time.sleep alternatives for trials?

Question

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.

Answer 1

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)