Find indices of elements for given difference

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Jayant chouragade
Jayant chouragade 2020-7-19
评论： madhan ravi 2020-7-19
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
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SilverSurfer
SilverSurfer 2020-7-19
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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采纳的回答

madhan ravi
madhan ravi 2020-7-19
编辑：madhan ravi 2020-7-19
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)

更多回答（3 个）

Bruno Luong
Bruno Luong 2020-7-19
编辑：Bruno Luong 2020-7-19
i = interp1(t, 1:length(t), 0:10:max(t), 'nearest', 'extrap');
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显示 7更早的评论
Bruno Luong
Bruno Luong 2020-7-19
Here is the evidence
>> sum('Barney:')==sum('the god')
ans =
logical
1

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dpb
dpb 2020-7-19
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.
Image Analyst
Image Analyst 2020-7-19
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 个评论
Bruno Luong
Bruno Luong 2020-7-19
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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