圖像處理 之 二維快速傅里葉變換（FFT2）

# -*- coding: utf-8 -*-
"""
Created on Sun Jul  8 21:05:51 2018

@author: Diko
"""

import numpy


def FFT_v1(Img,Wr):
    if Img.shape[0]==2:
        pic = numpy.zeros([2],dtype=complex)
        pic = pic*(1+0j)
        pic[0]=Img[0]+Img[1]*Wr[0]
        pic[1]=Img[0]-Img[1]*Wr[0]
        return pic
    else:
        pic = numpy.empty([Img.shape[0]],dtype=complex)
        pic[0:Img.shape[0]//2] = FFT_v1(Img[::2],Wr[::2])+Wr*FFT_v1(Img[1::2],Wr[::2])
        pic[Img.shape[0]//2:Img.shape[0]]=FFT_v1(Img[::2],Wr[::2])-Wr*FFT_v1(Img[1::2],Wr[::2])
        return pic;


def FFT_1d(Img):
    Wr = numpy.ones([Img.shape[0]//2])*[numpy.cos(2*numpy.pi*i/Img.shape[0])-1j*numpy.sin(2*numpy.pi*i/Img.shape[0]) for i in numpy.arange(Img.shape[0]/2)]
    return FFT_v1(Img,Wr)


def FFT_2d(Img):
    pic = numpy.zeros([Img.shape[0],Img.shape[1]],dtype=complex)
    for i in numpy.arange(Img.shape[0]):
        pic[:,i]=FFT_1d(Img[:,i])
    for i in numpy.arange(Img.shape[1]):
        pic[i,:]=FFT_1d(pic[i,:])
    return pic


import time
from skimage import io,data
if __name__ == "__main__":
    array = numpy.zeros([512],dtype=complex)
    array[0],array[1],array[2],array[3],array[4],array[5],array[6],array[7],array[8]=1,5,3,2,5,6,1,6,3
    
    img = data.camera()
    
    print("numpy.fft.fft2()函數計算結果：")
    t_s1=time.time()
    print(numpy.fft.fft2(img[:16,0:16]))
    t_e1=time.time()
    print("計算時間："+str(t_e1-t_s1))
    
    print("FFT_2d()函數的計算結果：")
    t_s2 = time.time()
    print(FFT_2d(img[:16,0:16]))
    t_e2 =time.time()
    print("計算時間："+str(t_e2-t_s2))
   # io.imshow(numpy.log(numpy.real(numpy.fft.fft2(img))))
    #io.imshow(numpy.real(FFT_2d(img[0:256,0:256])))
    #img = data.camera()
    #print(numpy.fft.fft2(img))
    #io.imshow(FFT_2d(img))
    
    

下面是與官方給出函數的比較：

numpy.fft.fft2()的結果：

自己實現的結果：

結果都一樣，但是時間卻多了一倍，並且當計算量越大，兩個函數之間的差距就越大，還需要繼續不斷優化。