Python 在整数列表中查找重复序列？

Question

I have a list of lists and each list has a repeating sequence. I'm trying to count the length of repeated sequence of integers in the list:

list_a = [111,0,3,1,111,0,3,1,111,0,3,1] 

list_b = [67,4,67,4,67,4,67,4,2,9,0]

list_c = [1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9,0,23,18,10]

Which would return:

list_a count = 4 (for [111,0,3,1])

list_b count = 2 (for [67,4])

list_c count = 10 (for [1,2,3,4,5,6,7,8,9,0])

Any advice or tips would be welcome. I'm trying to work it out with re.compile right now but, its not quite right.

Answer 1

Guess the sequence length by iterating through guesses between 2 and half the sequence length. If no pattern is discovered, return 1 by default.

def guess_seq_len(seq):
    guess = 1
    max_len = len(seq) / 2
    for x in range(2, max_len):
        if seq[0:x] == seq[x:2*x] :
            return x

    return guess

list_a = [111,0,3,1,111,0,3,1,111,0,3,1] 
list_b = [67,4,67,4,67,4,67,4,2,9,0]
list_c = [1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9,0,23,18,10]

print guess_seq_len(list_a)
print guess_seq_len(list_b)
print guess_seq_len(list_c)
print guess_seq_len(range(500))   # test of no repetition

This gives (as expected):

4
2
10
1

As requested, this alternative gives longest repeated sequence. Hence it will return 4 for list_b. The only change is guess = x instead of return x

def guess_seq_len(seq):
    guess = 1
    max_len = len(seq) / 2
    for x in range(2, max_len):
        if seq[0:x] == seq[x:2*x] :
            guess = x

    return guess

Answer 2

I took Maria 's faster and more stackoverflow-compliant answer and made it find the largest sequence first:

def guess_seq_len(seq, verbose=False):
    seq_len = 1
    initial_item = seq[0]
    butfirst_items = seq[1:]
    if initial_item in butfirst_items:
        first_match_idx = butfirst_items.index(initial_item)
        if verbose:
            print(f'"{initial_item}" was found at index 0 and index {first_match_idx}')
        max_seq_len = min(len(seq) - first_match_idx, first_match_idx)
        for seq_len in range(max_seq_len, 0, -1):
            if seq[:seq_len] == seq[first_match_idx:first_match_idx+seq_len]:
                if verbose:
                    print(f'A sequence length of {seq_len} was found at index {first_match_idx}')
                break
    
    return seq_len

Answer 3

This worked for me.

def repeated(L):
    '''Reduce the input list to a list of all repeated integers in the list.'''
    return [item for item in list(set(L)) if L.count(item) > 1]

def print_result(L, name):
    '''Print the output for one list.'''
    output = repeated(L)
    print '%s count = %i (for %s)' % (name, len(output), output)

list_a = [111, 0, 3, 1, 111, 0, 3, 1, 111, 0, 3, 1]
list_b = [67, 4, 67, 4, 67, 4, 67, 4, 2, 9, 0]
list_c = [
    1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2,
    3, 4, 5, 6, 7, 8, 9, 0, 23, 18, 10
]

print_result(list_a, 'list_a')
print_result(list_b, 'list_b')
print_result(list_c, 'list_c')

Python's set() function will transform a list to a set, a datatype that can only contain one of any given value, much like a set in algebra. I converted the input list to a set, and then back to a list, reducing the list to only its unique values. I then tested the original list for each of these values to see if it contained that value more than once. I returned a list of all of the duplicates. The rest of the code is just for demonstration purposes, to show that it works.

Edit: Syntax highlighting didn't like the apostrophe in my docstring.