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import datetime |
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from collections import Counter |
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3
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4
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5
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def most_frequent_days(year): |
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# since the week repeats in the whole year, we only need to find out |
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# incompleted weekdays in the beginning and the end of the year. |
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9
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# incompleted weekdays in the beginning |
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week_list = [ |
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'', |
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12
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'Monday', |
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'Tuesday', |
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'Wednesday', |
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15
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'Thursday', |
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'Friday', |
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'Saturday', |
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'Sunday', |
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] |
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weekdays = [] |
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for i in range(1, 8): |
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weekday = datetime.datetime(year, 1, i).isoweekday() |
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if weekday == 1: |
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# no incompleted week in beginning |
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break |
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weekdays.append(weekday) |
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# incompleted weekdays in the end |
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for i in range(7): |
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weekday = datetime.datetime(year, 12, 31 - i).isoweekday() |
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if weekday == 7: |
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# no incompleted week in beginning |
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break |
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weekdays.append(weekday) |
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36
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# the most common elements |
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ret = [] |
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week_counter = Counter(weekdays).most_common() |
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ret.append(week_counter[0][0]) |
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for i in week_counter[1:]: |
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if i[1] == week_counter[0][1]: |
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ret.append(i[0]) |
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else: |
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break |
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ret = sorted(ret) |
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return [week_list[i] for i in ret] |
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48
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49
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if __name__ == '__main__': # pragma: no cover |
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# These "asserts" using only for self-checking and not necessary for |
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# auto-testing |
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52
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assert most_frequent_days(2399) == ['Friday'], "1st example" |
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assert most_frequent_days(1152) == ['Tuesday', 'Wednesday'], "2nd example" |
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54
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assert most_frequent_days(56) == ['Saturday', 'Sunday'], "3rd example" |
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assert most_frequent_days(2909) == ['Tuesday'], "4th example" |
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56
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KenMercusLai /
checkio
