from math import pi
class Circle(object):
'Circle(x,y,r)'
def __init__(self, x=0, y=0, r=1):
self._r = r
self._x = x
self._y = y
def __repr__(self):
return 'Circle({},{},{})'.\
format(self.getx(), self.gety(),\
self.getr())
#silly, but has a point: str can be different from repr
def __str__(self):
return 'hello world'
def __contains__(self, item):
'point in circle'
px, py = item
return (self.getx() - px)**2 + \
(self.gety() - py)**2 < self.getr()**2
def getr(self):
'radius'
return self._r
def getx(self):
'x'
self._lst.append(self._x)
return self._x
def gety(self):
'y'
self._lst.append(self._y)
return self._y
def setr(self,r):
'set r'
self._r = r
def setx(self,x):
'set x'
self._x = x
def sety(self,y):
'set y'
self._y = y
def move(self,x,y):
self._x += x
self._y += y
def concentric(self, d):
d = self._list
def area(self):
'area of circle'
return (self.getr())**2*pi
def circumference(self):
'circumference of circle'
return 2*self.getr()*pi
我的问题措辞有点笨拙,但我要做的是检查两个不同的圈子是否具有相同的中心(x,y)
。我认为解决这个问题的最简单方法是将2个点输入到列表中,但我不确定如何比较2个列表,因为每次尝试我的代码时它都会将所有内容添加到同一个列表中
答案 0 :(得分:2)
将以下方法添加到Circle
班级。
def equal_center(self, other):
'check if another circle has same center'
return (self._x == other._x) & (self._y == other._y)
<强>用法强>
C1 = Circle(3, 5, 8)
C2 = Circle(3, 5, 10)
C3 = Circle(3, 2, 1)
C1.equal_center(C2) # True
C1.equal_center(C3) # False
答案 1 :(得分:0)
我建议创建一个带有两个圆形对象的函数,如果坐标相同或不相同,则通过比较每个对象的x
和y
值来返回:
def same_center(circle_1, circle_2):
if circle_1.getx() == circle_2.getx() and circle_1.gety() == circle_2.gety():
return True
else:
return False
此解决方案比使用列表容易得多,并且应该易于实现。
答案 2 :(得分:0)
如果您有两个班级实例......
a = Circle(0,0,1)
b = Circle(0,0,1)
您可以将它们添加到圈子列表中......
circles = [a,b]
循环浏览列表,检查其值......
for i in circles:
for j in filter(lambda x : x != i, circles):
if i._x == j._x and i._y == j._y:
return True #two circles have same center
这应该适用于该类的n个实例,但如果只有两个要检查
if a._x == b._x and a._y == a._y:
return True