Index exceeds matrix dimensions.

3 次查看（过去 30 天）
Sara
Sara 2012-6-20
for the code below I received the error index exceeds.. I do not know how can I fix it.
x = zeros( 2,length(v0) ) ;
for k = 1:length(v0)
A = [ factor1(:,k) factor2(:,k)];
b = -rhs(:,k);
x(:,k) = A\b ;
//end of for
length v0 = 2501
A\b is 2 by 2 matrix but x(:,k) is 2 by 2501.
  3 个评论
显示 1更早的评论
Sara
Sara 2012-6-20
the sizes of factor1 and factor2 are 2 by 2501.
by the line x (:,k) = A\b , it has error.
Walter Roberson
Walter Roberson 2012-6-20
That doesn't sound like an "index exceeds" error, that sounds like an error about RHS having more non-singleton dimensions than the LHS has.
Please show the exact error.
If A\b is a 2 x 2 matrix, then were you only expecting a 2 x 1 matrix as the result of the A\b operation? Or you only care about one of the rows or one of the columns? Or you want to store all 4 results and do that in a single column ?

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

per isakson
per isakson 2012-6-20
Do you use the debugger?
  1. Put a break on the line that causes the error or
  2. Tools | Debug | Stop if Error
  3. Run the code
  4. At the break inspect the values of the index
  2 个评论
Walter Roberson
Walter Roberson 2012-6-20
A zero index would generate the error about indices needing to be positive integers, not the error about exceeding matrix dimensions.
per isakson
per isakson 2012-6-20
Thanks Walter; fixed. I try to make the point that better asking the debugger than Answer.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Creating and Concatenating Matrices 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by