手动生成ifft曲线以将其用于预测

Question

fft_tr = np.fft.fft(demand)
psd = (fft_tr * np.conj(fft_tr))/len(fft_tr)
ampl = (np.sqrt(fft_tr * np.conj(fft_tr))/(len(fft_tr)/2)).real
freq = 1/(1*len(train_df.demand)) * np.arange(len(train_df.demand))
l = np.arange(0, np.floor(len(freq)/2), dtype = "int")
df = pd.DataFrame(zip(freq,psd.real[l],fft_tr.real[l], fft_tr.imag[l],fft_tr[l],ampl[l]),columns = ["freq","PSD_r","fft_real","fft_imag","fft_tr","ampl"])
df["Theta"] = (df["fft_real"]/df["fft_imag"]).apply(lambda x : math.atan(x))


df1 = df.loc[df.PSD_r > 15000,["freq","Theta","ampl"]].reset_index().reset_index(drop = True, inplace=False)

def construct_Waves(time_series, v):
    
    ''' this function will construct the waves given a time series 
    along with the list of freq and phase which is theta'''

    return v[2] * np.sin(2*np.pi*time_series*v[0]-v[1])

v = np.zeros(len(time))
for i in range(1,len(df1)):
    x = list(df1.loc[i,["freq","Theta","ampl"]])
    x1 = np.array(construct_Waves(time+1,x))
    v = v+x1

plt.figure(figsize=(20,4))
plt.plot(v)

usp = sharing？手动ifft。从FFT结果。

这是我得到的图。实际的IFFT曲线就是这样。

可以帮助我手动获取实际曲线，以便我可以将其扩展到将来的时间段。