Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 1-by-20.

23 次查看（过去 30 天）
Maddy
Maddy 2022-9-25
回答： Walter Roberson 2022-11-6
Unable to perform assignment because the size of the left
side is 1-by-1 and the size of the right side is 1-by-20.
  15 个评论
显示 13更早的评论
Nilanshu Ranjan
Nilanshu Ranjan 2022-9-28
for i = 1 : s-2
R(i,i) = 1-b0;
R(i+1,i) = b01(t);
R(i+2,i) = b02(t);
end
Here b0 is a 1×20 vector that you are trying to assign to R(i,i) which is a 1×1 type. You perhaps want R to be a 3-D matrix of dimensions s×s×20 to be able to assign b0 to R(i,i) .
If you do not require R to be a 3-D matrix, you may want to use a specific value from the vector b0 as b0(index).
Also see that you have used b01(t) and b02(t) which are 1×1 double types and indexing them with a number greater than 1 will throw error.
Maddy
Maddy 2022-11-6
@Nilanshu Ranjan
Hello Brother, I understnad what you have said.
R is not a 3-D matrix. It is a 100*100 matrix having same values in each column in a descending loop manner. and t is a 1*2000 matrix having different values in each column.
The values from t matrix are varying. Depend on these varying values i need variation in R matrix too. But the main problem is dimension are not the same. what should i need to do to match them.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Walter Roberson
Walter Roberson 2022-11-6
t = 1:20; % Number of time steps
That is a vector of size 1 x 20.
b1(t) = b01*(t) + 0.01*(t);
b2(t) = b02*(t) + 0.001*(t);
Since t is 1 x 20 and b01 and b02 and 0.01 and 0.001 are scalars, then b1 and b2 will become scalars of size 1 x 20.
Beware: the b1(t) on the left hand side hints that you are thinking of those two lines as if they are defining formulas relating any input t value to an output value. However, in MATLAB, when you have an indexed variable on the left hand side of the assignment, the only time you can be defining a formula is if you are using the Symbolic Toolbox, such as
syms t
b1(t) = b01*(t) + 0.01*(t)
which would create a symbolic function relating the symbolic variable t to an output.
When t is numeric, as it is in your situation, then b1(t) on output means the same as indexing b1 at the values stored in t. If you had, for example, defined t = linspace(0, 20, 50) then that would include non-integers and assigning to b1(t) would be an error because of the non-integer indices.
You should typically be assigning to an unindexed variable in such numeric situations, such as
b1 = b01 * t + 0.01 * t;
and if you have the case where b1 is already initialized and you want to write to the beginning of it, you should tend more towards.
b1(1:length(t)) = b01 * t + 0.01 * t;
Continuing:
b0 = 1 - b1 - b2; % Density of state changing
with b1 and b2 both being 1 x 20, b0 is 1 x 20.
Then in your loop,
for i = 1 : s-2
R(i,i) = 1-b0;
you appear to be wanting to write 1-b0 into diagonal entries in R. But remember b0 is a 1 x 20 array, and R(i,i) is a single output location. 20 values will not fit into one value.
You have a few possibilities at this point:
  • you can select one of the b0 values to use for the calculation. For example, R(i,i) = 1-b0(i) . The difficulty with that is that the maximum value of i is s-2 which is 100-2 which is 98, but b0 only has 20 elements
  • you can extend R into the third dimension, writing to R(i,i,:) . Your comments seem to imply that you do not want to do this.
  • You could make R into a cell array and write to R{i,i} so that each location could store a 1 x 20 vector. But if you do that, you would have trouble doing mathematics with R
I have to wonder if you are trying to initialize the upper left corner diagnonals of R?? If so then
for i = 1 : length(t)
R(i,i) = 1-b0(i);
R(i+1,i) = b01(i);
R(i+2,i) = b02(i);
end

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Matrix Indexing 的更多信息

标签

产品


版本

R2021b

Community Treasure Hunt

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

Start Hunting!

Translated by