How to find if a cell reaches an average value in a specific amount of instances.

I have a cell array thats made up of different vectors. What I want to do is eveluate each vector in the cell to determine if at any point in the vector a mean of a say 10 entries falls within +or - 10 of a value?

[C B]

can you Please eloobrate what ipnut and your expected output?

like this

input should be

 A= {[11 23 44 99] [111 223 434 939] [1111 2223 433 9339]}

[Jan]

I do not undestand this part: "determine if at any point in the vector a mean of a say 10 entries falls within +or - 10 of a value".

Answer 1

[Image Analyst]

Try this:

A= {[11 23 44 99] [111 223 434 939] [1111 2223 433 9339]}
targetValue = 426; % Whatever.
for k = 1 : length(A)
    thisVector = A{k};
    fprintf('\nVector #%d is', k); disp(thisVector);
    thisMean = mean(thisVector);
    fprintf('   The mean for vector #%d is %f.\n', k, thisMean);
    difference = abs(thisMean - targetValue);
    percentDifference = 100 * difference / targetValue;
    fprintf('   The difference of the mean from the target value of %.1f is %.2f = %.2f%%.\n', targetValue, difference, percentDifference);
    if percentDifference < 10
        % It's within 10% of the target value.
        fprintf('Yay!!!!  Cell #%d has a mean value that is within 10 percent of %.2f.\n', k, targetValue)
    end
end

You'll see

Vector #1 is 11 23 44 99

The mean for vector #1 is 44.250000.

The difference of the mean from the target value of 426.0 is 381.75 = 89.61%.

Vector #2 is 111 223 434 939

The mean for vector #2 is 426.750000.

The difference of the mean from the target value of 426.0 is 0.75 = 0.18%.

Yay!!!! Cell #2 has a mean value that is within 10 percent of 426.00.

Vector #3 is 1111 2223 433 9339

The mean for vector #3 is 3276.500000.

The difference of the mean from the target value of 426.0 is 2850.50 = 669.13%.

Obviously the code could be shortened a lot by taking out the informative messages.

[Brooks Lane]

This worked great! Thank you for the help!