python - 计算 def f(x) 评估次数的简单方法？

Question

我正在尝试计算f(x)评估的次数，而不必过多地更改我的代码，看起来应该不是很困难，但我似乎无法弄清楚。

def f (x):
    f = 12*x**5-45*x**4+40*x**3+5
    return f
def bounding():
    d=.1
    x=6
    n=0

while(n<50):
    Lb=x-d
    n+=1
    Ub=x+d
    if f(Lb)>=f(x) and f(Ub)<=f(x):
        x=x+d           
    elif f(Lb)<=f(x) and f(Ub)>=f(x):
        x=x-d           
    elif f(Lb)>=f(x) and f(Ub)>=f(x):
        print("Lower bound:",Lb,"Upperbound:",Ub)
        break
    print (n)
bounding()

Answer 1

基于装饰器的解决方案，您可以将其应用于您想要的任何功能......

def count(fn):
        def wrapper(*args, **kwargs):
            wrapper.called+= 1
            return fn(*args, **kwargs)
        wrapper.called= 0
        wrapper.__name__= fn.__name__
        return wrapper

@count
def test():
    print "something"

test()

print test.called #will print 1

Answer 2

class F:
    count = 0
    def __call__(self, x):
        self.count += 1
        return 12*x**5-45*x**4+40*x**3+5

f = F()

从这里开始和以前一样，计数由 给出f.count。测试:)

>>> f = F()
>>> f(1)
12
>>> f(2)
-11
>>> f.count
2
>>> f(2)
-11
>>> f.count
3