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.

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.

[Dyuman Joshi]

How can you assign 20 values to a single place holder only?

[Maddy]

Thats what i am asking.. is there any solution for it or it is impossible?

[Dyuman Joshi]

It depends. What are you trying to do? Show some data and give an example.

[Torsten]

We don't see the context within your code where the error appears.

So we can't decide how it could possibly be avoided.

But you can't assign an array with 20 elements to a scalar with only 1 element - that's for sure.

[Maddy]

clc
close all;
clear all;

b01 = 0.011 ; % Density of 1state/1time
b02 = 0.001; % Density of 2state/1time
t = 1:20; % Number of time steps
s = 100; % Number of states
b1(t) = b01*(t) + 0.01*(t);
b2(t) = b02*(t) + 0.001*(t);
b0 = 1 - b1 - b2; % Density of state changing
eps = 0.001; % Epsilon
dis = 100; % draw step of interval
dds = 500; % draw step of distributions
q = 60; % X length of parameters

t/dis;

for j = 1 : s
    
    L(j) = j-1;
    
end

% Determination of startvector
p  = zeros(s,1);
p(1) = 0;
p(2) = 0.0006;
p(3) = 0.0024;
p(4) = 0.0079;
p(5) = 0.0224;
p(6) = 0.0540;
p(7) = 0.1109;
p(8) = 0.1942;
p(9) = 0.2897;
p(10) = 0.3683;
p(11) = 0.3989; 
p(12) = 0.3683;
p(13) = 0.2897;
p(14) = 0.1942;
p(15) = 0.1109;
p(16) = 0.0540;
p(17) = 0.0224;
p(18) = 0.0079;
p(19) = 0.0024;
p(20) = 0.0006;
p(21) = 0;

% Determination of Markov matrix
R = zeros(s,s);

for i = 1 : s-2
    
    R(i,i) = 1-b0;
    R(i+1,i) = b01(t);
    R(i+2,i) = b02(t);
    
end   

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.

R(s-1,s-1) = 1-b0;
R(s,s-1) = b01(t);
R(s,s) = 1-b0;

P(:,1) = p;

for i = 1 : t
    
    Z(i) = i-1;
    pn = R * p;
    P(:,i+1) = pn;
    p = pn;
    
end
Z(t+1) = t;

for j = 1 : t/dis %Columns vector
    w(j) = j * dis;  
    for i = 1 : s   %Rows vector
        if P(i,j*dis) > eps
            min(j)= i;
        end
        
    end
end


for j = 1 : t/dis %Columns vector
    for i = s : -1 : 1   %Rows vector
        if P(i,j*dis) > eps
            max(j)= i;
        end
        
    end
end

figure(1)
hold on
plot(w,min,'k','LineWidth',2);
plot(w,max,'k','LineWidth',2);
legend('Minimum','Maximum')
xlabel('time') 
ylabel('ⴄ') 
hold off 

figure(2)
plot (L,P(:,1),'k','LineWidth',2)
ylim([0 0.4])
xlim([0 q])
legend('t = 0')
xlabel('ⴄ') 
ylabel('f(ⴄ)')

figure(3)
plot (L,P(:,20),'k','LineWidth',2)
ylim([0 0.4])
xlim([0 q])
legend('t = 20')
xlabel('ⴄ') 
ylabel('f(ⴄ)')

figure(4)
subplot(1,1,1)
plot (L,P(:,20),'k','LineWidth',2)
ylim([0 0.4])
xlim([0 q])
legend('t = 20')
xlabel('ⴄ') 
ylabel('f(ⴄ)')
subplot(1,1,2)
plot (L,P(:,1),'k','LineWidth',2)
ylim([0 0.4])
xlim([0 q])
legend('t = 0')
xlabel('ⴄ') 
ylabel('f(ⴄ)')

figure(5)
hold on
plot (L,P(:,1),'k','LineWidth',2)
plot (L,P(:,20),'k','LineWidth',2)
legend('t = 0','t = 20')
ylim([0 0.4])
xlim([0 q])
xlabel('ⴄ') 
ylabel('f(ⴄ)')

[Maddy]

welcome back @Torsten

[Torsten]

Hello Maddy, maybe you could explain what you try to do where the error in your code occurs ?

[Maddy]

@Torsten I am trying to get minimum and maximum values of vectors and plot them.

Can you please check the code and tell me how to resolve the error.

[Torsten]

I can't since I don't know what you try to do in these lines:

b1(t) = b01*(t) + 0.01*(t);
b2(t) = b02*(t) + 0.001*(t);
b0 = 1 - b1 - b2; % Density of state changing

and

for i = 1 : s-2
    
    R(i,i) = 1-b0;
    R(i+1,i) = b01(t);
    R(i+2,i) = b02(t);
    
end   

R(s-1,s-1) = 1-b0;
R(s,s-1) = b01(t);
R(s,s) = 1-b0;

[Maddy]

Can you please give me your contact/e-mail id.So i can disucss further. @Torsten

[Rik]

If you can't explain it clearly in a comment here, why do you think you can in an email?

[Torsten]

@Maddy

Sorry, but discussion with me can only happen here in the forum.

[Maddy]

Sure, I understand @Torsten

[Nilanshu Ranjan]

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]

@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.

Answer 1

[Walter Roberson]

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