Python class Vector 的创建，pythonvector,创建一个向量的class
创建一个向量的class，可以让vector相加，相减， 乘以一个系数， 或者清空成为(0，0，0)from math import sqrtclass Vector: “”” A Vector is a 3-tuple of (x,y,z) coordinates. “”” def __init__(self,x,y,z): self._x = x self._y = y self._z = z def __repr__(self): return ‘{%.3g} ,{%.3g} ,{%.3g}’%(self._x, self._y, self._z) def __str__(self): return ‘({},{},{})’.format(self._x, self._y, self._z) def __add__(self,other): return Vector(self._x + other._x, self._y + other._y, self._z + other._z) def __sub__(self,other): return Vector(self._x – other._x, self._y – other._y, self._z – other._z) def norm(self): result = sqrt(self._x**2 + self._y**2 + self._z**2) return result def __mul__(self,other): return Vector(self._x * other, self * other, self._z * other) def x(self): return self._x def y(self): return self._y def z(self): return self._z def clear(self): return Vector(0 ,0 ,0)
以下是我test的结果v1 = Vector(1, 1, 1)v2 = Vector(2, 2, 2)v3 = Vector(3, 3, 3)assert str(v1) == ‘(1,1,1)’assert str(v2) == ‘(2,2,2)’assert Vector(1, 2, 3) == Vector(1, 2, 3)assert Vector(1, 2, 3) != Vector(1.001, 2.001, 3.001)—————————————————————————AssertionError Traceback (most recent call last) in ()—-> 1 assert Vector(1, 2, 3) == Vector(1, 2, 3) 2 assert Vector(1, 2, 3) != Vector(1.001, 2.001, 3.001)AssertionError: assert v1 + v2 == v3—————————————————————————AssertionError Traceback (most recent call last) in ()—-> 1 assert v1 + v2 == v3 AssertionError: assert v3 – v2 == v1—————————————————————————AssertionError Traceback (most recent call last) in ()—-> 1 assert v3 – v2 == v1AssertionError: assert v3 == v1 * 3#Error 信息太多 不复制出来了assert round(v2.norm(), 10) == round(sqrt(12), 10)v1.clear()assert v1 == Vector(0,0,0)—————————————————————————AssertionError Traceback (most recent call last) in () 1 v1.clear()—-> 2 assert v1 == Vector(0,0,0)AssertionError:
问题到底出在哪里？
比如向量相加，我print v1+v2 是(3，3，3)v3 也是(3，3，3)，为什么还是assertion error？
你需要定义 __eq__：def __eq__(self, a): return self._x == a._x and self._y == a._y and self._z == a._z
原因在此：The behavior of the default equality comparison, that instances with different identities are always unequal, may be in contrast to what types will need that have a sensible definition of object value and value-based equality.
via https://docs.python.org/3/ref…
也就是说，没定义 __eq__ 时，v1 == v2 相当于 id(v1) == id(v2)，当然是 False 了。
编橙之家文章，