如何高效检查嵌套Timestamp列表中是否存在3个时间戳在指定范围？

我有一个嵌套列表，每个子列表是长度4到250的Timestamp序列。我已经知道怎么检查两个Timestamp是否在指定时间范围内，但不知道怎么检查是否存在3个Timestamp处于该范围？

附上我写的代码，这段代码冗余低效，还存在索引越界问题，求更优实现方案：

master = [Timestamp('2014-03-11 00:00:00'), Timestamp('2014-03-12 00:00:00'), Timestamp('2014-03-13 00:00:00'), Timestamp('2014-05-07 00:00:00'), Timestamp('2014-05-11 00:00:00'), Timestamp('2014-05-28 00:00:00'), Timestamp('2014-05-29 00:00:00')]
[Timestamp('2014-03-09 00:00:00'), Timestamp('2014-04-02 00:00:00'), Timestamp('2014-04-15 00:00:00'), Timestamp('2014-04-23 00:00:00'), Timestamp('2014-05-01 00:00:00')]
[Timestamp('2013-08-09 00:00:00'), Timestamp('2013-08-13 00:00:00'), Timestamp('2013-09-05 00:00:00'), Timestamp('2013-09-11 00:00:00'), Timestamp('2013-09-24 00:00:00')]
[Timestamp('2014-04-02 00:00:00'), Timestamp('2014-04-11 00:00:00'), Timestamp('2014-04-30 00:00:00'), Timestamp('2014-05-05 00:00:00'), Timestamp('2014-05-15 00:00:00'), Timestamp('2014-05-21 00:00:00')]
[Timestamp('2012-11-16 00:00:00'), Timestamp('2012-11-28 00:00:00'), Timestamp('2012-12-09 00:00:00'), Timestamp('2012-12-10 00:00:00'), Timestamp('2012-12-22 00:00:00'), Timestamp('2013-01-14 00:00:00'), Timestamp('2013-01-31 00:00:00')]
[Timestamp('2012-11-23 00:00:00'), Timestamp('2013-01-01 00:00:00'), Timestamp('2013-01-02 00:00:00'), Timestamp('2013-01-07 00:00:00'), Timestamp('2013-01-17 00:00:00')]
final_result = []
for stamp in master:
    sub_list = []
    stamp = stamp.sort(reverse = True)
    x1 = stamp[0] - stamp[1]
    x2 = stamp[1] - stamp[2]
    if x1+x2 <=7:
         sub_list.append(1)
         continue
    else:
         sub_list.append(0)
if 1 in sub_list:
    final_result.append(1)
else:
    final_result.append(0)

核心思路 #

要检查是否存在3个Timestamp在指定时间范围（比如7天）内，最高效的方式是先对每个子列表的Timestamp排序，然后用滑动窗口检查连续3个Timestamp的首尾时间差是否≤指定阈值。排序后只需检查连续元素即可——因为如果存在非连续的三个元素满足条件，其中必然有连续的三个也满足（排序后时间呈递增/递减顺序）。

代码实现 #

import pandas as pd

# 整理嵌套列表（原代码中仅第一个列表存入master，其余未正确赋值）
timestamp_lists = [
    [pd.Timestamp('2014-03-11 00:00:00'), pd.Timestamp('2014-03-12 00:00:00'), pd.Timestamp('2014-03-13 00:00:00'), pd.Timestamp('2014-05-07 00:00:00'), pd.Timestamp('2014-05-11 00:00:00'), pd.Timestamp('2014-05-28 00:00:00'), pd.Timestamp('2014-05-29 00:00:00')],
    [pd.Timestamp('2014-03-09 00:00:00'), pd.Timestamp('2014-04-02 00:00:00'), pd.Timestamp('2014-04-15 00:00:00'), pd.Timestamp('2014-04-23 00:00:00'), pd.Timestamp('2014-05-01 00:00:00')],
    [pd.Timestamp('2013-08-09 00:00:00'), pd.Timestamp('2013-08-13 00:00:00'), pd.Timestamp('2013-09-05 00:00:00'), pd.Timestamp('2013-09-11 00:00:00'), pd.Timestamp('2013-09-24 00:00:00')],
    [pd.Timestamp('2014-04-02 00:00:00'), pd.Timestamp('2014-04-11 00:00:00'), pd.Timestamp('2014-04-30 00:00:00'), pd.Timestamp('2014-05-05 00:00:00'), pd.Timestamp('2014-05-15 00:00:00'), pd.Timestamp('2014-05-21 00:00:00')],
    [pd.Timestamp('2012-11-16 00:00:00'), pd.Timestamp('2012-11-28 00:00:00'), pd.Timestamp('2012-12-09 00:00:00'), pd.Timestamp('2012-12-10 00:00:00'), pd.Timestamp('2012-12-22 00:00:00'), pd.Timestamp('2013-01-14 00:00:00'), pd.Timestamp('2013-01-31 00:00:00')],
    [pd.Timestamp('2012-11-23 00:00:00'), pd.Timestamp('2013-01-01 00:00:00'), pd.Timestamp('2013-01-02 00:00:00'), pd.Timestamp('2013-01-07 00:00:00'), pd.Timestamp('2013-01-17 00:00:00')]
]

def has_three_within_range(timestamps, max_days=7):
    # 排序Timestamp序列，默认升序
    sorted_ts = sorted(timestamps)
    # 遍历连续三个元素的窗口，避免索引越界
    for i in range(len(sorted_ts) - 2):
        # 计算第一个和第三个Timestamp的时间差
        delta = sorted_ts[i+2] - sorted_ts[i]
        if delta.days <= max_days:
            return 1
    return 0

# 生成最终结果
final_result = [has_three_within_range(ts_list) for ts_list in timestamp_lists]
print(final_result)

关键优化点 #

• 修复原代码错误：原代码中sort()是原地修改方法，返回None，导致后续索引操作报错；同时修正了子列表未正确存入嵌套列表的问题。
• 降低时间复杂度：排序时间为O(n log n)，遍历检查为O(n)，远低于暴力枚举所有三元组的O(n³)，适合子列表长度较大的场景。
• 避免索引越界：遍历范围控制在len(sorted_ts)-2，确保i+2不会超出列表边界，兼容4到250长度的所有子列表。

内容的提问来源于stack exchange，提问作者Fernando Araiza