numpy vectorized way to count non-zero bits in array of integers

Question

I have an array of integers:

[int1, int2, ..., intn]

I want to count how many non-zero bits are in the binary representation of these integers.

For example:

bin(123) -> 0b1111011, there are 6 non-zero bits

Of course I can loop over list of integers, use bin() and count('1') functions, but I'm looking for vectorized way to do it.

Answer 1

Assuming your array is a, you can simply do:

np.unpackbits(a.view('uint8')).sum()

example:

a = np.array([123, 44], dtype=np.uint8)
#bin(a) is [0b1111011, 0b101100]
np.unpackbits(a.view('uint8')).sum()
#9

---

Comparison using benchit:

#@Ehsan's solution
def m1(a):
  return np.unpackbits(a.view('uint8')).sum()

#@Valdi_Bo's solution
def m2(a):
  return sum([ bin(n).count('1') for n in a ])

in_ = [np.random.randint(100000,size=(n)) for n in [10,100,1000,10000,100000]]

m1 is significantly faster.