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

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Sim
Sim 2022-11-18
评论： Jan 2022-11-18
Ran in:
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.

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采纳的回答

Jan
Jan 2022-11-18
编辑：Jan 2022-11-18
Ran in:
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]
  2 个评论
Sim
Sim 2022-11-18
编辑：Sim 2022-11-18
Amazing!!! Many thanks @Jan!!!
I did not exploit the full potential of "diff"...!!!
Jan
Jan 2022-11-18
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. :-)

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