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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.

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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.
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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.

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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

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