从某个2D点到2D点列表的最近点Python

Question

我有一个大约40个2D点（x和y）的列表。我想要一个从列表中的第一个点开始的脚本（例如，180,0）并找到从列表到该第一个点（180,0）的最近点。一旦找到最近点，它应该再次做同样的事情（所以最近点成为第一点），而不使用已经使用过的点（180,0）。这样做直到每一点都被使用了。通过这种方式，我的随机列表应该被排序成一个点列表，成为一个流畅的线路径。到目前为止我的代码看起来像这样：

def arrange_points(points):
    # arranges all 2D points in a fluent and consistent linepath order
    size = len(points) #number of elements in list
    di = [] #list containing all distances from start point
    start_point = points[0]

    for i in range (size):
        next_points = points[i] 

        dist = math.sqrt((next_points[0] - start_point[0])**2 + (next_points[1] - start_point[1])**2) #distance of start point to all other points
        di.append(dist)

        ln = min(filter(None, di)) # smallest distance from start point, except 0 
        f = di.index(ln) # finds place of ln in list di
        next_point = points[f] # retrieves corresponding point from points, that corresponds to ln
    return di
    return next_point

di = arrange_points(points)

# 0 and previous points cannot be taken

这就是我的linepath现在的样子：

这就是它应该是这样的：

绘制的点看起来像这样:(错误的顺序）所以基本上如果我从180,0开始并继续跟随最近的点（不让代码返回），它应该最终以正确的顺序结束列表。

有谁可以帮我处理我的代码？

Answer 1

IIUC，你可以这样做：

def get_nearest(points, coord):
    """Return closest point to coord from points"""
    dists = [(pow(point[0] - coord[0], 2) + pow(point[1] - coord[1], 2), point)
              for point in points]              # list of (dist, point) tuples
    nearest = min(dists)
    return nearest[1]  # return point only

next_point = points.pop(0)  # get first point
line_path = []
line_path.append(next_point)

while len(points) > 1:
    next_point = get_nearest(points, next_point)
    line_path.append(next_point)
    points.remove(next_point)

line_path.extend(points)  # add last point