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how can I multiply a vector by scalar?

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Parham Babakhani Dehkordi
Parham Babakhani Dehkordi 2016년 5월 19일
댓글: the cyclist 2016년 5월 19일
Hi, I have a very simple case. I do not know why there is an error which says, (Undefined operator '*' for input arguments of type 'cell').C1 is a [1*44998] and k=[1*300] vectors. my purpose is to calculate the vector of sss in which a scalar value of 2.7 is multiplied by vector C (which has a size [1*300]. any help would be appreciated.
k=find(c1==1);
time1=t(k);
i=diff(k);
j=find(i>1);
j=[0,j];
n=length(j);
for m=1:n-1;
I{m}=k(j(m)+1:j(m+1))
time{m}=t(I{m})
initial_time{m}=time{m}(1)
final_time{m}=time{m}(end)
end
I{m+1}=k(j(end)+1:end)
time{m+1}=t(I{m+1})
initial_time{m+1}=time{m+1}(1)
final_time{m+1}=time{m+1}(end)
C = cellfun(@minus,final_time,initial_time,'UniformOutput',false)
sss=C*2.7;

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the cyclist
the cyclist 2016년 5월 19일
편집: the cyclist 2016년 5월 19일
A cell array is a special type of "container", which you cannot do all operations on. You need to use a numeric array for numerical operations. Depending on the contents of C, you might just be able to do
C_num = cell2mat(C)
and carry out your operations on that.
  댓글 수: 2
Parham Babakhani Dehkordi
Parham Babakhani Dehkordi 2016년 5월 19일
well-done
the cyclist
the cyclist 2016년 5월 19일
You might want to back-track and see if C should have been a numerical array all along, rather than being stored in a cell array.

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추가 답변 (2개)

Jos (10584)
Jos (10584) 2016년 5월 19일
편집: Jos (10584) 2016년 5월 19일
C is a cell array. I think you want the content of a cell in C to be multiplied by 2.7. You can do this using cellfun
C = {[10 7 4], 100, 1:5} % a cellarray
C2 = cellfun(@(x) 2.7*x, C, 'un', 0)

dpb
dpb 2016년 5월 19일
You get the error because C isn't a vector but a cell containing a vector. You can work around it by
C=C{:}*2.7; % result is double vector
or
C={C{:}*2.7}; % result is cast back to a cell
W/o a way to run your code snippet easily I'm not positive but it's likely you could use logical addressing and avoid creating the cell array to begin with...

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