根据列表的数值差异比较列表的最有效方法？

Question

我正在处理仅包含数值数据的列表，并且希望将它们之间的数值差异进行比较：

primary_seq = [0, 2, 4, 5, 7, 8, 10]
    ...Code code code
primary_pattern = 2^2^1^2^1^2

sub_seq = [0, 2, 3, 5, 6]
    ...Code code code
sub_pattern = 2^1^2^1

鉴于此，我希望程序将sub_pattern与primary_pattern进行比较：

>>> primary_pattern >= sub_pattern
>>> True

所以基本上

221212  <--- Above seq.
 2121   <--- This is a section of the above seq.

Answer 1

您可以创建一个类来计算模式1 ^ 2 ^ 1 ^ 2并定义> =的含义（这似乎是字符串的in）。

class pattern:
  pattern=''
  def __init__(self,lst):
    pat=str([lst[i+1]-lst[i] for i in range(len(lst)-1)])[1:-1]
    self.pattern=pat.replace(', ','^')
  def __repr__(self): return  self.pattern
  def __ge__(self,other): return (other.pattern in self.pattern)

 >>>primary_pattern =pattern(primary_seq) 
 >>>sub_pattern = pattern(sub_seq)
 >>>sub_pattern
 2^1^2^1
 >>>primary_pattern >= sub_pattern
 True

Answer 2

假设您的意思是一个列表中的差异之和大于另一个列表中的> =，那么您可以将列表本身拉大并求和...

>>> primary_pattern = [abs(a-b) for a,b in zip(primary_seq[1:], primary_seq[:-1])]
>>> sub_pattern = [abs(a-b) for a,b in zip(sub_seq[1:], sub_seq[:-1])]
>>> sum(primary_pattern[1:-1]) >= sum(sub_pattern)
True

注意：如果序列是严格排序的，则不需要abs() 。