Efficient way to join elements under a conditional











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2
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I am solving a certain challenge given by my friend.



I want to print the first day of the year if it is not a leap year, while I want both first and second days if it is a leap year.



I want to know if there is an efficient way to rewrite this piece of code:



If[Mod[year,4]==0,DayName/@{{year,1,1},{year,1,2}},{DayName[{year,1,1}]}]





list-manipulation







share|improve this question






















  • As an aside: leap years are not years which are multiples of 4, the definition is slightly more complex
    – Lonidard
    5 hours ago















up vote
2
down vote

favorite












I am solving a certain challenge given by my friend.



I want to print the first day of the year if it is not a leap year, while I want both first and second days if it is a leap year.



I want to know if there is an efficient way to rewrite this piece of code:



If[Mod[year,4]==0,DayName/@{{year,1,1},{year,1,2}},{DayName[{year,1,1}]}]





list-manipulation







share|improve this question






















  • As an aside: leap years are not years which are multiples of 4, the definition is slightly more complex
    – Lonidard
    5 hours ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











I am solving a certain challenge given by my friend.



I want to print the first day of the year if it is not a leap year, while I want both first and second days if it is a leap year.



I want to know if there is an efficient way to rewrite this piece of code:



If[Mod[year,4]==0,DayName/@{{year,1,1},{year,1,2}},{DayName[{year,1,1}]}]





list-manipulation







share|improve this question













I am solving a certain challenge given by my friend.



I want to print the first day of the year if it is not a leap year, while I want both first and second days if it is a leap year.



I want to know if there is an efficient way to rewrite this piece of code:



If[Mod[year,4]==0,DayName/@{{year,1,1},{year,1,2}},{DayName[{year,1,1}]}]





list-manipulation




list-manipulation






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 8 hours ago









JustCurious

1264




1264












  • As an aside: leap years are not years which are multiples of 4, the definition is slightly more complex
    – Lonidard
    5 hours ago


















  • As an aside: leap years are not years which are multiples of 4, the definition is slightly more complex
    – Lonidard
    5 hours ago
















As an aside: leap years are not years which are multiples of 4, the definition is slightly more complex
– Lonidard
5 hours ago




As an aside: leap years are not years which are multiples of 4, the definition is slightly more complex
– Lonidard
5 hours ago










2 Answers
2






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oldest

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up vote
4
down vote













Here is an alternate solution not using If:



Table[ DayName[ {year, 1, x} ], {x, 1 + Boole[ LeapYearQ[{year} ]]} ]





share|improve this answer




























    up vote
    2
    down vote













    LeapYearQ should be more reliable, in particular since special rules apply if the year is divisible by 100 or 400.



    f = year [Function] If[
    
   LeapYearQ[{year, 1, 1}],
    
   DayName /@ {{year, 1, 1}, {year, 1, 2}},
    
   {DayName[{year, 1, 1}]}
    
   ]





    share|improve this answer





















    • Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
      – JustCurious
      7 hours ago










    • Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
      – Henrik Schumacher
      7 hours ago










    • I'm concerned about efficiency since it's a challenge. Nothing much.
      – JustCurious
      7 hours ago











    Your Answer





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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    up vote
    4
    down vote













    Here is an alternate solution not using If:



    Table[ DayName[ {year, 1, x} ], {x, 1 + Boole[ LeapYearQ[{year} ]]} ]





    share|improve this answer

























      up vote
      4
      down vote













      Here is an alternate solution not using If:



      Table[ DayName[ {year, 1, x} ], {x, 1 + Boole[ LeapYearQ[{year} ]]} ]





      share|improve this answer























        up vote
        4
        down vote










        up vote
        4
        down vote









        Here is an alternate solution not using If:



        Table[ DayName[ {year, 1, x} ], {x, 1 + Boole[ LeapYearQ[{year} ]]} ]





        share|improve this answer












        Here is an alternate solution not using If:



        Table[ DayName[ {year, 1, x} ], {x, 1 + Boole[ LeapYearQ[{year} ]]} ]






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 7 hours ago









        sakra

        2,4381328




        2,4381328






















            up vote
            2
            down vote













            LeapYearQ should be more reliable, in particular since special rules apply if the year is divisible by 100 or 400.



            f = year [Function] If[
            
   LeapYearQ[{year, 1, 1}],
            
   DayName /@ {{year, 1, 1}, {year, 1, 2}},
            
   {DayName[{year, 1, 1}]}
            
   ]





            share|improve this answer





















            • Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
              – JustCurious
              7 hours ago










            • Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
              – Henrik Schumacher
              7 hours ago










            • I'm concerned about efficiency since it's a challenge. Nothing much.
              – JustCurious
              7 hours ago















            up vote
            2
            down vote













            LeapYearQ should be more reliable, in particular since special rules apply if the year is divisible by 100 or 400.



            f = year [Function] If[
            
   LeapYearQ[{year, 1, 1}],
            
   DayName /@ {{year, 1, 1}, {year, 1, 2}},
            
   {DayName[{year, 1, 1}]}
            
   ]





            share|improve this answer





















            • Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
              – JustCurious
              7 hours ago










            • Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
              – Henrik Schumacher
              7 hours ago










            • I'm concerned about efficiency since it's a challenge. Nothing much.
              – JustCurious
              7 hours ago













            up vote
            2
            down vote










            up vote
            2
            down vote









            LeapYearQ should be more reliable, in particular since special rules apply if the year is divisible by 100 or 400.



            f = year [Function] If[
            
   LeapYearQ[{year, 1, 1}],
            
   DayName /@ {{year, 1, 1}, {year, 1, 2}},
            
   {DayName[{year, 1, 1}]}
            
   ]





            share|improve this answer












            LeapYearQ should be more reliable, in particular since special rules apply if the year is divisible by 100 or 400.



            f = year [Function] If[
            
   LeapYearQ[{year, 1, 1}],
            
   DayName /@ {{year, 1, 1}, {year, 1, 2}},
            
   {DayName[{year, 1, 1}]}
            
   ]






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 7 hours ago









            Henrik Schumacher

            46.4k466133




            46.4k466133












            • Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
              – JustCurious
              7 hours ago










            • Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
              – Henrik Schumacher
              7 hours ago










            • I'm concerned about efficiency since it's a challenge. Nothing much.
              – JustCurious
              7 hours ago


















            • Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
              – JustCurious
              7 hours ago










            • Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
              – Henrik Schumacher
              7 hours ago










            • I'm concerned about efficiency since it's a challenge. Nothing much.
              – JustCurious
              7 hours ago
















            Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
            – JustCurious
            7 hours ago




            Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
            – JustCurious
            7 hours ago












            Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
            – Henrik Schumacher
            7 hours ago




            Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
            – Henrik Schumacher
            7 hours ago












            I'm concerned about efficiency since it's a challenge. Nothing much.
            – JustCurious
            7 hours ago




            I'm concerned about efficiency since it's a challenge. Nothing much.
            – JustCurious
            7 hours ago


















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