How to simply the code

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Riccardo Tronconi
Riccardo Tronconi 2021-7-20
回答： Peter Perkins 2021-7-27
Dear guys I have addressed this code but I would appreciate to know if there is a more efficient way to compute this operation.
I have C = mx2 as input data in which:
Column 1 Column 2
'05-Mar-2021 11:41:37.000000000' 1.82586529734179
'05-Mar-2021 11:41:38.000000000' 1.77388482137334
'05-Mar-2021 11:41:39.000000000' 1.88714319565667
'05-Mar-2021 11:41:40.000000000' 2.29089559960008
'05-Mar-2021 11:41:41.000000000' 2.30346664313351
'05-Mar-2021 11:41:42.000000000' 1.87713583687103
'05-Mar-2021 11:41:43.000000000' 1.76028809274914
'05-Mar-2021 11:41:44.000000000' 2.29089559960008
'05-Mar-2021 11:41:45.000000000' 1.91069619238497
'05-Mar-2021 11:41:46.000000000' 1.87713583687103
'05-Mar-2021 11:41:47.000000000' 1.70193502387195
'05-Mar-2021 11:41:48.000000000' 1.95785425847213
'05-Mar-2021 11:41:49.000000000' 1.87823181906136
I loop over each row and I stay inside a while loop if the value in the second column is lower than 2. Once, it is greater than the threshold I exit and it is computed the difference between the first timestamp to the last one to find the duration.
is it possibile to simply the code?
  2 个评论
Yongjian Feng
Yongjian Feng 2021-7-20
You want to post your code here, so others can review?
Jan
Jan 2021-7-20
Is your data a cell array or a table?

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采纳的回答

Jan
Jan 2021-7-20
编辑：Jan 2021-7-20
With some guessing:
Value = cat(1, C{:, 2});
Index = find(Value >= 2, 1, 'first') - 1;
if ~isempty(Index) % [EDITED]
Time1 = C{1, 1};
Time2 = C{index, 1};
Duration = datetime(Time2) - datetime(Time1)
else
% What do you want to do, if no element matchs?
end
  4 个评论
显示 2更早的评论
Jan
Jan 2021-7-20
@Riccardo Tronconi: I've inserted a check, if no value is found.
The shown procedure should work for an arbitrary number of rows.

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更多回答（1 个）

Peter Perkins
Peter Perkins 2021-7-27
"I'm working with cell arrays": Riccardo, you don't want to be doing that, especially not with 100000 rows. Or using a loop for that matter.
I think the following does what you want, though I'm not actually 100% sure what that is.
>> t = datetime({'05-Mar-2021 11:41:37.000000000'
'05-Mar-2021 11:41:38.000000000'
'05-Mar-2021 11:41:39.000000000'
'05-Mar-2021 11:41:40.000000000'
'05-Mar-2021 11:41:41.000000000'
'05-Mar-2021 11:41:42.000000000'
'05-Mar-2021 11:41:43.000000000'
'05-Mar-2021 11:41:44.000000000'
'05-Mar-2021 11:41:45.000000000'
'05-Mar-2021 11:41:46.000000000'
'05-Mar-2021 11:41:47.000000000'
'05-Mar-2021 11:41:48.000000000'
'05-Mar-2021 11:41:49.000000000'});
>> x = [1.82586529734179
1.77388482137334
1.88714319565667
2.29089559960008
2.30346664313351
1.87713583687103
1.76028809274914
2.29089559960008
1.91069619238497
1.87713583687103
1.70193502387195
1.95785425847213
1.87823181906136];
>> t = table(t,x)
t =
13×2 table
t x
____________________ ______
05-Mar-2021 11:41:37 1.8259
05-Mar-2021 11:41:38 1.7739
05-Mar-2021 11:41:39 1.8871
05-Mar-2021 11:41:40 2.2909
05-Mar-2021 11:41:41 2.3035
05-Mar-2021 11:41:42 1.8771
05-Mar-2021 11:41:43 1.7603
05-Mar-2021 11:41:44 2.2909
05-Mar-2021 11:41:45 1.9107
05-Mar-2021 11:41:46 1.8771
05-Mar-2021 11:41:47 1.7019
05-Mar-2021 11:41:48 1.9579
05-Mar-2021 11:41:49 1.8782
>> t.run = cumsum(x >= 2)
t =
13×3 table
t x run
____________________ ______ ___
05-Mar-2021 11:41:37 1.8259 0
05-Mar-2021 11:41:38 1.7739 0
05-Mar-2021 11:41:39 1.8871 0
05-Mar-2021 11:41:40 2.2909 1
05-Mar-2021 11:41:41 2.3035 2
05-Mar-2021 11:41:42 1.8771 2
05-Mar-2021 11:41:43 1.7603 2
05-Mar-2021 11:41:44 2.2909 3
05-Mar-2021 11:41:45 1.9107 3
05-Mar-2021 11:41:46 1.8771 3
05-Mar-2021 11:41:47 1.7019 3
05-Mar-2021 11:41:48 1.9579 3
05-Mar-2021 11:41:49 1.8782 3
>> rowfun(@(t,x)t(end) - t(1), t, "InputVariables","t","GroupingVariables","run")
ans =
4×3 table
run GroupCount Var3
____ __________ ________
0 3 00:00:02
1 1 00:00:00
2 3 00:00:02
3 6 00:00:05
I've used a table, not a timetable because rowfun doesn't currently let you use the row times as a data variable.

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