我正在寻找一种有效的方法来计算（复杂）数学函数的结果。
现在，它看起来可与：
def f(x):
return x**2
def g(x):
if x is 0: return 1
return f(x)*5
def h(x):
return g(x)
res = [f(input[i]) for i in range(1000000)]
由于每个函数调用在Python中都代价高昂，因此如果f（x）与g（x）合并，则代码应该已经运行得更快。不幸的是，在那种情况下，g（x）函数的'return ...'行已经很长了。此外，当前实际上总共定义了6个函数，因此完整的公式占用几行。
那么，计算物理公式结果的巧妙方法是什么？
If you're concerned about the length of the
return ...line you could use intermediate values, e.g.:In this context, it is incorrect to use the
isoperator as inx is 0. It does not check for numerical equality, which is what==does. Theisoperator checks that the two operands refer to exactly the same object in memory, which happens to be true in this case because the Python interpreter is intelligently reusing number objects. It can lead to confusing errors, for example:In practice,
isis mainly used only to check if a valueis None.