Fast way of counting non-zero bits in positive integer

Python 3.10 introduces int.bit_count():

>>> n = 19
>>> bin(n)
'0b10011'
>>> n.bit_count()
3
>>> (-n).bit_count()
3

This is functionally equivalent to bin(n).count("1") but should be ~6 times faster. See also Issue29882.

For arbitrary-length integers, bin(n).count("1") is the fastest I could find in pure Python.

I tried adapting Óscar's and Adam's solutions to process the integer in 64-bit and 32-bit chunks, respectively. Both were at least ten times slower than bin(n).count("1") (the 32-bit version took about half again as much time).

On the other hand, gmpy popcount() took about 1/20th of the time of bin(n).count("1"). So if you can install gmpy, use that.

To answer a question in the comments, for bytes I'd use a lookup table. You can generate it at runtime:

counts = bytes(bin(x).count("1") for x in range(256))  # py2: use bytearray

Or just define it literally:

counts = (b'\x00\x01\x01\x02\x01\x02\x02\x03\x01\x02\x02\x03\x02\x03\x03\x04'
          b'\x01\x02\x02\x03\x02\x03\x03\x04\x02\x03\x03\x04\x03\x04\x04\x05'
          b'\x01\x02\x02\x03\x02\x03\x03\x04\x02\x03\x03\x04\x03\x04\x04\x05'
          b'\x02\x03\x03\x04\x03\x04\x04\x05\x03\x04\x04\x05\x04\x05\x05\x06'
          b'\x01\x02\x02\x03\x02\x03\x03\x04\x02\x03\x03\x04\x03\x04\x04\x05'
          b'\x02\x03\x03\x04\x03\x04\x04\x05\x03\x04\x04\x05\x04\x05\x05\x06'
          b'\x02\x03\x03\x04\x03\x04\x04\x05\x03\x04\x04\x05\x04\x05\x05\x06'
          b'\x03\x04\x04\x05\x04\x05\x05\x06\x04\x05\x05\x06\x05\x06\x06\x07'
          b'\x01\x02\x02\x03\x02\x03\x03\x04\x02\x03\x03\x04\x03\x04\x04\x05'
          b'\x02\x03\x03\x04\x03\x04\x04\x05\x03\x04\x04\x05\x04\x05\x05\x06'
          b'\x02\x03\x03\x04\x03\x04\x04\x05\x03\x04\x04\x05\x04\x05\x05\x06'
          b'\x03\x04\x04\x05\x04\x05\x05\x06\x04\x05\x05\x06\x05\x06\x06\x07'
          b'\x02\x03\x03\x04\x03\x04\x04\x05\x03\x04\x04\x05\x04\x05\x05\x06'
          b'\x03\x04\x04\x05\x04\x05\x05\x06\x04\x05\x05\x06\x05\x06\x06\x07'
          b'\x03\x04\x04\x05\x04\x05\x05\x06\x04\x05\x05\x06\x05\x06\x06\x07'
          b'\x04\x05\x05\x06\x05\x06\x06\x07\x05\x06\x06\x07\x06\x07\x07\x08')

Then it's counts[x] to get the number of 1 bits in x where 0 ≤ x ≤ 255.