Raymond Hettinger <[email protected]> added the comment:
FWIW, here are two approaches to getting an equi-distributed version of
int(n*random()) where 0 < n <= 2**53. The first mirrors the approach currently
in the code. The second approach makes fewer calls to random().
def rnd1(n):
assert 0 < n <= 2**53
N = 1 << (n-1).bit_length()
r = int(N * random())
while r >= n:
r = int(N * random())
return r
def rnd2(n, N=1<<53):
assert 0 < n <= N
NN = N - (N % n) + 0.0 # largest multiple of n <= N
r = N * random() # a float with an integral value
while r >= NN:
r = N * random()
return int(r) % n
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Python tracker <[email protected]>
<http://bugs.python.org/issue9025>
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