Unable to perform assignment because the left and right sides have a different number of elements.

Hey everyone,

I have some acceleration data of experiments saved as a vector acc_g.

1) I now want to count the number of values in acc_g that are above +/- 1.5 g acceleration, which gives the variable k.

2) Then I want to determine these values, which gives acc_peak.

3) And at last, I want to display the row number of these maximum values and save them in the vector times_peak_acc. So I compare the acc_peak with the entire acceleration vector acc_g.

I get an error of "Unable to perform assignment because the left and right sides have a different number of elements." for step 3. However, the code works when k < 32??? What do I do wrong?

The length of the acc_g vector is 16342 x 1, so it is not smaller than 32...

Here is my code:

And thanks a lot in advance!

1)

k = 0;

for i = 1:length(acc_g)

if abs(acc_g(i,4)) >= 1.5

k = k+1;

end

end

2)

acc_peak = maxk(acc_g,k,'ComparisonMethod','abs');

3)

for n = 1:k

times_acc_peak(n) = find(acc_g==acc_peak(n));

end

Answer 1

[Star Strider]

We will likely need to have the ‘acc_g’ matrix. (It can’t be a vector, since the reference to it in the first loop is to column 4.)

This runs without error —

acc_g = randn(1000,4);                                         % Create Missing Data

% 1)
k = 0;
for i = 1:length(acc_g)
    if abs(acc_g(i,4)) >= 1.5
        k = k+1;
    end
end
k                                                               % Display Result

k = 123

% 2)
acc_peak = maxk(acc_g,k,'ComparisonMethod','abs');
% 3)
for n = 1:k
    times_acc_peak(n) = find(acc_g==acc_peak(n));
end

times_acc_peak                                                  % Display Result

times_acc_peak = 1×123

549 988 229 946 146 254 945 350 671 571 682 714 475 314 238 970 226 716 934 125 932 550 165 753 438 628 297 806 748 500

Note that ‘acc_peak’ is a (kx4) matrix, so that could be a problem, however it wasn’t one here.

.

[Lisa Oswald]

Oh yes sorry, acc_g is a 16342 x 4 matrix, but I only need column 4 for the entire code. I deleted the reference to column 4 at some points (just to post the problem here) but forgot to delete it in the first step.

So I created a new variable acc_a4 now, which is the vector acc_g(:,4)

But the code still did not work...

acc_a4 = acc_g(:,4);

% count acceleration peaks over +/- 1.5 g
k_acc = 0;
for i = 1:length(acc_a4)
    if abs(acc_a4(i)) >= 1.5
        k_acc = k_acc+1;
    end
end

% find k maximum values of acceleration
acc_peak = maxk(acc_a4,k_acc,'ComparisonMethod','abs');
for n = 1:k_acc
    times_acc_peak(n) = find(acc_a4==acc_peak(n));
end

same error in the for loop...

[Star Strider]

As I mentioned in my Answer, it will likely be necessary to provide the file, preferably as a spreadsheet or text file.

If the code worked with my synthetic data and not with the actual data, that points to the actual data being the problem.

.

[Lisa Oswald]

here you go!

I just found out: that it stops the loop because at this iteration, there are 2 positions (6577 and 7100) for this value. so it can't save both of them. Do you know how to solve this problem? If I change the value to another value, the loop continues...

thanks a lot in advance!

[Star Strider]

My pleasure!

The problem is caused because in certain instances the find call is returning more than one result, so a reference to a scalar value ‘times_acc_peak(n)’ overwrote the destination element. That threw the error.

That occurred for the first time at ‘n=5’ and stopped the code (so there may be others). One way to deal with that is to use a cell array and then deal with a cell array of numeric indices later, however the best approach is to avoid the loop altogether and use the ismember function —

LD = load('acc_g.mat');
acc_g = LD.acc_g;

acc_a4 = acc_g(:,4);

% count acceleration peaks over +/- 1.5 g
k_acc = 0;
for i = 1:length(acc_a4)
    if abs(acc_a4(i)) >= 1.5
        k_acc = k_acc+1;
    end
end

% find k maximum values of acceleration
acc_peak = maxk(acc_a4,k_acc,'ComparisonMethod','abs');

Lv = ismember(acc_a4, acc_peak);                            % Logical Vector
times_acc_peak = find(Lv)                                   % Numeric Indices
CorrespondingMatrix = acc_g(Lv,:);                          % Array Corresponding To Rows Identified By 'ismember'

I am not certain what the intent is for the results of this code, so I also created the ‘CorrespondingMatrix’ array that returns all the columns of ‘acc_g’ that are indexed by the ‘Lv’ rows. If that is not necessary, just delete that assignment, however I assume the indices are going to be used elsewhere in the code, possible for this, and so decided to do it here.

.

[Lisa Oswald]

Wow thank you so much for your kind help! Now it works all fine :D

You made my day!!

[Star Strider]

As always, my pleasure!

Thank you!

.