What does this mean? "In an assignment A(I) = B, the number of elements in B and I must be the same"

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Amit
Amit 2013-4-2
p0=1.207;
t=0;
9=9.81;
i=0
for i=1:1000
y(i+1)=(t.*ydot(i))-(0.5*g*(t.^2));
p(i+1)=p0*(1-(2.333*10e-5)*y).^5;
t=t+dt
end
this is a small section of my code. y(i+1) works perfectly, but whenever i try and evaluate p(i+1), I keep getting, "In an assignment A(I) = B, the number of elements in B and I must be the same". All I want is there to be an array of p based on each value of y.
I would appreciate any help thanks.
  1 个评论
Jan
Jan 2013-4-2
编辑：Jan 2013-4-2
I strongly recommend to search for this frequently asked question in this forum. This question is answered at least 5 times per week. Searching by your own before letting others create an answer is efficient also: Google or the search of this forum find corresponding answers in less than a second.
Please format you code properly. This time I've done this for you, but when you follow the "? help" link, you will learn more tricks about formatting in this forum.

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采纳的回答

Mahdi
Mahdi 2013-4-2
In the p(i+1) section, the y that you are multiplying by is actually a matrix.
I think there is a mistake there because you probably just want the current element, so this should solve your problem:
p(i+1)=p0.*(1-2.33.*10e-5.*y(i+1)).^5;

更多回答（1 个）

Jan
Jan 2013-4-2
编辑：Jan 2013-4-2
The error message means, that you have a vector on the right, but a scalar on the left hand side for the assignment:
p(i+1) = p0*(1-(2.333*10e-5)*y).^5;
Here y is a vector, such that the expression of the right is a vector also. Perhaps you want (sorry, pure guessing, because you did not explain what the code shoudl achieve):
p(i+1) = p0 * (1 - 2.333e-5*y(i+1)) .^ 5;
"2.333e-5" is more efficient, because it does not need a multiplcation as in "2.333*10e-5".
Btw. please search for the term "pre-allocation" in this forum and the Matlab documentation. It helps to improve the speed dramatically.

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