Python：具有点数而不是epsilon的Ramer-Douglas-Peucker（RDP）算法

问题描述

限时删除！！

我想针对，目的是不使用epsilon，而是选择我想保留在最后的点数：

I would like to modify this following python script for RDP algorithm with the purpose of not using epsilon but to choose the number of points I want to keep at the final :

class DPAlgorithm():

    def distance(self,  a, b):
        return  sqrt((a[0] - b[0]) ** 2 + (a[1] - b[1]) ** 2)

    def point_line_distance(self,  point, start, end):
        if (start == end):
            return self.distance(point, start)
        else:
            n = abs(
                (end[0] - start[0]) * (start[1] - point[1]) - (start[0] - point[0]) * (end[1] - start[1])
            )
            d = sqrt(
                (end[0] - start[0]) ** 2 + (end[1] - start[1]) ** 2
            )
            return n / d

    def rdp(self, points, epsilon):
        """
        Reduces a series of points to a simplified version that loses detail, but
        maintains the general shape of the series.
        """
        dmax = 0.0
        index = 0
        i=1
        for i in range(1, len(points) - 1):
            d = self.point_line_distance(points[i], points[0], points[-1])
            if d > dmax :
                index = i
                dmax = d

        if dmax >= epsilon :
            results = self.rdp(points[:index+1], epsilon)[:-1] + self.rdp(points[index:], epsilon)
        else:
            results = [points[0], points[-1]]
        return results

我本着这种精神发现了一个Java脚本：

I found a Java script in that spirit : https://gist.github.com/msbarry/9152218

有人知道Python 3.X的版本吗？

Does anyone know a version for Python 3.X ?

谢谢
Momow

ThanksMomow

推荐答案

从上面的链接中将JS代码移植到Python [2.7]：

Ported JS code from the link above to Python [2.7]:

# -*- coding: utf-8 -*-

import math
import time


def timenow():
    return int(time.time() * 1000)

def sqr(x):
    return x*x

def distSquared(p1, p2):
    return sqr(p1[0] - p2[0]) + sqr(p1[1] - p2[1])

class Line(object):
    def __init__(self, p1, p2):
        self.p1 = p1
        self.p2 = p2
        self.lengthSquared = distSquared(self.p1, self.p2)

    def getRatio(self, point):
        segmentLength = self.lengthSquared
        if segmentLength == 0:
            return distSquared(point, p1);
        return ((point[0] - self.p1[0]) * (self.p2[0] - self.p1[0]) + \
        (point[1] - self.p1[1]) * (self.p2[1] - self.p1[1])) / segmentLength

    def distanceToSquared(self, point):
        t = self.getRatio(point)

        if t < 0:
            return distSquared(point, self.p1)
        if t > 1:
            return distSquared(point, self.p2)

        return distSquared(point, [
            self.p1[0] + t * (self.p2[0] - self.p1[0]),
            self.p1[1] + t * (self.p2[1] - self.p1[1])
        ])

    def distanceTo(self, point):
        return math.sqrt(self.distanceToSquared(point))


def simplifyDouglasPeucker(points, pointsToKeep):
    weights = []
    length = len(points)

    def douglasPeucker(start, end):
        if (end > start + 1):
            line = Line(points[start], points[end])
            maxDist = -1
            maxDistIndex = 0

            for i in range(start + 1, end):
                dist = line.distanceToSquared(points[i])
                if dist > maxDist:
                    maxDist = dist
                    maxDistIndex = i

            weights.insert(maxDistIndex, maxDist)

            douglasPeucker(start, maxDistIndex)
            douglasPeucker(maxDistIndex, end)

    douglasPeucker(0, length - 1)
    weights.insert(0, float("inf"))
    weights.append(float("inf"))

    weightsDescending = weights
    weightsDescending = sorted(weightsDescending, reverse=True)

    maxTolerance = weightsDescending[pointsToKeep - 1]
    result = [
        point for i, point in enumerate(points) if weights[i] >= maxTolerance
    ]

    return result

这篇关于Python：具有点数而不是epsilon的Ramer-Douglas-Peucker（RDP）算法的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！

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