问题
Given a MATLAB uint32 to be interpreted as a bit string, what is an efficient and concise way of counting how many nonzero bits are in the string?
I have a working, naive approach which loops over the bits, but that's too slow for my needs. (A C++ implementation using std::bitset count() runs almost instantly).
I've found a pretty nice page listing various bit counting techniques, but I'm hoping there is an easy MATLAB-esque way.
http://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetNaive
Update #1
Just implemented the Brian Kernighan algorithm as follows:
w = 0;
while ( bits > 0 )
bits = bitand( bits, bits-1 );
w = w + 1;
end
Performance is still crappy, over 10 seconds to compute just 4096^2 weight calculations. My C++ code using count() from std::bitset does this in subsecond time.
Update #2
Here is a table of run times for the techniques I've tried so far. I will update it as I get additional ideas/suggestions.
Vectorized Scheiner algorithm => 2.243511 sec Vectorized Naive bitget loop => 7.553345 sec Kernighan algorithm => 17.154692 sec length( find( bitget( val, 1:32 ) ) ) => 67.368278 sec nnz( bitget( val, 1:32 ) ) => 349.620259 sec Justin Scheiner's algorithm, unrolled loops => 370.846031 sec Justin Scheiner's algorithm => 398.786320 sec Naive bitget loop => 456.016731 sec sum(dec2bin(val) == '1') => 1069.851993 sec
Comment: The dec2bin() function in MATLAB seems to be very poorly implemented. It runs extremely slow.
Comment: The "Naive bitget loop" algorithm is implemented as follows:
w=0;
for i=1:32
if bitget( val, i ) == 1
w = w + 1;
end
end
Comment: The loop unrolled version of Scheiner's algorithm looks as follows:
function w=computeWeight( val )
w = val;
w = bitand(bitshift(w, -1), uint32(1431655765)) + ...
bitand(w, uint32(1431655765));
w = bitand(bitshift(w, -2), uint32(858993459)) + ...
bitand(w, uint32(858993459));
w = bitand(bitshift(w, -4), uint32(252645135)) + ...
bitand(w, uint32(252645135));
w = bitand(bitshift(w, -8), uint32(16711935)) + ...
bitand(w, uint32(16711935));
w = bitand(bitshift(w, -16), uint32(65535)) + ...
bitand(w, uint32(65535));
回答1:
I'd be interested to see how fast this solution is:
function r = count_bits(n)
shifts = [-1, -2, -4, -8, -16];
masks = [1431655765, 858993459, 252645135, 16711935, 65535];
r = n;
for i=1:5
r = bitand(bitshift(r, shifts(i)), masks(i)) + ...
bitand(r, masks(i));
end
Going back, I see that this is the 'parallel' solution given on the bithacks page.
回答2:
Unless this is a MATLAB implementation exercise, you might want to just take your fast C++ implementation and compile it as a mex function, once per target platform.
回答3:
EDIT: NEW SOLUTION
It appears that you want to repeat the calculation for every element in a 4096-by-4096 array of UINT32 values. If this is what you are doing, I think the fastest way to do it in MATLAB is to use the fact that BITGET is designed to operate on matrices of values. The code would look like this:
numArray = ...your 4096-by-4096 matrix of uint32 values...
w = zeros(4096,4096,'uint32');
for iBit = 1:32,
w = w+bitget(numArray,iBit);
end
If you want to make vectorized versions of some of the other algorithms, I believe BITAND is also designed to operate on matrices.
The old solution...
The easiest way I can think of is to use the DEC2BIN function, which gives you the binary representation (as a string) of a non-negative integer:
w = sum(dec2bin(num) == '1'); % Sums up the ones in the string
It's slow, but it's easy. =)
回答4:
Implemented the "Best 32 bit Algorithm" from the Stanford link at the top. The improved algorithm reduced processing time by 6%. Also optimized the segment size and found that 32K is stable and improves time by 15% over 4K. Expect 4Kx4K time to be 40% of Vectorized Scheiner Algorithm.
function w = Ham(w)
% Input uint32
% Output vector of Ham wts
for i=1:32768:length(w)
w(i:i+32767)=Ham_seg(w(i:i+32767));
end
end
% Segmentation gave reduced time by 50%
function w=Ham_seg(w)
%speed
b1=uint32(1431655765);
b2=uint32(858993459);
b3=uint32(252645135);
b7=uint32(63); % working orig binary mask
w = bitand(bitshift(w, -1), b1) + bitand(w, b1);
w = bitand(bitshift(w, -2), b2) + bitand(w, b2);
w =bitand(w+bitshift(w, -4),b3);
w =bitand(bitshift(w,-24)+bitshift(w,-16)+bitshift(w,-8)+w,b7);
end
回答5:
Did some timing comparisons on Matlab Cody. Determined a Segmented Modified Vectorized Scheiner gives optimimum performance.
Have >50% time reduction based on Cody 1.30 sec to 0.60 sec change for an L=4096*4096 vector.
function w = Ham(w)
% Input uint32
% Output vector of Ham wts
b1=uint32(1431655765); % evaluating saves 15% of time 1.30 to 1.1 sec
b2=uint32(858993459);
b3=uint32(252645135);
b4=uint32(16711935);
b5=uint32(65535);
for i=1:4096:length(w)
w(i:i+4095)=Ham_seg(w(i:i+4095),b1,b2,b3,b4,b5);
end
end
% Segmentation reduced time by 50%
function w=Ham_seg(w,b1,b2,b3,b4,b5)
% Passing variables or could evaluate b1:b5 here
w = bitand(bitshift(w, -1), b1) + bitand(w, b1);
w = bitand(bitshift(w, -2), b2) + bitand(w, b2);
w = bitand(bitshift(w, -4), b3) + bitand(w, b3);
w = bitand(bitshift(w, -8), b4) + bitand(w, b4);
w = bitand(bitshift(w, -16), b5) + bitand(w, b5);
end
vt=randi(2^32,[4096*4096,1])-1;
% for vt being uint32 the floor function gives unexpected values
tic
v=num_ones(mod(vt,65536)+1)+num_ones(floor(vt/65536)+1); % 0.85 sec
toc
% a corrected method is
v=num_ones(mod(vt,65536)+1)+num_ones(floor(double(vt)/65536)+1);
toc
回答6:
A fast approach is counting the bits in each byte using a lookup table, then summing these values; indeed, it's one of the approaches suggested on the web page given in the question. The nice thing about this approach is that both lookup and sum are vectorizable operations in MATLAB, so you can vectorize this approach and compute the hamming weight / number of set bits of a large number of bit strings simultaneously, very quickly. This approach is implemented in the bitcount submission on the MATLAB File Exchange.
回答7:
Try splitting the job into smaller parts. My guess is that if you want to process all data at once, matlab is trying to do each operation on all integers before taking successive steps and the processor's cache is invalidated with each step.
for i=1:4096,
«process bits(i,:)»
end
回答8:
I'm reviving an old thread here, but I ran across this problem and I wrote this little bit of code for it:
distance = sum(bitget(bits, 1:32));
Looks pretty concise, but I'm scared that bitget is implemented in O(n) bitshift operations. The code works for what I'm going, but my problem set doesn't rely on hamming weight.
回答9:
num_ones=uint8(zeros(intmax('uint32')/2^6,1));
% one time load of array not implemented here
tic
for i=1:4096*4096
%v=num_ones(rem(i,64)+1)+num_ones(floor(i/64)+1); % 1.24 sec
v=num_ones(mod(i,64)+1)+num_ones(floor(i/64)+1); % 1.20 sec
end
toc
tic
num_ones=uint8(zeros(65536,1));
for i=0:65535
num_ones(i+1)=length( find( bitget( i, 1:32 ) ) ) ;
end
toc
% 0.43 sec to load
% smaller array to initialize
% one time load of array
tic
for i=1:4096*4096
v=num_ones(mod(i,65536)+1)+num_ones(floor(i/65536)+1); % 0.95 sec
%v=num_ones(mod(i,65536)+1)+num_ones(bitshift(i,-16)+1); % 16 sec for 4K*1K
end
toc
%vectorized
tic
num_ones=uint8(zeros(65536,1));
for i=0:65535
num_ones(i+1)=length( find( bitget( i, 1:32 ) ) ) ;
end % 0.43 sec
toc
vt=randi(2^32,[4096*4096,1])-1;
tic
v=num_ones(mod(vt,65536)+1)+num_ones(floor(vt/65536)+1); % 0.85 sec
toc
来源：https://stackoverflow.com/questions/1024904/calculating-hamming-weight-efficiently-in-matlab