First clear solution for The Most Frequent Weekdays by Sim0000 - python coding challenges

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First solution in Clear category for The Most Frequent Weekdays by Sim0000

from calendar import weekday, isleap, day_name

def most_frequent_days(year):
    d = [weekday(year, 1, i) for i in range(1, 2 + isleap(year))]
    return [day_name[i] for i in sorted(d)]

    # Because 365 % 7 == 1, the most frequent weekday is the weekday
    # of the 1/1 of the year. In leap year case, because 366 % 7 == 2,
    # both 1/1 and 1/2 are the most frequent.



if __name__ == '__main__':
    # These "asserts" using only for self-checking and not necessary for auto-testing
    assert most_frequent_days(2399) ==  ['Friday'], "1st example"
    assert most_frequent_days(1152) == ['Tuesday', 'Wednesday'], "2nd example"
    assert most_frequent_days(56) == ['Saturday', 'Sunday'], "3rd example"
    assert most_frequent_days(2909) == ['Tuesday'], "4th example"

March 22, 2016

Comments:

freeman_lex on March 23, 2016, 7:29 p.m.
Hey, bro. Same as mine!)

south on March 25, 2016, 2:46 p.m.
I'm not understanding why people are importing isleap, whether it's a leap year or not has no outcome on the answer. We just need to look at the first and last week's and find the most frequent days.

lezeroq on March 31, 2016, 2:22 p.m.
Imagine that you don't have any functions in standard library like ```datetime.isoweekday``` and you have to write an efficient function to calculate most frequent day of the year. First most likely you will write an efficient function that calculate a day of a week by date. And most likely it will be something like this (It is taken from the ceil's solutions): <pre class='brush: python'>def zeller(_y,m,d): m+=1 if m<4: _y-=1;m+=12 y=_y//100 # this is about leap year z=_y%100 # and this is about leap year return (5*y+z+y//4+z//4+13*m//5+d-1)%7 # y//4 + z//4 is also about leap year </pre> Each call of that function adds some cost to ```mostfrequentdays``` function. And minimal cost is to call this function only once. Second optional function checks leap year <pre class='brush: python'>def leap(y): if y%400==0: return True if y%100==0: return False return y%4==0 </pre> So my first point (but not most important): Call of ```leap``` function costs less than ```zeller``` function, so for efficiency better to use each function once than use first function more than once. The second point: Function ```zeller``` already contains "knowledge" about leap year. And the last one: There are many way to solve this task. Most dump way is to calculate number for each day of a week in a year and choose most frequent without using leap function. For me much easier to understand solution like this than any other. Especially with good comments. My solution is almost the same.

Master of the Time Stone

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