How to select a time from one array which is close to the time in another array ?

I have 2 time arrays (input)

A = [{'05:10:34'} {'05:42:52'} {'06:52:17'} {'07:34:36'} {'07:44:37'} {'08:44:35'}]

B = [{'05:07:46'} {'05:14:43'} {'05:18:25'} {'05:40:04'} {'05:46:02'} {'06:00:32'} {'06:29:36'} {'06:49:31'} {'07:31:48'} {'07:41:49'} {'07:47:48'}]

need to select time of B which is closest to time in A (but time difference between A and B should not be more than 15 min).

and the time in B that has least time difference from A should be the desired output as below:

c = [{'05:07:46'} {'05:40:04'} {'06:49:31'} {'07:31:48'} {'07:41:49'}]

Answer 1

[Jan]

Prefer to use modern datetime objects instead of cellstrings containing datestr values:

A = [{'05:10:34'} {'05:42:52'} {'06:52:17'} {'07:34:36'} {'07:44:37'} {'08:44:35'}]; 
B = [{'05:07:46'} {'05:14:43'} {'05:18:25'} {'05:40:04'} {'05:46:02'} {'06:00:32'}, ...
     {'06:29:36'} {'06:49:31'} {'07:31:48'} {'07:41:49'} {'07:47:48'}];

a = datetime(A);
b = datetime(B);
c = NaT(size(a));
for k = 1:numel(A)
    [dist, index] = min(abs(a(k) - b));
    if minutes(dist) < 15
        c(k) = b(index);
    end
end

If you really need the olde datestr elements:

a = datenum(A);
b = datenum(B);
c = cell(size(a));
for k = 1:numel(A)
    [dist, index] = min(abs(a(k) - b));
    if dist < 1 / 96  % Explicitly: 15*60/86400
        c{k} = B{index};
    end
end

[Feral]

Thanks ! the code is perfectly fine , but how to find the indices of A when the time is <15 min.

[Jan]

@Feral: The code does find the indices already, but does not store it. If you want to collect the indices also, simply add:

iA = nan(size(a));
for k = 1:numel(A)
    [dist, index] = min(abs(a(k) - b));
    if minutes(dist) < 15
        c(k)  = b(index);
        iA(k) = index;
    end
end

[Feral]

@Jan: I did the same before asking the previous question but the code gives me the index of B i.e [1,4,8,9,10,NaN] what I want is index of A which would be [1,2,3,4,5,NaN].

[Jan]

@Freal: Ah, I understand. Then replace:

iA(k) = index;

by

iA(k) = k;

It is not hard, isn't it? The loop runs over the elements of A, so the loop index is the wanted index relalted to A also.

[Feral]

Yes! exaclty it. Thanks alot.

Answer 2

[Peter Perkins]

N loops needed for this. Use durations amd 'nearest' interpolation. Actually, this problem is probably ill-posed, because if these are times of day, they'd better be within the same day. If they are not, you are in troubhle. If they are, try this:

>> A = [{'05:10:34'} {'05:42:52'} {'06:52:17'} {'07:34:36'} {'07:44:37'} {'08:44:35'}];
>> B = [{'05:07:46'} {'05:14:43'} {'05:18:25'} {'05:40:04'} {'05:46:02'} {'06:00:32'} {'06:29:36'} {'06:49:31'} {'07:31:48'} {'07:41:49'} {'07:47:48'}];
>> A = duration(A)
A = 
  1×6 duration array
   05:10:34   05:42:52   06:52:17   07:34:36   07:44:37   08:44:35
>> B = duration(B)
B = 
  1×11 duration array
   05:07:46   05:14:43   05:18:25   05:40:04   05:46:02   06:00:32   06:29:36   06:49:31   07:31:48   07:41:49   07:47:48
>> B2 = [seconds(0) B seconds(86400)] % make extrapolation work
B2 = 
  1×13 duration array
Columns 1 through 11
       0 sec   18466 sec   18883 sec   19105 sec   20404 sec   20762 sec   21632 sec   23376 sec   24571 sec   27108 sec   27709 sec
Columns 12 through 13
   28068 sec   86400 sec
>> closestBIndex = interp1(B2,[1 1:length(B) length(B)],A,'nearest')
closestBIndex =
     1     4     8     9    10    11
>> closestBTime = B(closestBIndex)
closestBTime = 
  1×6 duration array
   05:07:46   05:40:04   06:49:31   07:31:48   07:41:49   07:47:48
>> diffToClosestBTime = closestBTime - A
diffToClosestBTime = 
  1×6 duration array
   -00:02:48   -00:02:48   -00:02:46   -00:02:48   -00:02:48   -00:56:47
>> closestBIndexWithin15 = abs(diffToClosestBTime) < minutes(15)
closestBIndexWithin15 =
  1×6 logical array
   1   1   1   1   1   0
>> B(closestBIndexWithin15)
ans = 
  1×5 duration array
   05:07:46   05:14:43   05:18:25   05:40:04   05:46:02