将构造函数参数传递给生成默认值的函数失败

Question

我对Python非常陌生，所以我认为我正在做一些非常错误的事情，但我没有看到什么，Google也没有帮助过这么多。这有什么问题？

def lookup_permille(name): 
    # TODO: implement a permille lookup table 
    return 0 

def lookup_known_product(name): 
    # TODO: implement a known product lookup table 
    return 0 

class ProductEntry: 
    def __init__(self, source, name, price, volume, permille = lookup_permille(name), known_product_id = lookup_known_product(name), category = 0): 
     self.source = source 
     self.name = name 
     self.price = price 
     self.volume = volume 
     self.permille = permille 
     self.price_per_permille = self.permille/self.price; 
     self.category = category 
     self.known_product_id = known_product_id 

调用的ProductEntry构造失败：定义函数时

def __init__(self, source, name, price, volume, permille = lookup_permille(name), known_product_id = lookup_known_product(name), category = 0): 
NameError: name 'name' is not defined 

Answer 1

定义默认参数的表达式求值，而不是当它被调用。在定义__init__时，name不存在，因此它不能用于表达式中以计算默认参数。

执行此类操作的常用方法是将备用值作为默认参数，并将其替换为您在函数体内实际需要的任何值。

def __init__(self, source, name, price, volume, 
      permille=None, known_product_id=None, category=0): 
    if permille is None: 
     permille = lookup_permille(name) 
    if known_product_id is None: 
     known_product_id = lookup_known_product(name) 
    ...