concatonate time axis using a loop
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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 件のコメント
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;
回答 (1 件)
Marc Jakobi
2016 年 10 月 5 日
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 件のコメント
Jordan Gallacher
2016 年 10 月 5 日
Marc Jakobi
2016 年 10 月 5 日
編集済み: Marc Jakobi
2016 年 10 月 5 日
Then I would I would replace
idx = find(ismember(t, 0));
with
idx = find([0, diff(t)] <= 0);
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