写一个代码,功能如下:1.从三维点集中拟合出多条直线;2.找出长度前3的直线,并返回
时间: 2024-06-04 20:12:39 浏览: 228
import numpy as np
from sklearn.cluster import KMeans
from sklearn.decomposition import PCA
from scipy.spatial.distance import cdist
def fit_lines(points, num_lines):
"""
从三维点集中拟合出多条直线
:param points: 三维点集,形状为(n, 3),n为点的个数
:param num_lines: 需要拟合的直线数量
:return: 拟合后的直线列表,每条直线由两个点表示
"""
# 使用PCA将三维点集降到二维
pca = PCA(n_components=2)
points_2d = pca.fit_transform(points)
# 使用KMeans将点集聚类
kmeans = KMeans(n_clusters=num_lines)
kmeans.fit(points_2d)
# 从每个簇中找到距离最远的两个点,作为直线的端点
lines = []
for i in range(num_lines):
cluster_points = points_2d[kmeans.labels_ == i]
distances = cdist(cluster_points, cluster_points)
max_distance_indices = np.unravel_index(distances.argmax(), distances.shape)
p1, p2 = cluster_points[max_distance_indices[0]], cluster_points[max_distance_indices[1]]
p1_3d = pca.inverse_transform(np.hstack([p1, 0]))
p2_3d = pca.inverse_transform(np.hstack([p2, 0]))
lines.append((p1_3d, p2_3d))
return lines
def top_n_lines(lines, n):
"""
找出长度前n的直线,并返回
:param lines: 直线列表,每条直线由两个点表示
:param n: 需要返回的直线数量
:return: 长度前n的直线列表,每条直线由两个点表示
"""
# 计算每条直线的长度
line_lengths = [np.linalg.norm(p1 - p2) for p1, p2 in lines]
# 找到长度前n的直线
indices = np.argsort(line_lengths)[-n:]
return [lines[i] for i in indices]
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