Question

嘿，我陷入了一种情况，我有条件内部循环。如果条件满足，我希望输出延迟10秒。而不是期望的输出，我同时获得所有值，然后最后的值重复10秒延迟。下面是代码

import threading
import time
a=[2, 3, 4, 10, 12]
b=2
for x in a:
    if x >2:
        def delayfunction():
            print(x,"is not ok")
        threading.Timer(10, delayfunction).start()
        delayfunction()
    else:
        print(x," is less than equal to 2")

输出是：

2  is less than equal to 2
3 is not ok
4 is not ok
10 is not ok
12 is not ok
12 is not ok
12 is not ok
12 is not ok
12 is not ok

如果能在这里得到一些帮助，我将非常感激。感谢

Answer 1

问题在于你的范围。在定时器之后，延迟功能将打印当前的x值，而不是定时器启动时的x值。

你需要像这样传递x作为参数：

import threading
import time
a=[2, 3, 4, 10, 12]
b=2
for x in a:
    if x >2:
        def delayfunction(current_x):
            print(current_x,"is not ok")
        threading.Timer(10, delayfunction, [x]).start()
        delayfunction(x)
    else:
        print(x," is less than equal to 2")

输出将是：

2  is less than equal to 2
3 is not ok
4 is not ok
10 is not ok
12 is not ok
3 is not ok
4 is not ok
10 is not ok
12 is not ok

如果您不想在计时器之前输出，只是不要在if语句中调用延迟函数。

实际上，threading.Timer将在10次借调（作为第一个参数给出）之后调用你的函数（作为第二个参数给出）

import threading
import time
a=[2, 3, 4, 10, 12]
b=2
for x in a:
    if x >2:
        def delayfunction(current_x):
            print(current_x,"is not ok")
        threading.Timer(10, delayfunction, [x]).start()
    else:
        print(x," is less than equal to 2")

将输出：

2  is less than equal to 2 # immediatly
3 is not ok                # before 10 second
4 is not ok                # before 10 second
10 is not ok               # before 10 second
12 is not ok               # before 10 second

threading.timer只打印for循环的最后一个值

1 个答案: