Variable change at every time step

for i = 1 :1000
a(:,i) = b ;% assume b size 800 X 1 vector
% but b size changes at random iteration (for loop) depends on the problem i mean it size
% decrease so how to store b values in a as it showing error
end
Error:
Unable to perform assignment because the size of the left side is 1-by-800 and
the size of the right side is 797-by-1.
Other case :
for i = 1 :1000
a(:,i) = b'; % assume b size 800 X 1 vector
% but b size changes at random iteration (for loop) depends on the problem i mean it size
% decrease so how to store b values in a as it showing error
end
Unable to perform assignment because the indices on the left side are not
compatible with the size of the right side.
I can understand what error showing but how to tackle this problem. Any help appreciated .
Thanks in advance.

4 comentarios

Geoff Hayes
Geoff Hayes el 11 de En. de 2022
@RAKESH KUMAR TOTA - can a be a cell array instead since the b at each iteration can be of different sizes?
RAKESH KUMAR TOTA
RAKESH KUMAR TOTA el 11 de En. de 2022
I have no issue 'a' can be of any data type . I need to use 'a' further to investigate and plot results. thanks
VBBV
VBBV el 11 de En. de 2022
Can you paste the actual error that you faced. Everything in red color text shown at command window
RAKESH KUMAR TOTA
RAKESH KUMAR TOTA el 11 de En. de 2022
Unable to perform assignment because the size of the left side is 1-by-800 and the size of the right side is 797-by-1. for first case type
Unable to perform assignment because the indices on the left side are not
compatible with the size of the right side. for second case type

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KSSV
KSSV el 11 de En. de 2022
When you are not sure of size b, you can save them into a cell array.
a = cell(1000,1) ;
for i = 1 :1000
a{i} = b ;
end
You can access any cell using a{1}, a{2},....a{i},...a{1000}.

5 comentarios

RAKESH KUMAR TOTA
RAKESH KUMAR TOTA el 11 de En. de 2022
Sorry i forgot to mention i need to store 'a' for intance of size 800X1 at every iteration of for loop even though b size decreasing by assigning all zeros (decreased size) to the rest of 'a' compensating to maintain original size (800 X1). Actually in 'b' certain position become zero we eliminate in the next iteration and store it in 'a' but 'a' size should not change.
I appreciate your response thats the good way to represent to assign 'b' to 'a' so easy to convert using formulae cell2mat for further analysis.
Any idea in this direction please.
If the size of the b is always less than or equal 800x1....you may follow the below:
a = zeros(800,1000) ;
for i = 1 :1000
temp = b ;
a(1:length(b),i) = b ;
end
Thanks a lot.
Just a question.
Is it better to use
a = []; % or
a = zeros(800,10000); % if we dont know exactly how many iteration it will take
% for better programming.
KSSV
KSSV el 11 de En. de 2022
Initializing with dimensions is suggested.
RAKESH KUMAR TOTA
RAKESH KUMAR TOTA el 11 de En. de 2022
okay thank you.

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