python Graham求凸包并画图
python写Graham没有c++那么好写,但是python画图简单。只需要用matplotlib里的pyplot,c++画图太难了。
Graham算法写起来比较简单,只需要想办法对最小点和其他的点所连成的直线,与x轴正半轴的夹角进行排序,然后其他的就直接套用Graham算法模板就好了,因为c++可以重载排序函数sort,不用计算角度(用其他的数学方法),但是python不行(也许是我不知道而已,菜)。
python必须要在结构体里面加上角度这个变量,然后才能按照角度排序。排好序后就变得容易了,用stack栈存放答案,算完答案后,用scatter(散点图)画出点,用plt(折线图)画边界就好了。
import matplotlib.pyplot as plt import math import numpy as np class Node: def __init__(self): self.x = 0 self.y = 0 self.angel = 0 #和最左下的点连成的直线,与x轴正半轴的夹角大小 #按照角度从小到大排序 def cmp(x): return x.angel def bottom_point(points): min_index = 0 n = len(points) #先判断y坐标,找出y坐标最小的点,x坐标最小的点 for i in range(1, n): if points[i].y < points[min_index].y or (points[i].y == points[min_index].y and points[i].x < points[min_index].x): min_index = i return min_index #计算角度 def calc_angel(vec): norm = math.sqrt(vec[0] * vec[0] + vec[1] * vec[1]) if norm == 0: return 0 angel = math.acos(vec[0]/norm) if vec[1] >= 0: return angel else: return math.pi * 2 - angel def multi(v1, v2): return v1[0] * v2[1] - v1[1] * v2[0] point = [] n = 30 #生成30个点的坐标,n可以修改 for i in range(n): temp = Node() temp.x = np.random.randint(1, 100) temp.y = np.random.randint(1, 100) point.append(temp) index = bottom_point(point) for i in range(n): if i == index: continue #计算每个点和point[index]所连成的直线与x轴正半轴的夹角 vector = [point[i].x - point[index].x, point[i].y - point[index].y] #vector是向量 point[i].angel = calc_angel(vector) #排序 point.sort(key=cmp) #答案存入栈中 stack = [] stack.append(point[0]) stack.append(point[1]) #for循环更新答案 for i in range(2, n): L = len(stack) top = stack[L - 1] next_top = stack[L - 2] vec1 = [point[i].x - next_top.x, point[i].y - next_top.y] vec2 = [top.x - next_top.x, top.y - next_top.y] #一定要大于等于零,因为可能在一条直线上 while multi(vec1, vec2) >= 0: stack.pop() L = len(stack) top = stack[L - 1] next_top = stack[L - 2] vec1 = [point[i].x - next_top.x, point[i].y - next_top.y] vec2 = [top.x - next_top.x, top.y - next_top.y] stack.append(point[i]) #画出图像 for p in point: plt.scatter(p.x, p.y, marker='o', c='g') L = len(stack) for i in range(L-1): plt.plot([stack[i].x, stack[i+1].x], [stack[i].y, stack[i+1].y], c='r') plt.plot([stack[0].x, stack[L-1].x], [stack[0].y, stack[L-1].y], c='r') plt.show()
Python 找到凸包 Convex hulls
图形学可以说经常遇到这东西了,这里给出一个库函数的实现
from scipy.spatial import ConvexHull points = np.random.rand(10, 2) # 30 random points in 2-D hull = ConvexHull(points) import matplotlib.pyplot as plt plt.plot(points[:,0], points[:,1], 'o') for simplex in hull.simplices: plt.plot(points[simplex,0], points[simplex,1], 'k-') plt.show()
以上为个人经验,希望能给大家一个参考,也希望大家多多支持自学编程网。
- 本文固定链接: https://zxbcw.cn/post/214074/
- 转载请注明:必须在正文中标注并保留原文链接
- QQ群: PHP高手阵营官方总群(344148542)
- QQ群: Yii2.0开发(304864863)