How to create an if loop with conditions

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GB92_R
GB92_R 2022년 8월 28일
댓글: GB92_R 2022년 8월 28일
Hi,
in my Time analysis i need to divide my vector 'pupil diameter' in 3 moments and for each one i have to calculate the median
and i did it in this way
%% TIME SPAN ANALYSIS
time_1 = median(diametre(1:33000));
time_2 = median(diametre(33001:66000));
time_3 = median(diametre(66001:99000));
but not all my subject have vector length which reaches 99000 timestamps so i created a loop where i wrote:
for i=1:length(list)
time_1 = median(diametre(1:33000));
time_2 = median(diametre(33001:66000));
if diametre > 66000
time_3 = median(diametre(66001:99000));
end
list is the list of my participant. When it comes to calculate time_3 then it tell me that the matrix exceed the dimension so it means that the loop doesnt work despite i check the length of the signal in all my participant before choosing where to cut
can you please help me write it correctly?
thank you very much in advance

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Walter Roberson
Walter Roberson 2022년 8월 28일
if numel(diametre) > 66000
  댓글 수: 5
Walter Roberson
Walter Roberson 2022년 8월 28일
for i=1:length(list)
What is list ?
time_1 = median(diametre(1:33000));
You do not appear to be using the value of i so it is not clear why you are looping there.
If list were empty, then after the for loop, time_1 would be undefined.
Be careful, when diametre is small, time_3 would be undefined.
GB92_R
GB92_R 2022년 8월 28일
yes, thank you
it works !

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