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

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Jyothi Alugolu on 8 Feb 2017
Commented: Jyothi Alugolu on 15 Feb 2017
Hello, I have a cell of binary elements with 0's and 1's of size 1*1624..now i have to convert every 8 binary elements from 1624 elements into a decimal number and store in an array..so,the final output must be of size 1*203(each element in 203 must consists of the value of each 8 binary elements...can anyone help me please
Stephen23 on 8 Feb 2017

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;
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Jyothi Alugolu on 15 Feb 2017
one more question...i have 4 binary(0's and 1's ) cells of sizes 1*1120,1*1344,1*868,1*812...now, i need to split or do partition on each cell i.e the entire cell must divide into each 8 bits,so that output of cell's must be of size first cell:8*140, second cell: 8*168, third cell must be 8*109..109 because 108 columns contains 108*8=864 binary numbers,and there will remain 4 binary number's..these 4 binary numbers must store in another column i.e 109th column..so third cell size must be 8*109..and fourth cell size must be 8*102 ...and finally i need to calculate decimal value for each splitted file..in case of 3rd cell,the 109th column contains 4 elements,these 4 elements also must convert into decimal value...final decimal vector must contain size of 1*140,1*168,1*109,1*102....

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
Jyothi Alugolu on 10 Feb 2017
if suppose, if the binary text file with 0's and 1's is of size 1*1624.now i need to reshape this 1624 binary elements into 16 *N (N=1624),then i will have 16*101 cell and 8 bits will be remaining..these 8 bits must store in another cell..so,finally the output must be 16*102..since 101 columns will have each 16 bit values and 102th column must have remaining 8 bits..now i need to convert these 102 column values into decimal values...the final output must be a row vector of size 1*102..how to do this?

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)'))}'
Jyothi Alugolu on 8 Feb 2017
can you explain it properly??

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
Jan on 8 Feb 2017
You can omit the sum(), when you use a matrix multiplication.

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