how to replace the elements row by rows instead of column by column in matrix

A =[ 0 0 3 3 3 0 0 3 0 0; 0 0 0 3 3 3 0 3 3 0]
[rows,colms ] = size(A)
for i = 1:rows
    for j = 1:colms
            index-1 = find(A==3,1,'first')
            index_2 = find(A==3,1,'last')
                    If     A(i,j)=3 & A(i,j)==index_1
                           A(i,index_1:index_2) = A(i,index_1:index_2) +1
                    end 
    end 
end

it gives me 5th and 18th indices while i want to get row wise like first should be 3rd and last should be 6th.

please help me in resolving this problem.

warm regards in advance.

[madhan ravi]

Show how your expected result should look like.

[M.S. Khan]

Mr. Madhan Ravi, i want the first 3 in the rows as the lowest element and the last one as the heighest element in each row. and want to add 1 .e.g.

0 0 3 3 3 0 0 3 0 0 => 0 0 4 4 1 1 4 0 0 (i want to add 1 from lowest 3 to highest 3)

0 0 0 3 3 3 0 3 3 0 => 0 0 0 4 4 4 1 4 4 0

Answer 1

[TADA]

find(A==3,1,'first')
find(A==3,1,'last')

These lines find linear indices not [row, col] subsets. Linear indices go along all the rows of the first column, then on to the second column and so on.

These two lines also disregard i and j completely, so they always give the absolute first and last linear indices in the entire matrix each iteration (5 and 18).

I dont know what exactly you're trying to achieve, but maybe you need to compare only current row:

index_1 = find(A(i,:)==3,1,'first');
index_2 = find(A(i,:)==3,1,'last');

Look here for explanation on array indexing in Matlab

if A(i,j)=3 & A(i,j)==index_1

This condition compares the value of A(i,j) to 3 and to the first index which equals 3, that makes little sence to me, but i may be missing your intent

If you explain with more detail what you are trying to do, we may be able to help you get to the right solution

[M.S. Khan]

Dear TADA, i have already explained. Thanks for your time.

This condition (if A(i,j)=3 & A(i,j)==index_1) only gives the first one in the row but the problem is with the heighest one in the same row. Thanks for your feedback. Regards

[TADA]

B = A == 3;
for i = 1:size(A,1)
    ind1 = find(B(i,:),1,'first');   
    ind2 = find(B(i,:),1,'last');
    A(i,ind1:ind2) = A(i,ind1:ind2) + 1;
end

[M.S. Khan]

Mr. TADA, thanks for all your guidance. God bless you. Warm Regards.

i am very thankful to all community friends who provided me their best feedback.

My special prayers to all for sharing their professional knowlege and cooperation.

[TADA]

Cheers mohammad :-)

[M.S. Khan]

Cheers TADA. Heartiest Greetings!!!

Answer 2

[Bruno Luong]

f = @(A)cumsum(A==3,2)>0;
A = A + f(A).*fliplr(f(fliplr(A)))

[M.S. Khan]

Mr. Bruno, thanks for guidance. i couldn't understand the functions and code you have used. could you plz explain to me.

[TADA]

Very elegant +1

[Bruno Luong]

"could you plz explain to me."

Sure here is step by step for single row input:

> A = [0 0 3 3 3 0 0 3 0 0].

This will put 1 at the place where there is element with value == 3, so 0 before the first 3 on the left

>> A==3

ans =

  1×10 logical array

   0   0   1   1   1   0   0   1   0   0

When I apply cumsum to this, after the first 3 element values of the ouput are >= 1. (it actually increase by 1 when it meets a 3)

>> cumsum(A==3)

ans =

     0     0     1     2     3     3     3     4     4     4

I need array of 0s, but then 1s starting from the most left 3, so I do logical commparison

>> cumsum(A==3)>0

ans =

  1×10 logical array

   0   0   1   1   1   1   1   1   1   1

The three steps are combined, I put in a anonymous function

   f = @(A)cumsum(A==3,2)>0;
   mask1 = f(A)

Now I want to do the exact same trick but runiing from right-to-left. I simply flip the input array, apply f(), then flip the output back

> mask2 = flip(f(flip(A)))

mask2 =

  1×10 logical array

   1   1   1   1   1   1   1   1   0   0

Meaning I have array with 0s on the right of the last 3 and 1s on the left.

The product of mask1 and mask2 gives

>> mask1.*mask2

ans =

     0     0     1     1     1     1     1     1     0     0

provides array with 0s on the right of the last 3, 0 on the left of the first 3, and 1s in between them.

I then simply add them to the original array A to get the desired result.

[M.S. Khan]

Dear Bruno, its so complicated. its blowing above my head.

Thanks for your time.

Regards

Answer 3

[KALYAN ACHARJYA]

A=[0 0 3 3 3 0 0 3 0 0; 0 0 0 3 3 3 0 3 3 0]
[rows colm]=size(A);
B=zeros(rows,colm);
for i=1:rows
B(i,i+2:end-3+i)=1;
end
result=A+B

Command Window:

A =
     0     0     3     3     3     0     0     3     0     0
     0     0     0     3     3     3     0     3     3     0
result =
     0     0     4     4     4     1     1     4     0     0
     0     0     0     4     4     4     1     4     4     0
>> 

[M.S. Khan]

Thanks Mr. Kalyan for kind contribution and support