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Find indices of elements for given difference

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Hi,
I have an incrementing time vector from 0 to 500 ms . Increment in time is not constant. I want to find indices every ~10 ms . E.g
t=[0, 1 ,3,4,7,10,13,15,16,19,20, 23,25,27,31...........500ms];
Then I would like to find indices of 10,20,31 ...., that will be 6th, 11th,15th.
Is this possible without loop.
thanks
jayant

  1 Comment

SilverSurfer
SilverSurfer on 19 Jul 2020
If you know in advance which numbers you need to identify you can use find function.
Here there is a suggestion for finding multiple elements.
t=[0,1,3,4,7,10,13,15,16,19,20,23,25,27,31];
num = [10,20,31];
c = ismember(t,num);
indexes = find(c);

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Accepted Answer

madhan ravi
madhan ravi on 19 Jul 2020
Edited: madhan ravi on 19 Jul 2020
Nearest element after or equal to the boundary:
Dt = t - (10:10:max(t)).';
Dt(Dt<0) = inf;
[~, Indices] = min(Dt,[],2)
Wanted = t(Indices)
Nearest elements before or equal it crosses boundary:
Dt = t - (10:10:max(t)).';
Dt(Dt>0) = -inf;
[~, Indices] = max(Dt,[],2)
Wanted = t(Indices)

  1 Comment

madhan ravi
madhan ravi on 19 Jul 2020
Use
Dt = bsxfun(@minus, t, (10:10max(t)).') % if you’re using version prior to 2016b

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More Answers (3)

Bruno Luong
Bruno Luong on 19 Jul 2020
Edited: Bruno Luong on 19 Jul 2020
i = interp1(t, 1:length(t), 0:10:max(t), 'nearest', 'extrap');

  9 Comments

Show 6 older comments
Bruno Luong
Bruno Luong on 19 Jul 2020
Here is the evidence
>> sum('Barney:')==sum('the god')
ans =
logical
1
madhan ravi
madhan ravi on 19 Jul 2020
😂 , a good sense of humour after all.

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dpb
dpb on 19 Jul 2020
find and/or ismember will only return EXACT matches -- will NOT return something "on or about" a 10 ms interval.
Two possibilities come to mind
  1. ismembertol to find within some defined tolerance about the target, or
  2. interp1 with 'nearest' option
The second will return something for every input in range; the first may not find something if the spacing is such there isn't one within the given tolerance--or could potentially return more than one if the tolerance is too large.

  1 Comment

dpb
dpb on 19 Jul 2020
Possibly simply because 0 being first element wasn't hard to find... :)

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Image Analyst
Image Analyst on 19 Jul 2020
Here's one way to record the index and time of when the times first cross "10" boundaries:
t = sort(randperm(500, 200)) % Sample data
times = [0,0];
counter = 1;
for k = 0 : 10 : max(t)
index = find(t >= k, 1, 'first'); % Find where it crosses multiple of 10 for the first time.
if ~isempty(index)
times(counter, 1) = index; % Log index
times(counter, 2) = t(index); % Log the actual time.
counter = counter + 1;
end
end
times % Show in command window.

  1 Comment

Bruno Luong
Bruno Luong on 19 Jul 2020
The same can be achieved without for-loop by using INTERP1 with 'NEXT' method in recent MATLAB realeases (just change 'nearest' in my anser to 'next'), or a combo of HISTC/ACCUMARRAY on older MATLAB.

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