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

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Mostafa Salah 2020-5-14

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https://ww2.mathworks.cn/matlabcentral/answers/525572-i-need-to-split-sequance-of-binary-numbers-to-groups-with-step-log2m

编辑： Ameer Hamza 2020-5-16

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

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Ameer Hamza 2020-5-14

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/525572-i-need-to-split-sequance-of-binary-numbers-to-groups-with-step-log2m#answer_432526

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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)

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Mostafa Salah 2020-5-16

在 MATLAB Online 中打开

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 2020-5-16

编辑：Ameer Hamza 2020-5-16

在 MATLAB Online 中打开

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