i have a single row vector contains 0s and 1s and want to convert every 3 bits into integer how can i do that??

i have a single row vector contains 0s and 1s and want to convert every 3 bits in this vector into integer how can i do that??

i have single row vector contains 0s and 1s values and i need every 3 bits in this vector convert to decimal number

for example

x=[0 1 0 1 1 1 ]

then i want this single row to be converted to decimel every 3 of them to decimal [010],[111]

Answer 1

[Mathieu NOE]

hello

see below

hope it helps

% dummy data
nbits = 3;
samples = 8;
x = round(rand(1,nbits*samples));
nbits_vect = 2.^((nbits-1:-1:0));


%%%% main loop %%%%
iter = fix((length(x)-nbits)/nbits +1);

for ci=1:iter
    start_index = 1+(ci-1)*nbits;
    stop_index = min(start_index+ nbits-1,length(x));
    int_data(ci) = sum(nbits_vect.*x(start_index:stop_index));  % your interger data  

end
figure(1),
plot(int_data,'r'); % your interger data 

[Mathieu NOE]

Forgot to say that x in my code is the single row vector contains 0s and 1s

[Aly Khafagy]

to be more clear, i meant that i have single row vector contains 0s and 1s values and i need every 3 bits in this vector convert to decimal number

do you understand me??

for example

x=[0 1 0 1 1 1 ]

then i want this single row to be converted to decimel every 3 of them to decimal [010],[111]

do you understand me??

[Mathieu NOE]

hello again

yes I understand you

the 3 consecutives bits (like [010]) is what is done with x(start_index:stop_index)

and if you want to have the corresponding integer (as on your original post) this is int_data

[010] = 2 (integer) , so int_data = 2

I do this in a for loop until the end of your input data is reached

[Aly Khafagy]

i am sorry for annoying you but i will be grateful if you can explain to me the code as i dont understand it.

[Aly Khafagy]

hello Mathew, i need to know what is meant by nbits_vect ?? i cannot understand this variable.

[Bruno Luong]

it's equivalent to the sequence

[100,10,1] but in base 2.

Answer 2

[Bruno Luong]

x=[0 1 0 1 1 1 ];
bin2dec(char(reshape(x,3,[])+'0')')

ans = 2×1

2 7

Answer 3

[Voss]

You say you want [010,111] as a result:

x=[0 1 0 1 1 1 ];

xx = reshape(x,3,[]).';
y = zeros(1,size(xx,1));
for ii = 1:size(xx,1)
    y(ii) = polyval(xx(ii,:),10);
end

disp(y)

    10   111

If you want [2,7] as a result:

for ii = 1:size(xx,1)
    y(ii) = polyval(xx(ii,:),2);
end

disp(y)

     2     7