A very fast way to (i) concatenate and (ii) calculate the difference of elements?

A very fast way to (i) concatenate and (ii) calculate the difference of elements ?

Here below my case / example:

a1 = datetime([
   '2022-10-27 00:22:50.000'
   '2022-10-27 05:29:45.000'
   '2022-10-27 05:32:19.400'
   '2022-10-27 05:36:44.100'
   '2022-10-27 05:39:26.600'
   '2022-10-27 05:43:18.200'
   '2022-10-27 05:49:17.400'
   '2022-10-27 05:55:27.300'
   '2022-10-27 05:58:07.000'
   '2022-10-27 06:17:13.800'
   '2022-10-27 06:41:28.700'
   '2022-10-27 07:03:06.000'
   '2022-10-27 07:09:49.800'
   '2022-10-27 07:17:39.700'
   '2022-10-27 07:35:09.000'
   '2022-10-27 07:42:33.600'
   '2022-10-27 07:46:50.500'
   '2022-10-27 08:07:02.700'
   '2022-10-27 08:26:29.600'
   '2022-10-27 08:45:03.500'
   '2022-10-27 08:48:09.700'
   '2022-10-27 08:53:57.000'
   '2022-10-27 08:59:43.400'
   '2022-10-27 09:13:24.100'
   '2022-10-27 09:19:25.000'
   '2022-10-27 09:26:35.000'
   '2022-10-27 09:29:54.000'
   '2022-10-27 09:46:45.700'
   '2022-10-27 10:10:59.900'
   '2022-10-27 10:29:04.600'
   '2022-10-27 10:31:50.100'
   '2022-10-27 10:37:45.300'
   '2022-10-27 10:42:35.800'
   '2022-10-27 10:58:42.300'
   '2022-10-27 11:03:48.900'
   '2022-10-27 11:10:44.700'
   '2022-10-27 11:13:11.100'
   '2022-10-27 11:31:25.100'
   '2022-10-27 11:56:01.200'
   '2022-10-27 12:19:25.300'
   '2022-10-27 12:25:12.500'
   '2022-10-27 12:30:16.900'
   '2022-10-27 12:49:19.100'
   '2022-10-27 12:55:42.700'
   '2022-10-27 12:58:18.100'
   '2022-10-27 13:16:06.300'
   '2022-10-27 14:00:44.700'
   '2022-10-27 14:04:14.200'
   '2022-10-27 14:11:54.600'
   '2022-10-27 14:17:10.000'
   '2022-10-27 14:42:03.300'
   '2022-10-27 14:45:22.100'
   '2022-10-27 15:02:07.900'
   '2022-10-27 15:25:39.600'
   '2022-10-27 15:44:58.300'
   '2022-10-27 15:48:35.800'
   '2022-10-27 15:54:23.100'
   '2022-10-27 16:00:17.300'
   '2022-10-27 16:19:23.700'
   '2022-10-27 16:27:30.800'
   '2022-10-27 16:31:30.600'
   '2022-10-27 16:52:09.700'
   '2022-10-27 17:16:08.800'
   '2022-10-27 18:01:39.400'
   '2022-10-27 18:08:24.800'
   '2022-10-27 18:17:44.500'
   '2022-10-27 18:25:02.500'
   '2022-10-27 18:27:45.700'
   '2022-10-27 18:48:32.000'
   '2022-10-27 19:02:01.700'
   '2022-10-27 19:24:08.300'
   '2022-10-27 19:30:01.400'
   '2022-10-27 19:43:02.200'
   '2022-10-27 19:48:50.300'
   '2022-10-27 19:55:41.900'
   '2022-10-27 19:58:23.700'
   '2022-10-27 20:17:15.800'
   '2022-10-27 20:28:47.600'
   '2022-10-27 20:49:36.900'
   '2022-10-27 20:53:02.700'
   '2022-10-27 21:07:12.100'
   '2022-10-27 21:09:47.800'
   '2022-10-27 21:50:42.300'
   '2022-10-27 22:07:34.100'
   '2022-10-27 22:09:18.800']);



a2 = [25
    82
    34
     9
    75
    76
    10
    33
    81
    26
    25
    34
     9
    75
    10
    33
    81
    26
    25
    82
    34
     9
    75
    76
    10
    33
    81
    26
    25
    81
    34
     9
    75
    76
    10
    33
    81
    26
    25
    34
     9
    75
    10
    33
    81
    26
    82
    34
     9
    75
    33
    81
    26
    25
    82
    34
     9
    75
    10
    33
    81
    26
    25
    34
     9
    10
    33
    81
    26
    25
    34
     9
    76
    10
    33
    81
    26
    25
    82
    34
    33
    81
    34
    33
    81];

a = table(a1,a2,'VariableNames',{'time','value'});

% Any way that could be faster than this one ?
tic
diff_times = diff([a.time vertcat(a.time(2:end),datetime('0001-01-01 00:00:00.000'))],1,2);
diff_times(end) = NaN;
toc

Elapsed time is 0.091898 seconds.

Answer 1

[Jan]

a1 = datetime([
   '2022-10-27 00:22:50.000'
   '2022-10-27 05:29:45.000'
   '2022-10-27 05:32:19.400'
   '2022-10-27 05:36:44.100'
   '2022-10-27 05:39:26.600'
   '2022-10-27 05:43:18.200'
   '2022-10-27 05:49:17.400'
   '2022-10-27 05:55:27.300'
   '2022-10-27 05:58:07.000'
   '2022-10-27 06:17:13.800'
   '2022-10-27 06:41:28.700'
   '2022-10-27 07:03:06.000'
   '2022-10-27 07:09:49.800'
   '2022-10-27 07:17:39.700'
   '2022-10-27 07:35:09.000'
   '2022-10-27 07:42:33.600'
   '2022-10-27 07:46:50.500'
   '2022-10-27 08:07:02.700'
   '2022-10-27 08:26:29.600'
   '2022-10-27 08:45:03.500'
   '2022-10-27 08:48:09.700'
   '2022-10-27 08:53:57.000'
   '2022-10-27 08:59:43.400'
   '2022-10-27 09:13:24.100'
   '2022-10-27 09:19:25.000'
   '2022-10-27 09:26:35.000'
   '2022-10-27 09:29:54.000'
   '2022-10-27 09:46:45.700'
   '2022-10-27 10:10:59.900'
   '2022-10-27 10:29:04.600'
   '2022-10-27 10:31:50.100'
   '2022-10-27 10:37:45.300'
   '2022-10-27 10:42:35.800'
   '2022-10-27 10:58:42.300'
   '2022-10-27 11:03:48.900'
   '2022-10-27 11:10:44.700'
   '2022-10-27 11:13:11.100'
   '2022-10-27 11:31:25.100'
   '2022-10-27 11:56:01.200'
   '2022-10-27 12:19:25.300'
   '2022-10-27 12:25:12.500'
   '2022-10-27 12:30:16.900'
   '2022-10-27 12:49:19.100'
   '2022-10-27 12:55:42.700'
   '2022-10-27 12:58:18.100'
   '2022-10-27 13:16:06.300'
   '2022-10-27 14:00:44.700'
   '2022-10-27 14:04:14.200'
   '2022-10-27 14:11:54.600'
   '2022-10-27 14:17:10.000'
   '2022-10-27 14:42:03.300'
   '2022-10-27 14:45:22.100'
   '2022-10-27 15:02:07.900'
   '2022-10-27 15:25:39.600'
   '2022-10-27 15:44:58.300'
   '2022-10-27 15:48:35.800'
   '2022-10-27 15:54:23.100'
   '2022-10-27 16:00:17.300'
   '2022-10-27 16:19:23.700'
   '2022-10-27 16:27:30.800'
   '2022-10-27 16:31:30.600'
   '2022-10-27 16:52:09.700'
   '2022-10-27 17:16:08.800'
   '2022-10-27 18:01:39.400'
   '2022-10-27 18:08:24.800'
   '2022-10-27 18:17:44.500'
   '2022-10-27 18:25:02.500'
   '2022-10-27 18:27:45.700'
   '2022-10-27 18:48:32.000'
   '2022-10-27 19:02:01.700'
   '2022-10-27 19:24:08.300'
   '2022-10-27 19:30:01.400'
   '2022-10-27 19:43:02.200'
   '2022-10-27 19:48:50.300'
   '2022-10-27 19:55:41.900'
   '2022-10-27 19:58:23.700'
   '2022-10-27 20:17:15.800'
   '2022-10-27 20:28:47.600'
   '2022-10-27 20:49:36.900'
   '2022-10-27 20:53:02.700'
   '2022-10-27 21:07:12.100'
   '2022-10-27 21:09:47.800'
   '2022-10-27 21:50:42.300'
   '2022-10-27 22:07:34.100'
   '2022-10-27 22:09:18.800']);

a2 = rand(85, 1);  % This value is not required for the question
a  = table(a1, a2, 'VariableNames', {'time', 'value'});

tic
diff_times = diff([a.time vertcat(a.time(2:end),datetime('0001-01-01 00:00:00.000'))],1,2);
diff_times(end) = NaN;
toc

Elapsed time is 0.032950 seconds.

% Easier and faster:
tic
diff_times_2 = [diff(a.time); NaN];
toc

Elapsed time is 0.005685 seconds.

isequaln(diff_times, diff_times_2)  % Same result:

ans = logical
   1

You have the a vector x and want to determine the difference between neighboring elements. Then diff is the perfect solution already. Apply it directly. You code is slower because it uses an indirection:

diff([x, [x(2:end); 0]], 1, 2)
% Is the same as:
x - [x(2:end); 0]
% is the same as:
[diff(x); 0]

[Sim]

Amazing!!! Many thanks @Jan!!!

I did not exploit the full potential of "diff"...!!!

[Jan]

You are welcome. It took me some time to understand, how you have used diff, because it was the correct command for this job. For future questions I suggest, that you explain the purpose of a command also. Most of all, the English sentence, which describes the procedure exactly, is almost the Matlab code already. :-)