How to extract elements of each column 1 by 1 in matrix and transform it to a column vector ?

Hello everyone!

i want to ask you if it is possible to do what i'm trying to explain or not!

at first i have a binary matrix for example (6*4) as follow :

M=
[1     0     0     0    
 0     1     0     0     
 0     1     0     0    
 0     0     0     1    
 0     0     0     1     
 0     0     0     1  ]     

in each column i want to verify :

if "1 " is the first element in column (only "1" i don't consider "0") so i try to fill a new column vector V in its first element equal to "0"

if "1" is the second element in column so the second element in V is "k : it is a value"

if "1" is the third element in column so the third element in V is "2*k"

......

as a result i want a vector as shown below

V =
[ 0     
    0      
    k        
    0       
    k         
    2*k]

I hope all is clear.

Thanks in advance.

[Dyuman Joshi]

So, if 1 is the ith element in a column, you want the value to be (i-1)*k ?

or 1:(i-1)*k

And all the values joined with a 0 in between them?

[Safia]

@Dyuman Joshi Hi! if 1 is the ith element in a column, i want the value to be (i-1)*k

i don't inderstand this : "And all the values joined with a 0 in between them?"

[Dyuman Joshi]

In your example -

For 1st column the value will be 0

For 2nd column the value will be k

For 3rd column there is no 1 (what would be the expected output/value here?)

For 4th column the value will be 3k

What is the expected here?

And are you looking for the first position of 1 in the column only?

[Safia]

at first i'm looking for rows :

for each row i see "1" ,

if it is the first element in column so the result in V = 0

if it is the second element in column so the result in V = k

if it is the third element in column so the result in V = 2* k

......

[Dyuman Joshi]

So, from my understanding, for the example you have given

In the first row, 1 appears at 1st column so value is 0

In the second row, 1 appears at 2nd column so the value is k

same for 3rd row

4th, 5th and 6th row, value = 3k

[Safia]

@Dyuman Joshi i will try to explain more:

In the first row, 1 appears as the 1st element of column so value is 0

In the second row, 1 appears at 1st element of column so the value is 0

In the third row, 1 appears at 2nd element of column so the value is k

In the fourth row, 1 appears at 1st element of column so the value is 0

In the fifth row, 1 appears at 2nd element of column so the value is k

In the sixth row, 1 appears at 3rd element of column so the value is 2*k

i consider just non zero element , just '1'

[Dyuman Joshi]

"In the second row, 1 appears at 1st element of column so the value is 0"

Second row is [0 1 0 0]

Can you explain your statement regarding this - '1 appears at 1st element of column'

What I can observe is, in the 2nd row, 1 appears as the 2nd element.

[Safia]

"What I can observe is, in the 2nd row, 1 appears as the 2nd element".

in the 2nd element in the row, but when i see the position of "1" in column , it the first non zero element right?

Answer 1

[Dyuman Joshi]

Apologies, it took quite some time and questions to understand.

Code for the 1st non-zero element in a row -

M=[1     0     0     0    
 0     1     0     0     
 0     1     0     0    
 0     0     0     1    
 0     0     0     1     
 0     0     0     1];
v=zeros(size(M,1),1);
for i=1:size(M,1)
    c=find(M(i,:),1);
    v(i,1)=nnz(M(1:i-1,c));
end
v

v = 6×1

0 0 1 0 1 2

You can multiply the vector 'v' with any arbitary value of k.

[Safia]

Good! thank you so much for your help!

[Dyuman Joshi]

You are welcome!