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 Commento

Adam
Adam il 5 Ott 2016
Modificato: Adam il 5 Ott 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;

Risposte (1)

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 Commenti

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 il 5 Ott 2016
Modificato: Marc Jakobi il 5 Ott 2016
Then I would I would replace
idx = find(ismember(t, 0));
with
idx = find([0, diff(t)] <= 0);

Questa domanda è chiusa.

Chiuso:

il 20 Ago 2021

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