Unable to perform assignment because the size of the left side is 1-by-10000 and the size of the right side is 1-by-20000. how to solve ??

clearvars;close all;clc;
count=0;

k=10e3:10e3:100e3;
for i=1:1:10
p(count+1,:)=rand(1,k(i));
count=count+1;
end

[Stephen23]

MATLAB arrays must be rectangular (i.e. cannot be "ragged"), so this allocation will not work because the number of elements changes on each iteration:

p(count+1,:)=rand(1,k(i));

You might be able to use a cell array.

[pavan kumar]

thank u sir..

please suggest me how to use a cell array ??

Answer 1

[Walter Roberson]

You are increasing the number of values generated for each element of the loop. How do you expect to store that continually increasing number of values as rows of a matrix? A matrix cannot have a different number of columns for each row.

[pavan kumar]

thank u sir.

sir, how can i store differernt coloumns in a same matrix ??

[Walter Roberson]

k = 10e3:10e3:100e3;
p = [];
for i = 1:1:10
    if size(p,1) < k(i)
        p(end:k(i),1:i) = nan;  %expands rows and columns
    end
    p(1:k(i), i) = rand(k(i), 1);
    p(k(i):end, i) = nan;
end

This is not as efficient as it could be. More efficient would be to figure out the largest size ahead of time:

k = 10e3:10e3:100e3;
maxk = max(k);
N = 10;
p = nan(maxk, N);
for i = 1:1:N
    p(1:k(i), i) = rand(k(i), 1);
end

Both of these pad the short columns with NaN to signal missing data.

Both versions are coded to account for the possibility that you might have shorter columns after longer columns but must still pad with nan properly.