用 python 实现小波变换的简单示例

最近工作需要，看了一下小波变换方面的东西，用python实现了一个简单的小波变换类，将来可以用在工作中。

 

简单说几句原理，小波变换类似于傅里叶变换，都是把函数用一组正交基函数展开，选取不同的基函数给出不同的变换。例如傅里叶变换，选择的是sin和cos，或者exp(ikx)这种复指数函数；而小波变换，选取基函数的方式更加灵活，可以根据要处理的数据的特点（比如某一段上信息量比较多），在不同尺度上采用不同的频宽来对已知信号进行分解，从而尽可能保留多一点信息，同时又避免了原始傅里叶变换的大计算量。以下计算采用的是haar基，它把函数分为2段（A1和B1，但第一次不分），对第一段内相邻的2个采样点进行变换（只考虑A1），变换矩阵为

sqrt(0.5)       sqrt(0.5)

sqrt(0.5)        -sqrt(0.5)

变换完之后，再把第一段（A1）分为两段，同样对相邻的点进行变换，直到无法再分。

 

下面直接上代码

Wavelet.py

import math

class wave:
    def __init__(self):
        M_SQRT1_2 = math.sqrt(0.5)
        self.h1 = [M_SQRT1_2, M_SQRT1_2]
        self.g1 = [M_SQRT1_2, -M_SQRT1_2]
        self.h2 = [M_SQRT1_2, M_SQRT1_2]
        self.g2 = [M_SQRT1_2, -M_SQRT1_2]
        self.nc = 2
        self.offset = 0

    def __del__(self):
        return

class Wavelet:
    def __init__(self, n):
        self._haar_centered_Init()
        self._scratch = []
        for i in range(0,n):
            self._scratch.append(0.0)
        return
        
    def __del__(self):
        return
        
    def transform_inverse(self, list, stride):
        self._wavelet_transform(list, stride, -1)
        return
        
    def transform_forward(self, list, stride):
        self._wavelet_transform(list, stride, 1)
        return
        
    def _haarInit(self):
        self._wave = wave()
        self._wave.offset = 0
        return
        
    def _haar_centered_Init(self):
        self._wave = wave()
        self._wave.offset = 1
        return
        
    def _wavelet_transform(self, list, stride, dir):
        n = len(list)
        if (len(self._scratch) < n):
            print("not enough workspace provided")
            exit()
        if (not self._ispower2(n)):
            print("the list size is not a power of 2")
            exit()
        
        if (n < 2):
            return

        if (dir == 1):  # 正变换
            i = n
            while(i >= 2):
                self._step(list, stride, i, dir)
                i = i>>1
             
        if (dir == -1):   # 逆变换
            i = 2
            while(i <= n):
                self._step(list, stride, i, dir)
                i = i << 1
        return
        
    def _ispower2(self, n):
        power = math.log(n,2)
        intpow = int(power)
        intn = math.pow(2,intpow)
        if (abs(n - intn) > 1e-6):
            return False
        else:
            return True
            
    def _step(self, list, stride, n, dir):
        for i in range(0, len(self._scratch)):
            self._scratch[i] = 0.0
        
        nmod = self._wave.nc * n
        nmod -= self._wave.offset
        n1 = n - 1
        nh = n >> 1
        
        if (dir == 1):  # 正变换
            ii = 0
            i = 0
            while (i < n):
                h = 0
                g = 0
                ni = i + nmod
                for k in range(0, self._wave.nc):
                    jf = n1 & (ni + k)
                    h += self._wave.h1[k] * list[stride*jf]
                    g += self._wave.g1[k] * list[stride*jf]
                self._scratch[ii] += h
                self._scratch[ii + nh] += g
                i += 2
                ii += 1
        
        if (dir == -1):    # 逆变换
            ii = 0
            i = 0
            while (i < n):
                ai = list[stride*ii]
                ai1 = list[stride*(ii+nh)]
                ni = i + nmod
                for k in range(0, self._wave.nc):
                    jf = n1 & (ni + k)
                    self._scratch[jf] += self._wave.h2[k] * ai + self._wave.g2[k] * ai1
                i += 2
                ii += 1
                
        for i in range(0, n):
            list[stride*i] = self._scratch[i]

 

测试代码如下：

test.py

import math
import Wavelet

waveletn = 256
waveletnc = 20   #保留的分量数
wavelettest = Wavelet.Wavelet(waveletn)
waveletorigindata = []
waveletdata = []
for i in range(0, waveletn):
    waveletorigindata.append(math.sin(i)*math.exp(-math.pow((i-100)/50,2))+1)
    waveletdata.append(waveletorigindata[-1])
    
wavelettest.transform_forward(waveletdata, 1)
newdata = sorted(waveletdata, key = lambda ele: abs(ele), reverse=True)
for i in range(waveletnc, waveletn):   # 筛选出前 waveletnc个分量保留
    for j in range(0, waveletn):
        if (abs(newdata[i] - waveletdata[j]) < 1e-6):
            waveletdata[j] = 0.0
            break
    
wavelettest.transform_inverse(waveletdata, 1)
waveleterr = 0.0
for i in range(0, waveletn):
    print(waveletorigindata[i], ",", waveletdata[i])
    waveleterr += abs(waveletorigindata[i] - waveletdata[i])/abs(waveletorigindata[i])
print("error: ", waveleterr/waveletn)

 

当waveletnc = 20时，可得到下图，误差大约为2.1

 

当waveletnc = 100时，则为下图，误差大约为0.04

 

当waveletnc = 200时，得到下图，误差大约为0.0005