python画布.坐标（）积累内存

Question

我用tkinter做了一个简单的模拟时钟来研究python的GUIing。代码如下：

from tkinter import *
from datetime import datetime
import threading
from math import cos, sin, pi

root = Tk()
root.geometry('400x400')
pause = True

def sec_to_coord(second):
    x = cos((second - 15) * pi / 30)
    y = sin((-second - 15) * pi / 30)
    return x,y

def update():
    global pause
    if pause: return
    threading.Timer(1, update).start()
    now = datetime.now()
    x,y = sec_to_coord(now.second)
    canv.coords(line, 200, 200, x * 100 + 200, y * 100 + 200) # drains RAM
    x,y = sec_to_coord(now.minute + now.second / 60)
    canv.coords(minhand, 200, 200, x * 90 + 200, y * 90 + 200) # drains RAM
    x,y = sec_to_coord(now.hour * 5 + now.minute / 12)
    canv.coords(hourhand, 200, 200, x * 50 + 200, y * 50 + 200) # drains RAM
    print(':'.join([str(now.hour), str(now.minute), str(now.second)]))

def start():
    global pause
    if not pause: return
    pause = False
    print('pause =', pause)
    update()

def halt():
    global pause
    if not pause: 
        pause = True
        print('pause =', pause)

def on_closing():
    halt()
    print('BUY-BUY!')
    root.destroy()

canv = Canvas(root, width=400, height=400)
canv.pack()
line = canv.create_line(200, 200, 200, 100)
minhand = canv.create_line(200, 200, 200, 110, width = 3)
hourhand = canv.create_line(200, 200, 200, 150, width= 7)
ticks = []
for i in range(60):
    x,y = sec_to_coord(i)
    if i % 5 == 0:
        ticks.append(canv.create_line(x*90+200, y*90+200, \
            x*100+200, y*100+200, width=3))
    else:
        ticks.append(canv.create_line(x*95+200, y*95+200, \
            x*100+200, y*100+200, width=3))

root.protocol('WM_DELETE_WINDOW', on_closing)
start()
root.mainloop()

问题是：在滴答滴答的时候，时钟积累了RAM。通过对部分代码进行注释，我发现canv.coord()中的def update():函数负责。 拜托，有人能告诉我密码有什么问题吗？你知道吗

Answer 1

谢谢你，特里！！！解决了问题！我只需要按照你的建议更改一行root.after(1000, update)而不是threading.Timer(1, update).start()，RAM累积就消失了！你知道吗