how to count the number of consecutive rows less than a given sum?

I have the following kind of data:

Data=[1; 2; 3; 8; 9; 10; 20; 30]

I want to count the number of consecutive rows which the difference between the first and the last element is less than or equal to 5.

For the example above, it would result in 3 counts:

-rows 1, 2 and 3 = 3

-rows 3 and 8 = 2

-rows 8, 9 and 10 = 3

In the end, I’d like to create a matrix in which the first column indicates the first row of the counting and the second column indicates the sum achieved:

Result =[1 3; 3 2; 8 3]

Many thanks in advance.

Flavio

[Image Analyst]

For case 2, rows 3 and 8, the rows are not consecutive, which your subject line said is a requirement. Why do you have separated rows? Row 3 is not next to row 8.

For your last case you mention rows 8, 9, and 10. However your "Data" has only 8 rows! There is no row 9 or row 10!

This question needs a complete rewording.

Perhaps you can use rangefilt() in the Image Processing Toolbox.

[Guillaume]

@IA, I think the user meant element of value 3 and 8 (which would be row 4 and 5).

@Flavio, your wording is indeed poor. When we say row 8, we mean the 8th row, which in your case is the element with value 30.

[Image Analyst]

OK, for rows 6,7, & 8, the values are 10,20,& 30. So "the difference between the first and the last element" would be the difference from the element at row 6, which is 10, and the element at row 8, which is 30 to give a difference of 20. The requirement was that the difference "is less than or equal to 5." So how does Flavio say that there are 3 elements? I see zero elements because there are no elements within 5 of each other in those rows.

[Flavio Silva]

Thank you all for your supporting.

Indeed, my wording led to a misunderstanding... Instead of "row", I meant "element". So, the idea is to get an output such as:

-Rows 1-3: there are 3 elements which the difference between the first (1) and the last (3) is =<5

-Rows 3-4: there are 2 elements which the difference between the first (3) and the last (8) is =<5

-Rows 4-6: there are 3 elements which the difference between the first (8) and the last (10) is =<5

The code provided by Stephen works smoothly for the example above. However, when I increase the data set, for example: Data = [1; 2; 3; 8; 9; 10; 20; 30; 33; 38; 39; 65; 89; 90; 1054]; it gets an error: "Subscripted assignment dimension mismatch.

Error in i30a (line 7) out(:,2) = 1+C-R % count elements"

Actually my data set is much larger (c.a. 20000 rows) and consists of timestamps (DD-MM-YYYY HH:MM)

Any ideas?

Again, thank you very much for your kind attention, and apologies for my poor code wording...

[Stephen23]

The problem is that the variable out still exists from the last time that you ran my code. You should clear the variables from the MATLAB workspace and then run the code again.

Alternatively change the last two rows of code to this:

out = Data(R);      % first element
out(:,2) = 1+C-R;   % count elements

Answer 1

[Stephen23]

Assuming that the data is monotonically increasing then this is easy using vectorized code:

Data = [1; 2; 3; 8; 9; 10; 20; 30];

X = bsxfun(@minus,Data(:).',Data(:));
Y = fliplr(0<X & X<=5);
Z = fliplr(Y & cumsum(Y,1)==1 & cumsum(Y,2)==1);
[R,C] = find(Z);
out = Data(R);      % first element
out(:,2) = 1+C-R;   % count elements

The code produces this output:

>> out
out =
   1   3
   3   2
   8   3

EIDT to make out allocation replace existing out.

[Stephen23]

Your new Data also works fine with my code:

>> clear  
>> Data = [1; 2; 3; 8; 9; 10; 20; 30; 33; 38; 39; 65; 89; 90; 1054];

and run the code again:

>> out
out =

      1    3
      3    2
      8    3
     30    2
     33    2
     38    2
     89    2

[Flavio Silva]

Thank you for your help! The code runs properly.