Find indices of elements for given difference

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

[SilverSurfer]

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);

Answer 1

[madhan ravi]

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)

[madhan ravi]

Use

 Dt = bsxfun(@minus, t, (10:10max(t)).') % if you’re using version prior to 2016b

Answer 2

[Bruno Luong]

i = interp1(t, 1:length(t), 0:10:max(t), 'nearest', 'extrap');

[madhan ravi]

It’s wrong. This gives index one as well.

[Bruno Luong]

So why 1 is wrong? (assuming t starts from 0 as stated by OP)

[madhan ravi]

“I want to find indices every ~10 ms”

[Bruno Luong]

Every ~10 ms meaning (in english) is that the difference between 2 consecutive samples is about 10 ms, NOT necessary it must start at 10 ms.

[madhan ravi]

Thank you for the “English lessons” ;) I wonder why the OP didn’t include it in the expected result then xD.

[Bruno Luong]

We do not need OP to confirm whereas t=0 must be considered, since

>> why(0)
Barney wanted it that way.

Who are we to argue with Barney? Clear now?

[madhan ravi]

Who’s Barney? A God xD?

[Bruno Luong]

Here is the evidence

>> sum('Barney:')==sum('the god')

ans =

  logical

   1

[madhan ravi]

😂 , a good sense of humour after all.

Answer 3

[dpb]

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.

[dpb]

Possibly simply because 0 being first element wasn't hard to find... :)

Answer 4

[Image Analyst]

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.

[Bruno Luong]

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.