for loop vector multiplication

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jonas paludan
jonas paludan 2017-3-12
编辑： Jan 2017-3-12
i have this assigment:
Create a function that computes the cost of each item and returns it as a vector. Find an elegant and simple way to write the code using what you have learned about matrix and vector operations.
here's what i have come up with so far which is a not so elegant solution and only works for a specific size of matrix. in my script it has have more than 1 column and less than 4 columns:
function itemCost=computeItemCost(resourceItemMatrix,resourceCost)
if length(resourceItemMatrix)>1
A=sum(resourceItemMatrix(:,1).*reshape(resourceCost,length(resourceCost),1));
itemCost=A;
if length(resourceItemMatrix)>1
B=sum(resourceItemMatrix(:,2).*reshape(resourceCost,length(resourceCost),1));
C=sum(resourceItemMatrix(:,3).*reshape(resourceCost,length(resourceCost),1));
itemCost=[A,B,C];
if length(resourceItemMatrix)>3
D=sum(resourceItemMatrix(:,4).*reshape(resourceCost,length(resourceCost),1));
itemCost=[A,B,C,D];
end
end
end
end
I know there is a much simpler way to do this(perhaps a for-loop?), but just keep running my head against a wall when i try. help would be much appreciated.
  1 个评论
Jan
Jan 2017-3-12
"less than 4 columns"? 4 columns are accepted also, but 2 columns are not.

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采纳的回答

Jan
Jan 2017-3-12
编辑：Jan 2017-3-12
Do you mean:
function itemCost = computeItemCost(resourceItemMatrix, resourceCost)
itemCost = resourceCost(:).' * resourceItemMatrix;
end
You check for length(resourceItemMatrix)>1 twice in your code. If resourceItemMatrix is a matrix, length() might not do what you expect. Prefer:
if size(resourceItemMatrix, 2) > 1
A simplification of your original code:
function itemCost = computeItemCost(resourceItemMatrix, resourceCost)
switch size(resourceItemMatrix, 2)
case {0, 2}
error('Unexpected input.'); % ???
case {1, 3, 4}
itemCost = resourceCost(:).' * resourceItemMatrix;
otherwise
itemCost = resourceCost(:).' * resourceItemMatrix(:, 1:4);
end
end
Or according to the text of the question:
function itemCost = computeItemCost(resourceItemMatrix, resourceCost)
n = size(resourceItemMatrix, 2);
if n >= 1 && n <= 4
itemCost = resourceCost(:).' * resourceItemMatrix;
else
error('Unexpected input.'); % Or whatever you need.
end
end

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