How to split data in reference to certain values from one column and organise into a cell array?- The logical indices in position 1 contain a true value outside of the array bounds. ERROR

I have a large data selection of 1037803 points for each test and 3 repeats. The data has 12 columns that indicate various different parameters. Column 6 indicates the speed of which the test ran. In total there was 100 different speeds. The first speed is 31.4159261390256 and then it increases by 31.4159 each time up to the last speed of 3141.59. Therefore, i want to split the data into a cell array that includes 100 rows, one for each speed with the data for each test next to it.

I have used this;

for n= 0:100 % 100 different speeds
    ans = n;
    indexes=((New_Data{1,1}(:,6)> (31+31.4159*n)& (New_Data{1,1}(:,6)<(32.4159+31.4159*n))) ) ; %values are not completly fixed and change by some degree
    speeds{n+1,1} = num2str(31+31.4159*n); %first column = the speed values
    speeds{n+1,2} = New_Data{:,1}(indexes,:); %test 1
    speeds{n+1,3} = New_Data{:,2}(indexes,:); %test 2
    speeds{n+1,4} = New_Data{:,3}(indexes,:); %test 3
 end

This works well untill New Data {:,3} for the last speed, where it appears blank (figire 1) . I have checked my data and the data is same length and has the final speed so the original data is defo not the issue (figure 2).

I keep getting the following error.

The logical indices in position 1 contain a true value outside of the array bounds.

Error in dat_proc2 (line 58)

speeds{n+1,4} = New_Data{:,3}(indexes,:);

But if i just look at data 3 (test 3) alone it seems to work but as soon as i look at it with the other 2 tests i get this error.

I think it may be the indexing. If anyone can help me solve this problem that would be great.

[Walter Roberson]

Your search bounds are strange. The lower bound is 31 + 10*pi*n and the upper end is 29 + 10*pi*n + pi . It would make more sense if the upper bound were 31 + 10*pi*(n+1) or 31 + 10*pi*n + pi .

[Walter Roberson]

It seems as if New_Data{1,1} has more rows than New_Data{1,3} does.

[Sufia Fatima]

Hi They defo have the same amount of data and rows thats the first thing i checked. The thing is it works when i add just two sets of data but when i add a 3rd it just doesnt work.

[Sufia Fatima]

I changed the upper bounds to what you suggested i still get the same error.

[Sufia Fatima]

Also since i am splitting it according to specific values from specific columns surly the amount of data shouldnt be an issue? If it is do you how i can bypass the problem. So when i do split the data its always just dependent on the specific column values.

[Walter Roberson]

At the comman prompt command

dbstop if error

and then run. When it stops with the error message, show us

  size(New_Data{1,1})
  size(New_Data{1,2})
  size(New_Data{1,3})
  find(indexes,1,'last')

[Sufia Fatima]

Hi so inputted the following and thats what i got. you right there is a slight diff in length of the data but is there a way for it to ignore the length and just focus on splitting the data according to speed.

>> dbstop if error
>> size(New_Data{1,1})
  size(New_Data{1,2})
  size(New_Data{1,3})
  find(indexes,1,'last')

ans =

     1037803          12


ans =

     1037803          12


ans =

     1037800          12


ans =

     1037803

[Walter Roberson]

L3 = min( size(New_Data{:,3},1), length(indexes) );
speeds{n+1,4} = New_Data{:,3}(indexes(1:L3),:); %test 3

[Sufia Fatima]

Hi Walter, Thank you that worked very well. I was wondering if you could explain what each part does. Thank you agin.

[Walter Roberson]

size(New_Data{:,3},1) is the number of rows in New_Data{:,3} .

length(indices) is the number of entries in the logical mask.

min() of the two gives the smaller of the two values.

indexes(1:L3) is therefore indexing only to the shorter of the two.

So if there are more entries in indexes than there are rows, then only up to the number of actual rows will be considered. If there are more rows then there are entries in indexes, then only up to the number of entries in indexes will be considered.

Note: in the case where there are more rows than entries in indices, you end up indexing with a logical vector that is shorter than what is being indexed. MATLAB defines the result as-if the logical vector were padded out with false to reach the size of what is being indexed.

[Sufia Fatima]

Thank You for all your help, this makes sense.