Krdytkyu

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"

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

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 "-:--"

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"

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)