I'm learning how to use Python with the Raspberry Pi. I successfully followed the tutorial for how to have a Python script run on the Pi to check for new email and turn on an LED if any new messages are in an inbox. I wanted to give myself a challenge by adding a feature to blink a secondary LED to indicate the number of waiting messages. I played around for a while and got a script working, but would like to know if I implemented this correctly.

Am I using threading correctly? Is there an easier way to check a feed every n seconds, and if new messages are present, then continually run blink() until the next feed check?

#!/user/bin/env python

from threading import Thread, Event
from time import sleep
import feedparser
import RPi.GPIO as GPIO

BIG_LED = 23

USERNAME = "user@gmail.com"
PASSWORD = "gzvioCNiAvFvKoqY"

class Repeat(Thread):
    def __init__(self,delay,function,*args,**kwargs):
        self.abort = Event()
        self.delay = delay
        self.args = args
        self.kwargs = kwargs
        self.function = function
    def stop(self):
    def run(self):
        while not self.abort.isSet():

def blink(count):
    for x in range(0, count):
        print "blink", x + 1
        GPIO.output(BLUE_LED, True)
        GPIO.output(BLUE_LED, False)
    sleep(1) #pause between blinks

    while True:

        print "Starting b"
        b = Repeat(1,blink,0)
        big_led = 0

        print "---> ---> Checking feed"
        messages = int(feedparser.parse("https://" + USERNAME + ":" + PASSWORD + "@mail.google.com/gmail/feed/atom")["feed"]["fullcount"])

        if messages >= 1:
            # turn on big LED
            print "Turning on Big LED"
            GPIO.output(BIG_LED, True)
            big_led = 1
            print "There are", messages, "unread email messages"
            b = Repeat(1,blink,messages)
            print "There are no new messages"
            if big_led == 1:
                print "Turning off Big LED"
                GPIO.output(BIG_LED, False)

        sleep(60) # check the feed every minute for new mail
        print "Stopping b"
except (KeyboardInterrupt, SystemExit):
    print 'Program Stopped Manually!'
  • 1
    \$\begingroup\$ I would replace Thread.__init__(self) with the super(Repeat, self).__init__() as it is a better way of calling base functions. \$\endgroup\$
    – flakes
    Apr 15, 2014 at 18:18
  • \$\begingroup\$ Also consider breaking up larger lines into smaller portions to keep on track with PEP8's "Limit all lines to a maximum of 79 characters" rule. And use the sting.format function when surrounding strings with more text. ie) "https://{}:{}@...".format(username,password) \$\endgroup\$
    – flakes
    Apr 15, 2014 at 18:29
  • \$\begingroup\$ Minor point: the try...except could quite possibly be better a try...finally. \$\endgroup\$
    – Veedrac
    Apr 16, 2014 at 15:38
  • \$\begingroup\$ Great suggestions! Thank you. I'm going to try Veedrac's structure below and look up what PEP8 says. I generally like to avoid writing long lines of code. \$\endgroup\$
    – backwardm
    Apr 17, 2014 at 22:10

1 Answer 1


That looks about right. Your thread doesn't do very much, though, so you might want to consider something like so (untested):

def blink(abort, delay, count):
    while not abort.isSet():
        for x in range(0, count):
            print "blink", x + 1
            GPIO.output(BLUE_LED, True)
            GPIO.output(BLUE_LED, False)
        sleep(1) #pause between blinks

Then later

thread_stop = Event()
b = threading.Thread.run(target=blink, args=(thread_stop, 1, 0))

which prevents need for the whole class. It's not a suggestion as much as a note that you did more than you needed to.

Note that your blink should really be using abort.wait and should cooperate with the other abort.wait to have sensible timings.

  • \$\begingroup\$ Thanks for your answer Veedrac. I'll give your suggestion a try. I didn't know about about.wait which seems like exactly what I was going for. Good to know I at least got the threading aspect right—even though it is doing very little. \$\endgroup\$
    – backwardm
    Apr 17, 2014 at 22:07
  • \$\begingroup\$ The start parameter of range defaults to zero, so you can write for x in range(count) which is the Python idiom for this scenario. \$\endgroup\$
    – Adam
    Apr 18, 2014 at 7:18

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