bit manipulation - Calculating Hamming weight efficiently in matlab

Question

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));

Answer 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.