Question

我有一个纬度/经度坐标列表，我想从中创建一条线：

import time
import random

class Point:
    def __init__(self, lng, lat, neighbours):
        self.x = lng
        self.y = lat
        self.neighbours = neighbours

def find_neighbours(point):
        '''Simulate neighbour search in kdtree'''
        print(f'Searching for {point} in the tree')
        time.sleep(1)
        neighbours = random.randint(0, 100)
        return neighbours

class Line:
    def __init__(self, point_A, point_B):
        self.point_A = point_A
        self.point_B = point_B
        self.line = [
            point_A, 
            point_B,
        ]

def get_points():
    return [
    [51.11453, 17.10942],
    [51.11441, 17.10941],
    [51.11349, 17.10779],
    [51.11367, 17.10733],
    [51.1143, 17.10648],
    [51.11493, 17.10553],
    [51.11471, 17.10519],
    [51.11434, 17.10506],
    [51.11376, 17.10462],
    [51.11372, 17.10433],
    [51.11353, 17.1042],
    [51.11388, 17.10168],
    [51.11386, 17.10108],
    [51.11362, 17.10098],
    [51.11177, 17.10099],
    [51.11169, 17.10105],
    [51.11169, 17.10129],
]

def make_lines_from_points(points):
    lines = []

    for index, point in enumerate(points):
        try:
            next_point = points[index + 1]
        except IndexError: 
            # last point reached, so no need to make a line
            next_point = None

        if next_point:
            point_A = Point(
                point[1],
                point[0],
                find_neighbours(point)
            )
            point_B = Point(
                next_point[1],
                next_point[0],
                find_neighbours(next_point)
            )
            line = Line(point_A, point_B)
            lines.append(line.line)
    return lines

def main():
    points = get_points()
    lines = make_lines_from_points(points)
    print(lines)

if __name__ == '__main__':
    main()

在这段代码中，我将每个坐标及其旁边的坐标转换为Point类的实例，然后从这两个点创建Line的实例。

在初始化每个Point实例之后，必须找到点的邻居并将其分配给实例的neighbours属性。这可能需要更长的时间。

考虑到这一点，我想改进for循环，以使B点在下一次迭代中成为A点，以避免再次计算同一点的邻居。

我该怎么做？

Answer 1

您可以尝试以下方法：

def make_lines_from_points(points):
    lines = []

    saved_neighbours = None
    for index, point in enumerate(points):
        try:
            next_point = points[index + 1]
        except IndexError: 
            # last point reached, so no need to make a line
            next_point = None

        if next_point:
            point_A = Point(
                point[1],
                point[0],
                saved_neighbours if saved_neighbours is not None else find_neighbours(point)
            )
            point_B = Point(
                next_point[1],
                next_point[0],
                find_neighbours(next_point)
            )
            line = Line(point_A, point_B)
            lines.append(line.line)
            saved_neighbours = point_B.neighbours
    return lines

Answer 2

def make_lines_from_points(points):
    lines = []

    point_A = None

    for index, point in enumerate(points):
        try:
            next_point = points[index + 1]
        except IndexError: 
            # last point reached, so no need to make a line
            next_point = None

        if next_point:
            if not point_A:
                point_A = Point(
                    point[1],
                    point[0],
                    find_neighbours(point)
                )
            point_B = Point(
                next_point[1],
                next_point[0],
                find_neighbours(next_point)
            )

            point_A = point_B

            line = Line(point_A, point_B)
            lines.append(line.line)
    return lines

将结果从一个“ for”循环迭代传递到下一个迭代？

2 个答案: