interpolación cúbica por piezas para la curva 3D en python

Question

Tengo una curva en el espacio 3D. Quiero usar una interpolación cúbica por partes que preserve la forma, similar a pchip en matlab. Investigué funciones provistas en scipy.interpolate, p. Ej. interp2d, pero las funciones funcionan para algunas estructuras de curva y no para los puntos de datos que tengo. ¿Alguna idea de cómo hacerlo?

Éstos son los puntos de datos:

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

Es posible que desee utilizar splprep() and splev(), como esto (explicación básica en los comentarios):

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() 

Produce:

que es bueno y suave, también esto funciona e para su conjunto completo de datos publicados.

Answer 2

Encontré an email de un compañero que también estaba buscando una versión nativa de Python de la función pchip() de Matlab. Adjuntó his code que desafortunadamente quiere descargar como 'attachment-001.bin'. Si guarda el archivo y lo cambia a pychip.py, encontrará que es exactamente lo que está pidiendo.

Answer 3

Scipy tiene pchip en scipy.interpolate --- pero, uhh, alguien se olvidó de incluirla en la documentación:

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) 

Para los datos en 3-D, simplemente interpolar cada una de las 3 coordenadas separado.

Answer 4

Aquí hay otra solución que hace lo que yo quería (preservación de forma).
El problema es que no hay una fórmula o ecuación clara para conectar los puntos. La respuesta está en crear un nuevo conjunto de datos que sea común a los diferentes puntos. Este nuevo conjunto de datos es la longitud a lo largo de la ruta (norma). Luego usamos interp1 para interpolar cada conjunto.

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() 

El resultado se muestra en la figura siguiente. la línea azul es la información de ajuste de la curva, mientras que los puntos rojos son el conjunto de datos original. Una cosa que noté con este enfoque, sin embargo, es que si el conjunto de datos es largo, entonces interpolate.interp1d no es eficiente y lleva mucho tiempo.