How do I take average of specific portions of data with overlap?

Question

Anthony 2017-1-23

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https://ww2.mathworks.cn/matlabcentral/answers/321640-how-do-i-take-average-of-specific-portions-of-data-with-overlap

评论： Anthony 2017-1-25

采纳的回答： Kelly Kearney

Flex1.xls

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Hi all,

I have a question regarding the EMG data I am working on. I have attached a file containing a sample data Excel sheet. What I need to do is compute the average of several "windows" for each data set, while also accounting for overlapped sections.For example, Window Size 1 = 1500 points & Overlap Size 1=750 points I then need the mean of each window as follows: [From 1 to 1500] then [750-2250] then [1500 – 3000] etc.

What is the best way to do this? I'm thinking some kind of for loop, but I don't know how exactly to account for overlap. Thank you so much in advance!

Below is the code I have right now to accomplish this. I need a more efficient way to implement this logic into my data set so that I can repeat it easily for various window sizes and overlaps

.
clc;
clear;
[d]=xlsread('Flex1.xls', 'Flex1', 'D5:D10000');
[t]=xlsread('Flex1.xls', 'Flex1', 'A5:A10000');
mean_d_1_1 = mean(d(1:1500))
mean_d_1_2 = mean(d(750:2250))
mean_d_1_3 = mean(d(1500:3000))
mean_d_1_4 = mean(d(2250:3750))
mean_d_1_5 = mean(d(3000:4500))
mean_d_1_6 = mean(d(3750:5250))
mean_d_1_7 = mean(d(4500:6000))
mean_d_1_8 = mean(d(5250:6750))
mean_d_1_9 = mean(d(6000:7500))
mean_d_1_10 = mean(d(6750:8250))
mean_d_1_11 = mean(d(7500:8694))
  % code
end

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Answer 1

Kelly Kearney 2017-1-23

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https://ww2.mathworks.cn/matlabcentral/answers/321640-how-do-i-take-average-of-specific-portions-of-data-with-overlap#answer_251721

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First comment: don't store values in variables with numbered names like that. It just makes a mess, and that's why arrays were invented.

So instead of mean_d_1_1, mean_d_1_2, etc., you can create a single array named mean_d and access specific elements: mean_d(1,1), mean_d(1,2), etc.

As for your loop, just look at your code and see which pieces are incrementing...

the first index (1, 750, 1500, 2250)
the second index (1500, 2250, 3000)

So:

d = rand(10000,1);
nslide = 750;
nwin = 1500;
n = ceil(length(d)/nslide);
mean_d = zeros(n,1);
for ii = 1:n
    idx1 = nslide*(ii-1) + 1;
    idx2 = min(nslide*ii,length(d));
    mean_d(ii) = mean(d(idx1:idx2));
end

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Kelly Kearney 2017-1-24

编辑：Kelly Kearney 2017-1-24

在 MATLAB Online 中打开

You're right, I wrote that out too quickly and used the wrong equation for idx2. Easy way to check this is to do some simple debugging; for example, print out the values of idx1 and idx2:

n = ceil(length(d)/nslide);
mean_d = zeros(n,1);
for ii = 1:n
    idx1 = nslide*(ii-1) + 1;
    idx2 = min(nslide*ii,length(d));
    mean_d(ii) = mean(d(idx1:idx2));
    fprintf('%d %d\n', idx1, idx2);
end

Results:

Oops, that's not right. I was using the slide size and not the window size. Let's try again:

for ii = 1:n
    idx1 = nslide*(ii-1) + 1;
    idx2 = min(idx1 + nwin - 1, length(d));
    mean_d(ii) = mean(d(idx1:idx2));
    fprintf('%d %d\n', idx1, idx2);
end

Results:

Much better. Though note that it's still not quite identical to the way you typed it out in your example... in that one, your first window was 1500 long and all the rest were 1501. If you really want it that way, you'll need to adjust the idx1 equation accordingly.

Anthony 2017-1-25