# Find index of first zero searching from left first column first row, then find index of first zero searching from last column last row

조회 수: 153(최근 30일)
monkey_matlab 2017년 10월 22일
편집: Cedric Wannaz 2017년 10월 22일
Hello,
In the code below, I am attempting to find the index of first zero searching from the left first column first row, going down the column, then find the index of first zero from last column last row searching up (as shown in the arrows below)?
In the picture above, rstart = 1, cstart = 1, rstop = 10, cstop = 20.
This is what I have so far:
n = 10;
m = 20;
M = randi([0 1], n,m);
for i = 1:1:n
for j = 1:1:m
if M(i,j) == 0
rstart = i;
cstart = j;
break
end
end
end
for ii = n:-1:1
for jj = m:-1:1
if M(ii,jj) == 0
rstop = ii;
cstop = jj;
break
end
end
end

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### 채택된 답변

Cedric Wannaz 2017년 10월 22일
편집: Cedric Wannaz 2017년 10월 22일
MATLAB stores data column first in memory. Accessing your 2D array linearly follows this structure. Evaluate
>> M(:)
and you will see a column vector that represents the content of the array read column per column. This means that you can FIND the linear index of these 0s using the 'first' (default) or 'last' options of FIND, by operating on M(:):
>> M = randi( 10, 4, 5 ) > 5
M =
4×5 logical array
1 1 1 1 0
1 0 1 0 1
0 0 0 1 1
1 1 1 0 1
>> linId = find( M(:) == 0, 1 )
linId =
3
>> linId = find( M(:) == 0, 1, 'last' )
linId =
17
What remains is to convert linear indices back to subscripts, and you can do this using IND2SUB. For example:
>> [r,c] = ind2sub( size( M ), find( M(:) == 0, 1, 'last' ))
r =
1
c =
5

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