Asked by M.S. Khan
on 20 Jul 2019

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.

Answer by TADA
on 20 Jul 2019

Accepted Answer

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');

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
on 21 Jul 2019

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
on 21 Jul 2019

Cheers mohammad :-)

M.S. Khan
on 21 Jul 2019

Cheers TADA. Heartiest Greetings!!!

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Answer by Bruno Luong
on 20 Jul 2019

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

A = A + f(A).*fliplr(f(fliplr(A)))

TADA
on 21 Jul 2019

Very elegant +1

Bruno Luong
on 21 Jul 2019

"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
on 23 Jul 2019

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

Thanks for your time.

Regards

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Answer by KALYAN ACHARJYA
on 20 Jul 2019

Edited by KALYAN ACHARJYA
on 20 Jul 2019

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

>>

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## 2 Comments

## madhan ravi (view profile)

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## M.S. Khan (view profile)

## Direct link to this comment

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