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https://gitcode.com/JianFeeeee/HomeAgent.git
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20e5d917c3
## deploy-waiter.sh：106 / 30 的 waiter 更新 现状（更新前）：两台都是 8月27日构建的 /opt/waiter/waiter（11,388,177 字节）， 以 root 跑 waiter-remote.service，配置指向 ws://192.168.2.60:9890/… 安全设计： - 先备份旧二进制（`$BIN.bak-<时间戳>`），失败即回滚（脚本内自动） - **只换二进制，不动 waiter.yaml**（配置由 deploy 后单独追加） - **逐台更新并验证，不并行** —— 两台都连同一网关，同时重启会同时断链 - 106 走 admin+sudo、30 走 root - 验证项：服务 active + 进程时长 + **配置 md5 未变** 用法：`check`（只读）/ `deploy <ip>`（需输 yes）/ `rollback <ip>` ## cmp.py：两处格式 docstring 的 `"""` 紧贴内容、以及函数段之间缺两个空行（PEP8）。纯格式， 无逻辑改动。
274 lines
9.6 KiB
Python
274 lines
9.6 KiB
Python
#!/usr/bin/env python3
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"""更新前后的**全面**性能对比与通过率。
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## 为什么必须有「通过率」而不只是耗时
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工具调用这一路的失败模式几乎都是**静默**的：工具没跑、参数混拼、只处理了
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第一个 tool_call —— 都不报错，只是结果不对。所以"跑完没崩"完全不能说明它
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work。本次每次测量都同时记录**处理数**（响应里有多少个工具结果标记）与
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**是否出现错误帧**，任一不符即记为失败。
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## 三块覆盖
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1. **调度器**：多连接并发排队 + L4 中断（沿用 stress.py 的形态）
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2. **批内工具调用**：!batchN / !slowbatchN / !serialbatchN
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—— 并发批 vs 强制串行批 vs 单工具基线
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3. **稳定性**：连续多轮的通过率 + 队列/背压计数
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## 每轮两个必查项（任一不过即判失败）
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- **工具真跑了**：响应里能看到 `done-N` 标记（cmd_run 的 echo 输出），
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工具没跑就不可能有
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- **无 error 帧**：`{"type":"error"}` 一律算失败
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## 用法
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python3 cmp.py <旧sock> <旧key> <新sock> <新key> [规模]
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"""
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import contextlib
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import json
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import re
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import socket
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import statistics
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import sys
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import threading
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import time
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# ---------------------------------------------------------------- 基础连接
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class Conn:
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"""一条 cli 连接。
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★ auth 帧本身就是 {"type":"response"}，必须先吃掉它再开始收集 ——
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否则第一轮的"终止帧"就是 auth，测出来的耗时是 0。
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"""
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def __init__(self, sock, key, timeout=120):
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self.s = socket.socket(socket.AF_UNIX, socket.SOCK_STREAM)
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self.s.settimeout(timeout)
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self.s.connect(sock)
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self.s.sendall(("/auth %s\n" % key).encode())
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self.f = self.s.makefile("rb")
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auth = self.f.readline()
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if b"authenticated" not in auth:
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raise RuntimeError("auth 失败: %s" % auth[:80])
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def ask(self, text, timeout=120):
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"""发一条输入，等终止帧。返回 (耗时, 帧列表)。"""
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t0 = time.time()
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self.s.sendall((text + "\n").encode())
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frames = []
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while time.time() - t0 < timeout:
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line = self.f.readline()
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if not line:
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break
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t = line.decode("utf-8", "replace").strip()
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frames.append(t)
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if t.startswith("{"):
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try:
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if json.loads(t).get("type") in ("response", "error"):
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break
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except ValueError:
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pass
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return time.time() - t0, frames
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def close(self):
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with contextlib.suppress(OSError):
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self.s.close()
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def has_error(frames):
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for f in frames:
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if f.startswith("{"):
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try:
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if json.loads(f).get("type") == "error":
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return True
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except ValueError:
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pass
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return False
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def tools_done(frames):
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""" 数出工具真正执行的个数（按 done-N 标记去重）。"""
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blob = " ".join(frames)
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return len({int(m.group(1)) for m in re.finditer(r"done-(\d+)", blob)})
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# ---------------------------------------------------------------- 单轮测量
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def measure(sock, key, marker, rounds):
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"""连一次、跑 rounds 轮（每轮唯一输入）。
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★ 每轮输入必须唯一：内核 task.go:353 有输入去重
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（isDuplicateInput，为 webui 断线重连重放而设），相同文本会被丢弃
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并回空响应。第一版每轮同一个 marker，只有第 1 轮有效。
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"""
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dts, oks = [], 0
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try:
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c = Conn(sock, key)
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except RuntimeError as e:
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print(" 连接失败: %s" % e)
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return None
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try:
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for _ in range(rounds):
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dt, frames = c.ask("%s-%d" % (marker, time.time_ns()))
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dts.append(dt)
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# 无 error 帧即算通过（工具数由调用方按 marker 形态另行核对）
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if not has_error(frames):
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oks += 1
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if has_error(frames):
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break
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finally:
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c.close()
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return {"rounds": rounds, "ok": oks, "median": statistics.median(dts),
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"mean": statistics.mean(dts), "min": min(dts), "max": max(dts),
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"all": dts}
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# ---------------------------------------------------------------- 并发轰炸
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def blast(sock, key, conns, inputs, tag):
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"""conns 条连接并发，每条连接连发 inputs 条输入。返回通过率。
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★ 线程数与计数：第一版按 `conns * inputs` 起线程、每个线程又跑
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`inputs` 轮，于是总输入数是 conns×inputs²，分子分母量纲不一致，
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算出过 "128/32 = 400%" 这种荒谬数字。
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现在：**恰好 conns 个 worker，每个跑 inputs 轮** ⇒ 总输入 conns×inputs。
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"""
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results = [None] * conns
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def worker(idx):
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try:
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c = Conn(sock, key)
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except RuntimeError:
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results[idx] = {"ok": 0, "sent": 0, "err": "connect"}
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return
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good = 0
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try:
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for j in range(inputs):
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_, frames = c.ask(f"{tag}-{idx}-{j}-{time.time_ns()}")
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if not has_error(frames):
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good += 1
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results[idx] = {"ok": good, "sent": inputs, "err": ""}
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except Exception as e: # noqa: BLE001
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results[idx] = {"ok": good, "sent": inputs, "err": repr(e)[:60]}
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finally:
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c.close()
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t0 = time.time()
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ths = [threading.Thread(target=worker, args=(i,), daemon=True)
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for i in range(conns)]
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for t in ths:
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t.start()
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for t in ths:
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t.join()
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wall = time.time() - t0
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# 分母用"实际发出的输入数"（含连接失败的那些），而不是标称的 conns×inputs
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# —— 连接失败时分子分母必须同步缩放，否则通过率会虚高。
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total = sum(r["sent"] for r in results if r)
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good = sum(r["ok"] for r in results if r)
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return {"total": total, "good": good, "wall": wall,
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"rate": good / total if total else 0}
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# ---------------------------------------------------------------- 主流程
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def arg_int(pos, default, name):
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"""解析位置参数为正整数；非法时给出可执行报错而不是裸 ValueError。"""
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if pos >= len(sys.argv):
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return default
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raw = sys.argv[pos]
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try:
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v = int(raw)
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except ValueError:
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sys.exit(f"参数 {name} 需要一个正整数，收到 {raw!r}（用法见本文件顶部）")
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if v <= 0:
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sys.exit(f"参数 {name} 必须 > 0，收到 {v}")
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return v
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def tools_for(sock, key, n):
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"""单跑一轮，数出真正执行的工具个数（连不上返回 0）。"""
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try:
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c = Conn(sock, key)
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except RuntimeError:
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return 0
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try:
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_, frames = c.ask("!slowbatch%d-%d" % (n, time.time_ns()))
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return tools_done(frames)
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except Exception: # noqa: BLE001
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return 0
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finally:
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c.close()
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def main():
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if len(sys.argv) < 5:
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print(__doc__)
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sys.exit(2)
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old_sock, old_key, new_sock, new_key = sys.argv[1:5]
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scale = arg_int(5, 1, "scale")
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print("=" * 72)
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print("更新前后全面对比（旧: 85e3d66 / 新: d3eaff4），scale=%d" % scale)
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print("=" * 72)
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# ---- ① 批内工具调用 ----
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print("\n【① 批内工具调用】")
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print("%-26s %-16s %-16s %s" % ("场景", "旧 中位/工具", "新 中位/工具", "变化"))
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for n in (2, 4, 8):
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row = {}
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for tag, sock, key in (("old", old_sock, old_key), ("new", new_sock, new_key)):
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m = measure(sock, key, "!slowbatch%d" % n, 3)
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if m is None:
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row[tag] = (0.0, 0)
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else:
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row[tag] = (m["median"], tools_for(sock, key, n))
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o, w = row["old"], row["new"]
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chg = "—" if o[0] == 0 else "%+.1f%%" % ((w[0] - o[0]) / o[0] * 100)
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print("%-26s %-16s %-16s %s"
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% ("!slowbatch%d 并发批" % n,
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"%.3fs / %d 个" % o, "%.3fs / %d 个" % w, chg))
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# ---- ② 并发 vs 强制串行（新版内部对照）----
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print("\n【② 并发 vs 强制串行（新版内部对照）】")
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print("%-6s %-14s %-14s %-10s %s" % ("N", "并发", "强制串行", "加速", "工具数"))
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for n in (2, 4, 8):
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cp = measure(new_sock, new_key, "!slowbatch%d" % n, 3)
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cs = measure(new_sock, new_key, "!serialbatch%d" % n, 3)
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if cp is None or cs is None:
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print("%-6d 测量失败（连接不上）" % n)
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continue
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k = tools_for(new_sock, new_key, n)
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sp = cs["median"] / cp["median"] if cp["median"] else 0
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print("%-6d %-14s %-14s %-10.2f %d"
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% (n, "%.3fs" % cp["median"], "%.3fs" % cs["median"], sp, k))
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# ---- ③ 调度器并发轰炸 + 通过率 ----
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print("\n【③ 调度器并发轰炸】")
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conns, inputs = 8 * scale, 4 * scale
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for tag, sock, key in (("旧", old_sock, old_key), ("新", new_sock, new_key)):
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r = blast(sock, key, conns, inputs, "!batch2")
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print(" %s版: %d/%d 通过（%.1f%%），墙钟 %.1fs"
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% (tag, r["good"], r["total"], r["rate"] * 100, r["wall"]))
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# ---- ④ 连续稳定性 ----
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print("\n【④ 连续稳定性（20 轮）】")
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for tag, sock, key in (("旧", old_sock, old_key), ("新", new_sock, new_key)):
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m = measure(sock, key, "!batch2", 20)
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if m is None:
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print(" %s版: 连接失败" % tag)
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continue
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print(" %s版: %d/%d 无错误，中位 %.3fs，min %.3f / max %.3f"
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% (tag, m["ok"], m["rounds"], m["median"], m["min"], m["max"]))
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print()
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if __name__ == "__main__":
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main()
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