How to manipulate nested cell arrays using for loop?

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Jung BC 2016-3-10

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评论： Jung BC 2016-3-16

Hi, I have a cell array of size 1x316 with each cell different sizes.In each cell, there is two columns of data: first is speed and second is time.Both are double numeric in types. Firstly, i loaded that cell array in a structure 's1'.Its variable is called ''all_cords''.Now, using for loop, i am trying to do some arithmatics here.i need to calculate pause time variable in a cell array of size 1x316.

I need to check in each cell, if there is 0's in first column i.e speed field,if there is single 0, then pause time will be its time value and if there are multiple 0's then pause time will be tn-t0.

Thanks in advance!

Here is my codes:-

s1 = load('cell_speed_data.mat'); 
tpause =   cell(1,length(s1.all_cords));
pause_times = cell(1,length(s1.all_cords));
for i = 1:length(s1.all_cords)
      if find(s1.all_cords{i}(:,1)==0)              
      tpause{i} = deal(s1.all_cords{i}(:,1));    
                                              
      pause_times{i} = max(tpause{i}(:,1))-min(tpause{i}(:,1)); 
                                                                  
      end                                                        
      continue;                                       
      end;

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Chad Greene 2016-3-12

This question is unclear. Can you provide some sample data and describe exactly what you are trying to do?

Guillaume 2016-3-16

From a dimensional analysis point of view, there is some inconsistency in the result that you want.

If there is only one time where the speed is zero, you want as an output the time at which it occurs. Hence your output is a time.

If there is more than one time where the speed is zero, you want the difference between the first and last time, so your output is a duration.

Even if both have the same unit (time), they don't represent the same thing, so how is that output going to be useful?

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Answer 1

Guillaume 2016-3-16

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See my comment on your question about the mismatch in what the output represents. I'm creating two outputs here, pause_start which is the time at which the pause starts, and pause_duration which is the duration of the pause.

pause_start = cell(size(s1.all_coords)); %safer than length
pause_duration = cell(size(s1.all_coords)); %safer than length
for iter = 1:numel(s1.all_coords) %safer than length as well
   ispause = s1.all_coord{iter}(:, 1) == 0; %no need for find, use logical array
   pause_times = s1.all_coords{iter}(ispause, 2);
   if ~isempty(pause_times) 
      pause_start{iter} = pause_times(1)
      pause_duration{iter} = pause_times(end) - pause_times(1);
   end
end

Note that the pause duration ignores cases where speed goes back up between start and end of the pause, as per your question. To me that doesn't sound correct but maybe it doesn't happen with your data, so it does not matter.

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Jung BC 2016-3-16

编辑：Jung BC 2016-3-16

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Hi Guillaume, Thanks for your answer.I should describe my problem more in details here now,then you may understand it more.

The sample data of first cell of s1.all_coords looks like this:-

44.88 0.0043

28.94 0.0040

0 0.0044

0 0.0037

0 0.0042

10.83 0.00415

35.03 0.00432

57.87 0.00400

24.65 0.00432

0 0.00402

8.88 0.00406

Now, in above sample data in the cell, i need to compute pause time variable everytime where it detects 0's in the speed column.

So lets say, for above sample,

first pause time = p1 = 0.0044(highest among 3 consecutives)-0.0037(lowest)

second pause time= p2 = 0.00402(since its single 0's)

similar this way, i need to store all pause time values from every cells and store in a new cell array of size 1x316.

In your solution, i find only one pause time value from entire cell.I need to store all the pause times of cell.

e.g. cell_1 : p1, p2, p3 ,p4,.... . . .

      .
      cell_316: p1, p2, p3, p4, and so on.

Hope you got my problem much broad way this time.

Guillaume 2016-3-16

编辑：Guillaume 2016-3-16

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Ok, so the problem is a bit more complicated. You want to identify runs of 0 in the speed column. diff and find on the logical array will do that.

I still think that using the time when the run is of length 1 is not consistent (and makes the code more complicated since you need to special-case it).

pause_duration = cell(size(s1.all_coords)); %safer than length
for iter = 1:numel(s1.all_coords) %safer than length as well
   ispause = s1.all_coord{iter}(:, 1) == 0; %a logical array of 0 and 1
   transitions = find(diff([0 ispause 0])); %location of transitions from 0 to 1 and 1 to 0 in logical array
   %because of the prepending and appending of 0, transition is 
   %guaranteed to have an even numbers of elements and to start
   %with the index of a transition from 0 to 1 (i.e. a beginning
   %of a run).
   runstarts = transitions(1:2:end);
   runends = transitions(2:2:end) - 1; 
   durations = cell(1, numel(runstarts));
   for run = 1 : numel(runstarts)
      runtimes = s1.all_coords{iter}(runstarts(run):runends(run), 2);
      if numel(runtimes) == 1 %special case when single 0
          durations{run} = runtimes;  %but it's not a duration!
      else
          durations{run} = max(runtimes) - min(runtimes);
      end
      pause_duration{iter} = durations;
   end
end

Guillaume 2016-3-16

编辑：Guillaume 2016-3-16

在 MATLAB Online 中打开

Oh, of course, ispause is a column vector. Change the transitions line to:

transitions = find(diff([0; ispause; 0]));

I've edited my answer and fixed another bug later on. There may still be some more bugs. Code is untested.

Jung BC 2016-3-16

Thanks a lot Guilaume,finally it's working well.

I really appreciate your help.One last request to you,

How can i extract minimum, maximum, and average

values from the final output- the cell array- pause_duration{}

and save them in different cell

arrays?

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