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I have some algorithm including a while loop::

while (condition)

.......................................................

.......................................................

.......% do something and return a result array B...

end

Example: I MUST run the code (while...end). For instance, my code will loop 4 time to satify the condition. Therefore, I will get the result of each loop is {B1,B2,B3,B4}

Let's say:

-Loop 1: B1=[1 2 3 4 5....9];

-Loop 2: B2=[10 11 12....15];

-Loop 3: B3=[16 17 18 19] ;

-Loop 4: B4=[20 21 22....30];

The question is: from the result matrix Bi (at each loop):

1/ How to create array index_Bi according to the number of element in array Bi as follows:?

index_B1= [1 1 1 .....1] {9 element 1}

index_B2= [2 2 2.....2] {6 element 2}

index_B3= [3 3 3 3] {4 element 3}

index_B4= [4 4 4.....4] {11 element 4}

_ Note: the number of array "index_Bi" depend on the number of loop (i). In this example, Assumed that the number of loop i=4._

2/ After that, How to add two array (Bi & index_Bi) to matrix A, we will have the result:

A=[ B1 index_B1

B2 index_B2

B3 index_B3

B4 index_B4]

I hope the result of matrix A will be as follows:

A=[ 1 1

2 1

3 1

...1

9 1

10 2

11 2

....2

15 2

16 3

17 3

18 3

19 3

20 4

21 4

.....

30 4]

In the above example, I assume the algorithm is just loop 4 time (i=4). { But actually, For my real data, "the while loop" may be loop up to 100 times (or more). Therefore, i may be equal or larger than 100}*

Andrei Bobrov
on 1 Sep 2017

B = [];

ii = 1;

x = [9 6 4 11];

add1 = 0;

while ii <= 4

add1 = add1(end) + (1:x(ii));

B = [B;{add1}];

ii = ii + 1;

end

A = [[B{:}]',repelem((1:numel(B))',cellfun(@numel,B))];

Stephen Cobeldick
on 1 Sep 2017

Edited: Stephen Cobeldick
on 1 Sep 2017

You can do this very simply inside the loop, there is no need to use cell arrays, slow cellfun, or repelem.

out = NaN(0,2);

itr = 1;

while ...

tmp = ... your code here

tmp = tmp(:);

tmp(:,2) = itr;

out = [out;tmp];

itr = itr + 1;

end

Walter Roberson
on 1 Sep 2017

Edited: Walter Roberson
on 1 Sep 2017

index_B4= [4 4 4;.....4] {11 element 4}

The above is not possible. The space between the first two "4" tells us that you have multiple columns, and the ';' after the third "4" tells us that you have multiple rows. In MATLAB, each row must be the same size. Therefore, the total number of elements must be divisible by the number of rows. But the total number of elements is 11, which is a prime number; to have that work out, you would need to have only one column per row -- but we see that there a multiple columns per row, so the total number of elements cannot be prime. If there are multiple rows and multiple columns in an array, then the total number of elements must be a composite number, not a prime.

The creating of 11 copies of element 4 is easy:

repmat(4, 1, 11)

However, this requires that you somehow know that the proper length is 11.

Is the length of each subsection somehow known? Is there something particular about the first element of each subsection (for example if the first element were always negative but the others were always positive) ?

Walter Roberson
on 1 Sep 2017

Your restrictions make it impossible to solve the problem.

Suppose, for example, that I told you that the following list of numbers was generated in bursts, and I asked you to identify each burst:

1 5 10 7 9 2 4 3 3 6 8 7 8 7 7 3 1

I can offer the additional information that no burst was negative length, and that no burst had length more than 16383. What were the burst boundaries?

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