把主线程那一行剔掉再排

（每道题开头都有同一段：那张线程表 PSL，和把它切成字典列表的 rows()——每个字典有 pid / tid / nlwp / stat / cpu / comm 六项。）

找「算账」里最忙的线程：先剔掉主线程行（TID 等于 PID），再按 %CPU 排。看剔前剔后各是谁：

PSL = '''PID  TID  NLWP STAT  %CPU  COMM
1    1    1    Ss    0.0   bash
12   12   1    S     0.0   收数据.sh
17   17   4    Ssl   49.8  算账
17   18   4    Rsl   49.6  算账-核对
17   19   4    Ssl   0.0   算账-写盘
17   20   4    Ssl   0.1   算账-心跳
21   21   2    Ssl   0.0   通知
21   22   2    Ssl   0.0   通知-等待'''

def rows(text):
    out = []
    for line in text.split("\n")[1:]:
        pid, tid, nlwp, stat, cpu, comm = line.split()
        out.append({"pid": int(pid), "tid": int(tid), "nlwp": int(nlwp), "stat": stat, "cpu": float(cpu), "comm": comm})
    return out

r = [x for x in rows(PSL) if x["pid"] == 17]
before = max(r, key=lambda x: x["cpu"])["comm"]
workers = [x for x in r if x["tid"] != x["pid"]]
after = max(workers, key=lambda x: x["cpu"])["comm"]
print(before + "/" + after)