Azzera filtri
Azzera filtri

find first line that satisfies condition; skip next few; then find next line that satisfies condition

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I have a long column of #'s. I want to find and replace values >4.8 with 10000 so that later code can find these spots. This works ok:
stim_time = find(TTL_column>4.8);
TTL_column(stim_time) = 10000;
However, there are ~3-6 points all grouped together that have almost the same time stamp (e.g., 15870,15871,15872,15873). I only want to replace the first one of these with the #10000 (e.g., only the value at 15870 = 10000). I am totally lost on how I'd get this code to replace only the first spot in each group and not the whole group, since each group has an unequal number of points. Any help would be greatly appreciated!
stim_time =
15870
15871
15872
15873
49863
49864
49865
49866
49867

Risposta accettata

Matt J
Matt J il 24 Mar 2019
Modificato: Matt J il 24 Mar 2019
idx = diff([0;TTL_column(:)>4.8])>0;
TTL_column(idx) = 10000;
  3 Commenti
Kathleen Hupfeld
Kathleen Hupfeld il 25 Mar 2019
or a way to count how many substitutions were made that's more efficient than this? :)
total = 0;
for i = 1 : length(idx)
if idx(i)==1
total=total+1;
end
end
total

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