Question

points = array

def get_nearest_point(x, y, n_points):

    dist = []
    val = []
    j = (x, y)
    for z in range(n_points):
        for i in range(len(points)):
            dist += [Distance(points[i][0],points[i][1],x,y)]
            a = dist.index(min(dist))
        val.append(points[a])
        dist.remove(min(dist))
        return val

输出：

get_nearest_point（50，50,3）

[[54，57]]

我想返回类似[[54,57]，[56,78]，[78,90]]

Answer 1

您实际上想做的是找到给定数组中最小3个元素的索引。您可以使用numpy.argsort方法轻松完成此操作

无论如何，这是您的numpy代码：

import numpy as np

def get_nearest_point(x, y, n):

    dist = [Distance(point[0], point[1], x, y) for point in points]
    indices = np.argsort(dist)
    return [points[i] for i in indices[:n]]

Answer 2

欢迎来到该网站！

您编写的代码的问题在于return语句位于循环内部-而是将其移至循环外部。这样，您可以在返回之前将所有三个值附加到val上。

Answer 3

在这里我们计算每个点的距离，然后按距离对它进行排序，然后得到前n个点。

def get_nearest_point(x, y, n_points):
    points_with_dist = []  # store here pairs (dist, point)
    for point in points:
        points_with_dist.append((
            Distance(point[0], point[1], x, y),
            point
        ))
    result_with_dist = sorted(points_with_dist, key=lambda struct: struct[0])[:n_points]
    result = [p[1] for p in result_with_dist]  # we need only points
    return result

我想遍历for循环以返回三点

3 个答案: