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How to convert binary elements of a cell into decimal number

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Hello, I have a cell of binary elements with 0's and 1's of size 1* i have to convert every 8 binary elements from 1624 elements into a decimal number and store in an,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

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;
Jyothi Alugolu
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*, 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 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 case of 3rd cell,the 109th column contains 4 elements,these 4 elements also must convert into decimal decimal vector must contain size of 1*140,1*168,1*109,1*102....

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More Answers (3)

Thibaut Jacqmin
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}]);
Thibaut Jacqmin
Thibaut Jacqmin on 8 Feb 2017
You don't have to write 203 in the for loop. Indeed size(a, 2) already equals to 203 after reshaping.
Jyothi Alugolu
Jyothi Alugolu on 10 Feb 2017
if suppose, if the binary text file with 0's and 1's is of size 1* 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,finally the output must be 16*102..since 101 columns will have each 16 bit values and 102th column must have remaining 8 i need to convert these 102 column values into decimal values...the final output must be a row vector of size 1* to do this?

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Alexandra Harkai
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, ...};
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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