如何在Python中手动复现ifft2？问题排查与修正

我尝试找出可累加还原原始二维实数据的FFT2正弦波集合，参考了一个一维FFT实现示例（感谢@mark snyder）。手动实现ifft2时遇到两个现象：

• 仅保留第一维度计算时，第一列数据始终能匹配原始数据
• 疑惑为何需要对dimT的输入减1才能对齐相位

以下是最初的尝试代码：

import pandas as pd
import numpy as np
from scipy.fftpack import fftfreq, fft2
import cmath
df=pd.DataFrame(    { 'Ang':[  0.091, -0.141, -0.114], 
                      'Con':[ -0.139, -0.259, -0.573],
                      'EAC':[  0.016, -0.106, -0.044]   } ,
                       columns=['Ang', 'Con', 'EAC']      ,
                       index=[2011,2012,2013]               )

my_fft = fft2(df)
T      , B      =len(df)       , len(df.columns)
freqsT , freqsB = fftfreq(T,1) , fftfreq(B,1)
def ifft2Manually(option):
    recombine      = np.zeros((T,B)) #recombineine   
    for t in list(range(T)):   
        for b in list(range(B)):    
            coef  = abs(my_fft[t][b])/(T*B)
            tfreq , bfreq = freqsT[t] , freqsB[b]
            ps    = cmath.phase(my_fft[t][b])
            dimT =  coef*np.cos(tfreq*2*np.pi*( np.array([df.index]).T - 1  )   + ps  )   
            dimB =  coef*np.cos(bfreq*2*np.pi*( np.array(list(range(B)))    )   + ps  )
            if option==1:
                sinusoid = dimT + dimB
            else: 
                sinusoid=  dimT #exclude dimB here 
            recombine=sinusoid +recombine
            #print('\nt={}\nb={}\ndimT={}\ndimB={}\nsinusoid={}'.format(t,b,dimT,dimB,sinusoid))
    if option==1:
        print('\n nothing matches when both dimensions used') 
    else: 
        print('\n first column matches when ONLY the first dimension (dimT) is used') 
    print('\n original data:\n{} \n\n my attempt to match:\n{}'.format(df, recombine )) 

ifft2Manually(option=1)
ifft2Manually(option=2)

修正后的可行代码 #

以下代码已能正常运行。主要错误在于：之前是将两个维度的余弦计算结果相加，正确的做法应该是把维度项放在cos()函数内部相加。希望这个示例能帮助遇到类似问题的人。

import pandas as pd
import numpy as np
from scipy.fftpack import fftfreq, fft2
import cmath

row=2 # arbitrary
col=3 # arbitrary
df=pd.DataFrame(    { 'A':[  0.09, -0.12, -0.11], 
                      'B':[ -0.13,  0.29, -0.57],
                      'C':[  0.01, -0.16, -0.04],
                      'D':[  0.10,  0.65,  0.46]   } ,
                        columns=['A', 'B', 'C','D']   , 
                        index=[2011,2012,2013]               )

print('here is the df:\n', df)
actual = np.array(df)[row][col] 
my_fft2 = fft2(df)
T      , B      =df.shape[0]      , df.shape[1]
mysum= 0 
for t in range(T):   
    for b in range(B): 
        coef  = abs(my_fft2[t][b])/(T*B)
        bfreq , tfreq =  fftfreq(B,1)[b] , fftfreq(T,1)[t]
        ps    = cmath.phase(my_fft2[t][b])
        mysum   += coef*np.cos( 2*np.pi*(tfreq*row + bfreq*col) + ps )   
print('\n replicate value at row={}, col={}'.format( row, col))
print('\n yhat   :', mysum) 
print(' actual :'  , actual ) 

内容的提问来源于stack exchange，提问作者bri85