Python测试点是否在矩形中


问题内容

我是python新手,仍然在学习绳索,但是我希望有更多经验的人可以帮助我。

我正在尝试编写以下Python脚本:

  1. 创造四个点
  2. 创建四个矩形
  3. 检查每个点是否在任何矩形中,然后将结果写到输出文件中。

问题涉及两个数据结构Point和Rectangle类。我已经开始创建Point类和Rectangle类。矩形类应包含根据随机模块的随机方法创建的相关数据集。从我的尝试中可以看出,我到处都是,但是我使用#comments来尝试获得自己想做的事情。

我的具体问题是:
1)如何使此脚本正常工作?
2)我缺少哪些变量或函数来生成随机矩形,并查看这些矩形中是否有特定点?

## 1. Declare the Point class
class Point:
    def __init__(self,x = 0.0, y = 0.0): 
        self.x = x 
        self.y = y
    pass
## 2. Declare the Rectangle class 
class Rectangle: 
    def __int__(self): ## A rectangle can be determined aby (minX, maxX) (minY, maxY) 
        self.minX = self.minY = 0.0 
        self.maxX = self.maxY = 1.0 
    def contains(self, point): ## add code to check if a point is within a rectangle 
        """Return true if a point is inside the rectangle."""
        # Determine if a point is inside a given polygon or not
        # Polygon is a list of (x,y) pairs. This function
        # returns True or False. 
    def point_in_poly(x,y,poly):
        n = len(poly)
        inside = False
        p1x,p1y = poly[0]
        for i in range(n+1):
            p2x,p2y = poly[i % n]
            if y > min(p1y,p2y):
                if y <= max(p1y,p2y):
                    if x <= max(p1x,p2x):
                        if p1y != p2y:
                            xints = (y-p1y)*(p2x-p1x)/(p2y-p1y)+p1x
                    if p1x == p2x or x <= xints:
                        inside = not inside
            p1x,p1y = p2x,p2y
        return inside
## 3. Generate four points 
##define a Point list to keep four points 
points = []
##add codes to generate four points and append to the points list
polygon = [(0,10),(10,10),(10,0),(0,0)]
point_x = 5
point_y = 5

## 4. Generate four rectangles 
##define a Rectangle list 
rects = [] 
for i in range(4):
    rectangle = Rectangle() 
    ## Generate x 
    x1 = random.random() 
    x2 = random.random() 
    ## make sure minX != maxX 
    while(x1 == x2): 
        x1 = random.random() 
    if x1<x2: 
        rectangle.minX=x1 
        rectangle.maxX=x2 
    elif x1>x2:
        rectangle.minX=x2
        rectangle.maxX=x1
    rects.append(rectangle)
    ## Develop codes to generate y values below 
    ## make sure minY != maxY 
    while(y1 == y2):
        y1 = random.random()
    if y1<y2:
        rectangle.minY=y1
        rectangle.maxY=y2
    elif y1>y2:
        recetangle.minY=y2
        racetangle.maxY=y1
    ## add to the list 
    rects.append(rectangle)

## 5. Add code to check which point is in which rectangle 
resultList = [] ## And use a list to keep the results 
for i in range(4):
    for j in range(4):
        if points[i] in rectangle[j]:
            print i

# write the results to file
f=open('Code5_4_1_Results.txt','w') 
for result in resultList:
    f.write(result+'\n') 
f.close()

问题答案:

这是非常简单的数学。给定一个具有点(x1,y1)和(x2,y2)的矩形,并假设x1 < x2y1 < y2(如果没有,您可以交换它们),则点(x,y)在该矩形中,如果x1 < x < x2 and y1 < y < y2。由于可以将Python比较运算符链接在一起,因此这甚至是有效的Python代码,它们都应该产生正确的结果(在其他语言中,您必须编写,如x1 < x and x < x2等等)。

如果需要,可以使用<=代替<。使用<=手段意味着矩形边界上的点(例如,点(x1,y1))被视为在其内部,而使用<手段意味着此类点在其外部。