使用numpy计算二维闭合多边形的面积

Question

我有一个二维numpy数组，我想计算一下下面的等式。有没有一种类似numpy (矢量化)的方法来实现这种评估，而不只是迭代这些值？

额外的问题，有没有一种方法(再次矢量化)将乘积的每个结果保存在一个新的一维numpy数组中？提前感谢

顺便说一句，这个公式是从这个网站找到的：https://www.seas.upenn.edu/~ese502/lab-content/extra_materials/Polygon%20Area%20and%20Centroid.pdf

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Answer 1

您可以按如下方式使用numpy.roll：

import numpy as np

N = 5
a = np.random.rand(N, 2)
x, y = a[:,0], a[:,1]

area = 1/2 * np.sum(x*np.roll(y, 1) - y*np.roll(x, 1))

Answer 2

下面是SQF语言的一个实现：

private _n0 = count(_polygon);
private _surface = 0;

for "_i0" from 0 to (_n0 - 1) do {
    private _p1 = _polygon select _i0;
    private _p2 = _polygon select ((_i0 + 1) % _n0);
    private _p1x = _p1 select 0;
    private _p1y = _p1 select 1;
    private _p2x = _p2 select 0;
    private _p2y = _p2 select 1;
    _surface = _surface + ((_p1x * _p2y) - (_p2x * _p1y)) / 2;

你只需要把它翻译成python即可。

Answer 3

einsum可以方便地实现鞋带公式

A  # --- sample array representing a polygon ordered clockwise
array([[  0.00,   0.00],
       [  0.00,  10.00],
       [ 10.00,  10.00],
       [ 10.00,   0.00],
       [  0.00,   0.00]])

# --- can be implemented in a list comprehension of for loop

def _bit_area_(a):
    """Mini e_area, used by `areas` and `centroids`."""
    x0, y1 = (a.T)[:, 1:]
    x1, y0 = (a.T)[:, :-1]
    e0 = np.einsum('...i,...i->...i', x0, y0)
    e1 = np.einsum('...i,...i->...i', x1, y1)
    return np.sum((e0 - e1) * 0.5)
    

_bit_area_(A)
100.0  # --- result for a 10 x 10 unit square,