Efficient way to join elements under a conditional
Clash Royale CLAN TAG#URR8PPP
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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
asked Dec 3 at 20:02
Exp ikx
1285
add a comment |
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
asked Dec 3 at 20:02
Exp ikx
1285
1
As an aside: leap years are not years which are multiples of 4, the definition is slightly more complex
– Lonidard
Dec 3 at 22:27
add a comment |
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
asked Dec 3 at 20:02
Exp ikx
1285
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
asked Dec 3 at 20:02
Exp ikx
1285
asked Dec 3 at 20:02
Exp ikx
1285
asked Dec 3 at 20:02
Exp ikx
1285
asked Dec 3 at 20:02
Exp ikx
1285
asked Dec 3 at 20:02
Exp ikx
1285
1285
1
As an aside: leap years are not years which are multiples of 4, the definition is slightly more complex
– Lonidard
Dec 3 at 22:27
add a comment |
1
As an aside: leap years are not years which are multiples of 4, the definition is slightly more complex
– Lonidard
Dec 3 at 22:27
1
1
As an aside: leap years are not years which are multiples of 4, the definition is slightly more complex
– Lonidard
Dec 3 at 22:27
As an aside: leap years are not years which are multiples of 4, the definition is slightly more complex
– Lonidard
Dec 3 at 22:27
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Here is an alternate solution not using If:
Table[ DayName[ year, 1, x ], x, 1 + Boole[ LeapYearQ[year ]] ]
answered Dec 3 at 21:01
sakra
2,4531328
add a comment |
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]
]
answered Dec 3 at 20:07
Henrik Schumacher
47.2k466134
Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
– Exp ikx
Dec 3 at 20:12
Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
– Henrik Schumacher
Dec 3 at 20:14
I'm concerned about efficiency since it's a challenge. Nothing much.
– Exp ikx
Dec 3 at 20:16
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Here is an alternate solution not using If:
Table[ DayName[ year, 1, x ], x, 1 + Boole[ LeapYearQ[year ]] ]
answered Dec 3 at 21:01
sakra
2,4531328
add a comment |
up vote
4
down vote
accepted
Here is an alternate solution not using If:
Table[ DayName[ year, 1, x ], x, 1 + Boole[ LeapYearQ[year ]] ]
answered Dec 3 at 21:01
sakra
2,4531328
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Here is an alternate solution not using If:
Table[ DayName[ year, 1, x ], x, 1 + Boole[ LeapYearQ[year ]] ]
answered Dec 3 at 21:01
sakra
2,4531328
Here is an alternate solution not using If:
Table[ DayName[ year, 1, x ], x, 1 + Boole[ LeapYearQ[year ]] ]
answered Dec 3 at 21:01
sakra
2,4531328
answered Dec 3 at 21:01
sakra
2,4531328
answered Dec 3 at 21:01
sakra
2,4531328
answered Dec 3 at 21:01
sakra
2,4531328
2,4531328
add a comment |
add a comment |
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]
]
answered Dec 3 at 20:07
Henrik Schumacher
47.2k466134
Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
– Exp ikx
Dec 3 at 20:12
Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
– Henrik Schumacher
Dec 3 at 20:14
I'm concerned about efficiency since it's a challenge. Nothing much.
– Exp ikx
Dec 3 at 20:16
add a comment |
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]
]
answered Dec 3 at 20:07
Henrik Schumacher
47.2k466134
Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
– Exp ikx
Dec 3 at 20:12
Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
– Henrik Schumacher
Dec 3 at 20:14
I'm concerned about efficiency since it's a challenge. Nothing much.
– Exp ikx
Dec 3 at 20:16
add a comment |
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]
]
answered Dec 3 at 20:07
Henrik Schumacher
47.2k466134
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]
]
answered Dec 3 at 20:07
Henrik Schumacher
47.2k466134
answered Dec 3 at 20:07
Henrik Schumacher
47.2k466134
answered Dec 3 at 20:07
Henrik Schumacher
47.2k466134
answered Dec 3 at 20:07
Henrik Schumacher
47.2k466134
47.2k466134
Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
– Exp ikx
Dec 3 at 20:12
Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
– Henrik Schumacher
Dec 3 at 20:14
I'm concerned about efficiency since it's a challenge. Nothing much.
– Exp ikx
Dec 3 at 20:16
add a comment |
Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
– Exp ikx
Dec 3 at 20:12
Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
– Henrik Schumacher
Dec 3 at 20:14
I'm concerned about efficiency since it's a challenge. Nothing much.
– Exp ikx
Dec 3 at 20:16
Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
– Exp ikx
Dec 3 at 20:12
Thanks for that tip. But is there still a better way to write the rest of the code? I mean specifically for DayName part.
– Exp ikx
Dec 3 at 20:12
Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
– Henrik Schumacher
Dec 3 at 20:14
Hm. What does "better" mean? Are you concerned about efficiency? Why? How many years do you want to test this way?
– Henrik Schumacher
Dec 3 at 20:14
I'm concerned about efficiency since it's a challenge. Nothing much.
– Exp ikx
Dec 3 at 20:16
I'm concerned about efficiency since it's a challenge. Nothing much.
– Exp ikx
Dec 3 at 20:16
add a comment |
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As an aside: leap years are not years which are multiples of 4, the definition is slightly more complex
– Lonidard
Dec 3 at 22:27