鞋带配方的最快方法 [英] Fastest way to Shoelace formula

问题描述

我做了一个用鞋带方式计算面积多边形的函数.

I have made a function who calculate area polygon with Shoelace way.

这很完美，但现在我想知道是否没有更快的方法来获得相同的结果.我想知道，因为这个函数必须在处理有很多坐标的多边形时工作得更快.

That's works perfectly but right now I wonder if there is not a faster way to have the same result. I want to know that because this function must work faster with polygon with a lot of coordinates.

我的功能:

def shoelace_formula(polygonBoundary, absoluteValue = True):
    nbCoordinates = len(polygonBoundary)
    nbSegment = nbCoordinates - 1

    l = [(polygonBoundary[i+1][0] - polygonBoundary[i][0]) * (polygonBoundary[i+1][1] + polygonBoundary[i][1]) for i in xrange(nbSegment)]

    if absoluteValue:
        return abs(sum(l) / 2.)
    else:
        return sum(l) / 2.

我的多边形:

polygonBoundary = ((5, 0), (6, 4), (4, 5), (1, 5), (1, 0))

结果:

22.

有什么想法吗?

我尝试使用 Numpy :这是最快的，但你必须先转换你的坐标.

I try with Numpy : It's speedest but you have to convert your coordinates first.

import numpy as np
x, y = zip(*polygonBoundary)

def shoelace_formula_3(x, y, absoluteValue = True):

    result = 0.5 * np.array(np.dot(x, np.roll(y, 1)) - np.dot(y, np.roll(x, 1)))
    if absoluteValue:
        return abs(result)
    else:
        return result

推荐答案

对我来说，最快的方法是使用 numpy，你必须在鞋带方法中发送一个 (x,y) 坐标的 numpy 数组作为参数:

For me the fastest way would be using numpy where you have to send a numpy array of (x,y) cordinates as an argument in shoelace method:

import numpy as np
def shoelace(x_y):
    x_y = np.array(x_y)
    x_y = x_y.reshape(-1,2)

    x = x_y[:,0]
    y = x_y[:,1]

    S1 = np.sum(x*np.roll(y,-1))
    S2 = np.sum(y*np.roll(x,-1))

    area = .5*np.absolute(S1 - S2)

    return area

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