私は次の投稿を参照しています：期間の発見にscipy.signal.spectral.lombscargleを使用する
ある場合には正しい答えが出ていることに気づきます。
sin（x）の頻度、つまり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)
以下を印刷します。結構です。私は推測する。lombscargle結果をで割る理由は、2piラジアンを周波数に変換する必要があるためです。（f=ラジアン/2pi）
1/2pi = 0.159154943092
Frequency = 0.159154943092
ただし、次の場合はうまくいかないようです。
sin（2x）の頻度、つまり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)
以下を印刷しています。
1/pi = 0.318309886184
Frequency = 0.0780862900972
間違っているようです。私が逃したステップはありますか？
