有关 Python3 的 重载运算符 与 引用问题

有关 Python3 的 重载运算符 与 引用问题

假设我有这样一个需求：

给定三点：A、B、C；其中A与B坐标已知，C点坐标 = A点 + B点
当A点坐标发生变化时，C点也应该发生变化。

我是这样写的：

class Point:
    def __init__(self, coordinate:tuple):
        self.x = coordinate[0]
        self.y = coordinate[1]
    def __repr__(self):
        return "(%.2f, %.2f)" % (self.x, self.y)
    def __add__(self, other:Point):
        if isinstance(other, Point):
            x = self.x + other.x
            y = self.y + other.y
            return Point(coordinate=(x,y))

然后定义A与B的坐标 以及C的关系：

A = Point((1,1))
B = Point((4,1))
C = A + B

此时C点的坐标应该为：(5,2)

但我进行如下操作后

A.x=5

C点的坐标并没有发生变化...原因我知道，因为__add__给C返回的是一个新的Point对象。

我想请教各位，如何设计，才能让C点的坐标根据A点/B点的坐标变化而变化呢？

回答：

把C和AB关联起来就可以了

class Point:
    def __init__(self, coordinate:tuple,point = None,point2 = None):
        self._point = point
        self._point2 = point2
        if self._point is None or self._point2 is None:
            self._point,self._point2 = coordinate
    def _simple(self):
        return not isinstance(self._point, Point)
    def _reset(self,x,y):
        self._point = x
        self._point2 = y
    def x(self,vx = None):
        if vx is not None: self._reset(vx,self.y())
        return self._point if self._simple() else self._point.x() + self._point2.x()
    def y(self,vy = None):
        if vy is not None: self._reset(self.x(),vy)
        return self._point2 if self._simple() else self._point.y() + self._point2.y()
    def __add__(self, other):
        return Point(None, self, other) if isinstance(other, Point) else self
    def __repr__(self):
        return "(%.2f, %.2f)" % (self.x(), self.y())
A = Point((1,1))
B = Point((4,1))
C = A + B
print(C) #(5,2)
A.x(5)
print(C)  #(9,2)

回答：

class Point:
    def __init__(self, coordinate=None, parent_points=None):
        if coordinate is None and parent_points is None:
            raise ValueError("coordinate and parent_points cannot be both None")
        self.parent_points = parent_points
        self.coordinate = coordinate
    def __getattr__(self, item):
        if item not in ('x', 'y'):
            raise AttributeError("Point only has x and y")
        if self.parent_points:
            return sum(i.x if item == 'x' else i.y for i in self.parent_points)
        else:
            return self.coordinate[0 if item == 'x' else 1]
    @property
    def x(self):
        return self.__getattr__('x')
    @x.setter
    def x(self, n):
        if self.coordinate is None:
            raise Exception("this point is associated with other point, so it cannot be modified")
        self.coordinate[0] = n
    @property
    def y(self):
        return self.__getattr__('y')
    @y.setter
    def y(self, n):
        if self.coordinate is None:
            raise Exception("this point is associated with other point, so it cannot be modified")
        self.coordinate[1] = n
    def __repr__(self):
        return "(%.2f, %.2f)" % (self.x, self.y)
    def __add__(self, other):
        return Point(parent_points=[self, other])
if __name__ == '__main__':
    a = Point([1, 1])
    b = Point([4, 1])
    c = a + b
    d = a + b + c
    print("Now c is {}".format(c))
    print("Now d is {}".format(d))
    a.x = 5
    b.y = 2
    print("Now c is {}".format(c))
    print("Now d is {}".format(d))

以上是 有关 Python3 的 重载运算符 与 引用问题 的全部内容， 来源链接： utcz.com/a/160167.html

回到顶部