问题
In the time tests shown below, I found that Skyfield takes several hundred microseconds up to a millisecond to return obj.at(jd).position.km for a single time value in jd, but the incremental cost for longer JulianDate objects (a list of points in time) is only about one microsecond per point. I see similar speeds using Jplephem and with two different ephemerides.
My question here is: if I want to random-access points in time, for example as a slave to an external Runge-Kutta routine which uses its own variable stepsize, is there a way I can do this faster within python (without having to learn to compile code)?
I understand this is not at all the typical way Skyfield is intended to be used. Normally we'd load a JulianDate object with a long list of time points and then calculate them at once, and probably do that a few times, not thousands of times (or more), the way an orbit integrator might do.
Workaround: I can imagine a work-around where I build my own NumPy database by running Skyfield once using a JulianDate object with fine time granularity, then writing my own Runge-Kutta routine which changes step sizes up and down by discrete amounts such that the timesteps always correspond directly to the striding of NumPy array.
Or I could even try re-interpolating. I am not doing highly precise calculations so a simple NumPy or SciPy 2nd order might be fine.
Ultimately I'd like to try integrating the path of objects under the influence of the gravity field of the solar system (e.g. deep-space satellite, comet, asteroid). When looking for an orbit solution one might try millions of starting state vectors in 6D phase space. I know I should be using things like ob.at(jd).observe(large_body).position.km method because gravity travels at the speed of light like everything else. This seems to cost substantial time since (I'm guessing) it's an iterative calculation ("Let's see... where would Jupiter have been such that I feel it's gravity right NOW"). But let's peel the cosmic onion one layer at a time.
Figure 1. Skyfield and JPLephem performance on my laptop for different length JulianDate objects, for de405 and de421. They are all about the same - (very) roughly about a half-millisecond for the first point and a microsecond for each additional point. Also the very first point to be calculated when the script runs (Earth (blue) with len(jd) = 1) has an additional millisecond artifact.
Earth and Moon are slower because it is a two-step calculation internally (the Earth-Moon Barycenter plus the individual orbits about the Barycenter). Mercury may be slower because it moves so fast compared to the ephemeris time steps that it requires more coefficients in the (costly) Chebyshev interpolation?
SCRIPT FOR SKYFIELD DATA the JPLephem script is farther down
import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import load, JulianDate
import time
ephem = 'de421.bsp'
ephem = 'de405.bsp'
de = load(ephem)
earth = de['earth']
moon = de['moon']
earth_barycenter = de['earth barycenter']
mercury = de['mercury']
jupiter = de['jupiter barycenter']
pluto = de['pluto barycenter']
things = [ earth, moon, earth_barycenter, mercury, jupiter, pluto ]
names = ['earth', 'moon', 'earth barycenter', 'mercury', 'jupiter', 'pluto']
ntimes = [i*10**n for n in range(5) for i in [1, 2, 5]]
years = [np.zeros(1)] + [np.linspace(0, 100, n) for n in ntimes[1:]] # 100 years
microsecs = []
for y in years:
jd = JulianDate(utc=(1900 + y, 1, 1))
mics = []
for thing in things:
tstart = time.clock()
answer = thing.at(jd).position.km
mics.append(1E+06 * (time.clock() - tstart))
microsecs.append(mics)
microsecs = np.array(microsecs).T
many = [len(y) for y in years]
fig = plt.figure()
ax = plt.subplot(111, xlabel='length of JD object',
ylabel='microseconds',
title='time for thing.at(jd).position.km with ' + ephem )
for item in ([ax.title, ax.xaxis.label, ax.yaxis.label] +
ax.get_xticklabels() + ax.get_yticklabels()):
item.set_fontsize(item.get_fontsize() + 4) # http://stackoverflow.com/a/14971193/3904031
for name, mics in zip(names, microsecs):
ax.plot(many, mics, lw=2, label=name)
plt.legend(loc='upper left', shadow=False, fontsize='x-large')
plt.xscale('log')
plt.yscale('log')
plt.savefig("skyfield speed test " + ephem.split('.')[0])
plt.show()
SCRIPT FOR JPLEPHEM DATA the Skyfield script is above
import numpy as np
import matplotlib.pyplot as plt
from jplephem.spk import SPK
import time
ephem = 'de421.bsp'
ephem = 'de405.bsp'
kernel = SPK.open(ephem)
jd_1900_01_01 = 2415020.5004882407
ntimes = [i*10**n for n in range(5) for i in [1, 2, 5]]
years = [np.zeros(1)] + [np.linspace(0, 100, n) for n in ntimes[1:]] # 100 years
barytup = (0, 3)
earthtup = (3, 399)
# moontup = (3, 301)
microsecs = []
for y in years:
mics = []
#for thing in things:
jd = jd_1900_01_01 + y * 365.25 # roughly, it doesn't matter here
tstart = time.clock()
answer = kernel[earthtup].compute(jd) + kernel[barytup].compute(jd)
mics.append(1E+06 * (time.clock() - tstart))
microsecs.append(mics)
microsecs = np.array(microsecs)
many = [len(y) for y in years]
fig = plt.figure()
ax = plt.subplot(111, xlabel='length of JD object',
ylabel='microseconds',
title='time for jplephem [0,3] and [3,399] with ' + ephem )
# from here: http://stackoverflow.com/a/14971193/3904031
for item in ([ax.title, ax.xaxis.label, ax.yaxis.label] +
ax.get_xticklabels() + ax.get_yticklabels()):
item.set_fontsize(item.get_fontsize() + 4)
#for name, mics in zip(names, microsecs):
ax.plot(many, microsecs, lw=2, label='earth')
plt.legend(loc='upper left', shadow=False, fontsize='x-large')
plt.xscale('log')
plt.yscale('log')
plt.ylim(1E+02, 1E+06)
plt.savefig("jplephem speed test " + ephem.split('.')[0])
plt.show()
来源：https://stackoverflow.com/questions/35358401/integration-of-orbits-with-solar-system-gravity-fields-from-skyfield-speed-iss