如何在python中进行第二次插值 [英] How to do a second interpolation in python

问题描述

我使用 numpy进行了第一次插值.polyfit()和

I did my first interpolation with numpy.polyfit() and numpy.polyval() for 50 longitude values for a full satellite orbit.

现在，我只想看一个0-4.5度经度的窗口并进行第二次插值，以便在该窗口中有6,000个经度点.

Now, I just want to look at a window of 0-4.5 degrees longitude and do a second interpolation so that I have 6,000 points for longitude in the window.

我需要使用第一个插值中的等式/曲线来创建第二个插值，因为窗口范围内只有一个点.我不确定如何进行第二次插值.

I need to use the equation/curve from the first interpolation to create the second one because there is only one point in the window range. I'm not sure how to do the second interpolation.

输入:

lon = [-109.73105744378498, -104.28690174554579, -99.2435132929552, -94.48533149079628, -89.91054414962821, -85.42671400689177, -80.94616150449806, -76.38135021210172, -71.6402674905218, -66.62178379632216, -61.21120467960157, -55.27684029674759, -48.66970878028004, -41.23083703244677, -32.813881865289346, -23.332386757370532, -12.832819226213942, -1.5659455609661785, 10.008077792630402, 21.33116444634303, 31.92601575632583, 41.51883213364072, 50.04498630545507, 57.58103957109249, 64.26993028992476, 70.2708323505337, 75.73441871754586, 80.7944079829813, 85.56734813043659, 90.1558676264546, 94.65309120129724, 99.14730128118617, 103.72658922048785, 108.48349841714494, 113.51966824008079, 118.95024882101737, 124.9072309203375, 131.5395221402974, 139.00523971191907, 147.44847902856114, 156.95146022590976, 167.46163867248032, 178.72228750873975, -169.72898181991064, -158.44642409799974, -147.8993300787564, -138.35373014113995, -129.86955508919888, -122.36868103811106, -115.70852432245486]

myOrbitJ2000Time = [ 20027712.,  20027713.,  20027714.,  20027715.,  20027716.,
        20027717.,  20027718.,  20027719.,  20027720.,  20027721.,
        20027722.,  20027723.,  20027724.,  20027725.,  20027726.,
        20027727.,  20027728.,  20027729.,  20027730.,  20027731.,
        20027732.,  20027733.,  20027734.,  20027735.,  20027736.,
        20027737.,  20027738.,  20027739.,  20027740.,  20027741.,
        20027742.,  20027743.,  20027744.,  20027745.,  20027746.,
        20027747.,  20027748.,  20027749.,  20027750.,  20027751.,
        20027752.,  20027753.,  20027754.,  20027755.,  20027756.,
        20027757.,  20027758.,  20027759.,  20027760.,  20027761.]

代码:

deg = 30 #polynomial degree for fit
fittime = myOrbitJ2000Time - myOrbitJ2000Time[0]

'Longitude Interpolation'
fitLon = np.polyfit(fittime, lon, deg)   #gets fit coefficients
polyval_lon = np.polyval(fitLon,fittime) #interp.s to get actual values


'Get Longitude values for a window of 0-4.5 deg Longitude'
lonwindow =[]

for i in range(len(polyval_lon)):
    if 0 < polyval_lon[i] < 4.5:         # get lon vals in window
        lonwindow.append(polyval_lon[i]) #append lon vals

lonwindow = np.array(lonwindow)

推荐答案

首先，使用旧时间(x轴)值和插值经度(y轴)值生成多项式拟合系数.

First, generate the polynomial fit coefficients using the old time (x-axis) values, and interpolated longitude (y-axis) values.

import numpy as np  
import matplotlib.pyplot as plt

poly_deg = 3 #degree of the polynomial fit
polynomial_fit_coeff = np.polyfit(original_times, interp_lon, poly_deg)

接下来，使用np.linspace()根据窗口中的期望点数生成任意时间值.

Next, use np.linspace() to generate arbitrary time values based on the number of desire points in the window.

start = 0
stop = 4
num_points = 6000
arbitrary_time = np.linspace(start, stop, num_points)

最后，使用拟合系数和任意时间来获取实际的内插经度(y轴)值并绘图.

Finally, use the fit coefficients and the arbitrary time to get the actual interpolated longitude (y-axis) values and plot.

lon_intrp_2 = np.polyval(polynomial_fit_coeff, arbitrary_time)

plt.plot(arbitrary_time, lon_intrp_2, 'r') #interpolated window as a red curve
plt.plot(myOrbitJ2000Time, lon, '.') #original data plotted as points

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