Changing elements of vector with matrix

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Michael Clausen
Michael Clausen le 24 Juin 2020
Commenté : Michael Clausen le 24 Juin 2020
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

Réponse acceptée

Stephen23
Stephen23 le 24 Juin 2020
Modifié(e) : Stephen23 le 24 Juin 2020
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
  3 commentaires
Stephen23
Stephen23 le 24 Juin 2020
"I was hoping to do the manuvre without a loop"
See my edited answer.
Michael Clausen
Michael Clausen le 24 Juin 2020
Thanks :-)

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Plus de réponses (2)

Ashish Azad
Ashish Azad le 24 Juin 2020
Modifié(e) : Ashish Azad le 24 Juin 2020
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
  2 commentaires
Stephen23
Stephen23 le 24 Juin 2020
Modifié(e) : Stephen23 le 24 Juin 2020
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 le 24 Juin 2020
Yeah truly said Stephen, size would be robust option

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Alan Stevens
Alan Stevens le 24 Juin 2020
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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