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

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M.S. Khan on 20 Jul 2019
Commented: M.S. Khan on 23 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.

madhan ravi on 20 Jul 2019
Show how your expected result should look like.
M.S. Khan on 20 Jul 2019
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

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 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.
M.S. Khan on 21 Jul 2019

Bruno Luong on 20 Jul 2019
f = @(A)cumsum(A==3,2)>0;
A = A + f(A).*fliplr(f(fliplr(A)))

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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;
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
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.
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.
Regards

KALYAN ACHARJYA on 20 Jul 2019
Edited: 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
>>

#### 1 Comment

M.S. Khan on 21 Jul 2019
Thanks Mr. Kalyan for kind contribution and support