OverflowError:数学范围错误-指数Python
OverflowError:数学范围错误-指数Python
提问于 2017-09-17 04:44:28
我试图使用python创建一个简单的模拟退火搜索，但是在使用math.exp计算指数时总是会出现溢出错误。
这是我在python中的代码：
import random
import math
def getF(x1, x2):
t1 = (4 - (2.1 * (x1 ** 2) + (x1 ** 4) / 3)) * (x1 ** 2)
t2 = x1 * x2
t3 = (-4 + (4 * (x2 ** 2))) * (x2 ** 2)
t = t1 + t2 + t3
return t
def getP(dE, t):
return math.exp((-1*dE)/t)
def getNewRandInRange(x):
newX = x + random.randint(-5, 5)
while (newX > 10) or (newX < -10):
newX = x + random.randint(-5, 5)
return newX
initState1 = random.randint(-10, 10)
initState2 = random.randint(-10, 10)
currentState1 = initState1
currentState2 = initState2
BSF = getF(currentState1, currentState2)
T = 1000
Tmin = 1
while T > Tmin:
print("T= %f" %T)
newState1 = getNewRandInRange(currentState1)
newState2 = getNewRandInRange(currentState2)
currentF = getF(currentState1, currentState2)
newF = getF(newState1, newState2)
print("Current X1= %f" % currentState1)
print("Current X2= %f" % currentState2)
print("New X1= %f" % newState1)
print("New X2= %f" % newState2)
dE = currentF - newF
print ("delta E: %f" %dE)
if dE > 0:
currentState1 = newState1
currentState2 = newState2
BSF = getF(newState1, newState2)
else:
randNumber = random.uniform(0, 1)
p = getP(dE, T)
if (randNumber < p):
currentState1 = newState1
currentState2 = newState2
print("BSF: %f" %BSF)
print("\n\n\n")
T = T * 0.9
print(BSF) #final output错误信息：
Traceback (most recent call last):
return math.exp((-1*dE)/t)
OverflowError: math range error我尝试使用try和catch，但是它不会返回指数，它会对结果产生问题，我也在谷歌搜索，但没有找到任何符合我的要求的解决方案。
之前谢谢你！
回答 1
Stack Overflow用户
回答已采纳
发布于 2017-09-17 04:59:34
异常OverflowError
当算术运算的结果太大而无法表示时引发。对于长整数(它宁愿提高MemoryError而不是放弃)和大多数使用普通整数返回长整数的操作，都不会发生这种情况。由于C语言中浮点异常处理缺乏标准化，大多数浮点操作也没有被检查。参考
你试着计算一个大的数字(大于710)，这超出了双倍的范围。
您可以像这样使用try/except来处理它：
def getP(dE, t):
try:
return math.exp((-1*dE)/t)
except:
return -1 # or anything else :D您可以在Python的代码中找到这个注释：
/*
* For the sake of simplicity and correctness, we impose an artificial
* limit on ndigits, the total number of hex digits in the coefficient
* The limit is chosen to ensure that, writing exp for the exponent,
*
* (1) if exp > LONG_MAX/2 then the value of the hex string is
* guaranteed to overflow (provided it's nonzero)
*
* (2) if exp < LONG_MIN/2 then the value of the hex string is
* guaranteed to underflow to 0.
*
* (3) if LONG_MIN/2 <= exp <= LONG_MAX/2 then there's no danger of
* overflow in the calculation of exp and top_exp below.
*
* More specifically, ndigits is assumed to satisfy the following
* inequalities:
*
* 4*ndigits <= DBL_MIN_EXP - DBL_MANT_DIG - LONG_MIN/2
* 4*ndigits <= LONG_MAX/2 + 1 - DBL_MAX_EXP
*
* If either of these inequalities is not satisfied, a ValueError is
* raised. Otherwise, write x for the value of the hex string, and
* assume x is nonzero. Then
*
* 2**(exp-4*ndigits) <= |x| < 2**(exp+4*ndigits).
*
* Now if exp > LONG_MAX/2 then:
*
* exp - 4*ndigits >= LONG_MAX/2 + 1 - (LONG_MAX/2 + 1 - DBL_MAX_EXP)
* = DBL_MAX_EXP
*
* so |x| >= 2**DBL_MAX_EXP, which is too large to be stored in C
* double, so overflows. If exp < LONG_MIN/2, then
*
* exp + 4*ndigits <= LONG_MIN/2 - 1 + (
* DBL_MIN_EXP - DBL_MANT_DIG - LONG_MIN/2)
* = DBL_MIN_EXP - DBL_MANT_DIG - 1
*
* and so |x| < 2**(DBL_MIN_EXP-DBL_MANT_DIG-1), hence underflows to 0
* when converted to a C double.
*
* It's easy to show that if LONG_MIN/2 <= exp <= LONG_MAX/2 then both
* exp+4*ndigits and exp-4*ndigits are within the range of a long.
*/不管怎样，你可以使用Decimal
import decimal
...
def getP(dE, t):
return decimal.Decimal((-1*dE)/t).exp()页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/46260731
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