Is there a one-line code for this?

% I would like to go from A = [1 1;2 2;3 3;4 4;5 5] 
%                      to A = [1 1;2 2;3 3;4 4;5 5;1 2;2 3;3 4;4 5;5 6];
A = [1 1;2 2;3 3;4 4;5 5];

% Multi-line approach
B = A;
B(:,2) = B(:,2) + 1;
A = [A;B];
clear B

% Is there a way to do this in one line of code? I have to do similar operations multiple times, 
% and would like to know if there is a way to do so. Otherwise, I will settle for a function.

% Note that the actual matrices are 1,000,000+ x 30 in size.
% Thank you

Answer 1

[Stephen23]

A = [1 1;2 2;3 3;4 4;5 5];
A = [A;A+(0:1)]

A = 10×2

1 1 2 2 3 3 4 4 5 5 1 2 2 3 3 4 4 5 5 6

[Mitsu]

Thank you, Stephen.

I suppose for a specific column column_no in a larger matrix, the simplest way is the following then?

A = [A;A+[zeros(1,column_no-1),1,zeros(1,size(A,2)-column_no)] ];

[Stephen23]

"the simplest way is the following then?"

Probably the "simplest way" is just as you show in your question.

If you want to avoid duplicated data, you could create B first then add one to the appropriate (part-)column.

[Mitsu]

Understood. Thank you.

Answer 2

[SUPRIYA]

% Define the differential equation dy/dx = f(x, y) 
% Example: dy/dx = (y^2-x^2)/(y^2+x^2) f = @(x, y) x + y; 

% Input initial and final conditions
x0 = input('Enter the initial value of x0: ');
y0 = input('Enter the initial value of y0: ');
xg = input('Enter the final value of xg: '); 
h = input('Enter the value of step size h: '); 

% Number of steps
n = (xg - x0) / h; 

% Runge-Kutta 2nd order iteration 
for i = 1:n
    k1 = h * f(x0, y0);
    k2 = h * f(x0 + h, y0 + k1); 
    
    k = (k1 + k2) / 2; 
    
    yg = y0 + k; % Update y 
    x0 = x0 + h; % Update x y0=yg; 
end 

% Display final result 
fprintf('The final value of y at x = %f is y = %f\n', x0, y0);

Is there a way to do entire code in one line

[Stephen23]

"Is there a way to do entire code in one line"

x0 = input('Enter the initial value of x0: '); y0 = input('Enter the initial value of y0: '); xg = input('Enter the final value of xg: '); h = input('Enter the value of step size h: '); n = (xg - x0) / h; for i = 1:n; k1 = h * f(x0, y0); k2 = h * f(x0 + h, y0 + k1); k = (k1 + k2) / 2; yg = y0 + k; x0 = x0 + h; y0=yg; end; fprintf('The final value of y at x = %f is y = %f\n', x0, y0);

[Dyuman Joshi]

Lol.