Python实现信号时域统计特征提取及回溯算法实战

from pandas import Series
import math
pstf_list=[]
def psfeatureTime(data,p1,p2):
 #均值
 df_mean=data[p1:p2].mean()
 #方差
 df_var=data[p1:p2].var()
 #标准差
 df_std=data[p1:p2].std()
 #均方根
 df_rms=math.sqrt(pow(df_mean,2) + pow(df_std,2))
 #偏度
 df_skew=data[p1:p2].skew()
 #峭度
 df_kurt=data[p1:p2].kurt()
 sum=0
 for i in range(p1,p2):
  sum+=math.sqrt(abs(data[i]))
 #波形因子
 df_boxing=df_rms / (abs(data[p1:p2]).mean())
 #峰值因子
 df_fengzhi=(max(data[p1:p2])) / df_rms
 #脉冲因子
 df_maichong=(max(data[p1:p2])) / (abs(data[p1:p2]).mean())
 #裕度因子
 df_yudu=(max(data[p1:p2])) / pow((sum/(p2-p1)),2)
 featuretime_list = [df_mean,df_rms,df_skew,df_kurt,df_boxing,df_fengzhi,df_maichong,df_yudu]
 return featuretime_list 

class Solution:
 def letterCombination(self,digits):
  
  phone={'2': ['a', 'b', 'c'],
     '3': ['d', 'e', 'f'],
     '4': ['g', 'h', 'i'],
     
     '5': ['j', 'k', 'l'],
     '6': ['m', 'n', 'o'],
     '7': ['p', 'q', 'r', 's'],
     '8': ['t', 'u', 'v'],
     '9': ['w', 'x', 'y', 'z']}
  
  res=[]#存放组合结果
  def backtrack(combination,next_digits):#回溯函数
   #combination目前已经产生的组合，next_digits:输入的下一个字符
   if len(next_digits)==0: #递归出口
    res.append(combination)
   else:
    for i in phone[next_digits[0]]:
     backtrack(combination+i,next_digits[1:]) #递归实现回溯
  if digits:
   backtrack('',digits) #初始化
  return res

class Solution:
 def permute(self,nums):
  
  res=[] #存放组合结果
  size=len(nums)
  
  def backtrack(combination,nums):
   #combination目前已经产生的组合，nums为剩下的数组
   #递归出口
   #递归的结束一定 要有return
   if len(combination)==size:
    res.append(combination) 
    return #注意
   for i in range(len(nums)):
     backtrack(combination+[nums[i]],nums[:i]+nums[i+1:]) #递归回溯
  
  backtrack([],nums)
  return res
    
if __name__=='__main__':
 nums = [1,2,3]
 solution=Solution()
 print(solution.permute(nums)) 

class Solution:
 def combinationArray(self,candidates,target):
  
  candidates.sort()
  res=[] #存放组合结果
  size=len(candidates)
  
  def backtrack(combination,cur_sum,j):
   #combination目前已经产生的组合,cur_sum当前计算和，j用于控制求和的查找范围起点
   #递归出口
   if cur_sum>target:
    return 
   if cur_sum==target:
    res.append(combination)
   for i in range(j,size): #j避免重复
    if cur_sum+candidates[i]>target: #约束函数(剪)
     break
    j=i
    backtrack(combination+[candidates[i]],cur_sum+candidates[i],j)#递归回溯
    
  backtrack([],0,0)
  return res
if __name__=='__main__':
 candidates = [2,3,6,7]
 target = 7
 solution=Solution()
 print(solution.combinationArray(candidates,target))

class Solution: 
 def solveNqueen(self,n):
  
  res=[] #存放结果组合，对于N皇后问题，这里存放的是其放在每一行对应的列下标  
  def backtrack(combination):
    if len(combination)==n:
     res.append(combination)
     return
    for j in range(n):
     if combination:
      #排除当前行，列和对应的两个对角线。
      if j not in combination and j!=combination[-1]+1 and j!=combination[-1]-1:#约束条件
       backtrack(combination+[j]) #递归回溯
      else:
       continue 
     else:
     backtrack(combination+[j])     
             
  backtrack([]) #回溯初始化
  
  #转化为需要的格式
  output=[["." * k + "Q" + "." * (n - k - 1) for k in i] for i in res] #列表生成器
  return output
  
if __name__=='__main__':
 n=4
 solution=Solution()
 print(solution.solveNqueen(n)) 

class Solution(object):
 def subsets(self, nums):
  """
  :type nums: List[int]
  :rtype: List[List[int]]
  """
  res=[]#存放组合结果
  size=len(nums)
  
  def backtrack(combination,nums):
   #combination目前已经产生的组合，nums为剩下的数组
   if len(combination)<=size:
    res.append(combination)
   #递归出口
   #递归的结束一定 要有return
   if len(combination)==size:
    return 
   
   for i in range(len(nums)):
    backtrack(combination+[nums[i]],nums[i+1:]) #递归回溯
   
  backtrack([],nums)
  return res
 
if __name__=='__main__':
 nums=[1,2,3]
 solution=Solution()
 print(solution.subsets(nums)) 

class Solution:
 def letterpermute(self,S):
  
  res=[]
  size=len(S)
  
  def backtrack(combination,S):
   
   if len(combination)==size:
    res.append(''.join(combination))
    return 
   
   for i in range(len(S)):
    if "a"<=S[i]<= "z" or "A"<=S[i]<= "Z":
     for j in range(2):
      if j==0:
       backtrack(combination+[S[i].lower()],S[i+1:])
      if j==1:
       backtrack(combination+[S[i].upper()],S[i+1:])
      
    else:
     backtrack(combination+[S[i]],S[i+1:])
     
     
  backtrack([],S)
  return res   
 
if __name__=='__main__':
 S=[i for i in "1B2"]
 solution=Solution()
 print(solution.letterpermute(S)) 

class Solution:
 def generateParenthesis(self,n):
  
  res=[] #存放组合结果
  def backtrack(combination,left,right):
   #combination目前已经产生的组合
   if len(combination)==2*n: #递归出口
    res.append(combination)
   #对于有效的括号，左边先出
   if left<n:
    backtrack(combination+'(',left+1,right)#递归实现回溯
   if right<left:
    backtrack(combination+')',left,right+1)#递归实现回溯
     
  backtrack('',0,0) #初始化
  return res 
if __name__=='__main__':
 n=3
 solution=Solution()
 print(solution.generateParenthesis(n))