concatonate time axis using a loop

I have a time axis which keeps resetting due to drop outs in the logging e.g.
t = 0,1,2,3,4,5,0,1,2,3,4,5,6,7,8,9,10,0,1,2,3,4,5,6,7,8,9,10,0,1,2 and so on....
What is the most efficient piece of code to generate the new time vector so that the zeros continue on from the last time value before the dropout.
t = 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30
Thanks!
Jordan.

1 Kommentar

Adam
Adam am 5 Okt. 2016
Bearbeitet: Adam am 5 Okt. 2016
I assume your real problem is a little more complex than the example you posted because to get that time vector you can just do
t = 0:numel(t) - 1;

Antworten (1)

Marc Jakobi
Marc Jakobi am 5 Okt. 2016

0 Stimmen

This should do it:
t = [0,1,2,3,4,5,0,1,2,3,4,5,6,7,8,9,10,0,1,2,3,4,5,6,7,8,9,10,0,1,2];
idx = find(ismember(t, 0));
for i = 2:length(idx)-1
t(idx(i):idx(i+1) - 1) = t(idx(i):idx(i+1) - 1) + idx(i) - 1;
end
t(idx(end):end) = t(idx(end):end) + idx(end) - 1;

2 Kommentare

Jordan Gallacher
Jordan Gallacher am 5 Okt. 2016
Hi Marc,
Thanks! What about if the zero was not exactly zero and say 4 microseconds i.e. just a value less than the last time value before the logging reset?
Thanks,
Jordan.
Marc Jakobi
Marc Jakobi am 5 Okt. 2016
Bearbeitet: Marc Jakobi am 5 Okt. 2016
Then I would I would replace
idx = find(ismember(t, 0));
with
idx = find([0, diff(t)] <= 0);

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am 4 Okt. 2016

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am 20 Aug. 2021

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