bitget for array (count no. of bits)
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Aravin
el 7 de Ag. de 2012
Respondida: Oliver Woodford
el 22 de Nov. de 2013
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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Matt Fig
el 7 de Ag. de 2012
Editada: Matt Fig
el 8 de Ag. de 2012
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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Matt Fig
el 8 de Ag. de 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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Oliver Woodford
el 22 de Nov. de 2013
I assume from your question that you actually want to do a vectorized bitcount, rather than a vectorized bitget. If that is correct then I suggest you use the bitcount submission on the File Exchange.
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