Code some excel formulas in Matlab, in a smart way

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Mikkel Ibsen 2017-9-11

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https://ww2.mathworks.cn/matlabcentral/answers/356071-code-some-excel-formulas-in-matlab-in-a-smart-way

编辑： José-Luis 2017-9-11

采纳的回答： Steven Lord

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Hey

I have a quick question, I have this excel ark where there is a column that has the following code:

q1 = 1/2 * x1 * y1
q2 = 1/2 * x2 * y2 + x1 * y1
q3 = 1/2 * x3 * y3 + x2 * y2 + x1 * y1
q4 = 1/2.............

and so on and so on.

How do I program this smart? I've tried to make a for loop, but I can't figure out how the make the 1/2 only stick to the new x-y values and not the others.

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Answer 1

Steven Lord 2017-9-11

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This sounds vaguely homework-like so I'm not going to give the answer just a hint. I assume x1, x2, etc. refer to elements of a vector x and similarly for y1, y2, etc. Let xy be x.*y. Take a look at the cumsum function applied to xy. It doesn't get you all the way to your answer, but it gets you most of the way there.

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Mikkel Ibsen 2017-9-11

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I've tried to work around some loops instead, but is the cumsum more efficient code?

This is what i have:

x = ([9;8;9;10;10]);
y = ([1;1;1;1;1]);
qadd = x.*y;
for i = 1:length(x)
    q(i,1) = 0.5*qadd(i)+sum(qadd(1:i-1));
end

How would that look like in a cumsom?

José-Luis 2017-9-11

编辑：José-Luis 2017-9-11

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 dummy = x.*y; %doesn't make much sense if y is a vector of ones 
 result = cumsum(dummy) - dummy./2;

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Answer 2

KL 2017-9-11

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x = 1:5;
y = 6:10;
xy = arrayfun(@(a,b) [x(1):a;y(1):b],x,y,'UniformOutput',false);
q = cellfun(@(c) sum([0.5 ones(size(c)-1)].*prod(c)),xy,'UniformOutput',false)

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Mikkel Ibsen 2017-9-11

I can see it works, but I have no idea what is going on, and at the same time the results are cells, which I don't understand to use on further coding, is there a way to make it into a double ?

Image Analyst 2017-9-11

There is a mat2cell() function. See the FAQ http://matlab.wikia.com/wiki/FAQ#What_is_a_cell_array.3F for an explanation of cells. However I think Steven's answer/hint is the more straightforward, less cryptic approach.

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