根据您的特定需求,将输入参数裁剪为exp()
. 如果您真的想在inf
溢出时退出,或者您想获得荒谬的巨大数字,那么其他答案将更合适。
def powellBadlyScaled(X):
f1 = 10**4 * X[0] * X[1] - 1
f2 = numpy.exp(-numpy.float(X[0])) + numpy.exp(-numpy.float(X[1])) - 1.0001
return f1 + f2
def powellBadlyScaled2(X):
f1 = 10**4 * X[0] * X[1] - 1
arg1 = -numpy.float(X[0])
arg2 = -numpy.float(X[1])
too_big = log(sys.float_info.max / 1000.0) # The 1000.0 puts a margin in to avoid overflow later
too_small = log(sys.float_info.min * 1000.0)
arg1 = max([min([arg1, too_big]), too_small])
arg2 = max([min([arg2, too_big]), too_small])
# print(' too_small = {}, too_big = {}'.format(too_small, too_big)) # Uncomment if you're curious
f2 = numpy.exp(arg1) + numpy.exp(arg2) - 1.0001
return f1 + f2
print('\nTest against overflow: ------------')
x = [-1e5, 0]
print('powellBadlyScaled({}) = {}'.format(x, powellBadlyScaled(x)))
print('powellBadlyScaled2({}) = {}'.format(x, powellBadlyScaled2(x)))
print('\nTest against underflow: ------------')
x = [0, 1e20]
print('powellBadlyScaled({}) = {}'.format(x, powellBadlyScaled(x)))
print('powellBadlyScaled2({}) = {}'.format(x, powellBadlyScaled2(x)))
结果:
Test against overflow: ------------
*** overflow encountered in exp
powellBadlyScaled([-100000.0, 0]) = inf
powellBadlyScaled2([-100000.0, 0]) = 1.79769313486e+305
Test against underflow: ------------
*** underflow encountered in exp
powellBadlyScaled([0, 1e+20]) = -1.0001
powellBadlyScaled2([0, 1e+20]) = -1.0001
请注意,powellBadlyScaled2
原始powellBadlyScaled
版本没有上溢/下溢,但修改后的版本给出了1.79769313486e+305
而不是inf
在其中一个测试中给出。我想有很多应用程序1.79769313486e+305
实际上是这样的inf
,这很好,甚至是首选,因为1.79769313486e+305
它是一个实数,inf
而不是一个实数。