Question

我为纬度，经度和＆amp;做了numpy.polyfit()卫星轨道的高度数据，并用numpy.polyval()进行插值（50点）。

现在，我想要一个窗口（经度0-4.5度）并进行更高分辨率的插值（6,000点）。我认为我需要使用第一个低分辨率拟合的拟合系数，以便为我的经度窗口进行插值，我不太清楚如何做到这一点。

输入：

lat = [27.755611104020687, 22.50661883405905, 17.083576087905502, 11.53891099628959, 5.916633366002468, 0.2555772624429494, -5.407902834141322, -11.037514984810027, -16.594621304857206, -22.03556688048686, -27.308475759820045, -32.34927891621322, -37.07690156937186, -41.38803163295967, -45.15306971601912, -48.21703193866987, -50.41165326774015, -51.58419672864487, -51.63883932997542, -50.57025116952513, -48.46557920053242, -45.47329014246061, -41.76143266388077, -37.48707787049647, -32.782653540783, -27.754184631685046, -22.48503337048438, -17.041097574740743, -11.475689837873944, -5.833592289780744, -0.1543286595142316, 5.525119007560692, 11.167878192881306, 16.73476477885508, 22.18160021405449, 27.455997555900108, 32.493386953033685, 37.21222272985329, 41.508824407948275, 45.25350232626601, 48.291788915858554, 50.45698534747271, 51.59925055739275, 51.62660832560593, 50.53733379179681, 48.420673231121725, 45.42531420150485, 41.71819693220144, 37.45473807165676, 32.76569228387106]

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]

alt = [374065.49207488785, 372510.1635949105, 371072.75959230476, 369836.3092635453, 368866.7921820211, 368209.0950216997, 367884.3703536549, 367888.97894243425, 368195.08833668986, 368752.88080031495, 369494.21701128664, 370337.49662954226, 371193.3839051864, 371971.0136622536, 372584.272228585, 372957.752022573, 373032.0104747458, 372767.8112563471, 372149.0940816824, 371184.49208500446, 369907.2992362557, 368373.8795969478, 366660.5935723809, 364859.4071422184, 363072.42955020745, 361405.69765685993, 359962.58417682414, 358837.24421522504, 358108.5277743581, 357834.7679493668, 358049.8054538341, 358760.531463618, 359946.1257064284, 361559.04646970675, 363527.70518032915, 365760.6377191965, 368151.8843206526, 370587.2165838985, 372950.8014553002, 375131.8814988529, 377031.06540952163, 378565.8596562773, 379675.13241518533, 380322.2707576381, 380496.8682141012, 380214.86538256245, 379517.14674525027, 378466.68079100474, 377144.36811517406, 375643.83731560566]

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]

'Latitude Interpolation'    
fitLat = np.polyfit(fittime, lat, deg)
polyval_lat = np.polyval(fitLat,fittime)

'Longitude Interpolation'
fitLon = np.polyfit(fittime, lon, deg)
polyval_lon = np.polyval(fitLon,fittime)

'Altitude Interpolation'
fitAlt = np.polyfit(fittime, alt, deg)
polyval_alt = np.polyval(fitAlt,fittime)


'Get Lat, Lon, & Alt values for a window of 0-4.5 deg Longitude'
lonwindow =[]
latwindow = []
altwindow = []
for i in range(len(polyval_lat)):
    if 0 < polyval_lon[i] < 4.5:         # get lon vals in window
        lonwindow.append(polyval_lon[i]) #append lon vals
        latwindow.append(polyval_lat[i]) #append corresponding lat vals
        altwindow.append(polyval_alt[i]) #append corresponding alt vals

lonwindow = np.array(lonwindow)

只是要清楚 - 问题是我在窗口范围内只有一个点，我想使用上一步中的插值/方程/曲线。那么我可以使用它再次插值并在我的窗口范围内生成6,000个点。

Answer 1

原帖回复here

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。

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（）根据窗口中的欲望点数生成任意时间值。

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得实际插值经度（y轴）值和绘图。

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

如何插入numpy.polyval和numpy.polyfit python

1 个答案: