How to create an if loop with conditions

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GB92_R
GB92_R am 28 Aug. 2022
Kommentiert: GB92_R am 28 Aug. 2022
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

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 28 Aug. 2022
if numel(diametre) > 66000
  5 Kommentare
Walter Roberson
Walter Roberson am 28 Aug. 2022
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 am 28 Aug. 2022
yes, thank you
it works !

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