Python 如何检查一个列表是否在另一个具有相同顺序的列表中

我的问题与典型的不同。假设我们有

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

我们想知道另一个列表是否在X中，顺序完全相同。比如说,

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

---

有人能给我一些建议吗？

如果您的数字总是三位数，那么有一个可能的解决方案：

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)

但是，容易出现错误，例如：

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


outputs True

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

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

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