Python，如何连接列表中连续的和相似的元素？

def join_consecutive_elements(lst):
    if not lst:
        return []

    result = []
    current_element = str(lst[0])
    for i in range(1, len(lst)):
        if lst[i] == lst[i - 1]:
            current_element += str(lst[i])
        else:
            result.append(current_element)
            current_element = str(lst[i])
    result.append(current_element)
    return result

# 示例
numbers = [1, 1, 2, 2, 2, 3, 4, 4]
print(join_consecutive_elements(numbers))  # 输出: ['11', '222', '3', '44']

from itertools import groupby

def join_consecutive_elements(lst):
    return [''.join(map(str, group)) for key, group in groupby(lst)]

# 示例
numbers = [1, 1, 2, 2, 2, 3, 4, 4]
print(join_consecutive_elements(numbers))  # 输出: ['11', '222', '3', '44']

def custom_similarity(a, b):
    # 这里是你的相似性逻辑
    return a == b

def join_similar_elements(lst, similarity_func):
    if not lst:
        return []

    result = []
    current_group = [lst[0]]
    for i in range(1, len(lst)):
        if similarity_func(lst[i], lst[i - 1]):
            current_group.append(lst[i])
        else:
            result.append(current_group)
            current_group = [lst[i]]
    result.append(current_group)
    return result

# 示例
strings = ['apple', 'apples', 'banana', 'bananas', 'cherry']
print(join_similar_elements(strings, lambda x, y: x.startswith(y[:3])))  # 输出: [['apple', 'apples'], ['banana', 'bananas'], ['cherry']]