Problem introducing user input into an array

Hey guys, im new to matlab and i have a problem. I have to create a simulator of an RZ signal coding.

I have made the part where the user adds the 1´s and 0´s but when i want to creat a array to assing values to the 1´s and 0´s i just dont ge it why its not working

heres the code

% ask the user
bny = input('Insira o codigo a codificar: ' ,'s');

   % Verifyif only 0´s and 1´s
   inv_c=regexp(bny,'[a-z_A-Z2-9]');
   if(~isempty(inv_c))
      disp(['O seu código não é válido: ' bny(inv_c) ' Valores não validos= ' num2str(inv_c)])
   else
      % its valid
      disp(['your code is : ', bny ])
   end
   for i = 1 : length(bny)
       if bny(i) == 1
            n(i)=3;
       else
            n(i)=-3;
       end
   end

[Walter Roberson]

Note: I recommend you change

inv_c=regexp(bny,'[a-z_A-Z2-9]');

to

inv_c = regexp(bny, '[^01]');

for example ';' is not one of the characters you listed and so your checking would permit through ';'.

[António Pereira]

thanks for the note!

Answer 1

[Stephen23]

Simpler:

bny = input('Insira o codigo a codificar: ' ,'s');
idx = ismember(bny,'01');
assert(all(idx),'Input must contain only 0s and 1s, but it contains %s',bny(~idx))
fprintf('Your code is: %s\n',bny)
V = [-3,3];
n = V(bny-'/')

[António Pereira]

That code is giving me a error, "Index exceeds the number of array elements (2)."

[Stephen23]

@António Pereira: I fixed the last line of code and tested it:

Insira o codigo a codificar: 1101101
Your code is: 1101101
n =
   3   3  -3   3   3  -3   3

[António Pereira]

Thanks a lot!!! Not only it looks nicer as it works nicely

Answer 2

[Walter Roberson]

You are reading in characters. You have

if bny(i) == 1

but this is comparing the character in bny(i) to the number 1. The character is '0' or '1' which are not numeric 0 or 1 -- they are numeric 48 and 49.

You can vectorize your code.

    n = 3 * ((bny - '0') * 2 - 1)

or somewhat more obscurely,

    n = bny * 6 - 291;

[António Pereira]

Can you explain why do that way??

[Walter Roberson]

bny is '0' or '1'. When you subtract '0' then you get numeric 0 or 1. Multiply by 2 to get numeric 0 or 2. Subtract one from that to get numeric -1 or +1 . Multiply by 3 to get -3 or +3.

The second form with the 291 is an optimization of the first one. '0' is equal to 48 and 291 is 48*6 - 3 -- i.e., (6 * bny - 6 * '0' - 3) which is the result of expanding 3 * ((bny - '0') * 2 - 1)

[António Pereira]

I got it, someone showed me a way to make it work, but thanks anyway for your time