Python列表循环相同性验证代码解读及特性咨询

Alright, let's walk through this code to unpack what "circularly identical" means and how this implementation checks for that property between two lists.

What is "Circularly Identical"? #

From this code's logic, two lists are circularly identical if you can shift one list by moving some number of elements from the start to the end, and end up with the other list.

For example:

• [1, 2, 3, 4] and [3, 4, 1, 2] are circularly identical (shift the first 2 elements to the end).
• ["a", "b", "c"] and ["b", "c", "a"] are circularly identical (shift the first element to the end).
• [1, 2] and [2, 1] are circularly identical (shift 1 element from start to end).
• But [1, 2, 3] and [1, 3, 2] are not circularly identical—you can't get one from the other via a simple end-to-end shift.

Code Breakdown #

Let's go through each function one by one:

1. extract(e, l) #

This helper function finds all the indexes where element e appears in list l.

• It initializes an empty list t to store indexes.
• It loops through each element in l, tracking the current index with j.
• Whenever it finds an element equal to e, it adds the current index j to t.
• Finally, it returns the list of all matching indexes.

Example: extract(2, [2, 1, 2, 3]) returns [0, 2].

2. construct(a, l) #

This function creates a circularly shifted version of list l, starting at index a.

• It first takes all elements from index a to the end of l and adds them to t.
• Then it takes all elements from the start of l up to (but not including) index a and appends them to t.
• The result is the original list shifted so that element at index a is now the first element.

Example: construct(2, [1, 2, 3, 4]) returns [3, 4, 1, 2].

3. verif(l1, l2) #

This is the main function that checks if l1 and l2 are circularly identical.

• It starts with test = False (assuming they're not identical until proven otherwise).
• First, it checks if the first element of l1 exists in l2. If not, it immediately returns False—since if they were circularly identical, every element of l1 would have to be in l2, including the first one.
• If the first element is present, it uses extract() to get all indexes in l2 where this element appears.
• For each of those indexes, it uses construct() to create a shifted version of l2 starting at that index. If any of these shifted lists matches l1, it sets test = True and breaks out of the loop early.
• Finally, it returns the test boolean.

Edge Cases to Note #

• If l1 and l2 are identical (zero shift), this function will return True (since shifting at index 0 gives the original list).
• If l1 has duplicate elements (like [2, 1, 2]), the function checks all possible starting positions where the first element of l1 appears in l2—which is correct, because there might be multiple valid shifts.
• If either list is empty? Well, this code would throw an error when accessing l1[0], so you'd need to add a check for empty lists if that's a possible input.

内容的提问来源于stack exchange，提问作者amine0