最近工作需要,看了一下小波變換方面的東西,用python實(shí)現(xiàn)了一個(gè)簡單的小波變換類,將來可以用在工作中。

簡單說幾句原理,小波變換類似于傅里葉變換,都是把函數(shù)用一組正交基函數(shù)展開,選取不同的基函數(shù)給出不同的變換。例如傅里葉變換,選擇的是sin和cos,或者exp(ikx)這種復(fù)指數(shù)函數(shù);而小波變換,選取基函數(shù)的方式更加靈活,可以根據(jù)要處理的數(shù)據(jù)的特點(diǎn)(比如某一段上信息量比較多),在不同尺度上采用不同的頻寬來對已知信號(hào)進(jìn)行分解,從而盡可能保留多一點(diǎn)信息,同時(shí)又避免了原始傅里葉變換的大計(jì)算量。以下計(jì)算采用的是haar基,它把函數(shù)分為2段(A1和B1,但第一次不分),對第一段內(nèi)相鄰的2個(gè)采樣點(diǎn)進(jìn)行變換(只考慮A1),變換矩陣為
sqrt(0.5) sqrt(0.5)
sqrt(0.5) -sqrt(0.5)
變換完之后,再把第一段(A1)分為兩段,同樣對相鄰的點(diǎn)進(jìn)行變換,直到無法再分。
下面直接上代碼
Wavelet.py
import math
class wave:
def __init__(self):
M_SQRT1_2 = math.sqrt(0.5)
self.h2 = [M_SQRT1_2, M_SQRT1_2]
self.g1 = [M_SQRT1_2, -M_SQRT1_2]
self.h3 = [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): # 正變換
i = n
while(i >= 2):
self._step(list, stride, i, dir)
i = i>>1
if (dir == -1): # 逆變換
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): # 正變換
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.h2[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): # 逆變換
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.h3[k] * ai + self._wave.g2[k] * ai1
i += 2
ii += 1
for i in range(0, n):
list[stride*i] = self._scratch[i]
新聞名稱:詳解python實(shí)現(xiàn)小波變換的一個(gè)簡單例子-創(chuàng)新互聯(lián)
本文網(wǎng)址:http://www.yijiale78.com/article34/dchjse.html
成都網(wǎng)站建設(shè)公司_創(chuàng)新互聯(lián),為您提供網(wǎng)站收錄、虛擬主機(jī)、軟件開發(fā)、網(wǎng)站策劃、App設(shè)計(jì)、網(wǎng)站改版
聲明:本網(wǎng)站發(fā)布的內(nèi)容(圖片、視頻和文字)以用戶投稿、用戶轉(zhuǎn)載內(nèi)容為主,如果涉及侵權(quán)請盡快告知,我們將會(huì)在第一時(shí)間刪除。文章觀點(diǎn)不代表本網(wǎng)站立場,如需處理請聯(lián)系客服。電話:028-86922220;郵箱:631063699@qq.com。內(nèi)容未經(jīng)允許不得轉(zhuǎn)載,或轉(zhuǎn)載時(shí)需注明來源: 創(chuàng)新互聯(lián)
猜你還喜歡下面的內(nèi)容