How to find vector with if loops and for loops

Given a vector u with the dimension 20x1, which consists of random variables from a uniform distribution (distributed on [0,1]) exists. Instructions: • Create a code in MATLAB, – which creates the vector u, – which contains both ’for loops’ and ’if loops’, – and for all values in the vector u indicates in which quarter the number lies.This is one of my mathlab final exam question, could you please help me,i just know that i should use "rand" function

[Torsten]

Hello Jack, back again ?

https://de.mathworks.com/matlabcentral/answers/1740320-how-to-create-if-loop?s_tid=srchtitle

[Ömer Durkut]

No,I am not him

[Torsten]

Ok, then - under the link given - you get the hints the forum is willing to give to solve this assignment.

[Ömer Durkut]

Thanks a lot

Answer 1

[Image Analyst]

Are you allowed to turn in other people's solutions as your own for your final exam question?

The first 5 values of the 20-element u lie in the first quarter, the next 5 in the second quarter, the third 5 in the third quarter, and the final 5 elements in indexes 16-20 obviously lie in the last quarter. But I think they want you to use a for loop

u = rand( % You said you got this
for k = 1 : length(u)
    if u(k)...........
        More code
    end
end

You can either store the quarter that the number lives in, in a vector called quarter, or maybe you just want to print out the quarter it's in using fprintf(). Or maybe you want to do both.

[Ömer Durkut]

n = 20;
r = zeros(n,1);
for i=1:n
    r(i) = rand;
    if r(i) < 0.25
        quarter(i) = ("erstes quartal")
    elseif
       r(i) >=0.25 && r(i) < 0.5;
        quarter(i) = ("zweites quartal")
    elseif
       r(i) >=0.5 && r(i) < 0.75
        quarter(i) =("drittes quartal")
    else
       r(i) >=0.75 && r(i) < 1;
        quarter(i) =("viertes quartal")

    end
end


% this is the code of torsten,we are allowed to use open source therefore
% ıt wont be a problem to turn in the answer. but there is a problem with
% the first elseif command what could it be?

[Image Analyst]

Do you want to identify the numbers that are in the quarters of your actual numbers? Or do you want the quarters to be quarters of the theoretically possible distribution?

[Ömer Durkut]

The second one

[Torsten]

@Ömer Durkut

Just use the numbers 1,2,3 and 4 instead of your assignments to "quarter(i)".

The last "else" doesn't need a condition:

else
  quarter(i) = 4;
end

[Ömer Durkut]

@Torsten thanks

[Ömer Durkut]

u = rand(20,1);
for k = 1 : length(u)
    if u(k) <= 0.25
        quarter(k):1;
    elseif u(k) > 0.25 && u(k) <= 0.50
        quarter(k):2;
    elseif u(k) > 0.50 && u(k) <= 0.75
        quarter(k):3;
    else
        quarter(k):4;
        
    end
end

%ı wrote this but it gives untitled4 warning what could it be?

[Ömer Durkut]

there is something ı am missing

[Image Analyst]

Do it this way:

n = 20;
r = rand(20,1);
for i=1:n
    if r(i) < 0.25
        quarter(i) = 1;
    elseif r(i) >=0.25 && r(i) < 0.5
        quarter(i) = 2;
    elseif r(i) >=0.5 && r(i) < 0.75
        quarter(i) = 3;
    else
        r(i) >=0.75 && r(i) < 1;
        quarter(i) = 4;
    end
end
quarter

[Ömer Durkut]

thanks a lot,you've been so helpful to me :)

[Torsten]

@Ömer Durkut

Delete the line

r(i) >=0.75 && r(i) < 1;

in the else-section.

[Ömer Durkut]

thanks a lot

[Image Analyst]

OK, if this works, come back and tell us and click the "Accept this answer" link. Thanks in advance. 🙂

n = 20;
r = rand(20,1);
for i=1:n
    if r(i) < 0.25
        quarter(i) = 1;
    elseif r(i) >=0.25 && r(i) < 0.5
        quarter(i) = 2;
    elseif r(i) >=0.5 && r(i) < 0.75
        quarter(i) = 3;
    else
        quarter(i) = 4;
    end
end
quarter

[Jan]

    if r(i) < 0.25
        quarter(i) = 1;
    elseif r(i) >=0.25 && r(i) < 0.5
        %  ^^^^^^^^^^^^^^  This was excluded already

[Image Analyst]

@Jan, true. I just basically took the OP's code but an experienced programmer would do it more like this:

n = 20;
r = rand(n,1);
for k = 1 : numel(r)
    if r(k) < 0.25
        quarter(k) = 1;
    elseif r(k) < 0.50
        quarter(k) = 2;
    elseif r(k) < 0.75
        quarter(k) = 3;
    else % r(k) >= 0.75
        quarter(k) = 4;
    end
end
quarter

[Torsten]

I like if the order of the conditions in if-statements can be changed.

This is not the case in this simplified version because

if r(k) < 0.75
    quarter(k) = 3;
elseif r(k) < 0.25
    quarter(k) = 1;
elseif r(k) < 0.5
    quarter(k) = 2;
else 
    quarter(k) = 4;
end

would create chaos.