threading.timer只打印for循環的最后一個值

Question

嘿，我陷入了一種情況，我有條件內循環。 如果條件滿足，我希望輸出延遲10秒。 而不是期望的輸出，我同時獲得所有值，然后最后的值重復10秒延遲。 下面是代碼

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")

輸出是：

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

如果能在這里得到一些幫助，我將非常感激。 謝謝

Answer 1

問題是你的范圍。 在定時器之后，延遲功能將打印當前的x值，而不是定時器啟動時的x值。

您需要將x作為參數傳遞，如下所示：

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")

輸出將是：

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

如果您不想在計時器之前輸出，只是不要在if語句中調用延遲函數。

實際上，threading.Timer會在10次seconde之后調用你的函數（作為第二個參數給出）（作為第一個參數給出）

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")

將輸出：

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