Filling Rand matrix with specific numbers

Good evening,

Lets say i have a=rand(1500,1000) and I need to fill 1:300 numbers with ones, 300:600 with 0.75, 900:1200 with 0.5 and so on.. And then show as a picture. I feeling stuck on this one:/

[the cyclist]

Sounds like homework. What have you tried?

Also, why are you starting with matrix filled with random values, and then overwriting them with non-random values? (Or maybe I misunderstand what mean.)

[darova]

• I need to fill 1:300 numbers with ones

You have 2D matrix. Where are those numbers?

[Image Analyst]

a(1, 1:300) = 1; % Set first 300 columns in row 1 to be 1.

Also look at the handy functions repelem() and repmat().

[Deividas Silgalis]

Something between homework and self-learning. Teacher gave me but it's not for grades.

I tried:

matrix0=rand(1500,1000);
matrix0(1:300, 1:1000) = 1;
matrix0(301:600, 1:1000) = 0.75;
matrix0(601:900, 1:1000) = 0.5;
matrix0(901:1200, 1:1000) = 0.25;
matrix0(1200:1500, 1:1000) = 0;
matrix0(1200:1500, 1:1000) = 0;

image(matrix0, 'CDataMapping', 'scaled');

And it works well.

Is there any way to make code smaller/not so clumsy?

[the cyclist]

Note that these lines

matrix0(1200:1500, 1:1000) = 0;
matrix0(1200:1500, 1:1000) = 0;

should be

matrix0(1201:1500, 1:1000) = 0;
matrix0(1201:1500, 1:1000) = 0;

[Image Analyst]

or, in one line of code and using the "end" keyword:

matrix0(1201:end, :) = 0;

I don't know that your code is clumsy. Sometimes if there are just a small handful of things to do it's often more readable and maintainable just to spell it out like you did. Anyone can instantly see what's going on there, which may not be the case for a more cryptic specialized function call, and it might be just as many lines, if not fewer than constructing some loop to do the same thing.

[the cyclist]

lol ... I somehow missed that those two lines of code were identical. :-)

Answer 1

[darova]

Use for loops

matrix0=rand(1500,1000);
val = [1:-0.25:0];
for i = 1:5
    ii = (i-1)*300+1 : i*300;
    matrix0(ii,:) = val(i);
end
image(matrix0, 'CDataMapping', 'scaled');

or more elegant

val = repmat(1:-0.25:0,[300 1]);            % copy 300 rows
% val(:) - make one column of 1500 elements (300x5)
val = repmat(val(:),[1 1000]);
image(val)

colon

repmat

Answer 2

[Image Analyst]

I don't know that your code is clumsy. Sometimes if there are just a small handful of things to do it's often more readable and maintainable just to spell it out like you did. Anyone can instantly see what's going on there, which may not be the case for a more cryptic specialized function call, and it might be just as many lines, if not fewer than constructing some loop to do the same thing.

If you'd really like to do it with MATLAB functions, and not explicity/directly like you did, then you can do this:

% Define the values we want to put into the matrix.
verticalProfile = [1; 0.75; 0.5; 0.25; 0]
% Replicate these vertically the full height of 1500 rows
% by duplicating each number 300 times.
column1 = repelem(verticalProfile, 300);
% Copy this line all the way across 1000 columns.
matrix0 = repmat(column1, [1, 1000]);

Not including comments, that's 3 lines of code instead of your 7 lines of code. But it's not as readable as yours since now someone has to know that repelem() and repmat() do. I think it definitely looks more cryptic.

Answer 3

[Deividas Silgalis]

Thanks you all for your answers. It helped me alot!