[英]Precise math.log(x, 2) in Python
In python, I need to get the rounded down logarithm of positive integers for base 2, including big numbers. 在python中,我需要得到基数为2的正整数的向下舍入,包括大数。
However, since floating point math is used, I might get bad results, for example: 但是,由于使用了浮点数学,我可能会得到不好的结果,例如:
>>> import math
>>> int(math.log(281474976710655, 2))
48
However: 然而:
>>> 2 ** 48
281474976710656
So the correct result, rounded down, should be 47. 因此,向下舍入的正确结果应为47。
How can I get the correct value? 我怎样才能得到正确的价值?
In Python 3, integers have a .bit_length
method, so you should use that to get your rounded down base 2 logarithm. 在Python 3中,整数有一个
.bit_length
方法,所以你应该使用它来得到你的舍入基数2对数。
Here's a short demo: 这是一个简短的演示:
m = 2 ** 1000
for n in (281474976710655, m-1, m, m+1):
a = n.bit_length() - 1
b = 2 ** a
print(a, b <= n < 2 * b)
output 产量
47 True
999 True
1000 True
1000 True
In python 3 int
s even have an efficient .bit_length()
method! 在python 3
int
甚至有一个高效的.bit_length()
方法!
>>> (281474976710655).bit_length()
48
>>> (281474976710656).bit_length()
49
In python 2, instead of using floating point math, count the number of bits: 在python 2中,不使用浮点数学,而是计算位数:
def log2(n):
assert n >= 1
return len(bin(n)) - 3 # bin() returns a string starting with '0b'
( Edited following this comment ) (在此评论后 编辑 )
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