How to simply the code

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?

[Yongjian Feng]

You want to post your code here, so others can review?

[Jan]

Is your data a cell array or a table?

Answer 1

[Jan]

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

[Riccardo Tronconi]

I'm working with cell arrays

[Riccardo Tronconi]

Thanks but I have two issues:

1- C{1, 1}; may not be less than two. Should I use a find also for it?

2- To do this procedure for n-rows, let s say 100 000 how to do that?

[Jan]

@Riccardo Tronconi: I've inserted a check, if no value is found.

The shown procedure should work for an arbitrary number of rows.

[Riccardo Tronconi]

thanks!!

Answer 2

[Peter Perkins]

"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.