I need to count the number of trailing and leading zeros in a numpy uint64 variable, so right now I'm doing it like this:
# n > 0
n = np.uint64(100)
s = np.binary_repr(n)
trail_zeros = len(s) - len(s.rstrip('0'))
lead_zeros = 64 - len(s)
Is there a better way of doing this, without using strings? The priority is speed. Thank you!
For numbers in [0,2**63), we can use some arithmetic operations to get the leading and trailing zeros in their binary formats and hence skip the string manipulations -
def get_leading_trailing_zeros(n):
a = (2**np.arange(64) & n)
lead_zeros = 64-a.argmax()-1
if n==0:
trail_zeros = 1
else:
trail_zeros = (a==0).argmin()
return lead_zeros,trail_zeros
I am not sure about the speed of the following code. But you can certainly do it like this, without using strings.
n = np.uint64(100)
i=1
while((n>>np.uint64(i))%2==0):
i+=1
trail_zeros=i
You right shift the value n until you get an odd number. Number of right shifts done is equal to trail_zeros.
lead_zeros = int(64-np.ceil(np.log2(n)))
Because len(s) is equal to ceil(log2(n)). This is pure arithmetic operation so it can be perfectly vectorized by numpy and much faster than writing your own loop.
n = np.uint64(0).
trail_zeros = 63-lead_zeros-s.rfind('1')seems to be a little faster than your solution, still strings thon = np.uint64(0)?