data (in two columns, data.x and data.y) simultaneously, with the aim of averaging the data within these intervals so I can visualise a clearer scatter plot (see original plot below). I have provided a simplified idea of what I need in code below, However, I am running into the issue of the index not being an integer.
In the code below, I am trying to return the three averages of [1,2,3], [4,5,6] and [7,8,9] in each xmeans and ymeans array.
Would appreciate any tips on the best way to do this.
% test data
x = [1;2;3;4;5;6;7;8;9];
y = [1;2;3;4;5;6;7;8;9];
% means of storage of averages at each interval
xmeans = [];
ymeans = [];
% calculate uniform interval
distance_range = length(x);
interval = distance_range/3;
s = min(x);
while (s >= min(x)) && (s <= (max(x)+1))
for k = s : (s+interval)
xmeans(s) = mean(x(k:(k+interval)-1));
ymeans(s) = mean(y(k:(k+interval)-1));
end
s = s+interval
end

 Accepted Answer

When s >= 7 the code overwrites the array.
The indices are otherwise all integers greater than 0, so the problem with them not being integers doesn’t appear.

10 Comments

Yes, I see that. So my question now is how do I return correct averages for [1,2,3], [4,5,6] and [7,8,9] for the x and y data.
This code fills both the xmeans and ymeans array with [5,0,0,8,0,0,8], and I am unsure as to why.
You need to set the appropriate limits on the subscripts.
However there may be easier ways to do what you want, specifically the movmean function (introduced in R2016a).
If you want to take only the means of specific blocks (for example blocks of 3) of the vectors, one way is:
x = [1;2;3;4;5;6;7;8;9];
y = [1;2;3;4;5;6;7;8;9];
xmeans = mean(reshape(x, 3, []));
ymeans = mean(reshape(y, 3, []));
producing:
xmeans =
2 5 8
ymeans =
2 5 8
.
I think this will do the trick! I basically want to segment both axis for averaging, just like this!
Thank you very much for your help and quick responses.
My pleasure!
If my Answer helped you solve your problem, please Accept it!
.
Hi, this should be marked as an accepted answer, but I am not being given the option to accept it for some reason!
Thank you for the help.
As always, my pleasure!
I will notify MathWorks about the problem.
(Answers Dev) I have figured out the issue and emailed Robyn how to accept the answer.
Accepted the answer. Thanks for the help everyone!
Robyn Seery — Thank you!
Rena Berman — Thank you for helping with this!

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