Efficient way to join elements under a conditional












2












$begingroup$


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









$endgroup$








  • 1




    $begingroup$
    As an aside: leap years are not years which are multiples of 4, the definition is slightly more complex
    $endgroup$
    – Lonidard
    Dec 3 '18 at 22:27
















2












$begingroup$


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









$endgroup$








  • 1




    $begingroup$
    As an aside: leap years are not years which are multiples of 4, the definition is slightly more complex
    $endgroup$
    – Lonidard
    Dec 3 '18 at 22:27














2












2








2





$begingroup$


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









$endgroup$




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 Dec 3 '18 at 20:02









Exp ikxExp ikx

1486




1486








  • 1




    $begingroup$
    As an aside: leap years are not years which are multiples of 4, the definition is slightly more complex
    $endgroup$
    – Lonidard
    Dec 3 '18 at 22:27














  • 1




    $begingroup$
    As an aside: leap years are not years which are multiples of 4, the definition is slightly more complex
    $endgroup$
    – Lonidard
    Dec 3 '18 at 22:27








1




1




$begingroup$
As an aside: leap years are not years which are multiples of 4, the definition is slightly more complex
$endgroup$
– Lonidard
Dec 3 '18 at 22:27




$begingroup$
As an aside: leap years are not years which are multiples of 4, the definition is slightly more complex
$endgroup$
– Lonidard
Dec 3 '18 at 22:27










2 Answers
2






active

oldest

votes


















4












$begingroup$

Here is an alternate solution not using If:



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





share|improve this answer









$endgroup$





















    2












    $begingroup$

    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









    $endgroup$













    • $begingroup$
      Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
      $endgroup$
      – Exp ikx
      Dec 3 '18 at 20:12










    • $begingroup$
      Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
      $endgroup$
      – Henrik Schumacher
      Dec 3 '18 at 20:14










    • $begingroup$
      I'm concerned about efficiency since it's a challenge. Nothing much.
      $endgroup$
      – Exp ikx
      Dec 3 '18 at 20:16











    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Here is an alternate solution not using If:



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





    share|improve this answer









    $endgroup$


















      4












      $begingroup$

      Here is an alternate solution not using If:



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





      share|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Here is an alternate solution not using If:



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





        share|improve this answer









        $endgroup$



        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 Dec 3 '18 at 21:01









        sakrasakra

        2,4931428




        2,4931428























            2












            $begingroup$

            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









            $endgroup$













            • $begingroup$
              Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
              $endgroup$
              – Exp ikx
              Dec 3 '18 at 20:12










            • $begingroup$
              Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
              $endgroup$
              – Henrik Schumacher
              Dec 3 '18 at 20:14










            • $begingroup$
              I'm concerned about efficiency since it's a challenge. Nothing much.
              $endgroup$
              – Exp ikx
              Dec 3 '18 at 20:16
















            2












            $begingroup$

            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









            $endgroup$













            • $begingroup$
              Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
              $endgroup$
              – Exp ikx
              Dec 3 '18 at 20:12










            • $begingroup$
              Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
              $endgroup$
              – Henrik Schumacher
              Dec 3 '18 at 20:14










            • $begingroup$
              I'm concerned about efficiency since it's a challenge. Nothing much.
              $endgroup$
              – Exp ikx
              Dec 3 '18 at 20:16














            2












            2








            2





            $begingroup$

            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









            $endgroup$



            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 Dec 3 '18 at 20:07









            Henrik SchumacherHenrik Schumacher

            50.7k469145




            50.7k469145












            • $begingroup$
              Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
              $endgroup$
              – Exp ikx
              Dec 3 '18 at 20:12










            • $begingroup$
              Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
              $endgroup$
              – Henrik Schumacher
              Dec 3 '18 at 20:14










            • $begingroup$
              I'm concerned about efficiency since it's a challenge. Nothing much.
              $endgroup$
              – Exp ikx
              Dec 3 '18 at 20:16


















            • $begingroup$
              Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
              $endgroup$
              – Exp ikx
              Dec 3 '18 at 20:12










            • $begingroup$
              Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
              $endgroup$
              – Henrik Schumacher
              Dec 3 '18 at 20:14










            • $begingroup$
              I'm concerned about efficiency since it's a challenge. Nothing much.
              $endgroup$
              – Exp ikx
              Dec 3 '18 at 20:16
















            $begingroup$
            Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
            $endgroup$
            – Exp ikx
            Dec 3 '18 at 20:12




            $begingroup$
            Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
            $endgroup$
            – Exp ikx
            Dec 3 '18 at 20:12












            $begingroup$
            Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
            $endgroup$
            – Henrik Schumacher
            Dec 3 '18 at 20:14




            $begingroup$
            Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
            $endgroup$
            – Henrik Schumacher
            Dec 3 '18 at 20:14












            $begingroup$
            I'm concerned about efficiency since it's a challenge. Nothing much.
            $endgroup$
            – Exp ikx
            Dec 3 '18 at 20:16




            $begingroup$
            I'm concerned about efficiency since it's a challenge. Nothing much.
            $endgroup$
            – Exp ikx
            Dec 3 '18 at 20:16


















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