试图在另一个函数中使用一个函数的返回值，但得到的却是未定义的变量

Question

我有以下代码：

class Pet(object):

    def __init__(self,name=""):
        self.name = name 
        self.kind = "Unknown"
        self.toys = []  
    def add_toys(self,toys):
        new_list = []
        for toy in self.toys:
            if toy not in new_list:
                new_list.append(toy)   
        return new_list
    def __str__(self):
        toys_list = add_toys(self,toys)  
        if self.toys == []:
            return "{} is a {} that has no toys".format(self.name,self.kind)
        else:
            return "{} is a {} that has the following toys: {}".format(self.name,self.kind,toys_list)     

在函数add_toys()中，我有返回值new_list。 我想在函数__ str__中使用该返回值，并将其定义为toys_list。 然而，当我写toys_list = add_toys(self, toys)时，它说：

add_toys is an undefined variable

Answer 1

你的add_toys方法不好，你没有使用toys参数，它不应该返回任何东西，应该是这样的

class Pet:
    # __init__ is OK

    def add_toys(self, *toys):
        for toy in toys:
            if toy not in self.toys:
                self.toys.append(toy)

    def __str__(self):
        if not self.toys:
            return "{} is a {} that has no toys".format(self.name, self.kind)
        else:
            return "{} is a {} that has the following toys: {}".format(self.name, self.kind, self.toys)

像

p = Pet("Foo")
p.add_toys("ball")
p.add_toys("plate", "frisbee")
print(p) # Foo is a Unknown that has the following toys: ['ball', 'plate', 'frisbee']

---

您可以直接使用set

class Pet:

    def __init__(self, name, kind="Unknown"):
        self.name = name
        self.kind = kind
        self.toys = set()

    def add_toys(self, *toys):
        self.toys.update(toys)