python 实现快速傅里叶变换（一维）

第一版

仿照 c 语言，有很多丑陋的 for 循环

def __fft(sig, inv):
    N = sig.size # 原始长度，如果不是二的幂，补零
    bit=0 # 表示信号长度的最小比特数
    while ((1<<bit)<N):
        bit += 1;
    N = pow(2,bit) # 不小于 N 的最小二次幂
    sig = np.hstack((sig,np.zeros((sig.shape[0],N-sig.shape[1]),dtype=complex))) # 补零
    
    # 重新排序
    rev = [0 for i in range(N)]
    for i in range(N):
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1))
        if(i<rev[i]):
            temp = sig[0,i]
            sig[0,i] = sig[0,rev[i]]
            sig[0,rev[i]] = temp
    
    # 蝶形运算       
    freq = sig.astype(complex)
    for mid in [pow(2,i) for i in range(bit)]:
        for i in range(0,N,2*mid):
            for j in range(mid):
                x = freq[0,i+j]
                y = freq[0,i+j+mid]*np.exp(-1j*inv*np.pi*(i+j)/mid)
                freq[0,i+j] = x+y
                freq[0,i+j+mid] = x-y
    if inv == -1:
        return freq/N
    return freq

测试

n = 16
sig = np.matrix([ np.random.random() for i in range(pow(2,n))]) # 随机信号
cov2 = fft(sig) # numpy.fft 得到的结果
freq = __fft(sig,1) 
print('distance from numpy.fft: ',np.linalg.norm(freq[0,:cov2.size]-cov2))
back = __fft(freq,-1)  # 逆变换
print('reconstruction error: ',np.linalg.norm(sig-back[0,:cov2.size]))

结果

distance from numpy.fft:  2.5842860851574793e-07
reconstruction error:  1.0518201158802643e-09

第二版

上面第一版实在是太丑陋了，因为 python 根本不适合跑 for 循环

而且，在 python 数组中，可以利用切片非常方便地取出 奇数列 和 偶数列

用递归简化代码

def dft(sig):
    N = sig.size
    V = np.matrix([[np.exp(-1j*2*np.pi*v*y/N) for v in range(N)] for y in range(N)])
    return sig.dot(V)

def FFT(x):
    """A recursive implementation of the 1D Cooley-Tukey FFT"""
    x = np.asarray(x, dtype=float)
    N = x.shape[0]
    if N % 2 > 0:
        raise ValueError("size of x must be a power of 2")
    elif N <= 8:  # this cutoff should be optimized
        return dft(x)
    else:
        X_even = FFT(x[::2])
        X_odd = FFT(x[1::2])
        # print('size X_even',X_even.shape)
        factor = np.matrix(np.exp(-2j * np.pi * np.arange(N) / N))
        # print('size factor',factor.shape)
        return np.hstack([X_even + np.multiply(factor[0,:int(N/2)],X_odd),
                               X_even + np.multiply(factor[0,int(N/2):],X_odd)])

测试

n = 16
sig = [ np.random.random() for i in range(pow(2,n))]
cov2 = fft(sig)
freq = FFT(sig)
print('distance from numpy.fft: ',np.linalg.norm(freq-cov2))

distance from numpy.fft:  3.946983495917491e-11