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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