Alarm clock printing

Question

I know there must be an easier way to write this but I'm stuck in over-complicating mindset instead of just following the Zen of Python. Please help me simplify this.

Given a day of the week encoded as 0=Sun, 1=Mon, 2=Tue, ...6=Sat, and a boolean indicating if we are on vacation, return a string of the form "7:00" indicating when the alarm clock should ring. Weekdays, the alarm should be "7:00" and on the weekend it should be "10:00". Unless we are on vacation -- then on weekdays it should be "10:00" and weekends it should be "off".

alarm_clock(1, False) → '7:00'
alarm_clock(5, False) → '7:00'
alarm_clock(0, False) → '10:00'

def alarm_clock(day, vacation):

    weekend = "06"
    weekdays = "12345"
    if vacation:
        if str(day) in weekend:
            return "off"
        else:
            return "10:00"
    else:
        if str(day) in weekend:
            return "10:00"
        else:
            return "7:00"

Answer 1

I don't think it can get much simpler than this (Pythonic, easy to read, performance great enough to never be a bottleneck):

def alarm_clock(day, vacation):
    weekend = int(day) in (0, 6)
    if weekend and vacation:
        return 'off'
    elif weekend or vacation:
        return '10:00'
    return '7:00'

---

I came up with this after creating a weekend boolean value and then checking the return values alarm_clock should have:

return_values = {
    # (weekend, vacation): Return value,
    (True, True): 'off',
    (True, False): '10:00',
    (False, True): '10:00',
    (False, False): '7:00'
}

As you can see, if both are True (if weekend and vacation:), we should return 'off', and if one of them is True (if weekend or vacation:), we should return 10:00 regardless of which one. Else return 7:00

Answer 2

1. You don't use weekdays.
2. You can have two return statements. (Shown below).

This keeps the same logic, it just removes the need for so meany return statements.

def alarm_clock(day, vacation):
    weekend = "06"
    if vacation:
        return "off" if str(day) in weekend else "10:00"
    else:
        return "10:00" if str(day) in weekend else "7:00"

---

I would improve it further by adding a check, that you enter a number 0-6.

if not (0 <= day <= 6):
    return "-:--"

Answer 3

What about:

1. using 10:00 as default:
2. only check for weekend
3. you might replace (str(day) in weekend) by (0 == day %6) but it is harder to understand

Code:

def alarm_clock(day, vacation):
    weekend = "06"
    if vacation and (str(day) in weekend):
        return "off"
    else:
        if not (str(day) in weekend):
            return "7:00"
    return "10:00"

The bit more cryptic version:

def alarm_clock(day, vacation):
    if vacation and 0 == day % 6:
        return "off"
    else:
        if 0 != day % 6:
            return "7:00"
    return "10:00"

Answer 4

Building on @Joe Wallis' answer, I would shorten it as follows:

def alarm_clock(day, vacation):
    weekend = "06"

    times = {"weekend": "10:00", "weekday": "7:00"}
    if vacation:
        times = {"weekend": "off", "weekday": "10:00"}

    return times['weekend'] if str(day) in weekend else times['weekday']

Which could be further shortened to (detrimental to readability though):

def alarm_clock(day, vacation):
    times = {"weekend": "off", "weekday": "10:00"} if vacation \
            else {"weekend": "10:00", "weekday": "7:00"}

    return times['weekend'] if str(day) in "06" else times['weekday']

The advantages are that you have a dict with the weekend/weekday times, so you only need one generic return statement. The magic/hardcoded string in the further shortened version is a no-no though. Furthermore, you could extend the function to allow for custom times to be passed in, as such:

def alarm_clock(day, vacation, times={}):
    times = times.get('regular', {"weekend": "10:00", "weekday": "7:00"})
    if vacation:
        times = times.get('vacation', {"weekend": "off", "weekday": "10:00"})

    return times['weekend'] if str(day) in "06" else times['weekday']

You can then call it as such:

times = {'regular': {'weekend': "9:00", "weekday": "7:00"}, "vacation": {"weekend": "12:00", "weekday": "6:00"}}
alarm_clock(2, False, times)