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035c2cc4c5
86 个采样点 / 约 1.5 小时：
QUIC: 最低 1 最高 8 平均 7.5
HTTP/2: 最低 1 最高 2 平均 2.0
QUIC 塌陷（<= 2 条）只有 2 次，占 2.3%
塌陷时 HTTP/2 一直是 >= 1 条（2/2 = 100%），归零 0 次
QUIC 连接数分布：
8 条 76.7%
7 条 15.1%
6 条 3.5%
5 条 2.3%
1 条 2.3% <- 塌陷（1.5 小时里约 2 分钟）
一次典型塌陷的时间线也记下来了（watchdog 日志）：
重启把中断从「可能几分钟」缩到 26 秒。
目标「0 掉线」做不到 —— 4G 链路上 QUIC 会周期性塌陷。
但影响被压到 2.3% 的窗口，而且 HTTP/2 全程兜住。
教训：多行中文提交信息里有引号时，PowerShell 的 -m 参数会被吃掉。
一律写进文件再用 -F 提交。
130 lines
4.0 KiB
Python
130 lines
4.0 KiB
Python
#!/usr/bin/env python3
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# -*- coding: utf-8 -*-
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r'''分析连接数时间线 long2h_conns.tsv，找塌陷的规律。
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目的：回答两个问题
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1. QUIC 塌陷是不是周期性的？周期多长？
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2. 塌陷期间 HTTP/2 那半边是不是一直活着？（这是双协议兜底的价值）
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'''
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from __future__ import annotations
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from pathlib import Path
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p = Path("/root/long2h_conns.tsv")
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if not p.is_file():
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print(" 没有 %s" % p)
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raise SystemExit(0)
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raw = p.read_text(encoding="utf-8", errors="replace").splitlines()
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print(" 共 %d 行，前 3 行：" % len(raw))
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for line in raw[:3]:
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print(" %r" % line[:120])
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print()
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rows = []
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for line in raw:
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line = line.strip()
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if not line:
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continue
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parts = line.replace(",", "\t").split("\t")
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stamp = ""
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nums = []
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for x in parts:
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x = x.strip()
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if not x:
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continue
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if ":" in x and ("-" in x or len(x) <= 8) and not x.replace(":", "").isdigit():
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stamp = x
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else:
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try:
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nums.append(int(x))
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except ValueError:
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pass
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if nums:
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rows.append((stamp, nums))
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if not rows:
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print(" 解析不出数字")
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raise SystemExit(0)
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# 假设列是 [quic, h2] 或 [总连接, quic, h2] —— 按每行长度自适应
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width = max(len(n) for _s, n in rows)
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print(" 每行 %d 个数字（列）" % width)
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print(" 前 5 行解析结果：")
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for s, n in rows[:5]:
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print(" %-12s %s" % (s, n))
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print()
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# 尝试识别：如果一行有 2 列，当作 [quic, h2]；3 列当作 [total, quic, h2]
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lower = [n for _s, n in rows]
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if width == 2:
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quic = [n[0] for n in lower]
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h2 = [n[1] for n in lower]
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elif width >= 3:
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quic = [n[1] for n in lower]
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h2 = [n[2] for n in lower]
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else:
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print(" 列数不支持")
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raise SystemExit(0)
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total = len(quic)
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print(" ── 连接数统计（%d 个采样点）──" % total)
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print(" QUIC: 最低 %d 最高 %d 平均 %.1f"
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% (min(quic), max(quic), sum(quic) / total))
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print(" HTTP/2: 最低 %d 最高 %d 平均 %.1f"
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% (min(h2), max(h2), sum(h2) / total))
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# 塌陷 = QUIC <= 2
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collapse_idx = [i for i, q in enumerate(quic) if q <= 2]
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print()
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print(" ── QUIC 塌陷（<=2 条）──")
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print(" 出现 %d 次，占比 %.1f%%" % (len(collapse_idx), len(collapse_idx) * 100.0 / total))
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# 塌陷时 h2 还在吗
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if collapse_idx:
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h2_alive = sum(1 for i in collapse_idx if h2[i] >= 1)
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h2_zero = sum(1 for i in collapse_idx if h2[i] == 0)
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print(" 塌陷时 HTTP/2 >=1 条: %d 次（%.0f%%）"
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% (h2_alive, h2_alive * 100.0 / len(collapse_idx)))
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print(" 塌陷时 HTTP/2 也归零: %d 次" % h2_zero)
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print(" -> 这就是双协议兜底的价值：%s"
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% ("QUIC 塌了但 HTTP/2 还在扛" if h2_zero < h2_alive
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else "两个一起塌，兜底没起作用"))
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# 合并连续的塌陷区间，看周期
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spans = []
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start = None
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for i, q in enumerate(quic):
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if q <= 2 and start is None:
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start = i
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elif q > 2 and start is not None:
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spans.append((start, i - 1))
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start = None
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if start is not None:
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spans.append((start, total - 1))
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print()
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print(" ── 塌陷区间（连续采样点合并）──")
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print(" 共 %d 段" % len(spans))
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for a, b in spans[:15]:
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sa = rows[a][0] if rows[a][0] else "第%d点" % a
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sb = rows[b][0] if rows[b][0] else "第%d点" % b
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print(" %-10s -> %-10s 持续 %d 个采样点 (quic 最低 %d)"
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% (sa, sb, b - a + 1, min(quic[a:b + 1])))
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# 塌陷间隔（周期）
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if len(spans) >= 2:
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gaps = [spans[i + 1][0] - spans[i][0] for i in range(len(spans) - 1)]
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print()
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print(" ── 两次塌陷之间隔多少个采样点 ──")
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print(" %s" % gaps[:15])
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print(" 平均 %.1f 个采样点" % (sum(gaps) / len(gaps)))
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# 连接数分布
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from collections import Counter
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print()
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print(" ── QUIC 连接数分布 ──")
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for k in sorted(set(quic)):
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c = quic.count(k)
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print(" %2d 条: %4d 次 %5.1f%%" % (k, c, c * 100.0 / total))
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