Zachowująca kształt interpolacja sześcienna dla krzywej 3D w pytonie

Question

Mam krzywą w przestrzeni 3D. Chcę używać na nim zachowującej kształt, sześciennej interpolacji sześciennej, podobnej do pchip w programie Matlab. Badałem funkcje udostępniane w scipy.interpolate, np. interp2d, ale funkcje działają dla niektórych struktur krzywych, a nie dla posiadanych punktów danych. Wszelkie pomysły, jak to zrobić?

Oto punkty danych:

x,y,z 
0,0,0 
0,0,98.43 
0,0,196.85 
0,0,295.28 
0,0,393.7 
0,0,492.13 
0,0,590.55 
0,0,656.17 
0,0,688.98 
0,0,787.4 
0,0,885.83 
0,0,984.25 
0,0,1082.68 
0,0,1181.1 
0,0,1227.3 
0,0,1279.53 
0,0,1377.95 
0,0,1476.38 
0,0,1574.8 
0,0,1673.23 
0,0,1771.65 
0,0,1870.08 
0,0,1968.5 
0,0,2066.93 
0,0,2158.79 
0,0,2165.35 
0,0,2263.78 
0,0,2362.2 
0,0,2460.63 
0,0,2559.06 
0,0,2647.64 
-0.016,0.014,2657.48 
-1.926,1.744,2755.868 
-7.014,6.351,2854.041 
-15.274,13.83,2951.83 
-26.685,24.163,3049.031 
-41.231,37.333,3145.477 
-58.879,53.314,3240.966 
-79.6,72.076,3335.335 
-103.351,93.581,3428.386 
-130.09,117.793,3519.96 
-159.761,144.66,3609.864 
-192.315,174.136,3697.945 
-227.682,206.16,3784.018 
-254.441,230.39,3843.779 
-265.686,240.572,3868.036 
-304.369,275.598,3951.483 
-343.055,310.627,4034.938 
-358.167,324.311,4067.538 
-381.737,345.653,4118.384 
-420.424,380.683,4201.84 
-459.106,415.708,4285.286 
-497.793,450.738,4368.741 
-505.543,457.756,4385.461 
-509.077,460.955,4393.084 
-536.475,485.764,4452.188 
-575.162,520.793,4535.643 
-613.844,555.819,4619.09 
-624.393,565.371,4641.847 
-652.22,591.897,4702.235 
-689.427,631.754,4784.174 
-725.343,675.459,4864.702 
-759.909,722.939,4943.682 
-793.051,774.087,5020.95 
-809.609,801.943,5060.188 
-820.151,820.202,5085.314 
-824.889,828.407,5096.606 
-830.696,838.466,5110.448 
-846.896,867.72,5150.399 
-855.384,883.717,5172.081 
-884.958,939.456,5247.626 
-914.53,995.188,5323.163 
-944.104,1050.927,5398.708 
-973.675,1106.659,5474.246 
-1003.249,1162.398,5549.791 
-1032.821,1218.13,5625.328 
-1062.395,1273.869,5700.873 
-1091.966,1329.601,5776.411 
-1121.54,1385.34,5851.956 
-1151.112,1441.072,5927.493 
-1180.686,1496.811,6003.038 
-1210.257,1552.543,6078.576 
-1239.831,1608.282,6154.121 
-1269.403,1664.015,6229.658 
-1281.875,1687.521,6261.517 
-1298.67,1720.451,6304.797 
-1317.209,1760.009,6353.528 
-1326.229,1780.608,6377.639 
-1351.851,1844.711,6447.786 
-1375.462,1912.567,6515.035 
-1379.125,1923.997,6525.735 
-1397.002,1984.002,6579.217 
-1416.406,2058.808,6640.141 
-1433.629,2136.794,6697.655 
-1448.619,2217.744,6751.587 
-1453.008,2244.679,6768.334 
-1461.337,2301.426,6801.784 
-1471.745,2387.628,6848.122 
-1479.813,2476.093,6890.468 
-1485.519,2566.597,6928.713 
-1488.852,2658.874,6962.744 
-1489.801,2752.688,6992.481 
-1488.358,2847.765,7017.828 
-1484.534,2943.865,7038.72 
-1478.344,3040.704,7055.099 
-1469.806,3137.966,7066.915 
-1469.799,3138.035,7066.922 
-1458.925,3235.574,7074.155 
-1445.742,3333.07,7076.775 
-1444.753,3339.757,7076.785 
-1438.72,3380.321,7076.785 
-1431.268,3430.42,7076.785 
-1416.787,3527.779,7076.785 
-1402.308,3625.128,7076.785 
-1401.554,3630.192,7076.785 
-1387.827,3722.487,7076.785 
-1373.347,3819.836,7076.785 
-1358.866,3917.195,7076.785 
-1357.872,3923.882,7076.785 
-1353.32,3954.485,7076.785 
-1344.387,4014.544,7076.785 
-1329.906,4111.903,7076.785 
-1315.427,4209.252,7076.785 
-1300.946,4306.611,7076.785 
-1286.466,4403.96,7076.785 
-1271.985,4501.319,7076.785 
-1257.504,4598.678,7076.785 
-1243.025,4696.027,7076.785 
-1228.544,4793.386,7076.785 
-1214.065,4890.735,7076.785 
-1199.584,4988.094,7076.785 
-1185.104,5085.443,7076.785 
-1170.623,5182.802,7076.785 
-1156.144,5280.151,7076.785 
-1141.663,5377.51,7076.785 
-1127.183,5474.859,7076.785 
-1112.703,5572.218,7076.785 
-1098.223,5669.567,7076.785 
-1083.742,5766.926,7076.785 
-1069.263,5864.275,7076.785 
-1054.782,5961.634,7076.785 
-1040.302,6058.983,7076.785 
-1025.821,6156.342,7076.785 
-1011.342,6253.692,7076.785 
-996.861,6351.05,7076.785 
-982.382,6448.4,7076.785 
-967.901,6545.759,7076.785 
-953.421,6643.108,7076.785 
-938.94,6740.467,7076.785 
-924.461,6837.816,7076.785 
-909.98,6935.175,7076.785 
-895.499,7032.534,7076.785 
-895.234,7034.314,7076.785 
-883.075,7130.158,7076.785 
-874.925,7228.243,7076.785 
-871.062,7326.579,7076.785 
-871.491,7425,7076.785 
-876.213,7523.299,7076.785 
-885.218,7621.308,7076.785 
-898.489,7718.822,7076.785 
-916.003,7815.673,7076.785 
-937.722,7911.659,7076.785 
-963.61,8006.615,7076.785 
-993.613,8100.342,7076.785 
-1027.678,8192.681,7076.785 
-1065.735,8283.437,7076.785 
-1083.912,8323.221,7076.785 
-1107.12,8372.742,7076.785 
-1148.885,8461.861,7076.785 
-1190.655,8550.989,7076.785 
-1232.42,8640.108,7076.785 
-1274.19,8729.236,7076.785 
-1315.955,8818.354,7076.785 
-1357.724,8907.482,7076.785 
-1399.49,8996.601,7076.785 
-1441.259,9085.729,7076.785 
-1483.024,9174.848,7076.785 
-1486.296,9181.829,7076.785 
-1510.499,9231.861,7076.785 
-1526.189,9263.304,7076.785 
-1570.139,9351.377,7076.785 
-1614.085,9439.441,7076.785 
-1658.035,9527.514,7076.785 
-1701.98,9615.578,7076.785 
-1745.93,9703.651,7076.785 
-1789.876,9791.715,7076.785 
-1833.826,9879.788,7076.785 
-1877.771,9967.852,7076.785 
-1921.721,10055.925,7076.785 
-1965.667,10143.989,7076.785 
-2009.625,10232.059,7076.785 
-2053.585,10320.115,7076.785 
-2097.551,10408.18,7076.785 
-2141.512,10496.237,7076.785 
-2185.477,10584.302,7076.785 
-2229.438,10672.359,7076.785 
-2273.403,10760.424,7076.785 
-2317.364,10848.481,7076.785 
-2352.213,10918.285,7076.785 

Answer 1

Prawdopodobnie chcesz użyć splprep() and splev(), jak ten (podstawowa wyjaśnieniu w komentarzach):

import scipy 
from scipy import interpolate 
import numpy as np 

#This is your data, but we're 'zooming' into just 5 data points 
#because it'll provide a better visually illustration 
#also we need to transpose to get the data in the right format 
data = data[100:105].transpose() 

#now we get all the knots and info about the interpolated spline 
tck, u= interpolate.splprep(data) 
#here we generate the new interpolated dataset, 
#increase the resolution by increasing the spacing, 500 in this example 
new = interpolate.splev(np.linspace(0,1,500), tck) 

#now lets plot it! 
from mpl_toolkits.mplot3d import Axes3D 
import matplotlib.pyplot as plt 
fig = plt.figure() 
ax = Axes3D(fig) 
ax.plot(data[0], data[1], data[2], label='originalpoints', lw =2, c='Dodgerblue') 
ax.plot(new[0], new[1], new[2], label='fit', lw =2, c='red') 
ax.legend() 
plt.savefig('junk.png') 
plt.show() 

Produkuje:

Co jest ładne i gładkie, działa również na płetwy e dla pełnego opublikowanego zbioru danych.

Answer 2

Znalazłem an email od faceta, który również szukał macierzystej wersji Pythona funkcji pchip() Matlaba. Dołączył on his code, który niestety chce pobrać jako "attachment-001.bin". Jeśli zapiszesz plik i zmienisz jego nazwę na pychip.py, znajdziesz to, o co prosisz.

Answer 3

scipy ma pchip w scipy.interpolate --- ale uhh, ktoś zapomniał wymienić go w dokumentacji:

import numpy as np 
import matplotlib.pyplot as plt 
from scipy.interpolate import pchip 
x = np.linspace(0, 1, 20) 
y = np.random.rand(20) 
xi = np.linspace(0, 1, 200) 
yi = pchip(x, y)(xi) 
plt.plot(x, y, '.', xi, yi) 

Dla danych 3-D, po prostu interpolować każdej z 3 współrzędne oddzielnie.

Answer 4

Oto inne rozwiązanie, które robi to, co chciałem (zachowując kształt).
Problem polega na tym, że nie ma jasnej formuły ani równania do łączenia punktów. Odpowiedź leży w stworzeniu nowego zestawu danych, który jest wspólny dla różnych punktów. Ten nowy zbiór danych to długość wzdłuż ścieżki (norma). Następnie używamy interp1 do interpolacji każdego zestawu.

import numpy as np 
import matplotlib as mpl 
from matplotlib import pyplot as plt 
from mpl_toolkits.mplot3d import Axes3D 

# read data from a file 
# as x, y , z 

# create a new array to hold the norm (distance along path) 
npts = len(x) 
s = np.zeros(npts, dtype=float) 
for j in range(1, npts): 
    dx = x[j] - x[j-1] 
    dy = y[j] - y[j-1] 
    dz = z[j] - z[j-1] 
    vec = np.array([dx, dy, dz]) 
    s[j] = s[j-1] + np.linalg.norm(vec) 

# create a new data with finer coords 
xvec = np.linspace(s[0], s[-1], 5000) 

# Call the Scipy cubic spline interpolator 
from scipy.interpolate import interpolate 

# Create new interpolation function for each axis against the norm 
f1 = interpolate.interp1d(s, x, kind='cubic') 
f2 = interpolate.interp1d(s, y, kind='cubic') 
f3 = interpolate.interp1d(s, z, kind='cubic') 

# create new fine data curve based on xvec 
xs = f1(xvec) 
ys = f2(xvec) 
zs = f3(xvec) 

# now let's plot to compare 
#1st, reverse z-axis for plotting 
z = z*-1 
zs = zs*-1 

dpi = 75 
fig = plt.figure(dpi=dpi, facecolor = '#617f8a') 
threeDPlot = fig.add_subplot(111, projection='3d') 
fig.subplots_adjust(left=0.03, bottom=0.02, right=0.97, top=0.98) 
mpl.rcParams['legend.fontsize'] = 10 

threeDPlot.scatter(x, y, z, color='r') # Original data as a scatter plot 
threeDPlot.plot3D(xs, ys, zs, label='Curve Fit', color='b', linewidth=1) 
threeDPlot.legend() 
plt.show() 

Wynik jest pokazany na poniższym rysunku. niebieska linia to dane dopasowania krzywej, a czerwone kropki to oryginalny zestaw danych. Jedną rzeczą, którą zauważyłem przy takim podejściu, jest to, że jeśli zbiór danych jest długi, to interpolate.interp1d nie jest wydajne i zajmuje dużo czasu.