Peak Algılama Lomb-Scargle Yöntemini Kullanarak

Question

Lomb-Scargle yöntemini kullanarak verilerdeki pik değerleri bulan bir Python kodu çalıştırmaya çalışıyorum. aşağıda Bu yöntemi kullanarak

http://www.astropython.org/snippets/fast-lomb-scargle-algorithm32/

,

import lomb 
x = np.arange(10) 
y = np.sin(x) 
fx,fy, nout, jmax, prob = lomb.fasper(x,y, 6., 6.) 
print jmax 

sorunsuz, gayet iyi çalışıyor. Bu

df = pd.read_csv('extinct.csv',header=None) 
Y = pd.rolling_mean(df[0],window=5) 
fx,fy, nout, jmax, prob = lomb.fasper(np.array(Y.index),np.array(Y),6.,6.) 
print jmax 

görüntüler sadece 0. Farklı OFAC'nin, hifac değerleri geçen çalıştı, verilerin başka bir parça (veri aşağıda dökümü) üzerinde Ancak 8. yazdırır hiçbiri bana mantıklı değerleri verir.

Ana fonksiyon

""" 
from numpy import * 
from numpy.fft import * 

def __spread__(y, yy, n, x, m): 
    """ 
    Given an array yy(0:n-1), extirpolate (spread) a value y into 
    m actual array elements that best approximate the "fictional" 
    (i.e., possible noninteger) array element number x. The weights 
    used are coefficients of the Lagrange interpolating polynomial 
    Arguments: 
    y : 
    yy : 
    n : 
    x : 
    m : 
    Returns: 

    """ 
    nfac=[0,1,1,2,6,24,120,720,5040,40320,362880] 
    if m > 10. : 
    print 'factorial table too small in spread' 
    return 

    ix=long(x) 
    if x == float(ix): 
    yy[ix]=yy[ix]+y 
    else: 
    ilo = long(x-0.5*float(m)+1.0) 
    ilo = min(max(ilo , 1), n-m+1) 
    ihi = ilo+m-1 
    nden = nfac[m] 
    fac=x-ilo 
    for j in range(ilo+1,ihi+1): fac = fac*(x-j) 
    yy[ihi] = yy[ihi] + y*fac/(nden*(x-ihi)) 
    for j in range(ihi-1,ilo-1,-1): 
     nden=(nden/(j+1-ilo))*(j-ihi) 
     yy[j] = yy[j] + y*fac/(nden*(x-j)) 

def fasper(x,y,ofac,hifac, MACC=4): 
    """ function fasper 
    Given abscissas x (which need not be equally spaced) and ordinates 
    y, and given a desired oversampling factor ofac (a typical value 
    being 4 or larger). this routine creates an array wk1 with a 
    sequence of nout increasing frequencies (not angular frequencies) 
    up to hifac times the "average" Nyquist frequency, and creates 
    an array wk2 with the values of the Lomb normalized periodogram at 
    those frequencies. The arrays x and y are not altered. This 
    routine also returns jmax such that wk2(jmax) is the maximum 
    element in wk2, and prob, an estimate of the significance of that 
    maximum against the hypothesis of random noise. A small value of prob 
    indicates that a significant periodic signal is present. 

    Reference: 
    Press, W. H. & Rybicki, G. B. 1989 
    ApJ vol. 338, p. 277-280. 
    Fast algorithm for spectral analysis of unevenly sampled data 
    (1989ApJ...338..277P) 

    Arguments: 
     X : Abscissas array, (e.g. an array of times). 
     Y : Ordinates array, (e.g. corresponding counts). 
     Ofac : Oversampling factor. 
     Hifac : Hifac * "average" Nyquist frequency = highest frequency 
      for which values of the Lomb normalized periodogram will 
      be calculated. 

    Returns: 
     Wk1 : An array of Lomb periodogram frequencies. 
     Wk2 : An array of corresponding values of the Lomb periodogram. 
     Nout : Wk1 & Wk2 dimensions (number of calculated frequencies) 
     Jmax : The array index corresponding to the MAX(Wk2). 
     Prob : False Alarm Probability of the largest Periodogram value 
     MACC : Number of interpolation points per 1/4 cycle 
      of highest frequency 

    History: 
    02/23/2009, v1.0, MF 
     Translation of IDL code (orig. Numerical recipies) 
    """ 
    #Check dimensions of input arrays 
    n = long(len(x)) 
    if n != len(y): 
    print 'Incompatible arrays.' 
    return 

    nout = 0.5*ofac*hifac*n 
    nfreqt = long(ofac*hifac*n*MACC) #Size the FFT as next power 
    nfreq = 64L    # of 2 above nfreqt. 

    while nfreq < nfreqt: 
    nfreq = 2*nfreq 

    ndim = long(2*nfreq) 

    #Compute the mean, variance 
    ave = y.mean() 
    ##sample variance because the divisor is N-1 
    var = ((y-y.mean())**2).sum()/(len(y)-1) 
    # and range of the data. 
    xmin = x.min() 
    xmax = x.max() 
    xdif = xmax-xmin 

    #extirpolate the data into the workspaces 
    wk1 = zeros(ndim, dtype='complex') 
    wk2 = zeros(ndim, dtype='complex') 

    fac = ndim/(xdif*ofac) 
    fndim = ndim 
    ck = ((x-xmin)*fac) % fndim 
    ckk = (2.0*ck) % fndim 

    for j in range(0L, n): 
    __spread__(y[j]-ave,wk1,ndim,ck[j],MACC) 
    __spread__(1.0,wk2,ndim,ckk[j],MACC) 

    #Take the Fast Fourier Transforms 
    wk1 = ifft(wk1)*len(wk1) 
    wk2 = ifft(wk2)*len(wk1) 

    wk1 = wk1[1:nout+1] 
    wk2 = wk2[1:nout+1] 
    rwk1 = wk1.real 
    iwk1 = wk1.imag 
    rwk2 = wk2.real 
    iwk2 = wk2.imag 

    df = 1.0/(xdif*ofac) 

    #Compute the Lomb value for each frequency 
    hypo2 = 2.0 * abs(wk2) 
    hc2wt = rwk2/hypo2 
    hs2wt = iwk2/hypo2 

    cwt = sqrt(0.5+hc2wt) 
    swt = sign(hs2wt)*(sqrt(0.5-hc2wt)) 
    den = 0.5*n+hc2wt*rwk2+hs2wt*iwk2 
    cterm = (cwt*rwk1+swt*iwk1)**2./den 
    sterm = (cwt*iwk1-swt*rwk1)**2./(n-den) 

    wk1 = df*(arange(nout, dtype='float')+1.) 
    wk2 = (cterm+sterm)/(2.0*var) 
    pmax = wk2.max() 
    jmax = wk2.argmax() 


    #Significance estimation 
    #expy = exp(-wk2)   
    #effm = 2.0*(nout)/ofac  
    #sig = effm*expy 
    #ind = (sig > 0.01).nonzero() 
    #sig[ind] = 1.0-(1.0-expy[ind])**effm 

    #Estimate significance of largest peak value 
    expy = exp(-pmax)   
    effm = 2.0*(nout)/ofac  
    prob = effm*expy 

    if prob > 0.01: 
    prob = 1.0-(1.0-expy)**effm 

    return wk1,wk2,nout,jmax,prob 

def getSignificance(wk1, wk2, nout, ofac): 
    """ returns the peak false alarm probabilities 
    Hence the lower is the probability and the more significant is the peak 
    """ 
    expy = exp(-wk2)   
    effm = 2.0*(nout)/ofac  
    sig = effm*expy 
    ind = (sig > 0.01).nonzero() 
    sig[ind] = 1.0-(1.0-expy[ind])**effm 
    return sig 

Veri,

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1.62195121951 
1.77743902439 
1.77743902439 
4.42073170732 
4.42073170732 
4.42073170732 
4.42073170732 
4.42073170732 
4.42073170732 
4.42073170732 
3.95426829268 
3.95426829268 
2.71036585366 
2.71036585366 
2.71036585366 
2.71036585366 
2.71036585366 
1.77743902439 
2.86585365854 
3.02134146341 
2.86585365854 
2.86585365854 
3.17682926829 
3.17682926829 

Plot

Herhangi bir yardım mutluluk duyacağız,

Answer 1

Afte biraz kazı, AstroML yöntemi en iyi gibi görünüyor.

Anlamlılık seviyesi verir

import numpy as np 
from matplotlib import pyplot as plt 
from astroML.time_series import lomb_scargle, search_frequencies 
import pandas as pd 

df = pd.read_csv('extinct.csv',header=None) 
Y = df[0] 

dy = 0.5 + 0.5 * np.random.random(len(df)) 
omega = np.linspace(10, 100, 1000) 

sig = np.array([0.1, 0.01, 0.001]) 
PS, z = lomb_scargle(df.index, Y, dy, omega, generalized=True, significance=sig) 

plt.plot(omega,PS) 
plt.hold(True) 

xlim = (omega[0], omega[-1]) 
for zi, pi in zip(z, sig): 
    plt.plot(xlim, (zi, zi), ':k', lw=1) 
    plt.text(xlim[-1] - 0.001, zi - 0.02, "$%.1g$" % pi, ha='right', va='top') 
    plt.hold(True) 
plt.show() 

de grafikte gösterilir. LS'yi genelleştirdim ve düzgünleştirme kullanmıyordum.