Replace a zeros row with one under a condition very important for me

if Matrix A (M x N) like this matrix as input of function

   A =[0 0 1 0 1 0 1 0 1 0 1 0 1 0 0 1
       0 1 1 0 1 1 1 1 1 1 1 1 0 1 0 1
       1 1 0 1 0 0 0 0 0 0 0 1 1 1 1 1
       0 1 0 1 1 1 0 0 0 0 0 1 1 1 1 1
       0 1 1 1 1 0 1 1 0 1 1 1 1 1 0 1
       0 0 1 1 0 1 1 0 1 1 1 0 1 0 0 1
       1 0 1 0 1 0 1 1 1 1 0 1 1 1 0 1
       0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0
       1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
       0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1];

then the function do many things

• First one

count number of ones in each row and in each column , For example

the first row is [0 0 1 0 1 0 1 0 1 0 1 0 1 0 0 1] then the first row in the row_matrix will be [1 1 1 1 1 1 1]

the second row is [ 0 1 1 0 1 1 1 1 1 1 1 1 0 1 0 1] then the second row in row_matrix will be [2 8 1 1]

the last row is [0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1] then in the last row in row_matrix will be [5 1]

• this Number in row_matrix compute by if [1 1 0 1 0 1 1] then solution will be [2 1 2]
• And do the same thing for the Column in matrix A
• Second one

a) compute number of ones in each row in A and put save it in vector sum_row , for example

sum_row = [ 7 
            12
            8
            9
            12
            9
            11
            13
            16
            6 ]

b) compute number of ones in each column in A and put the result in sum_col like this

sum_column = [ 3 7 6 6 7 6 8 6 7 7 6 7 8 7 3 9]

• third one

a) count number of element in rows of row_matrix and put it in count_row like this example

the first row in row_matrix = [1 1 1 1 1 1 1] then the first row in count_row will be [7]

the second row in row_matrix = [2 8 1 1] then the second row in count_row will be [4]

and so on ...

b) do the same thing for the column

• the last thing

Generate a (M x N) zeros_matrix then do this equation If N == (sum_row(k) + (count_row(k) - 1) where k is number of row in each matrix , for example

example one :

the 7th row in A matrix is [1 0 1 0 1 0 1 1 1 1 0 1 1 1 0 1]

then the 7th row in row_matrix will be [1 1 1 4 3 1]

and the 7th row in sum_row will be [11]

and the 7th row in count_row will be [6]

then we apply the equation N== (sum_row(k) + (count_row(k) - 1))

N== (sum_row(7) + (count_row(7)-1) >>> 16 == (11 + (6-1) >> then 16 ==16 (True)

example 2 : the 9th row in A matrix is [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]

then the 9th row in row_matrix will be [16]

and the 9th row in sum_row will be [16]

and the 9th row in count_row will be [1]

then when we apply the equation

N== (sum_row(9) + (count_row(9)-1) >>> 16 == (16 + (1-1) >> then 16 ==16 (True)

then in 7th and 9th row in A empty_matrix will be like this

   empty_matrix  = [ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
                     0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
                     0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
                     0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
                     0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
                     0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
                     1 0 1 0 1 0 1 1 1 1 0 1 1 1 0 1
                     0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
                     1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
                     0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ]

and do the same thing for a column in this equation M==(sum_column(k) + (count_column(k) - 1)

*Note i need this function to cont my graduation project *

[Walter Roberson]

You appear to know what results you want, so it is not clear what the question is?

In a case such as this which is holding up your research, I would suggest just going ahead and coding it the obvious ways with looping. Get it working first and you can worry about optimization later.

[Azzi Abdelmalek]

@Firas, you can make your question brief and clear. Post a smaller example, avoid posting comments that are not important. Post the expected result, but explain clearly how to obtain it.

[Firas Al-Kharabsheh]

okay Azzi

[Rena Berman]

(Answers Dev) Restored Question.

Answer 1

[dpb]

On the first, I know there is a submittal on File Exchange to count runs; I think it may be from Bruno Luong...but, consider what happens if you diff a row or column and what that tells you about start/stop locations...it's not really very difficult problem.

The rest are pretty simple and looking through the doc and "Getting Started" tutorial should lead you there pretty quickly...

sum_r=sum(M,2);  % sum rows
sum_c=sum(M);    % sum columns

The count vector is simply length(eachRow) in the answer to Part 1.

The last is convoluted enough I don't have time to dig thru the explanation to the result but Walter's advice is good...

Answer 2

[Roger Stafford]

I'll just answer for your "First one" in getting "the same thing for the columns in A". Let x be a column in A.

s = cumsum(diff([0;x])==1);
t = (x==1);
a = accumarray(s(t),1);

The 'a' will be a column of the various contiguous sums. In your A example for the second column you would get a = [4;2].