根据匹配的列表项分隔符将Python列表拆分为列表的正确方法

big_list = [1, 2, 3, 'separator', 4, 5, 6, 'separator', 7, 8, 9]
separator = 'separator'

split_lists = [sublist for sublist in big_list if sublist != separator]
result = [split_lists[i:i + len(sublist)] for i in range(0, len(split_lists), len(sublist))]

import itertools

big_list = [1, 2, 3, 'separator', 4, 5, 6, 'separator', 7, 8, 9]
separator = 'separator'

# 先将列表中的元素标记为True或False
marked_list = [(item != separator, item) for item in big_list]

# 使用groupby分组
groups = itertools.groupby(marked_list, key=lambda x: x[0])

# 提取分组后的数据并转换为列表
result = [[item for is_sep, item in group] for is_sep, group in groups if not is_sep]

def split_list_by_separator(big_list, separator):
    # 过滤掉分隔符
    filtered_list = [item for item in big_list if item != separator]
    
    # 初始化结果列表和临时子列表
    result = []
    temp_sublist = []
    
    # 遍历过滤后的列表
    for item in filtered_list:
        # 如果当前元素加上分隔符的长度等于下一个元素的索引，则认为遇到了分隔符
        if len(temp_sublist) > 0 and (filtered_list.index(item) + len(separator)) == filtered_list.index(filtered_list[filtered_list.index(item) + 1]):
            result.append(temp_sublist)
            temp_sublist = []
        else:
            temp_sublist.append(item)
    
    # 添加最后一个子列表
    if temp_sublist:
        result.append(temp_sublist)
    
    return result

# 测试数据
big_list = [1, 2, 3, 'separator', 4, 5, 6, 'separator', 7, 8, 9]
separator = 'separator'

# 调用函数并打印结果
print(split_list_by_separator(big_list, separator))