[FIXED] time.sleep alternatives for trials?

Issue

Are there any alternatives for time.sleep on python where I can run a function while timeouts?

Because I have these codes:

time.sleep(10)
pyautogui.locateCenterOnScreen('back.PNG')
pyautogui.click(120, 320)

What it does is that it runs the pyautogui after 10 seconds.

What I want is to run the pyautogui at the start and have trials for 10 seconds. If the image is not located within 10 seconds, then it will return error.

Solution

something like this?

for _ in range(10):
    pos = pyautogui.locateCenterOnScreen('back.PNG')
    if pos: break
    time.sleep(1)

if not pos: raise Exception('Where is this image?!')

pyautogui.click(*pos)

Note: it may take more time than 10 seconds, you could play with times instead

A bit better if locateCenterOnScreen takes a while

from datetime import datetime
start = datetime.now()
pos = None
while (datetime.now() - start).total_seconds() < 10 and not pos:
    pos = pyautogui.locateCenterOnScreen('back.PNG')

if not pos: raise Exception('Where is the image!!??')

pyautogui.click(*pos)

Or wrapped in a function

def retry(action, max_secs=10):
    from datetime import datetime
    start = datetime.now()
    res = None
    while (datetime.now() - start).total_seconds() < max_secs:
        res = action()
        if res: return res
    raise Exception('Timeout!')

pos = retry(lambda: pyautogui.locateCenterOnScreen('back.PNG'))
pyautogui.click(*pos)

Even "better" raising the exception outside of the function:

def retry(action, max_secs=10):
    from datetime import datetime
    start = datetime.now()
    res = None
    while (datetime.now() - start).total_seconds() < max_secs:
        res = action()
        if res: break
    return res

pos = retry(lambda: pyautogui.locateCenterOnScreen('back.PNG'))
if not pos: raise Exception('Where is this image??!!')
pyautogui.click(*pos)

Answered By - Jonas