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

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M.S. Khan
M.S. Khan 2019-7-20
评论： M.S. Khan 2019-7-23
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.
  2 个评论
M.S. Khan
M.S. Khan 2019-7-20
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

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采纳的回答

TADA
TADA 2019-7-20
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
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更多回答（2 个）

Bruno Luong
Bruno Luong 2019-7-20
f = @(A)cumsum(A==3,2)>0;
A = A + f(A).*fliplr(f(fliplr(A)))
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Bruno Luong
Bruno Luong 2019-7-21
编辑：Bruno Luong 2019-7-21
"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.

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KALYAN ACHARJYA
KALYAN ACHARJYA 2019-7-20
编辑：KALYAN ACHARJYA 2019-7-20
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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