#!/usr/bin/env python3
# -*- coding: utf-8 -*-
r'''分析连接数时间线 long2h_conns.tsv，找塌陷的规律。

目的：回答两个问题
  1. QUIC 塌陷是不是周期性的？周期多长？
  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）
'''
from __future__ import annotations

from pathlib import Path

p = Path("/root/long2h_conns.tsv")
if not p.is_file():
    print("  没有 %s" % p)
    raise SystemExit(0)

raw = p.read_text(encoding="utf-8", errors="replace").splitlines()
print("  共 %d 行，前 3 行：" % len(raw))
for line in raw[:3]:
    print("    %r" % line[:120])
print()

rows = []
for line in raw:
    line = line.strip()
    if not line:
        continue
    parts = line.replace(",", "\t").split("\t")
    stamp = ""
    nums = []
    for x in parts:
        x = x.strip()
        if not x:
            continue
        if ":" in x and ("-" in x or len(x) <= 8) and not x.replace(":", "").isdigit():
            stamp = x
        else:
            try:
                nums.append(int(x))
            except ValueError:
                pass
    if nums:
        rows.append((stamp, nums))

if not rows:
    print("  解析不出数字")
    raise SystemExit(0)

# 假设列是 [quic, h2] 或 [总连接, quic, h2] —— 按每行长度自适应
width = max(len(n) for _s, n in rows)
print("  每行 %d 个数字（列）" % width)
print("  前 5 行解析结果：")
for s, n in rows[:5]:
    print("    %-12s %s" % (s, n))
print()

# 尝试识别：如果一行有 2 列，当作 [quic, h2]；3 列当作 [total, quic, h2]
lower = [n for _s, n in rows]
if width == 2:
    quic = [n[0] for n in lower]
    h2 = [n[1] for n in lower]
elif width >= 3:
    quic = [n[1] for n in lower]
    h2 = [n[2] for n in lower]
else:
    print("  列数不支持")
    raise SystemExit(0)

total = len(quic)
print("  ── 连接数统计（%d 个采样点）──" % total)
print("    QUIC: 最低 %d  最高 %d  平均 %.1f"
      % (min(quic), max(quic), sum(quic) / total))
print("    HTTP/2: 最低 %d  最高 %d  平均 %.1f"
      % (min(h2), max(h2), sum(h2) / total))

# 塌陷 = QUIC <= 2
collapse_idx = [i for i, q in enumerate(quic) if q <= 2]
print()
print("  ── QUIC 塌陷（<=2 条）──")
print("    出现 %d 次，占比 %.1f%%" % (len(collapse_idx), len(collapse_idx) * 100.0 / total))

# 塌陷时 h2 还在吗
if collapse_idx:
    h2_alive = sum(1 for i in collapse_idx if h2[i] >= 1)
    h2_zero = sum(1 for i in collapse_idx if h2[i] == 0)
    print("    塌陷时 HTTP/2 >=1 条: %d 次（%.0f%%）"
          % (h2_alive, h2_alive * 100.0 / len(collapse_idx)))
    print("    塌陷时 HTTP/2 也归零: %d 次" % h2_zero)
    print("    -> 这就是双协议兜底的价值：%s"
          % ("QUIC 塌了但 HTTP/2 还在扛" if h2_zero < h2_alive
             else "两个一起塌，兜底没起作用"))

# 合并连续的塌陷区间，看周期
spans = []
start = None
for i, q in enumerate(quic):
    if q <= 2 and start is None:
        start = i
    elif q > 2 and start is not None:
        spans.append((start, i - 1))
        start = None
if start is not None:
    spans.append((start, total - 1))

print()
print("  ── 塌陷区间（连续采样点合并）──")
print("    共 %d 段" % len(spans))
for a, b in spans[:15]:
    sa = rows[a][0] if rows[a][0] else "第%d点" % a
    sb = rows[b][0] if rows[b][0] else "第%d点" % b
    print("    %-10s -> %-10s  持续 %d 个采样点  (quic 最低 %d)"
          % (sa, sb, b - a + 1, min(quic[a:b + 1])))

# 塌陷间隔（周期）
if len(spans) >= 2:
    gaps = [spans[i + 1][0] - spans[i][0] for i in range(len(spans) - 1)]
    print()
    print("  ── 两次塌陷之间隔多少个采样点 ──")
    print("    %s" % gaps[:15])
    print("    平均 %.1f 个采样点" % (sum(gaps) / len(gaps)))

# 连接数分布
from collections import Counter
print()
print("  ── QUIC 连接数分布 ──")
for k in sorted(set(quic)):
    c = quic.count(k)
    print("    %2d 条: %4d 次  %5.1f%%" % (k, c, c * 100.0 / total))