The DevOps Handbook
Development unless otherwise noted: a. Partially done work: b. Extra processes: c. Extra features: d. Task switching: e. Waiting: f. Motion: g. Defects: h. Nonstandard or manual work:i. Heroics: d of our code and environments—we do this by reserving cycles in each development interval, or by scheduling kaizen blitzes, which are periods when engineers self-organize into teams to work on fixing any0 码力 | 8 页 | 22.57 KB | 5 月前3The DevOps Handbook
rate) and throughput (deployment lead time, deployment frequency) e. ENABLE COORDINATION AND SCHEDULING OF CHANGES i. Loosely coupled architectures enable greater team independence and can likely succeed succeed with more informal coordination approaches like chat rooms ii. For deliberate change scheduling, gather representatives, not authorize changes, but to schedule and sequence. f. ENABLE PEER REVIEW0 码力 | 8 页 | 24.02 KB | 5 月前3Pomodoro Technique
• Turn off IMs or DND • No non-task related browsing • No bathroom breaks* • No conversations • No meetings • No eating • No preening How does it work? 1. Choose a task you’d like to get done 2. Set Set the Pomodoro for 25 minutes 3. Work on the task until the Pomodoro rings 4. When the Pomodoro rings, put a checkmark on a paper 5. Take a short break 6. Every 4 Pomodoros, take a longer break What activity requires you by monitoring how many Pomodoros you need in order to accomplish a specific task 2. Learn to protect your Pomodoro from the internal and external interruption 3. Make accurate estimations0 码力 | 3 页 | 289.16 KB | 5 月前3The Phoenix Project
percent divided by fifty percent, so one unit of time. Let’s call it one hour. So, on average, our task would wait in the queue for one hour before it gets worked. On the other hand, if a resource is or nine hours. In other words, our task would wait in queue nine times longer than if the resource were fifty percent idle.” I conclude, “So, for the Phoenix task, assuming we have seven handoffs, and0 码力 | 3 页 | 154.45 KB | 5 月前3python3学习手册
Lock()方法创建一个互斥锁, 互斥锁在同一时刻只允许一个线程访问共享数据 import threading import random num = 0 # 各线程之间可访问同一内存块,即可访问同一个全局变量 def task_thread(lock): global num # 获取锁,用于线程同步 lock.acquire() print("线程 {} 已获取 锁".format(threading Lock() t1 = threading.Thread(target=task_thread, args= (lock,)) t2 = threading.Thread(target=task_thread, args= (lock,)) t3 = threading.Thread(target=task_thread, args= (lock,)) t1.start()0 码力 | 213 页 | 3.53 MB | 1 年前3
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