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Tesco's Burger Relish Best Before End date number
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Tesco's Burger Relish Best Before End date number
Calculate the day number of the yearHow long until this date?Resolving the Date Format DisputeCalculate the day number of the yearLegalize Reversed DateWhat's the Date?Date OccurrencesConvert an Excel date code to a “date”Get the date of the nth day of week in a given year and monthMis-decode a dateOutput the date in the Mel calendar
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Given a date between 2010-01-01 and 2099-12-31 as three integers (please state your order if not [year,month,day]
), answer with Tesco's corresponding five-digit Burger Relish Best Before End date number.
The format is simply the last two digits of the year, followed immediately by three digits representing the day of that year, between 000 and 365.
Cookies if you can do this using only numeric operations and comparisons. You may put such code separately from your golfed code, if it isn't also the shortest code you have.
Test cases
[year,month,day]
→ BRBBED
:
[2010,1,1]
→ 10000
[2019,7,5]
→ 19185
[2020,5,20]
→ 20140
[2096,12,31]
→ 96365
[2099,12,31]
→ 99364
code-golf number date conversion
$endgroup$
|
show 2 more comments
$begingroup$
Given a date between 2010-01-01 and 2099-12-31 as three integers (please state your order if not [year,month,day]
), answer with Tesco's corresponding five-digit Burger Relish Best Before End date number.
The format is simply the last two digits of the year, followed immediately by three digits representing the day of that year, between 000 and 365.
Cookies if you can do this using only numeric operations and comparisons. You may put such code separately from your golfed code, if it isn't also the shortest code you have.
Test cases
[year,month,day]
→ BRBBED
:
[2010,1,1]
→ 10000
[2019,7,5]
→ 19185
[2020,5,20]
→ 20140
[2096,12,31]
→ 96365
[2099,12,31]
→ 99364
code-golf number date conversion
$endgroup$
$begingroup$
"Cookies if you can do this using only numeric operations." Good luck handling dates without e.g. comparison operators...
$endgroup$
– Erik the Outgolfer
7 hours ago
$begingroup$
@EriktheOutgolfer Addressed.
$endgroup$
– Adám
7 hours ago
$begingroup$
Closely related
$endgroup$
– Arnauld
7 hours ago
1
$begingroup$
@Arnauld But that's a do-X-without-Y with an input domain that makes things harder.
$endgroup$
– Adám
7 hours ago
$begingroup$
Oh indeed. I missed that rule in the linked challenge.
$endgroup$
– Arnauld
7 hours ago
|
show 2 more comments
$begingroup$
Given a date between 2010-01-01 and 2099-12-31 as three integers (please state your order if not [year,month,day]
), answer with Tesco's corresponding five-digit Burger Relish Best Before End date number.
The format is simply the last two digits of the year, followed immediately by three digits representing the day of that year, between 000 and 365.
Cookies if you can do this using only numeric operations and comparisons. You may put such code separately from your golfed code, if it isn't also the shortest code you have.
Test cases
[year,month,day]
→ BRBBED
:
[2010,1,1]
→ 10000
[2019,7,5]
→ 19185
[2020,5,20]
→ 20140
[2096,12,31]
→ 96365
[2099,12,31]
→ 99364
code-golf number date conversion
$endgroup$
Given a date between 2010-01-01 and 2099-12-31 as three integers (please state your order if not [year,month,day]
), answer with Tesco's corresponding five-digit Burger Relish Best Before End date number.
The format is simply the last two digits of the year, followed immediately by three digits representing the day of that year, between 000 and 365.
Cookies if you can do this using only numeric operations and comparisons. You may put such code separately from your golfed code, if it isn't also the shortest code you have.
Test cases
[year,month,day]
→ BRBBED
:
[2010,1,1]
→ 10000
[2019,7,5]
→ 19185
[2020,5,20]
→ 20140
[2096,12,31]
→ 96365
[2099,12,31]
→ 99364
code-golf number date conversion
code-golf number date conversion
edited 7 hours ago
Adám
asked 8 hours ago
AdámAdám
28.3k2 gold badges79 silver badges210 bronze badges
28.3k2 gold badges79 silver badges210 bronze badges
$begingroup$
"Cookies if you can do this using only numeric operations." Good luck handling dates without e.g. comparison operators...
$endgroup$
– Erik the Outgolfer
7 hours ago
$begingroup$
@EriktheOutgolfer Addressed.
$endgroup$
– Adám
7 hours ago
$begingroup$
Closely related
$endgroup$
– Arnauld
7 hours ago
1
$begingroup$
@Arnauld But that's a do-X-without-Y with an input domain that makes things harder.
$endgroup$
– Adám
7 hours ago
$begingroup$
Oh indeed. I missed that rule in the linked challenge.
$endgroup$
– Arnauld
7 hours ago
|
show 2 more comments
$begingroup$
"Cookies if you can do this using only numeric operations." Good luck handling dates without e.g. comparison operators...
$endgroup$
– Erik the Outgolfer
7 hours ago
$begingroup$
@EriktheOutgolfer Addressed.
$endgroup$
– Adám
7 hours ago
$begingroup$
Closely related
$endgroup$
– Arnauld
7 hours ago
1
$begingroup$
@Arnauld But that's a do-X-without-Y with an input domain that makes things harder.
$endgroup$
– Adám
7 hours ago
$begingroup$
Oh indeed. I missed that rule in the linked challenge.
$endgroup$
– Arnauld
7 hours ago
$begingroup$
"Cookies if you can do this using only numeric operations." Good luck handling dates without e.g. comparison operators...
$endgroup$
– Erik the Outgolfer
7 hours ago
$begingroup$
"Cookies if you can do this using only numeric operations." Good luck handling dates without e.g. comparison operators...
$endgroup$
– Erik the Outgolfer
7 hours ago
$begingroup$
@EriktheOutgolfer Addressed.
$endgroup$
– Adám
7 hours ago
$begingroup$
@EriktheOutgolfer Addressed.
$endgroup$
– Adám
7 hours ago
$begingroup$
Closely related
$endgroup$
– Arnauld
7 hours ago
$begingroup$
Closely related
$endgroup$
– Arnauld
7 hours ago
1
1
$begingroup$
@Arnauld But that's a do-X-without-Y with an input domain that makes things harder.
$endgroup$
– Adám
7 hours ago
$begingroup$
@Arnauld But that's a do-X-without-Y with an input domain that makes things harder.
$endgroup$
– Adám
7 hours ago
$begingroup$
Oh indeed. I missed that rule in the linked challenge.
$endgroup$
– Arnauld
7 hours ago
$begingroup$
Oh indeed. I missed that rule in the linked challenge.
$endgroup$
– Arnauld
7 hours ago
|
show 2 more comments
9 Answers
9
active
oldest
votes
$begingroup$
PowerShell, 74 70 67 bytes
param($y,$m,$d)-join"$y 00$((date "$m/$d/$y"|% d*r)-1)"[2,3+-3..-1]
Try it online!
Exactly what it says on the tin (but formatted weirdly). Takes in $y
ear, $m
onth, $d
ay, plucks out the last two digits of the $y
ear, gets a .NET datetime
object of the specified day, then gets the dayofyear
(with |% d*r
) thereof. Subtracts one to make it zero-indexed, then uses string formatting and slicing to make it three-padded (e.g., 000
instead of 0
), and then -join
s it all together into a single string with implicit output.
Note this is culture dependent due to the datetime
formatting. This works in en-us
, which is what TIO also uses.
-3 bytes thanks to mazzy
$endgroup$
$begingroup$
-1 byte
$endgroup$
– mazzy
5 hours ago
1
$begingroup$
@mazzy Don't need the parens in-join(...)
since it's a single string slice, then. So that's another -2. Thanks!
$endgroup$
– AdmBorkBork
5 hours ago
add a comment |
$begingroup$
PowerShell, 48 45 bytes
Port of Expired Data's answer for C#
$args[0]*1E3+($args-join'-'|date|% d*r)-2E6-1
Try it online!
PowerShell, 60 58 bytes
-2 bytes thanks @AdmBorkBork
-join"$args 00$(($args-join'-'|date|% d*r)-1)"[2,3+-3..-1]
Try it online!
Explanation:
- The script takes three integers
[year,month,day]
$args-join'-'
makes ISO date string likeYYYY-MM-DD
('YYYY-MM-DD'|date|% d*r)-1
calculatesdayOfYear
- the script makes a result string like
YYYY MM DD 00###
where###
isdayOfYear
- finally the script extracts chars from positions
[2,3+-3..-1]
of the result string and joins this chars to result
$endgroup$
1
$begingroup$
Clever use of$args
.
$endgroup$
– AdmBorkBork
5 hours ago
1
$begingroup$
Just like with my answer, you only have a single string slice, so you don't need parens in-join(...)
for -2 bytes. Try it online!
$endgroup$
– AdmBorkBork
5 hours ago
1
$begingroup$
Nice port! I knew there'd be a good way in PowerShell, but I didn't think it'd be that good
$endgroup$
– Expired Data
4 hours ago
add a comment |
$begingroup$
Python 2, 63 61 57 bytes
lambda y,m,d:y%100*1000+(153*m-2)/5+[m+1,1>y%4][2<m]+d-33
Try it online!
Returns the result as an integer.
Neil found that his Charcoal approach turned out to be lucky after some rearrangement.
Pure arithmetic and comparison solution (61 bytes):
lambda y,m,d:y%100*1000+30*m+m/2+(8<m<12)+~(0<y%4)*(2<m)+d-31
$endgroup$
$begingroup$
Seems to be a day off for January and February?
$endgroup$
– Neil
3 hours ago
$begingroup$
@Neil Oh LOL I just noticed I was subtracting the extra day February doesn't have even when that month hasn't passed yet. EDIT: should be fixed.
$endgroup$
– Erik the Outgolfer
3 hours ago
$begingroup$
I ported my Charcoal solution, and then golfed it down to 57 bytes:y%100*1000+(153*m-2)/5+[m+1,1>y%4][2<m]+d-33
. (Sadly this rearrangement has the same byte count in Charcoal.)
$endgroup$
– Neil
3 hours ago
$begingroup$
@Neil LOL Charcoal thinking beats Python thinking 10/10 I guess. :D I'll add it and see what I can further do, if I can.
$endgroup$
– Erik the Outgolfer
3 hours ago
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 88 78 73 71 59 57 bytes
a=>a[0]%100*1000+new DateTime(a[0],a[1],a[2]).DayOfYear-1
Try it online!
$endgroup$
add a comment |
$begingroup$
Python 3, 87 85 74 bytes
lambda y,m,d:sum([y*1000-2000029+28*m,3,y%4==0,3,2,3,2,3,3,2,3,2,3][:m])+d
Try it online!
This is using only numeric operations and comparisons.
$endgroup$
add a comment |
$begingroup$
Ruby, 40 bytes
->dTime.gm(*d).strftime('%y%j').to_i-1
Try it online!
So close to being a built-in format, but %j
is 1-indexed.
In a happier coincidence, though, Tesco Time (.gm
) is one character shorter than local time (.new
).
$endgroup$
add a comment |
$begingroup$
Charcoal, 33 bytes
IΣ⟦×φ﹪θ¹⁰⁰⊖ζ÷⊕×¹⁵³⊖η⁵⎇›η²⊖⊖¬﹪θ⁴⊖η
Try it online! Link is to verbose version of code. Try it online! Link includes test suite. I don't think Charcoal has any date functions, so here's a mathematical solution. Explanation:
×φ﹪θ¹⁰⁰ Year * 1000
⊖ζ Day - 1
÷⊕×¹⁵³⊖η⁵ Days in previous months
⎇›η² For months after Februrary
⊖⊖¬﹪θ⁴ Adjust for leap years else
⊖η Adjust for short Februrary
Σ⟦ Take the sum
I Cast to string for implicit print
$endgroup$
add a comment |
$begingroup$
JavaScript, 70 57 bytes
(y,m,d,U=Date.UTC)=>y%100*1e3+(U(y,m-1,d)-U(y,0,1))/864e5
Try it online!
$endgroup$
$begingroup$
59 bytes?
$endgroup$
– Arnauld
4 hours ago
$begingroup$
@Arnauld y%100*1e3 works because they have the same precedence.
$endgroup$
– Neil
4 hours ago
$begingroup$
Heh, I was overcomplicating it by using string concatenation when simple arithmetic is all that was needed. Thanks @Arnauld and @Neil!
$endgroup$
– darrylyeo
3 hours ago
$begingroup$
@darrylyeo something in the back of my head said -2000 was 1 byte too many but %100 was far too obvious!
$endgroup$
– Expired Data
2 hours ago
add a comment |
$begingroup$
Japt, 23 bytes
Takes input as 3 individual strings, assuming that's allowed.
¤+@ÐUTX ŶÐN Å}as ùT3 É
Try it
Very unhappy with this (there must be a better way to get the day of the year!) but it's been a long day. A port of darrylyeo's JS solution would be 2 bytes shorter but, for some reason, despite Japt being JavaScript, floating point inaccuracies abound.
$endgroup$
add a comment |
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9 Answers
9
active
oldest
votes
9 Answers
9
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
PowerShell, 74 70 67 bytes
param($y,$m,$d)-join"$y 00$((date "$m/$d/$y"|% d*r)-1)"[2,3+-3..-1]
Try it online!
Exactly what it says on the tin (but formatted weirdly). Takes in $y
ear, $m
onth, $d
ay, plucks out the last two digits of the $y
ear, gets a .NET datetime
object of the specified day, then gets the dayofyear
(with |% d*r
) thereof. Subtracts one to make it zero-indexed, then uses string formatting and slicing to make it three-padded (e.g., 000
instead of 0
), and then -join
s it all together into a single string with implicit output.
Note this is culture dependent due to the datetime
formatting. This works in en-us
, which is what TIO also uses.
-3 bytes thanks to mazzy
$endgroup$
$begingroup$
-1 byte
$endgroup$
– mazzy
5 hours ago
1
$begingroup$
@mazzy Don't need the parens in-join(...)
since it's a single string slice, then. So that's another -2. Thanks!
$endgroup$
– AdmBorkBork
5 hours ago
add a comment |
$begingroup$
PowerShell, 74 70 67 bytes
param($y,$m,$d)-join"$y 00$((date "$m/$d/$y"|% d*r)-1)"[2,3+-3..-1]
Try it online!
Exactly what it says on the tin (but formatted weirdly). Takes in $y
ear, $m
onth, $d
ay, plucks out the last two digits of the $y
ear, gets a .NET datetime
object of the specified day, then gets the dayofyear
(with |% d*r
) thereof. Subtracts one to make it zero-indexed, then uses string formatting and slicing to make it three-padded (e.g., 000
instead of 0
), and then -join
s it all together into a single string with implicit output.
Note this is culture dependent due to the datetime
formatting. This works in en-us
, which is what TIO also uses.
-3 bytes thanks to mazzy
$endgroup$
$begingroup$
-1 byte
$endgroup$
– mazzy
5 hours ago
1
$begingroup$
@mazzy Don't need the parens in-join(...)
since it's a single string slice, then. So that's another -2. Thanks!
$endgroup$
– AdmBorkBork
5 hours ago
add a comment |
$begingroup$
PowerShell, 74 70 67 bytes
param($y,$m,$d)-join"$y 00$((date "$m/$d/$y"|% d*r)-1)"[2,3+-3..-1]
Try it online!
Exactly what it says on the tin (but formatted weirdly). Takes in $y
ear, $m
onth, $d
ay, plucks out the last two digits of the $y
ear, gets a .NET datetime
object of the specified day, then gets the dayofyear
(with |% d*r
) thereof. Subtracts one to make it zero-indexed, then uses string formatting and slicing to make it three-padded (e.g., 000
instead of 0
), and then -join
s it all together into a single string with implicit output.
Note this is culture dependent due to the datetime
formatting. This works in en-us
, which is what TIO also uses.
-3 bytes thanks to mazzy
$endgroup$
PowerShell, 74 70 67 bytes
param($y,$m,$d)-join"$y 00$((date "$m/$d/$y"|% d*r)-1)"[2,3+-3..-1]
Try it online!
Exactly what it says on the tin (but formatted weirdly). Takes in $y
ear, $m
onth, $d
ay, plucks out the last two digits of the $y
ear, gets a .NET datetime
object of the specified day, then gets the dayofyear
(with |% d*r
) thereof. Subtracts one to make it zero-indexed, then uses string formatting and slicing to make it three-padded (e.g., 000
instead of 0
), and then -join
s it all together into a single string with implicit output.
Note this is culture dependent due to the datetime
formatting. This works in en-us
, which is what TIO also uses.
-3 bytes thanks to mazzy
edited 5 hours ago
answered 7 hours ago
AdmBorkBorkAdmBorkBork
28.4k4 gold badges70 silver badges245 bronze badges
28.4k4 gold badges70 silver badges245 bronze badges
$begingroup$
-1 byte
$endgroup$
– mazzy
5 hours ago
1
$begingroup$
@mazzy Don't need the parens in-join(...)
since it's a single string slice, then. So that's another -2. Thanks!
$endgroup$
– AdmBorkBork
5 hours ago
add a comment |
$begingroup$
-1 byte
$endgroup$
– mazzy
5 hours ago
1
$begingroup$
@mazzy Don't need the parens in-join(...)
since it's a single string slice, then. So that's another -2. Thanks!
$endgroup$
– AdmBorkBork
5 hours ago
$begingroup$
-1 byte
$endgroup$
– mazzy
5 hours ago
$begingroup$
-1 byte
$endgroup$
– mazzy
5 hours ago
1
1
$begingroup$
@mazzy Don't need the parens in
-join(...)
since it's a single string slice, then. So that's another -2. Thanks!$endgroup$
– AdmBorkBork
5 hours ago
$begingroup$
@mazzy Don't need the parens in
-join(...)
since it's a single string slice, then. So that's another -2. Thanks!$endgroup$
– AdmBorkBork
5 hours ago
add a comment |
$begingroup$
PowerShell, 48 45 bytes
Port of Expired Data's answer for C#
$args[0]*1E3+($args-join'-'|date|% d*r)-2E6-1
Try it online!
PowerShell, 60 58 bytes
-2 bytes thanks @AdmBorkBork
-join"$args 00$(($args-join'-'|date|% d*r)-1)"[2,3+-3..-1]
Try it online!
Explanation:
- The script takes three integers
[year,month,day]
$args-join'-'
makes ISO date string likeYYYY-MM-DD
('YYYY-MM-DD'|date|% d*r)-1
calculatesdayOfYear
- the script makes a result string like
YYYY MM DD 00###
where###
isdayOfYear
- finally the script extracts chars from positions
[2,3+-3..-1]
of the result string and joins this chars to result
$endgroup$
1
$begingroup$
Clever use of$args
.
$endgroup$
– AdmBorkBork
5 hours ago
1
$begingroup$
Just like with my answer, you only have a single string slice, so you don't need parens in-join(...)
for -2 bytes. Try it online!
$endgroup$
– AdmBorkBork
5 hours ago
1
$begingroup$
Nice port! I knew there'd be a good way in PowerShell, but I didn't think it'd be that good
$endgroup$
– Expired Data
4 hours ago
add a comment |
$begingroup$
PowerShell, 48 45 bytes
Port of Expired Data's answer for C#
$args[0]*1E3+($args-join'-'|date|% d*r)-2E6-1
Try it online!
PowerShell, 60 58 bytes
-2 bytes thanks @AdmBorkBork
-join"$args 00$(($args-join'-'|date|% d*r)-1)"[2,3+-3..-1]
Try it online!
Explanation:
- The script takes three integers
[year,month,day]
$args-join'-'
makes ISO date string likeYYYY-MM-DD
('YYYY-MM-DD'|date|% d*r)-1
calculatesdayOfYear
- the script makes a result string like
YYYY MM DD 00###
where###
isdayOfYear
- finally the script extracts chars from positions
[2,3+-3..-1]
of the result string and joins this chars to result
$endgroup$
1
$begingroup$
Clever use of$args
.
$endgroup$
– AdmBorkBork
5 hours ago
1
$begingroup$
Just like with my answer, you only have a single string slice, so you don't need parens in-join(...)
for -2 bytes. Try it online!
$endgroup$
– AdmBorkBork
5 hours ago
1
$begingroup$
Nice port! I knew there'd be a good way in PowerShell, but I didn't think it'd be that good
$endgroup$
– Expired Data
4 hours ago
add a comment |
$begingroup$
PowerShell, 48 45 bytes
Port of Expired Data's answer for C#
$args[0]*1E3+($args-join'-'|date|% d*r)-2E6-1
Try it online!
PowerShell, 60 58 bytes
-2 bytes thanks @AdmBorkBork
-join"$args 00$(($args-join'-'|date|% d*r)-1)"[2,3+-3..-1]
Try it online!
Explanation:
- The script takes three integers
[year,month,day]
$args-join'-'
makes ISO date string likeYYYY-MM-DD
('YYYY-MM-DD'|date|% d*r)-1
calculatesdayOfYear
- the script makes a result string like
YYYY MM DD 00###
where###
isdayOfYear
- finally the script extracts chars from positions
[2,3+-3..-1]
of the result string and joins this chars to result
$endgroup$
PowerShell, 48 45 bytes
Port of Expired Data's answer for C#
$args[0]*1E3+($args-join'-'|date|% d*r)-2E6-1
Try it online!
PowerShell, 60 58 bytes
-2 bytes thanks @AdmBorkBork
-join"$args 00$(($args-join'-'|date|% d*r)-1)"[2,3+-3..-1]
Try it online!
Explanation:
- The script takes three integers
[year,month,day]
$args-join'-'
makes ISO date string likeYYYY-MM-DD
('YYYY-MM-DD'|date|% d*r)-1
calculatesdayOfYear
- the script makes a result string like
YYYY MM DD 00###
where###
isdayOfYear
- finally the script extracts chars from positions
[2,3+-3..-1]
of the result string and joins this chars to result
edited 5 hours ago
answered 5 hours ago
mazzymazzy
3,5461 gold badge4 silver badges19 bronze badges
3,5461 gold badge4 silver badges19 bronze badges
1
$begingroup$
Clever use of$args
.
$endgroup$
– AdmBorkBork
5 hours ago
1
$begingroup$
Just like with my answer, you only have a single string slice, so you don't need parens in-join(...)
for -2 bytes. Try it online!
$endgroup$
– AdmBorkBork
5 hours ago
1
$begingroup$
Nice port! I knew there'd be a good way in PowerShell, but I didn't think it'd be that good
$endgroup$
– Expired Data
4 hours ago
add a comment |
1
$begingroup$
Clever use of$args
.
$endgroup$
– AdmBorkBork
5 hours ago
1
$begingroup$
Just like with my answer, you only have a single string slice, so you don't need parens in-join(...)
for -2 bytes. Try it online!
$endgroup$
– AdmBorkBork
5 hours ago
1
$begingroup$
Nice port! I knew there'd be a good way in PowerShell, but I didn't think it'd be that good
$endgroup$
– Expired Data
4 hours ago
1
1
$begingroup$
Clever use of
$args
.$endgroup$
– AdmBorkBork
5 hours ago
$begingroup$
Clever use of
$args
.$endgroup$
– AdmBorkBork
5 hours ago
1
1
$begingroup$
Just like with my answer, you only have a single string slice, so you don't need parens in
-join(...)
for -2 bytes. Try it online!$endgroup$
– AdmBorkBork
5 hours ago
$begingroup$
Just like with my answer, you only have a single string slice, so you don't need parens in
-join(...)
for -2 bytes. Try it online!$endgroup$
– AdmBorkBork
5 hours ago
1
1
$begingroup$
Nice port! I knew there'd be a good way in PowerShell, but I didn't think it'd be that good
$endgroup$
– Expired Data
4 hours ago
$begingroup$
Nice port! I knew there'd be a good way in PowerShell, but I didn't think it'd be that good
$endgroup$
– Expired Data
4 hours ago
add a comment |
$begingroup$
Python 2, 63 61 57 bytes
lambda y,m,d:y%100*1000+(153*m-2)/5+[m+1,1>y%4][2<m]+d-33
Try it online!
Returns the result as an integer.
Neil found that his Charcoal approach turned out to be lucky after some rearrangement.
Pure arithmetic and comparison solution (61 bytes):
lambda y,m,d:y%100*1000+30*m+m/2+(8<m<12)+~(0<y%4)*(2<m)+d-31
$endgroup$
$begingroup$
Seems to be a day off for January and February?
$endgroup$
– Neil
3 hours ago
$begingroup$
@Neil Oh LOL I just noticed I was subtracting the extra day February doesn't have even when that month hasn't passed yet. EDIT: should be fixed.
$endgroup$
– Erik the Outgolfer
3 hours ago
$begingroup$
I ported my Charcoal solution, and then golfed it down to 57 bytes:y%100*1000+(153*m-2)/5+[m+1,1>y%4][2<m]+d-33
. (Sadly this rearrangement has the same byte count in Charcoal.)
$endgroup$
– Neil
3 hours ago
$begingroup$
@Neil LOL Charcoal thinking beats Python thinking 10/10 I guess. :D I'll add it and see what I can further do, if I can.
$endgroup$
– Erik the Outgolfer
3 hours ago
add a comment |
$begingroup$
Python 2, 63 61 57 bytes
lambda y,m,d:y%100*1000+(153*m-2)/5+[m+1,1>y%4][2<m]+d-33
Try it online!
Returns the result as an integer.
Neil found that his Charcoal approach turned out to be lucky after some rearrangement.
Pure arithmetic and comparison solution (61 bytes):
lambda y,m,d:y%100*1000+30*m+m/2+(8<m<12)+~(0<y%4)*(2<m)+d-31
$endgroup$
$begingroup$
Seems to be a day off for January and February?
$endgroup$
– Neil
3 hours ago
$begingroup$
@Neil Oh LOL I just noticed I was subtracting the extra day February doesn't have even when that month hasn't passed yet. EDIT: should be fixed.
$endgroup$
– Erik the Outgolfer
3 hours ago
$begingroup$
I ported my Charcoal solution, and then golfed it down to 57 bytes:y%100*1000+(153*m-2)/5+[m+1,1>y%4][2<m]+d-33
. (Sadly this rearrangement has the same byte count in Charcoal.)
$endgroup$
– Neil
3 hours ago
$begingroup$
@Neil LOL Charcoal thinking beats Python thinking 10/10 I guess. :D I'll add it and see what I can further do, if I can.
$endgroup$
– Erik the Outgolfer
3 hours ago
add a comment |
$begingroup$
Python 2, 63 61 57 bytes
lambda y,m,d:y%100*1000+(153*m-2)/5+[m+1,1>y%4][2<m]+d-33
Try it online!
Returns the result as an integer.
Neil found that his Charcoal approach turned out to be lucky after some rearrangement.
Pure arithmetic and comparison solution (61 bytes):
lambda y,m,d:y%100*1000+30*m+m/2+(8<m<12)+~(0<y%4)*(2<m)+d-31
$endgroup$
Python 2, 63 61 57 bytes
lambda y,m,d:y%100*1000+(153*m-2)/5+[m+1,1>y%4][2<m]+d-33
Try it online!
Returns the result as an integer.
Neil found that his Charcoal approach turned out to be lucky after some rearrangement.
Pure arithmetic and comparison solution (61 bytes):
lambda y,m,d:y%100*1000+30*m+m/2+(8<m<12)+~(0<y%4)*(2<m)+d-31
edited 3 hours ago
answered 6 hours ago
Erik the OutgolferErik the Outgolfer
34.4k4 gold badges30 silver badges108 bronze badges
34.4k4 gold badges30 silver badges108 bronze badges
$begingroup$
Seems to be a day off for January and February?
$endgroup$
– Neil
3 hours ago
$begingroup$
@Neil Oh LOL I just noticed I was subtracting the extra day February doesn't have even when that month hasn't passed yet. EDIT: should be fixed.
$endgroup$
– Erik the Outgolfer
3 hours ago
$begingroup$
I ported my Charcoal solution, and then golfed it down to 57 bytes:y%100*1000+(153*m-2)/5+[m+1,1>y%4][2<m]+d-33
. (Sadly this rearrangement has the same byte count in Charcoal.)
$endgroup$
– Neil
3 hours ago
$begingroup$
@Neil LOL Charcoal thinking beats Python thinking 10/10 I guess. :D I'll add it and see what I can further do, if I can.
$endgroup$
– Erik the Outgolfer
3 hours ago
add a comment |
$begingroup$
Seems to be a day off for January and February?
$endgroup$
– Neil
3 hours ago
$begingroup$
@Neil Oh LOL I just noticed I was subtracting the extra day February doesn't have even when that month hasn't passed yet. EDIT: should be fixed.
$endgroup$
– Erik the Outgolfer
3 hours ago
$begingroup$
I ported my Charcoal solution, and then golfed it down to 57 bytes:y%100*1000+(153*m-2)/5+[m+1,1>y%4][2<m]+d-33
. (Sadly this rearrangement has the same byte count in Charcoal.)
$endgroup$
– Neil
3 hours ago
$begingroup$
@Neil LOL Charcoal thinking beats Python thinking 10/10 I guess. :D I'll add it and see what I can further do, if I can.
$endgroup$
– Erik the Outgolfer
3 hours ago
$begingroup$
Seems to be a day off for January and February?
$endgroup$
– Neil
3 hours ago
$begingroup$
Seems to be a day off for January and February?
$endgroup$
– Neil
3 hours ago
$begingroup$
@Neil Oh LOL I just noticed I was subtracting the extra day February doesn't have even when that month hasn't passed yet. EDIT: should be fixed.
$endgroup$
– Erik the Outgolfer
3 hours ago
$begingroup$
@Neil Oh LOL I just noticed I was subtracting the extra day February doesn't have even when that month hasn't passed yet. EDIT: should be fixed.
$endgroup$
– Erik the Outgolfer
3 hours ago
$begingroup$
I ported my Charcoal solution, and then golfed it down to 57 bytes:
y%100*1000+(153*m-2)/5+[m+1,1>y%4][2<m]+d-33
. (Sadly this rearrangement has the same byte count in Charcoal.)$endgroup$
– Neil
3 hours ago
$begingroup$
I ported my Charcoal solution, and then golfed it down to 57 bytes:
y%100*1000+(153*m-2)/5+[m+1,1>y%4][2<m]+d-33
. (Sadly this rearrangement has the same byte count in Charcoal.)$endgroup$
– Neil
3 hours ago
$begingroup$
@Neil LOL Charcoal thinking beats Python thinking 10/10 I guess. :D I'll add it and see what I can further do, if I can.
$endgroup$
– Erik the Outgolfer
3 hours ago
$begingroup$
@Neil LOL Charcoal thinking beats Python thinking 10/10 I guess. :D I'll add it and see what I can further do, if I can.
$endgroup$
– Erik the Outgolfer
3 hours ago
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 88 78 73 71 59 57 bytes
a=>a[0]%100*1000+new DateTime(a[0],a[1],a[2]).DayOfYear-1
Try it online!
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 88 78 73 71 59 57 bytes
a=>a[0]%100*1000+new DateTime(a[0],a[1],a[2]).DayOfYear-1
Try it online!
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 88 78 73 71 59 57 bytes
a=>a[0]%100*1000+new DateTime(a[0],a[1],a[2]).DayOfYear-1
Try it online!
$endgroup$
C# (Visual C# Interactive Compiler), 88 78 73 71 59 57 bytes
a=>a[0]%100*1000+new DateTime(a[0],a[1],a[2]).DayOfYear-1
Try it online!
edited 3 hours ago
answered 7 hours ago
Expired DataExpired Data
1,6934 silver badges23 bronze badges
1,6934 silver badges23 bronze badges
add a comment |
add a comment |
$begingroup$
Python 3, 87 85 74 bytes
lambda y,m,d:sum([y*1000-2000029+28*m,3,y%4==0,3,2,3,2,3,3,2,3,2,3][:m])+d
Try it online!
This is using only numeric operations and comparisons.
$endgroup$
add a comment |
$begingroup$
Python 3, 87 85 74 bytes
lambda y,m,d:sum([y*1000-2000029+28*m,3,y%4==0,3,2,3,2,3,3,2,3,2,3][:m])+d
Try it online!
This is using only numeric operations and comparisons.
$endgroup$
add a comment |
$begingroup$
Python 3, 87 85 74 bytes
lambda y,m,d:sum([y*1000-2000029+28*m,3,y%4==0,3,2,3,2,3,3,2,3,2,3][:m])+d
Try it online!
This is using only numeric operations and comparisons.
$endgroup$
Python 3, 87 85 74 bytes
lambda y,m,d:sum([y*1000-2000029+28*m,3,y%4==0,3,2,3,2,3,3,2,3,2,3][:m])+d
Try it online!
This is using only numeric operations and comparisons.
edited 47 mins ago
answered 5 hours ago
DatDat
5341 gold badge3 silver badges15 bronze badges
5341 gold badge3 silver badges15 bronze badges
add a comment |
add a comment |
$begingroup$
Ruby, 40 bytes
->dTime.gm(*d).strftime('%y%j').to_i-1
Try it online!
So close to being a built-in format, but %j
is 1-indexed.
In a happier coincidence, though, Tesco Time (.gm
) is one character shorter than local time (.new
).
$endgroup$
add a comment |
$begingroup$
Ruby, 40 bytes
->dTime.gm(*d).strftime('%y%j').to_i-1
Try it online!
So close to being a built-in format, but %j
is 1-indexed.
In a happier coincidence, though, Tesco Time (.gm
) is one character shorter than local time (.new
).
$endgroup$
add a comment |
$begingroup$
Ruby, 40 bytes
->dTime.gm(*d).strftime('%y%j').to_i-1
Try it online!
So close to being a built-in format, but %j
is 1-indexed.
In a happier coincidence, though, Tesco Time (.gm
) is one character shorter than local time (.new
).
$endgroup$
Ruby, 40 bytes
->dTime.gm(*d).strftime('%y%j').to_i-1
Try it online!
So close to being a built-in format, but %j
is 1-indexed.
In a happier coincidence, though, Tesco Time (.gm
) is one character shorter than local time (.new
).
answered 7 hours ago
histocrathistocrat
19.5k4 gold badges32 silver badges74 bronze badges
19.5k4 gold badges32 silver badges74 bronze badges
add a comment |
add a comment |
$begingroup$
Charcoal, 33 bytes
IΣ⟦×φ﹪θ¹⁰⁰⊖ζ÷⊕×¹⁵³⊖η⁵⎇›η²⊖⊖¬﹪θ⁴⊖η
Try it online! Link is to verbose version of code. Try it online! Link includes test suite. I don't think Charcoal has any date functions, so here's a mathematical solution. Explanation:
×φ﹪θ¹⁰⁰ Year * 1000
⊖ζ Day - 1
÷⊕×¹⁵³⊖η⁵ Days in previous months
⎇›η² For months after Februrary
⊖⊖¬﹪θ⁴ Adjust for leap years else
⊖η Adjust for short Februrary
Σ⟦ Take the sum
I Cast to string for implicit print
$endgroup$
add a comment |
$begingroup$
Charcoal, 33 bytes
IΣ⟦×φ﹪θ¹⁰⁰⊖ζ÷⊕×¹⁵³⊖η⁵⎇›η²⊖⊖¬﹪θ⁴⊖η
Try it online! Link is to verbose version of code. Try it online! Link includes test suite. I don't think Charcoal has any date functions, so here's a mathematical solution. Explanation:
×φ﹪θ¹⁰⁰ Year * 1000
⊖ζ Day - 1
÷⊕×¹⁵³⊖η⁵ Days in previous months
⎇›η² For months after Februrary
⊖⊖¬﹪θ⁴ Adjust for leap years else
⊖η Adjust for short Februrary
Σ⟦ Take the sum
I Cast to string for implicit print
$endgroup$
add a comment |
$begingroup$
Charcoal, 33 bytes
IΣ⟦×φ﹪θ¹⁰⁰⊖ζ÷⊕×¹⁵³⊖η⁵⎇›η²⊖⊖¬﹪θ⁴⊖η
Try it online! Link is to verbose version of code. Try it online! Link includes test suite. I don't think Charcoal has any date functions, so here's a mathematical solution. Explanation:
×φ﹪θ¹⁰⁰ Year * 1000
⊖ζ Day - 1
÷⊕×¹⁵³⊖η⁵ Days in previous months
⎇›η² For months after Februrary
⊖⊖¬﹪θ⁴ Adjust for leap years else
⊖η Adjust for short Februrary
Σ⟦ Take the sum
I Cast to string for implicit print
$endgroup$
Charcoal, 33 bytes
IΣ⟦×φ﹪θ¹⁰⁰⊖ζ÷⊕×¹⁵³⊖η⁵⎇›η²⊖⊖¬﹪θ⁴⊖η
Try it online! Link is to verbose version of code. Try it online! Link includes test suite. I don't think Charcoal has any date functions, so here's a mathematical solution. Explanation:
×φ﹪θ¹⁰⁰ Year * 1000
⊖ζ Day - 1
÷⊕×¹⁵³⊖η⁵ Days in previous months
⎇›η² For months after Februrary
⊖⊖¬﹪θ⁴ Adjust for leap years else
⊖η Adjust for short Februrary
Σ⟦ Take the sum
I Cast to string for implicit print
answered 3 hours ago
NeilNeil
85.9k8 gold badges46 silver badges183 bronze badges
85.9k8 gold badges46 silver badges183 bronze badges
add a comment |
add a comment |
$begingroup$
JavaScript, 70 57 bytes
(y,m,d,U=Date.UTC)=>y%100*1e3+(U(y,m-1,d)-U(y,0,1))/864e5
Try it online!
$endgroup$
$begingroup$
59 bytes?
$endgroup$
– Arnauld
4 hours ago
$begingroup$
@Arnauld y%100*1e3 works because they have the same precedence.
$endgroup$
– Neil
4 hours ago
$begingroup$
Heh, I was overcomplicating it by using string concatenation when simple arithmetic is all that was needed. Thanks @Arnauld and @Neil!
$endgroup$
– darrylyeo
3 hours ago
$begingroup$
@darrylyeo something in the back of my head said -2000 was 1 byte too many but %100 was far too obvious!
$endgroup$
– Expired Data
2 hours ago
add a comment |
$begingroup$
JavaScript, 70 57 bytes
(y,m,d,U=Date.UTC)=>y%100*1e3+(U(y,m-1,d)-U(y,0,1))/864e5
Try it online!
$endgroup$
$begingroup$
59 bytes?
$endgroup$
– Arnauld
4 hours ago
$begingroup$
@Arnauld y%100*1e3 works because they have the same precedence.
$endgroup$
– Neil
4 hours ago
$begingroup$
Heh, I was overcomplicating it by using string concatenation when simple arithmetic is all that was needed. Thanks @Arnauld and @Neil!
$endgroup$
– darrylyeo
3 hours ago
$begingroup$
@darrylyeo something in the back of my head said -2000 was 1 byte too many but %100 was far too obvious!
$endgroup$
– Expired Data
2 hours ago
add a comment |
$begingroup$
JavaScript, 70 57 bytes
(y,m,d,U=Date.UTC)=>y%100*1e3+(U(y,m-1,d)-U(y,0,1))/864e5
Try it online!
$endgroup$
JavaScript, 70 57 bytes
(y,m,d,U=Date.UTC)=>y%100*1e3+(U(y,m-1,d)-U(y,0,1))/864e5
Try it online!
edited 3 hours ago
answered 5 hours ago
darrylyeodarrylyeo
5,43410 silver badges34 bronze badges
5,43410 silver badges34 bronze badges
$begingroup$
59 bytes?
$endgroup$
– Arnauld
4 hours ago
$begingroup$
@Arnauld y%100*1e3 works because they have the same precedence.
$endgroup$
– Neil
4 hours ago
$begingroup$
Heh, I was overcomplicating it by using string concatenation when simple arithmetic is all that was needed. Thanks @Arnauld and @Neil!
$endgroup$
– darrylyeo
3 hours ago
$begingroup$
@darrylyeo something in the back of my head said -2000 was 1 byte too many but %100 was far too obvious!
$endgroup$
– Expired Data
2 hours ago
add a comment |
$begingroup$
59 bytes?
$endgroup$
– Arnauld
4 hours ago
$begingroup$
@Arnauld y%100*1e3 works because they have the same precedence.
$endgroup$
– Neil
4 hours ago
$begingroup$
Heh, I was overcomplicating it by using string concatenation when simple arithmetic is all that was needed. Thanks @Arnauld and @Neil!
$endgroup$
– darrylyeo
3 hours ago
$begingroup$
@darrylyeo something in the back of my head said -2000 was 1 byte too many but %100 was far too obvious!
$endgroup$
– Expired Data
2 hours ago
$begingroup$
59 bytes?
$endgroup$
– Arnauld
4 hours ago
$begingroup$
59 bytes?
$endgroup$
– Arnauld
4 hours ago
$begingroup$
@Arnauld y%100*1e3 works because they have the same precedence.
$endgroup$
– Neil
4 hours ago
$begingroup$
@Arnauld y%100*1e3 works because they have the same precedence.
$endgroup$
– Neil
4 hours ago
$begingroup$
Heh, I was overcomplicating it by using string concatenation when simple arithmetic is all that was needed. Thanks @Arnauld and @Neil!
$endgroup$
– darrylyeo
3 hours ago
$begingroup$
Heh, I was overcomplicating it by using string concatenation when simple arithmetic is all that was needed. Thanks @Arnauld and @Neil!
$endgroup$
– darrylyeo
3 hours ago
$begingroup$
@darrylyeo something in the back of my head said -2000 was 1 byte too many but %100 was far too obvious!
$endgroup$
– Expired Data
2 hours ago
$begingroup$
@darrylyeo something in the back of my head said -2000 was 1 byte too many but %100 was far too obvious!
$endgroup$
– Expired Data
2 hours ago
add a comment |
$begingroup$
Japt, 23 bytes
Takes input as 3 individual strings, assuming that's allowed.
¤+@ÐUTX ŶÐN Å}as ùT3 É
Try it
Very unhappy with this (there must be a better way to get the day of the year!) but it's been a long day. A port of darrylyeo's JS solution would be 2 bytes shorter but, for some reason, despite Japt being JavaScript, floating point inaccuracies abound.
$endgroup$
add a comment |
$begingroup$
Japt, 23 bytes
Takes input as 3 individual strings, assuming that's allowed.
¤+@ÐUTX ŶÐN Å}as ùT3 É
Try it
Very unhappy with this (there must be a better way to get the day of the year!) but it's been a long day. A port of darrylyeo's JS solution would be 2 bytes shorter but, for some reason, despite Japt being JavaScript, floating point inaccuracies abound.
$endgroup$
add a comment |
$begingroup$
Japt, 23 bytes
Takes input as 3 individual strings, assuming that's allowed.
¤+@ÐUTX ŶÐN Å}as ùT3 É
Try it
Very unhappy with this (there must be a better way to get the day of the year!) but it's been a long day. A port of darrylyeo's JS solution would be 2 bytes shorter but, for some reason, despite Japt being JavaScript, floating point inaccuracies abound.
$endgroup$
Japt, 23 bytes
Takes input as 3 individual strings, assuming that's allowed.
¤+@ÐUTX ŶÐN Å}as ùT3 É
Try it
Very unhappy with this (there must be a better way to get the day of the year!) but it's been a long day. A port of darrylyeo's JS solution would be 2 bytes shorter but, for some reason, despite Japt being JavaScript, floating point inaccuracies abound.
answered 1 hour ago
ShaggyShaggy
20.1k3 gold badges20 silver badges68 bronze badges
20.1k3 gold badges20 silver badges68 bronze badges
add a comment |
add a comment |
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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"Cookies if you can do this using only numeric operations." Good luck handling dates without e.g. comparison operators...
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– Erik the Outgolfer
7 hours ago
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@EriktheOutgolfer Addressed.
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– Adám
7 hours ago
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Closely related
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– Arnauld
7 hours ago
1
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@Arnauld But that's a do-X-without-Y with an input domain that makes things harder.
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– Adám
7 hours ago
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Oh indeed. I missed that rule in the linked challenge.
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– Arnauld
7 hours ago