Subtract each column of matrix until -3 is finished
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For example, i have a matrix a = [0 1 2 3 4]. How do I subtract each column by 1?
a = [0 3 2 6 4]
b = [0 2 1 2 3]
remainder = size(a, 2) - sum(b)
The remainder is -3. I need to minus from b when b(i) ~= 0 so that at the end the b = [0 1 0 1 3].
Thank you.
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Respuesta aceptada
Fangjun Jiang
el 14 de Dic. de 2011
I am curious too. Let me guess.
a=0:4;
b=-3; % b is an integer, positive or negative
c=zeros(size(a));
ind=find(a,1);
c(ind:ind+abs(b)-1)=-1;
a=a+c
a =
0 0 1 2 4
Updated for a generic solution
b = [0 2 1 2 3]
remainder=-6;
ind=find(b);
while remainder<0 && ~isempty(ind)
if length(ind)<=abs(remainder)
b(ind)=b(ind)-1
remainder=remainder+length(ind)
ind=find(b);
else
for k=1:abs(remainder)
b(ind(k))=b(ind(k))-1
remainder=remainder+1
end
end
end
b =
0 2 1 2 3
b =
0 1 0 1 2
remainder =
-2
b =
0 0 0 1 2
remainder =
-1
b =
0 0 0 0 2
remainder =
0
9 comentarios
Fangjun Jiang
el 16 de Dic. de 2011
I am confused with the c and a. I suggest you ask a separate question and use variable name consistently. Also, explain the logic or step. If it's hard to explain, try to show your step and intermediate result if you do it manually, like using a pencil and a piece of paper.
Más respuestas (1)
Walter Roberson
el 14 de Dic. de 2011
Does the minus sign of -3 indicate the third from the start or the third from the end? Your example is ambiguous about that.
Third from the start:
a(1:abs(b)) = a(1:abs(b)) - 1;
Third from the end:
a(1:end-abs(b)+1) = a(1:end-abs(b)+1) - 1;
2 comentarios
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