python'da 3D eğrisi için şekil koruyan parçalı kübik enterpolasyon

Question

3B alanda bir eğrim var. Matlabda pchip'e benzer bir şekil koruyan parçalı kübik enterpolasyon kullanmak istiyorum. Scipy.interpolate, ör. interp2d, ancak işlevler, sahip olduğum veri noktalarını değil, bazı eğri yapıları için çalışır. Nasıl yapılacağı hakkında bir fikrin var mı?

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

Muhtemelen böyle splprep() and splev(), kullanmak istediğiniz

(yorumlarda temel explaination):

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() 

üretir:

İşte

veri noktalarıdır

Bu güzel ve pürüzsüz, aynı zamanda çalışır Yayınlanan veri kümeniz için e.

Answer 2

Matlab'ın pchip() işlevinin bir yerel Python sürümünü de arayan bir arkadaşımdan an email buldum. Maalesef 'attachment-001.bin' olarak indirmek isteyen his code'u ekledi. Dosyayı kaydedip pychip.py'ye yeniden adlandırırsanız, tam olarak ne istediğinizi bulacaksınız.

Answer 3

Scipy pchipscipy.interpolate içinde var --- ama, ahh, birisi belgelerinde listeleyebilmemiz unuttum:

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) 

3-D verileri için, sadece 3 her interpole ayrı ayrı koordine eder.

Answer 4

İşte istediğimi yapan başka bir çözüm (şekil koruyucu).
Sorun, noktaları birleştirecek net bir formül veya denklem olmamasıdır. Cevap, farklı noktalarda ortak olan yeni bir veri seti oluşturulmasında yatar. Bu yeni veri seti yol boyunca (norm) olan uzunluktur. Daha sonra her setin enterpolasyonu için interp1 kullanırız.

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() 

Sonuç, aşağıdaki şekilde gösterilmiştir. Mavi çizgi, eğri uydurma verileridir, kırmızı noktalar ise orijinal veri kümesidir. Bu yaklaşımla fark ettiğim bir şey, veri kümesi uzunsa, interpolate.interp1d'nin verimli olmaması ve uzun zaman almasıdır.