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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采纳的回答
Fangjun Jiang
2011-12-14
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 个评论
Fangjun Jiang
2011-12-16
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.
更多回答(1 个)
Walter Roberson
2011-12-14
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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