# How to convert binary elements of a cell into decimal number

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### Accepted Answer

Jan
on 8 Feb 2017

In = {0,1,1,1,0,0,1,0, 1,0,1,1,0,1,0,1}; % 1 x 16

M = reshape([In{:}], 8, []);

Out = [128, 64, 32, 16, 8, 4, 2, 1] * M;

### More Answers (3)

Thibaut Jacqmin
on 8 Feb 2017

Edited: Thibaut Jacqmin
on 8 Feb 2017

Here I create a cell of size 16 (and not 1624) as an example

a = {1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1};

Reshape the cell in a 8xN cell (N = 203 in your case)

a = reshape(a, [8, length(a)/8]);

Convert each column of 8 binary in a decimal number and store it in res

res = [];

for i = 1:size(a, 2)

res(end+1) = bi2de([a{:, i}]);

end

Alexandra Harkai
on 8 Feb 2017

Considering it's a 1*1624 numeric array you have in the cell A, you can utilise bin2dec to do it after converting the numeric values to a string representation with num2str:

{bin2dec(num2str(reshape(A{:}, 8, size(A{:},2)/8)'))}'

Guillaume
on 8 Feb 2017

Edited: Guillaume
on 8 Feb 2017

Question: Why is the data in a cell array when a normal matrix would work just as well and make the code simpler? That is instead of:

binaryvector = {1, 0, 0, 1, 0, 1, 1, ...};

Have

binaryvector = [1, 0, 0, 1, 0, 1, 1, ...];

You also haven't told us which bit (1st or 8th) is the LSB and MSB.

Anyway, another way, probably the fastest, to convert your array:

binaryvector = num2cell(randi([0 1], 1, 1624)); %create a cell array of random bits for demonstration only. Use your own data

binaryvector = cell2mat(binaryvector); %convert cell array into matrix as cell array is pointless and just makes manipulating the bits harder

%in version R2016b ONLY:

decimalvector = sum(2.^(7:-1:0)' .* reshape(binaryvector, 8, []));

%in version prior to R2016b:

decimalvector = sum(bsxfun(@times, 2.^(7:-1:0)', reshape(binaryvector, 8, [])));

The above assumes the LSB is the 8th bit, if it's the first bit replace the 7:-1:0 by 0:7

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