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 comentario

Adam
Adam el 5 de Oct. de 2016
Editada: Adam el 5 de Oct. de 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;

Respuestas (1)

Marc Jakobi
Marc Jakobi el 5 de Oct. de 2016
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 comentarios

Jordan Gallacher
Jordan Gallacher el 5 de Oct. de 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 el 5 de Oct. de 2016
Editada: Marc Jakobi el 5 de Oct. de 2016
Then I would I would replace
idx = find(ismember(t, 0));
with
idx = find([0, diff(t)] <= 0);

La pregunta está cerrada.

Preguntada:

el 4 de Oct. de 2016

Cerrada:

el 20 de Ag. de 2021

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