如何檢查列表是否在另一個列表中使用相同的順序python

Question

我的問題與典型的不同。假設我們有，

X = ['123', '456', '789'] 

而且，我們想找到另一個列表是否與以相同的順序X。例如，

A = ['123', '456'] 
# should return True since A in X with same order 
B = ['456', '123'] 
# should return False since elements in B are not in same order with X 
C = ['123', '789'] 
# should return False since elements in C are not adjacent in X 

任何人都可以給我任何想法嗎？

Answer 1

的Python 2：

def is_a_in_x(A, X): 
    for i in xrange(len(X) - len(A) + 1): 
    if A == X[i:i+len(A)]: return True 
    return False 

的Python 3：

def is_a_in_x(A, X): 
    for i in range(len(X) - len(A) + 1): 
    if A == X[i:i+len(A)]: return True 
    return False 

Answer 2

一個可能的解決辦法，如果你的號碼總是三位數：

x = ['123', '456', '789'] 

A = ['123', '456'] 
''.join(A) in ''.join(x) 

B = ['456', '123'] 
''.join(B) in ''.join(x) 

C = ['123', '789'] 
''.join(C) in ''.join(x) 

雖然，容易出現的錯誤，例如：

D = ['1', '23'] 
''.join(D) in ''.join(x) 


outputs True 

Answer 3

def foo(your_list, your_main_list): 
    list_len = len(your_list) 
    for i, item in enumerate(your_list): 
     if your_main_list[i] == item: 
      if i + 1 == list_len: 
       return True 
     else: 
      return False