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I created a function to generate and propagate a satellite orbit. Now, I want to save all my data in a .dat file. I'm not sure how many for loops I need or quite how to do it.

I want time, latitude, longitude, and altitude all on one line for each propagation step.

Code for data:

myOrbitJ2000Time = [1085.0, 2170.0, 3255.0, 4340.1, 5425.1]

lat = [48.5, 26.5, -28.8, -48.1, 0.1]

lon = [-144.1, -50.4, -1.6, 91.5, 152.8]

alt = [264779.5, 262446.1, 319661.8, 355717.3, 306129.0]

Desired Output in .dat file:

J2000 Time, Lat, Lon, Alt

1085.0, 48.6, -144.2, 264779.5

2170.0, 26.5, -50.4, 262446.1

3255.0, -28.7, -1.6, 319661.8

4340.1, -48.0, 91.5, 355717.2

5425.1, 0.1, 152.8, 06129.0

Full Code:

import orbital
from orbital import earth, KeplerianElements, plot
import matplotlib.pyplot as plt
import numpy as np
from astropy import time
from astropy.time import TimeDelta, Time
from astropy import units as u
from astropy import coordinates as coord


#def orbitPropandcoordTrans(orbitPineapple_J2000time, _ecc, _inc, _raan, _arg_pe, _meananom, meanMotion):
def orbitPropandcoordTrans(propNum, orbitPineapple_J2000time, _ecc, _inc, _raan, _arg_pe, _meananom, meanMotion):
        '''
        Create original orbit and run for 100 propagations (in total one whole orbit)
        in order to get xyz and time for each propagation step.
        The end goal is to plot the lat, lon, & alt data to see if it matches ISS groundtrace.
        '''
    import orbital
    from orbital import earth, KeplerianElements, plot
    import matplotlib.pyplot as plt
    import numpy as np
    from astropy import time
    from astropy.time import TimeDelta, Time
    from astropy import units as u
    from astropy import coordinates as coord

    'Calculate Avg. Period from Mean Motion'
    _avgPeriod = 86400 / meanMotion
    #print('_avgPeriod', _avgPeriod)

    ######
    #propNum = int(_avgPeriod)

    'Generate Orbit'
    #orbitPineapple = KeplerianElements.with_period(5560, body=earth, e=0.0004525, i=(np.deg2rad(51.6414)), raan=(np.deg2rad(247.1662)))
    orbitPineapple = KeplerianElements.with_period(_avgPeriod, body=earth, e=_ecc, i=(np.deg2rad(_inc)), raan=(np.deg2rad(_raan)), arg_pe=(np.deg2rad(_arg_pe)), M0=(np.deg2rad(_meananom))) #ref_epoch=   
    #plot(orbitPineapple)
    #plt.show()
    #print('')
    #print('')

    'Propagate Orbit and retrieve xyz'
    myOrbitX = []         #X Coordinate for propagated orbit step
    myOrbitY = []         #Y Coordinate for propagated orbit step
    myOrbitZ = []         #Z Coordinate for propagated orbit step
    myOrbitTime = []      #Time for each propagated orbit step
    myOrbitJ2000Time = [] #J2000 times
    #propNum = 100        #Number of propagations and Mean Anomaly size (one orbit 2pi/propNum)

    for i in range(propNum):
        orbitPineapple.propagate_anomaly_by(M=(2.0*np.pi/propNum)) #Propagate the orbit by the Mean Anomaly
        myOrbitX.append(orbitPineapple.r.x)                        #x vals
        myOrbitY.append(orbitPineapple.r.y)                        #y vals
        myOrbitZ.append(orbitPineapple.r.z)                        #z vals
        myOrbitTime.append(orbitPineapple_J2000time)               #J2000 time vals
        myOrbitJ2000Time.append(orbitPineapple.t)

        #plot(orbitPineapple)
    #print('X:', 'Length:', len(myOrbitX))
    #print(myOrbitX)
    #print('Y:', 'Length:',len(myOrbitY))
    #print(myOrbitY)
    #print('Z:', 'Length:', len(myOrbitZ))
    #print(myOrbitZ)
    #print('J2000 Time:', 'Length:',len(myOrbitTime))
    #print(myOrbitTime)


    '''Because the myOrbitTime is only the time between each step to be the sum of itself plus
    all the previous times. And then I need to convert that time from seconds after J2000 to UTC.'''
    myT = [] #UTC time list

    for i in range(propNum):
        myT.append((Time(2000, format='jyear') + TimeDelta(myOrbitTime[i]*u.s)).iso) #Convert time from J2000 to UTC
    #print('UTC Time List Length:', len(myT))
    #print('UTC Times:', myT)

    '''Now I have xyz and time for each propagation step and need to convert the coordinates from
    ECI to Lat, Lon, & Alt'''
    now = []     #UTC time at each propagation step
    xyz =[]      #Xyz coordinates from OrbitalPy initial orbit propagation
    cartrep = [] #Cartesian Representation
    gcrs = []    #Geocentric Celestial Reference System/Geocentric Equatorial Inertial, the default coord system of OrbitalPy
    itrs =[]     #International Terrestrial Reference System coordinates
    lat = []     #Longitude of the location, for the default ellipsoid
    lon = []     #Longitude of the location, for the default ellipsoid
    alt = []     #Height of the location, for the default ellipsoid


    for i in range(propNum):
        xyz = (myOrbitX[i], myOrbitY[i], myOrbitZ[i])                   #Xyz coord for each prop. step
        now = time.Time(myT[i])                                         #UTC time at each propagation step
        cartrep = coord.CartesianRepresentation(*xyz, unit=u.m)         #Add units of [m] to xyz
        gcrs = coord.GCRS(cartrep, obstime=time.Time(myT[i]))           #Let AstroPy know xyz is in GCRS
        itrs = gcrs.transform_to(coord.ITRS(obstime=time.Time(myT[i]))) #Convert GCRS to ITRS
        loc = coord.EarthLocation(*itrs.cartesian.xyz)                  #Get lat/lon/height from ITRS
        lat.append(loc.lat.value)                                       #Create latitude list
        lon.append(loc.lon.value)                                       #Create longitude list
        alt.append(loc.height.value)           

    #print('Current Time:', now)
    #print('')
    #print('UTC Time:')
    #print(myT)
    print('myOrbitJ2000Time', myOrbitJ2000Time)
    print('')
    print('Lat:')
    print('')
    print(lat)
    print('')
    print('Lon:')
    print(lon)
    print('')
    print('Alt:')
    print(alt)

orbitPropandcoordTrans(5, -34963095, 0.0073662, 51.5946, 154.8079, 103.6176, 257.3038, 15.92610159)

4

2 回答 2

1

为了回答您的一般问题,您可以从创建 Numpy 数组开始,而不是定义一堆 Python 列表(附加和使用这些列表很慢,尤其是在您扩大分辨率时处理大量值时)的列表。初始化 Numpy 数组时,通常需要指定数组的大小,但在这种情况下,这很容易,因为您知道需要多少次传播。例如:

>>> orbitX = np.empty(propNum, dtype=float)

创建一个空的 NumpypropNum浮点值数组(这里的“空”表示该数组没有用任何值初始化——这是创建新数组的最快方法,因为我们稍后将填充其中的所有值无论如何。

然后在您的循环中,而不是使用orbitX.append(x)您将分配给与当前刻度对应的数组中的值:orbitX[i] = x。其他情况也一样。

然后有几种输出数据的可能性,但是使用 AstropyTable提供了很大的灵活性。您可以创建一个包含您想要的列的表,例如:

>>> from astropy.table import Table
>>> table = Table([J2000, lat, lon, alt], names=('J2000', 'lat', 'lon', 'alt'))

拥有 Table 对象的好处是有大量的输出格式选项。例如在 Python 提示符下:

>>> table
    J2000          lat            lon            alt
   float64       float64        float64        float64
------------- -------------- -------------- -------------
1085.01128806  48.5487129749 -144.175054697 264779.500823
2170.02257613  26.5068122883 -50.3805485685 262446.085716
3255.03386419 -28.7915478311  -1.6090285674 319661.817451
4340.04515225 -48.0536526356  91.5416828221 355717.274021
5425.05644032 0.084422392655  152.811717713  306129.02576

要首先输出到文件,您必须考虑如何格式化数据。您可以考虑使用许多常见的数据格式,但这取决于数据的用途和使用者(“.dat 文件”在文件格式方面没有任何意义;或者更确切地说,它可以什么意思)。但是在您的问题中,您指出您想要的是所谓的“逗号分隔值”(CSV),其中数据的格式为列向下,行中的每个值用逗号分隔。该类Table可以很容易地输出 CSV(和任何变体):

>>> import sys
>>> table.write(sys.stdout, format='ascii.csv')  # Note: I'm just using sys.stdout for demonstration purposes; normally you would give a filename
J2000,lat,lon,alt
1085.011288063746,48.54871297493748,-144.17505469691633,264779.5008225624
2170.022576127492,26.506812288280788,-50.38054856853237,262446.0857159357
3255.0338641912376,-28.79154783108773,-1.6090285674024463,319661.81745081506
4340.045152254984,-48.05365263557444,91.54168282208444,355717.2740210131
5425.05644031873,0.08442239265500713,152.81171771323176,306129.0257600865

不过还有很多其他选择。例如,如果您希望数据在对齐的列中很好地格式化,您也可以这样做。你可以在这里阅读更多关于它的信息。(另外,我建议如果你想要一个纯文本文件格式,你可以试试ascii.ecsv,它的优点是它输出额外的元数据,可以很容易地读回 Astropy 表):

>>> table.write(sys.stdout, format='ascii.ecsv')
# %ECSV 0.9
# ---
# datatype:
# - {name: J2000, datatype: float64}
# - {name: lat, datatype: float64}
# - {name: lon, datatype: float64}
# - {name: alt, datatype: float64}
# schema: astropy-2.0
J2000 lat lon alt
1085.01128806 48.5487129749 -144.175054697 264779.500823
2170.02257613 26.5068122883 -50.3805485685 262446.085716
3255.03386419 -28.7915478311 -1.6090285674 319661.817451
4340.04515225 -48.0536526356 91.5416828221 355717.274021
5425.05644032 0.084422392655 152.811717713 306129.02576

我要注意的另一件事是 Astropy 中的许多对象除了单个值之外还可以对整个数组进行操作,这通常可以更有效,尤其是对于许多值。特别是,你有这个 Python 循环:

for i in range(propNum):
    xyz = (myOrbitX[i], myOrbitY[i], myOrbitZ[i])                   #Xyz coord for each prop. step
    now = time.Time(myT[i])                                         #UTC time at each propagation step
    cartrep = coord.CartesianRepresentation(*xyz, unit=u.m)         #Add units of [m] to xyz
    gcrs = coord.GCRS(cartrep, obstime=time.Time(myT[i]))           #Let AstroPy know xyz is in GCRS
    itrs = gcrs.transform_to(coord.ITRS(obstime=time.Time(myT[i]))) #Convert GCRS to ITRS
    loc = coord.EarthLocation(*itrs.cartesian.xyz)                  #Get lat/lon/height from ITRS
    lat.append(loc.lat.value)                                       #Create latitude list
    lon.append(loc.lon.value)                                       #Create longitude list
    alt.append(loc.height.value)

这可以通过将它们视为坐标数组来完全重写而无需显式循环。例如:

>>> times = time.Time(myT)  # All the times, not just a single one
>>> cartrep = coord.CartesianRepresentation(orbitX, orbitY, orbitZ, unit=u.m)  # Again, an array of coordinates
>>> gcrs = coord.GCRS(cartrep, obstime=times)
>>> itrs = gcrs.transform_to(coord.ITRS(obstime=ts))
>>> loc = coord.EarthLocation(*itrs.cartesian.xyz)  # I'm not sure this is the most efficient way to do this but I'm not an expert on the coordinates package

有了这个,我们也可以将所有坐标作为数组获取。例如:

>>> loc.lat
<Latitude [ 48.54871297, 26.50681229,-28.79154783,-48.05365264,
             0.08442239] deg>

因此,这通常是处理大量坐标值的更有效方式。同样,在原始代码中转换时,您可以创建一个数组并将其添加到初始时间myT,而不是遍历所有时间偏移量。TimeDelta

不幸的是,我不是该orbital软件包的专家,但它似乎并不像人们想要获得轨道上不同点的坐标那样简单。ISTM 应该有。

于 2017-10-14T17:18:28.717 回答
0

可能最简单的方法是使用zip()

data = zip(myOrbitJ2000Time, lat, lon, alt)

因此,“数据”将具有您要序列化的格式。不要忘记序列化您想要的标头。

于 2017-10-12T19:39:13.220 回答