详解python实现小波变换的一个简单例子

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])
 
Wavelet.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
 
Wavelet.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

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