我指的是以下帖子：使用 scipy.signal.spectral.lombscargle 进行经期发现
我意识到在某些情况下给出的答案是正确的。
sin(x) 的频率，即 1/(2* pi)
# imports the numerical array and scientific computing packages
import numpy as np
import scipy as sp
from scipy.signal import spectral
# generates 100 evenly spaced points between 1 and 1000
time = np.linspace(1, 1000, 100)
# computes the sine value of each of those points
mags = np.sin(time)
# scales the sine values so that the mean is 0 and the variance is 1 (the documentation specifies that this must be done)
scaled_mags = (mags-mags.mean())/mags.std()
# generates 1000 frequencies between 0.01 and 1
freqs = np.linspace(0.01, 1, 1000)
# computes the Lomb Scargle Periodogram of the time and scaled magnitudes using each frequency as a guess
periodogram = spectral.lombscargle(time, scaled_mags, freqs)
# returns the inverse of the frequence (i.e. the period) of the largest periodogram value
print "1/2pi = " + str(1/(2*np.pi))
print "Frequency = " + str(freqs[np.argmax(periodogram)] / 2.0 / np.pi)
打印以下内容。没问题。我猜。我们将lombscargle结果除以的原因2pi是，我们需要将弧度转换为频率。(f = 弧度 / 2pi)
1/2pi = 0.159154943092
Frequency = 0.159154943092
但是，对于以下情况，事情似乎出了问题。
sin(2x) 的频率，即 1/(pi)
# imports the numerical array and scientific computing packages
import numpy as np
import scipy as sp
from scipy.signal import spectral
# generates 100 evenly spaced points between 1 and 1000
time = np.linspace(1, 1000, 100)
# computes the sine value of each of those points
mags = np.sin(2 * time)
# scales the sine values so that the mean is 0 and the variance is 1 (the documentation specifies that this must be done)
scaled_mags = (mags-mags.mean())/mags.std()
# generates 1000 frequencies between 0.01 and 1
freqs = np.linspace(0.01, 1, 1000)
# computes the Lomb Scargle Periodogram of the time and scaled magnitudes using each frequency as a guess
periodogram = spectral.lombscargle(time, scaled_mags, freqs)
# returns the inverse of the frequence (i.e. the period) of the largest periodogram value
print "1/pi = " + str(1/(np.pi))
print "Frequency = " + str(freqs[np.argmax(periodogram)] / 2.0 / np.pi)
正在打印以下内容。
1/pi = 0.318309886184
Frequency = 0.0780862900972
似乎不正确。我错过了什么步骤？
