I have a question regarding a simple operation that I cant find an answer to. Hope that someone can help me.
I have a vector a:
a = zeros(1,10)
and a matrix b:
b = [1, 3; 6, 8]
I want to change elements of a into ones in accordance with the segments indicated by matrix b:
The result should be
1 1 1 0 0 1 1 1 0 0
When I try:
a(b(:,1):b(:,2)) = 1
I get
1 1 1 0 0 0 0 0 0 0
Best regards,
Michael

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Stephen23
Stephen23 2020년 6월 24일
편집: Stephen23 2020년 6월 24일

1 개 추천

No loop required:
>> v = 1:numel(a);
>> x = any(v>=b(:,1) & v<=b(:,2), 1); % requires MATLAB >=R2016b
>> a(x) = 1
a =
1 1 1 0 0 1 1 1 0 0
For earlier versions replace the logical comparisons with bsxfun.
Or just use one simple loop:
>> for k = 1:size(b,1), a(b(k,1):b(k,2)) = 1; end
>> a
a =
1 1 1 0 0 1 1 1 0 0

댓글 수: 3

Thank you for the quick reply, Ill try your loop.
I was hoping to do the manuvre without a loop, but I guess it is not possible?
Stephen23
Stephen23 2020년 6월 24일
"I was hoping to do the manuvre without a loop"
See my edited answer.
Thanks :-)

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추가 답변 (2개)

Ashish Azad
Ashish Azad 2020년 6월 24일
편집: Ashish Azad 2020년 6월 24일

1 개 추천

The syntax you are using is very ambiguous and will never work
Try
for i=1:length(b)
a(b(i,1):b(i,2))=1;
end
Let me know if this work

댓글 수: 2

Stephen23
Stephen23 2020년 6월 24일
편집: Stephen23 2020년 6월 24일
Do NOT use length for this code:
for i=1:length(b)
Consider what would happen if b only has one row.
The only robust solution is to use size and specify the dimension.
Ashish Azad
Ashish Azad 2020년 6월 24일
Yeah truly said Stephen, size would be robust option

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Alan Stevens
Alan Stevens 2020년 6월 24일

1 개 추천

One way as follows:
a =
0 0 0 0 0 0 0 0 0 0
>> b
b =
1 3
6 8
>> a([b(1,1):b(1,2) b(2,1):b(2,2)]) = 1
a =
1 1 1 0 0 1 1 1 0 0

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도움말 센터File Exchange에서 Loops and Conditional Statements에 대해 자세히 알아보기

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