正如标题所述,我想知道math.log2(x)
的时间复杂度是多少。我知道可以用O(1)复杂性在C中编写这样的函数,但是我找不到有关在python中实现该函数的任何信息。
谢谢!
[在Python C2的CPython实现中,log2被实现为以下C函数,在C之上的一层中处理错误报告并专门处理整数,但最终即使在整数情况下,它也是执行对数的以下代码。 >
如果可以使用逻辑C,则逻辑上基本上使用标准C log2函数,否则根据log计算log2。在任何情况下,它都是O(1),但是由于所有检查和消毒层的原因,它具有相对较高的恒定因子。
/*
log2: log to base 2.
Uses an algorithm that should:
(a) produce exact results for powers of 2, and
(b) give a monotonic log2 (for positive finite floats),
assuming that the system log is monotonic.
*/
static double
m_log2(double x)
{
if (!Py_IS_FINITE(x)) {
if (Py_IS_NAN(x))
return x; /* log2(nan) = nan */
else if (x > 0.0)
return x; /* log2(+inf) = +inf */
else {
errno = EDOM;
return Py_NAN; /* log2(-inf) = nan, invalid-operation */
}
}
if (x > 0.0) {
#ifdef HAVE_LOG2
return log2(x);
#else
double m;
int e;
m = frexp(x, &e);
/* We want log2(m * 2**e) == log(m) / log(2) + e. Care is needed when
* x is just greater than 1.0: in that case e is 1, log(m) is negative,
* and we get significant cancellation error from the addition of
* log(m) / log(2) to e. The slight rewrite of the expression below
* avoids this problem.
*/
if (x >= 1.0) {
return log(2.0 * m) / log(2.0) + (e - 1);
}
else {
return log(m) / log(2.0) + e;
}
#endif
}
else if (x == 0.0) {
errno = EDOM;
return -Py_HUGE_VAL; /* log2(0) = -inf, divide-by-zero */
}
else {
errno = EDOM;
return Py_NAN; /* log2(-inf) = nan, invalid-operation */
}
}