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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Risposta accettata
Fangjun Jiang
il 14 Dic 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 Commenti
Fangjun Jiang
il 16 Dic 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.
Più risposte (1)
Walter Roberson
il 14 Dic 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;
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