将Delaney三角剖分按值分组

问题描述

我正在尝试绘制来自熊猫df的delaunay三角剖分。我希望按Time对点进行分组。目前，尝试从第一个时间点开始绘制点时出现错误。

QhullError: QH6214 qhull input error: not enough points(2) to construct initial simplex (need 6)

While executing:  | qhull d Q12 Qt Qc Qz Qbb
Options selected for Qhull 2019.1.r 2019/06/21:
  run-id 768388270  delaunay  Q12-allow-wide  Qtriangulate  Qcoplanar-keep
  Qz-infinity-point  Qbbound-last  _pre-merge  _zero-centrum  Qinterior-keep
  _maxoutside  0

似乎只将这两个数组作为单个点传递。

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from scipy.spatial import delaunay

df = pd.DataFrame({
    'Time' : [1,1,2,2],'A_X' : [5,5,6,4,3,4],'A_Y' : [5,6],})

fig,ax = plt.subplots(figsize = (6,6))
ax.set_xlim(0,10)
ax.set_ylim(0,10)
ax.grid(False)

points_x1 = df.groupby("Time")["A_X"].agg(list).tolist()
points_y1 = df.groupby("Time")["A_Y"].agg(list).tolist()

points = list(zip(points_x1,points_y1))

tri = delaunay(points[0])

#plot triangulation
plt.triplot(points[:,0],points[:,1],tri.simplices)
plt.plot(points[:,'o')

解决方法

您可以利用apply方法来对Series进行操作。

def make_points(x):
    return np.array(list(zip(x['A_X'],x['A_Y'])))

c = df.groupby("Time").apply(make_points)

结果是每个时间段的点数组形状正确：

Time
1    [[5,5],[5,6],[6,5]]
2    [[4,[3,[4,6]]
dtype: object

最后，只需为每个时间段计算Delaunay triangulation并将其绘制即可：

fig,axe = plt.subplots()
for p in c:
    tri = Delaunay(p)
    axe.triplot(*p.T,tri.simplices)

您甚至可以在一个呼叫中完成该操作：

def make_triangulation(x):
    return Delaunay(np.array(list(zip(x['A_X'],x['A_Y']))))

c = df.groupby("Time").apply(make_triangulation)

fig,axe = plt.subplots()
for tri in c:
    axe.triplot(*tri.points.T,tri.simplices)

delaunay pandas python