In an assignment A(I) = B, the number of elements in B and I must be the same.

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Stuart
Stuart 2012 年 4 月 26 日
Hi, I know this question must have been asked a million times before but I can't find a problem quite like mine. I am running a loop to predict a trajectory where I calculate time t1 from an equation to give a constant. However this t1 varies with every iteration of the program, hence t1(i).
I then want to use this to get t(i) from
t(i)=0:1:(t1(i)-2);
However it is spitting out the statement as in the title. Can someone suggest a way around this?
Thanks.

採用された回答

Walter Roberson
Walter Roberson 2012 年 4 月 26 日
The expression 0:1:(t1(i)-2) creates a row vector. You cannot store a row vector into the space of a single numeric array element t(i) .
You could consider using cell arrays:
t{i} = 0:1:t1(i)-2;

その他の回答 (2 件)

Honglei Chen
Honglei Chen 2012 年 4 月 26 日
I don't quite know what your intention is, but the left side is a scalar and the right side in general is a vector, could be scalar, could be empty, so an error is thrown.

Daniel Shub
Daniel Shub 2012 年 4 月 26 日
You need to step back and think about the error. What is the size of
t(i)
From inspection and a little bit of MATLAB knowledge you should realize it will be the same size as i. Since you are likely looping over i, i will be a scalar (1x1) so t(i) will have 1 element independent from the value of i. In other words, on every iteration t(i) will have 1 element.
Okay now look at
0:1:(t1(i)-2)
How big do you think that is going to be? Do you think it is going to have exactly 1 element on every iteration? Do you see where the error is yet? If not lets consider
0:1:n
where n is equal to (t1(i)-2). This obviously has n+1 elements. So for your assignment to work (i.e., the number of element in B and I to be the same) n must be equal to 0, and in turn t1(i)-2 must be zero and t1(i) must be -2. Are you initializing t1 to be a vector of -2's?
  4 件のコメント
Stuart
Stuart 2012 年 4 月 26 日
Sorry I'm not being clear.
A projectile takes a different time to cover a ceratin distance depending on its velocity in the x and y directions.
I am running 5 iterations with 5 different angles of launch and so the times will be different.
I then use these times to calculate a lot of other things.
Does that make more sense?
I was expecting variables like 0:1:2, 0:1:2.4, 0:1:2.45 etc
Daniel Shub
Daniel Shub 2012 年 4 月 26 日
While that is not going to work because will all be the same array 0:1:2.

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