bitget for array (count no. of bits)
Asked by Aravin on 7 Aug 2012
Latest activity Commented on by Aravin on 8 Aug 2012
Hi everyone,
How to take bitget of an array. Let say:
a = 7; s = sum(bitget(a,1:8));
will return 3 as there are three bits on in number 7. If I have any array
a = [1 2 7 6];
then how can I do the similar bit count operation for each element in a.
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1 Answer
Answer by Matt Fig on 7 Aug 2012
Edited by Matt Fig on 8 Aug 2012
Accepted answer
ARRAYFUN might be useful here.
arrayfun(@(x) sum(bitget(x,1:8)),[1 2 7 6])
This is much faster for larger row vectors. D is the number of bits (8 in the above formula):
sum(rem(floor(bsxfun(@times,x',pow2(1-D:0))),2),2);
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Aravin on 8 Aug 2012
Thanks Teja,
in my case my X is uint8 array (matrix). and it returns me the error : ??? Error using ==> bsxfun Mixed integer class inputs are not supported.
I don't want to change my x into double. is it possible to continue with uint8 ??
Matt Fig on 8 Aug 2012
Note carefully that the algorithm Teja gave you will automatically return all bits. I understood you to want to specify the bits to sum up. Either way, whether you specify D or calculate it using
D = floor(log2(max(x(:)))) + 1;
you can simply use reshape like Teja did to work the algorithm on a matrix. As for the uint8 part, there is no need to change your actual matrix to double, just change the copy used by the algorithm. For example:
x = uint8(round(rand(10,20)*255)); % Your matrix: we preserve this. S = double(x(:)); % Only temporary. Overwritten below. D = floor(log2(max(S))) + 1; % Or specify D! (As you need.) S = reshape(sum(rem(floor(bsxfun(@times,S,pow2(1-D:0))),2),2),size(x));
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