I need to split sequance of binary numbers to groups with step log2m

I need to split sequance of binary numbers to groups with step log2m

for example i need to let user input sequance of bits in array then this squance divide into groups with step log2m and each group tansfer to decimal and put them in array again

Answer 1

[Ameer Hamza]

Run this example

x = '100100111100001111';
m = 8;
n = floor(log2(m));
x = [repmat('0', 1, ceil(numel(x)/n)*n-numel(x)) x]; % pad values so that digits
                                                     % can be evenly divided in
                                                     % groups of m
x = reshape(x, n, []).';
y = bin2dec(x)

[Mostafa Salah]

I need x = '100100111100001111'; array not string

[Ameer Hamza]

What is the difference?

[Mostafa Salah]

for example x=[0 1 0 1 0 0 ]

i need to divide it

[Ameer Hamza]

Try this

x = [1 0 0 1 0 1 1 1 1 0 0 0 0 1 1 1 1];
m = 8;
n = floor(log2(m));
% pad zeros so that length become multiple of n
if ~(round(numel(x)/n)==numel(x)/n)
    x = [zeros(1, ceil(numel(x)/n)*n-numel(x)) x]; %
end

x = reshape(x, n, []).';
y = x*(2.^(n-1:-1:0)).';% multiply with [2^(n-1) 2^(n-1) ... 2^1 2^0] to convert binary to decimal

[Mostafa Salah]

It is work but when you put m=8 thats mean take first 3 elements then convert to decimal but the answer is not correct

first two in array is not correct beacuse it must be 4 according to 100

[Mostafa Salah]

and can you put comment please to help me understand the code ?

[Ameer Hamza]

Because there are 17 elements in 'x', which is not evenly divided by 3, so I padded the array with 0 on the left side to make it length 18 (divisible of 3). So the actual value of 'x' is

x = [0 1 0 0 1 0 1 1 1 1 0 0 0 0 1 1 1 1];

I also added a few comments. The rest of the code is quite simple. Just run those lines and see the output.

[Mostafa Salah]

okay thank you ,you mean this code if i put number of elements less than number of divsion the code will increase zeros which is perfect . if i need to get input from user in array

xi=[] ;
num=input ('please input number of values you need to enter']; 
for i=1:num xii =  input('Input value ');
  xi(end+1)=xii; end

is this a good code to get input zeros and ones from user ?

[Ameer Hamza]

The correct way depends on your requirement. If you just want to ask for each digit at a time, then this method is correct. However, I will add a check so that if the user enters anything other than 0 or 1, it will discard that value and tell the user to enter the correct value.

while true
    num = input('please input number of values you need to enter: ');
    if isnumeric(num) && round(num)==num % check if it is integer
        break;
    else
        fprintf('Value must be an integer\n');
    end
end
xi = zeros(1, num);
for i=1:num 
    while true
        xii = input('Input value [0 or 1]: ');
        if isnumeric(xii) && any(xii==[0 1]) % check if it is integer
            break;
        else
            fprintf('Value must be 0 or 1\n');
        end
    end
    xi(i)=xii;
end

[Mostafa Salah]

great i need the last thing to connect this code with the another one what i mean the user input the number of zeros and ones and then divide them accoeding to log2M please thank you <3

x = [1 0 0 1 0 1 1 1 1 0 0 0 0 1 1 1 1];
m = 8;
n = floor(log2(m));
% pad zeros so that length become multiple of n
if ~(round(numel(x)/n)==numel(x)/n)
    x = [zeros(1, ceil(numel(x)/n)*n-numel(x)) x]; %
end

x = reshape(x, n, []).';
y = x*(2.^(n-1:-1:0)).';% multiply with [2^(n-1) 2^(n-1) ... 2^1 2^0] to convert binary to decimal

[Ameer Hamza]

while true
    num = input('please input number of values you need to enter: ');
    if isnumeric(num) && round(num)==num % check if it is integer
        break;
    else
        fprintf('Value must be an integer\n');
    end
end
x = zeros(1, num);
for i=1:num 
    while true
        xii = input('Input value [0 or 1]: ');
        if isnumeric(xii) && any(xii==[0 1]) % check if it is integer
            break;
        else
            fprintf('Value must be 0 or 1\n');
        end
    end
    x(i)=xii;
end

m = 8;
n = floor(log2(m));
% pad zeros so that length become multiple of n
if ~(round(numel(x)/n)==numel(x)/n)
    x = [zeros(1, ceil(numel(x)/n)*n-numel(x)) x]; %
end
x = reshape(x, n, []).';
y = x*(2.^(n-1:-1:0)).';% multiply with [2^(n-1) 2^(n-1) ... 2^1 2^0] to convert binary to decimal