在遍历列表中的元素时，如何在python中创建对象？

我试图将列表中的元素传递给类，以便在Python中创建对象。稍后，当我使用相同的列表尝试调用该对象时，我得到错误：“str”对象没有属性“name”

我已经使用Python有一段时间了，但对OOP来说还是新手。想知道这是否与对象的范围有关>

class SwimmingWithTheFishes:
    def __init__(self, typeofshark):
        self.name = typeofshark

    def __str__(self):
        return f"This is the method shark: {self.name}"

    def reporting(self, shark):
        name = shark.name
        print(f"This is a method shark: {name}")

    def print_return(self):
        return f'{self.name}'


def main():
    # sharklist = [{"name": "mako"}, {"name": "hammerhead"}, {"name": "greatwhite"}, {"name": "reef"}]
    sharklist = ["mako", "hammerhead", "greatwhite", "reef"]

    for typeofshark in sharklist:
        typeofshark = SwimmingWithTheFishes(typeofshark)
        print(f"Heavens above, that's no fish: {typeofshark.name}")
        typeofshark.reporting(typeofshark)

    for shark in sharklist:
        print(SwimmingWithTheFishes.print_return(shark))


if __name__ == "__main__":
    main() 

---

print\u return

是一种

swimingwiththefishes

的方法，因此您应该用

shark

实例化

swimingwiththefishes

，这样

self

将成为

swimingwiththefishes

的对象，以便

self.name

在

print\u return

中工作

更改：

print(SwimmingWithTheFishes.print_return(shark))

致：

---

当您在列表上迭代并分配给当前变量时，您不会更改列表中的值，只会更改该局部变量

例如

---

要修改列表，您应该创建一个新列表，因为如果您修改迭代的列表，可能会遇到问题。这个新列表可以称为类似于

sharks

——其中元素包含类实例

最后，你对方法也有误解。。。您不需要每次调用实例上的方法时都传入对对象的引用。method函数的

self

参数自动获取从中调用方法的实例的值

这使得最终代码：

class SwimmingWithTheFishes:
    def __init__(self, typeofshark):
        self.name = typeofshark

    def __str__(self):
        return f"I am a {self.name} shark."

    def reporting(self):
        print(f"This is a {self.name} shark method.") 


def main():
    # shark_types = [{"name": "mako"}, {"name": "hammerhead"}, {"name": "greatwhite"}, {"name": "reef"}]
    shark_types = ["mako", "hammerhead", "greatwhite", "reef"]
    sharks = []

    for type_ in shark_types:
        shark = SwimmingWithTheFishes(type_)
        sharks.append(shark)
        print(f"Heavens above, that's no fish: {shark.name}")
        shark.reporting()

    for shark in sharks:
        print(shark)


if __name__ == "__main__":
    main() 

---

其中：

Heavens above, that's no fish: mako
This is a mako shark method.
Heavens above, that's no fish: hammerhead
This is a hammerhead shark method.
Heavens above, that's no fish: greatwhite
This is a greatwhite shark method.
Heavens above, that's no fish: reef
This is a reef shark method.
I am a mako shark.
I am a hammerhead shark.
I am a greatwhite shark.
I am a reef shark.

---

sharklist

是一个字符串列表，因此

shark

是一个字符串。您正在将该字符串传递给

print\u return

，这显然无法工作。非常感谢，这很有意义。谢谢。

class SwimmingWithTheFishes:
    def __init__(self, typeofshark):
        self.name = typeofshark

    def __str__(self):
        return f"I am a {self.name} shark."

    def reporting(self):
        print(f"This is a {self.name} shark method.") 


def main():
    # shark_types = [{"name": "mako"}, {"name": "hammerhead"}, {"name": "greatwhite"}, {"name": "reef"}]
    shark_types = ["mako", "hammerhead", "greatwhite", "reef"]
    sharks = []

    for type_ in shark_types:
        shark = SwimmingWithTheFishes(type_)
        sharks.append(shark)
        print(f"Heavens above, that's no fish: {shark.name}")
        shark.reporting()

    for shark in sharks:
        print(shark)


if __name__ == "__main__":
    main() 

Heavens above, that's no fish: mako
This is a mako shark method.
Heavens above, that's no fish: hammerhead
This is a hammerhead shark method.
Heavens above, that's no fish: greatwhite
This is a greatwhite shark method.
Heavens above, that's no fish: reef
This is a reef shark method.
I am a mako shark.
I am a hammerhead shark.
I am a greatwhite shark.
I am a reef shark.