Convert YYYYMM to MMMYY

Question

Challenge is rather simple: given a date period in the format YYYYMM output it in the format MMMYY.

Rules:

The input will be a number or a string exactly 6 characters long, consisting only of digits.
Last two digits will be between 01 and 12.
Output must be in the form MMMYY, where MMM represents uppercase three-letter code for the month (below) and YY represents two last digits of the YYYY part of the input.

List of months with corresponding code:

MM    MMM
01    JAN
02    FEB
03    MAR
04    APR
05    MAY
06    JUN
07    JUL
08    AUG
09    SEP
10    OCT
11    NOV
12    DEC

Examples:

Input     Output
201604    APR16
200001    JAN00
000112    DEC01
123405    MAY34

This question is very well-balanced. Both manual parsing and date libraries end up being about the same, at least in Python. — jqkul, Commented Jun 22, 2016 at 20:32
Yesterday, I saw "Convert YYYYMM to MMMYY" on HNQ beside the SO logo. Now I see the same title beside the PCG logo. I was very confused :) — cat, Commented Jun 23, 2016 at 14:30

roblogic · Accepted Answer · 2024-06-11 11:29:41Z

Zsh + BSD/MacOS `date`, 40 bytes

<<<${(U)$(date -jf %Y%m 20${1:2} +%b%y)}

It works for BSD date.
GNU date (as used by TIO) does not support j or f options.

_{I will think about making a GNU version. A relevant quote from the GNU manual:}

_{Our units of temporal measurement, from seconds on up to months, are so complicated, asymmetrical and disjunctive so as to make coherent mental reckoning in time all but impossible. Indeed, had some tyrannical god contrived to enslave our minds to time, to make it all but impossible for us to escape subjection to sodden routines and unpleasant surprises, he could hardly have done better than handing down our present system. It is like a set of trapezoidal building blocks, with no vertical or horizontal surfaces, like a language in which the simplest thought demands ornate constructions, useless particles and lengthy circumlocutions. Unlike the more successful patterns of language and science, which enable us to face experience boldly or at least level-headedly, our system of temporal calculation silently and persistently encourages our terror of time.}
_{… It is as though architects had to measure length in feet, width in meters and height in ells; as though basic instruction manuals demanded a knowledge of five different languages. It is no wonder then that we often look into our own immediate past or future, last Tuesday or a week from Sunday, with feelings of helpless confusion. (Robert Grudin, Time and the Art of Living.)}

Oops, a bash solution is just 18 bytes ... will leave this here and rethink my life :P — roblogic, Commented Jun 20 at 10:06

south · Accepted Answer · 2024-06-12 00:07:10Z

J `types/datetime`, 52 bytes

2 3&{,~'MMM'toupper@fmtDate 1800 todayno@,1,~4".@}.]

A solution using one of J's libraries.

2 3&{,~'MMM'toupper@fmtDate 1800 todayno@,1,~4".@}.]
                                                   ]   NB. Input as string
                                             4   }.    NB. Drop 4 chars
                                              ".@      NB. Then eval as int
                                          1,~          NB. Append 1
                            1800         ,             NB. Prepend 1800
                                 todayno@              NB. Create date int
       'MMM'        fmtDate                            NB. Format the month as Jan,Feb,...
            toupper@                                   NB. Then uppercase it
2 3&{                                                  NB. Index the 3rd/4th char
     ,~                                                NB. Append it

Attempt This Online!

miles · Accepted Answer · 2016-06-22 22:47:12Z

J, 70 bytes

4(}.((,~(_3]\'JANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC'){~1-~".)~2&}.){.)]

Usage

   f =: 4(}.((,~(_3]\'JANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC'){~1-~".)~2&}.){.)]
   f '201604'
APR16
   f '200001'
JAN00
   f '000112'
DEC01
   f '123405'
MAY34

Explanation

4(}.((,~(_3]\'...'){~1-~".)~2&}.){.)] Input: s
                                    ] Identity function, gets the value s
4                                     The constant 4
                                 {.   Take the first 4 chars from s
                            2&}.      Drop the first 2 (Take the last 2) to get the year
  }.                                  Drop the first 4 chars from s to get the month
                        ".            Parse the month substring as a number
                     1-~              Subtract 1 from it
             '...'                    List of MMM month names
         _3]\                         Split the list into nonoverlapping sublists of size 3
      ,~                              Join the MMM month name with the year and return

Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Pyth, 39 bytes

Hexdump:

0000000: 2b 40 63 2e 22 41 59 12 56 0a 7c bd 93 e3 1c 07 +@c."AY.V.|.....
0000010: e3 d4 d9 ed 5b 49 02 cd b4 92 83 86 22 33 73 3e ....[I......"3s>
0000020: 32 7a 3a 7a 32 20 34                            2z:z2 4

Test suite.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jun 23, 2016 at 9:19

Leaky Nun

50.1k6 gold badges110 silver badges284 bronze badges

Add a comment |

Community · Accepted Answer · 2017-04-13 12:39:16Z

jq, 35 characters

(34 characters code + 1 character command-line option.)

(Just tried whether the ^ trick used by Digital Trauma in his Bash answer works in jq too. Works. Now you know who inspired the most important character of this answer. (The alternative is to use the ascii_upcase function.))

strptime("%Y%m")|strftime("%^b%y")

Sample run (Option -R used only in this sample to pass all test cases.)

bash-4.3$ jq -Rr 'strptime("%Y%m")|strftime("%^b%y")' <<END
201604
200001
000112
123405
END
APR16
JAN00
DEC01
MAY34

On-line test: (Passing -R through URL is not supported – so input passed as JSON string literal. Passing -r through URL is not supported – check Raw Output yourself.)

Community · Accepted Answer · 2020-06-17 09:04:33Z

Factor, 82 78 bytes

[ [ 2 tail* 10 base> month-abbreviation ] [ 4 head 2 tail ] bi append >upper ]

Eshplained:

[                    ! new anonymouse function block (quotation)
  [                  ! new quotation 
    2 tail*          ! "201604" -> "04"
    10 base>         ! "04"     -> 4
    month-abbreviation ! 4 -> "Apr"
  ]                  ! end quotation
  [                  ! new quotation
    4 head           ! "201604" -> "2016"
    2 tail           ! "2016"   -> "16" 
  ]                  ! end quotation
  bi                 ! bifurcate two quotations to TOS
  append             ! "Apr" "16" -> "Apr16"
  >upper             ! "Apr16"    -> "APR16"
]                    ! end quotation

Marco · Accepted Answer · 2016-06-25 14:31:31Z

1

PHP, 78 bytes

<?=fscanf(STDIN,"%4d%d",$y,$m)?strtoupper(date("My",mktime(0,0,0,$m,1,$y))):0;

The "year 2038 problem" may occur on some computers, as here. But not in others, as here.

edited Jun 25, 2016 at 14:31

answered Jun 23, 2016 at 16:08

Marco

6014 silver badges5 bronze badges

\$\begingroup\$ @Titus, please don't try to edit other peoples' posts if there are problems and insert your own solutions; instead, please comment on the answer OR create a new answer yourself. \$\endgroup\$
– Value Ink
Commented Jun 25, 2016 at 7:32

Add a comment |

JAL · Accepted Answer · 2016-06-28 18:59:56Z

Swift 2.2, 149 bytes

let f = NSDateFormatter(),g = NSDateFormatter();f.dateFormat = "yyyyMM";g.dateFormat = "MMMyy"
g.stringFromDate(f.dateFromString(i)!).uppercaseString

Trying to get this shorter than Kotlin... It's a shame NSDateFormatter doesn't have an initializer that sets its dateFormat. NSDateFormatter also does not have a default dateFormat value, causing additional losses.

Swift 3, 136 bytes

let f = DateFormatter(),g = DateFormatter();f.dateFormat = "yyyyMM";g.dateFormat = "MMMyy"
g.string(from: f.date(from: i)!).uppercased()

Thanks to the removal of the NS prefix on some classes, I was able to make the Swift 3 answer a little shorter. Still not shorter than Kotlin though...

Test function and cases:

import Foundation
import XCTest

func dateConverter(i: String) -> String? {
    let f = DateFormatter(),g = DateFormatter();f.dateFormat = "yyyyMM";g.dateFormat = "MMMyy"

    if let date = f.date(from: i) {
        return g.string(from: date).uppercased()
    }

    return nil
}

XCTAssert(dateConverter(i: "201604") == "APR16")
XCTAssert(dateConverter(i: "200001") == "JAN00")
XCTAssert(dateConverter(i: "000112") == "DEC01")
XCTAssert(dateConverter(i: "123405") == "MAY34")

Don’t you have to import Foundation in Swift 3? I’ve never used Swift 3 seriously, but I know you need that in Swift 5 at least. — Bbrk24, Commented Jun 22, 2023 at 17:33

a soft pillow · Accepted Answer · 2016-09-16 11:49:20Z

1

R, 154 150 114 112 bytes

Takes six digit input into "b", separates the first four digits from the last two digits, abbreviates the 2-digit month and makes it uppercase, and concatenates it with the 3rd and 4th digit.

Golfed:

function(b){h=substr;i=sprintf;o="%06d";cat(toupper(month.abb[as.numeric(h(i(o,b),5,6))]),h(i(o,b),3,4),sep="")}

Ungolfed:

function(b){
   h=substr;i=sprintf;o="%06d";

   cat(
      toupper(month.abb[as.numeric(h(i(o,b),5,6))]),
      h(i(o,b),3,4),
   sep="")
}

EDITS: replaced duplicitous names with variables; fixed me being stupid. -2 bytes by turning function anonymous (thanks, cat).

edited Sep 16, 2016 at 11:49

answered Jun 23, 2016 at 14:40

a soft pillow

1315 bronze badges

\$\begingroup\$ Nice answer! You can leave off the a= for an anonymous function \$\endgroup\$
– cat
Commented Jun 24, 2016 at 17:56
\$\begingroup\$ @cat I'm not too familiar with anonymous functions but wouldn't I need to add parentheses at the beginning and end of the function if I removed the a= ? Something like this: (function(m) {a=3;m*a})(10) \$\endgroup\$
– a soft pillow
Commented Jun 24, 2016 at 18:52
1

\$\begingroup\$ Nope, function(b){h=substr;i=sprintf;o="%06d";cat(toupper(month.abb[as.numeric(h(i(o,b),5,6))]),h(i(o,b),3,4),sep="")} is a function object all on its own. \$\endgroup\$
– cat
Commented Jun 24, 2016 at 19:17
\$\begingroup\$ This doesn't appear to work. on RStudio with R 3.2.3 (2015-12-10) I get Error in i(o, b) : invalid format '%06d'; use format %s for character objects \$\endgroup\$
– cat
Commented Jun 24, 2016 at 19:18
\$\begingroup\$ @cat Darn new versions messing everything up! I'm on RStudio running R version 3.1.1 (2014-07-10) and it works fine. How does this work here, should I get the new version and change up the code? Also thanks for the anonymous function tip! \$\endgroup\$
– a soft pillow
Commented Jun 24, 2016 at 19:44

| Show 3 more comments

Community · Accepted Answer · 2020-06-17 09:04:33Z

1

SmileBASIC, 85 bytes

DEF C D?MID$(@__JANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC,VAL(D[4]+D[4])*3,3);D[2];D[3]END

@__JANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC is equivalent to the string literal "@__JANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC". JAN starts at index 3, so the program can calculate the position as month*3 instead of (month-1)*3 or month*3-3.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Apr 6, 2018 at 17:27

12Me21

6,7211 gold badge22 silver badges40 bronze badges

Add a comment |

Romen · Accepted Answer · 2018-04-06 19:28:30Z

1

C# - 75 Bytes

s=>{return new System.DateTime(s/100,s%100,1).ToString("MMMyy").ToUpper();}

answered Apr 6, 2018 at 19:28

Romen

1716 bronze badges

Add a comment |

mkst · Accepted Answer · 2018-04-07 21:33:43Z

K4, 65 59 bytes

Solution:

@[13 3#"DECJANFEBMARAPRMAYJUNJULAUGSEPOCTNOV";.*|x],*x:2 4_

Examples:

q)k)@[13 3#"DECJANFEBMARAPRMAYJUNJULAUGSEPOCTNOV";.*|x],*x:2 4_"201604"
"APR16"
q)k)@[13 3#"DECJANFEBMARAPRMAYJUNJULAUGSEPOCTNOV";.*|x],*x:2 4_"200001"
"JAN00"
q)k)@[13 3#"DECJANFEBMARAPRMAYJUNJULAUGSEPOCTNOV";.*|x],*x:2 4_"000112"
"DEC01"
q)k)@[13 3#"DECJANFEBMARAPRMAYJUNJULAUGSEPOCTNOV";.*|x],*x:2 4_"123405"
"MAY34"

Explanation:

Reshape the months of the year so that 0 => DEC and 1 => JAN ... 12 => DEC.

@[13 3#"DECJANFEBMARAPRMAYJUNJULAUGSEPOCTNOV";.*|x],*x:2 4_ / the solution
                                                       2 4_ / cut input at indices 2 & 4, "201604" => ["16", "04"]
                                                     x:     / save as x
                                                    *       / take first
                                                   ,        / join with
@[                                           ;    ]         / apply @[list;index]
                                               *|x          / reverse (|), first (*) aka 'last'
                                              .             / value ("05" => 5)
       "DECJANFEBMARAPRMAYJUNJULAUGSEPOCTNOV"               / offset months of the year
  13 3#                                                     / reshape

Bonus:

60 byte version in K (oK):

((13 3#"DECJANFEBMARAPRMAYJUNJULAUGSEPOCTNOV")@.*|x),*x:2 4_

Try it online!

homersimpson · Accepted Answer · 2018-04-09 17:22:45Z

Caché ObjectScript - 65 bytes

Golfed

r i s d=$zcvt($zd($zdth(i_"01",8),6),"u") w $e(d,1,3)_$e(d,*-1,*)

Ungolfed

r i                            ; Read stdin and store into 'i'                      [200203]
   s d=                        ; Set the variable 'd' to the following...
       $zcvt(                  ; Convert the string
             $zd(              ; Parse the string as a caché date
                 $zdth(        ; Convert the date to horolog format (# days from 12/31/1840)
                       i_"01", ; Append the string "01" to variable 'i'             [20020301]
                       8       ; Use format #8 - "YYYYMMDD" (e.g. 20020301)
                      ),                                                            [58864,0]
                 6             ; Use format #6 - "Mmm [D]D YYYY" (e.g. Mar 1 2002)
                ),                                                                  [Mar 1 2002]
             "u"               ; Convert the string to uppercase
            )                                                                       [MAR 1 2002]
   w                           ; Write out the following...
     $e(d,1,3)                 ; Extract the substring from position 1 to 3 (inclusive) [MAR]
     _                         ; Concatenate
     $e(d,                     ; Extract the substring...
        *-1,                   ; ...from 1 character from the end of the string     [0]
        *                      ; ...to the last character                           [2]
       )                                                                            [02]
                                                                                    [MAR02]

Commentary

I don't think I've ever seen another Caché answer on this site. I'm only forced to know the language because of work (legacy code), but language itself is very powerful.

Some notes about the language:

Caché doesn't really care about whitespace, so r x r y is the same as r x \n r y
Each function is the shortname for the longer version:
- r == read
- s == set
- $zcvt == $zconvert
- $zd == $zdate
- $zdth == $zdateh
- $e == $extract
Arrays are 1-indexed

Unfortunately, there is no online interpreter for Caché. Probably either because it's too obscure or maybe a licensing thing.

If you're interested in the language, you can check out their docs on their very 90's looking website :)

Francisco Hahn · Accepted Answer · 2018-04-09 18:59:05Z

PHP, 146 Bytes.

Manipulating string

Code Try it online

function f($d){echo strtr(ltrim(substr($d,-2),'0'),array_combine(range(1,12),
str_split(JANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC,3))).substr($d,2,-2);}

Explanation

function f($d){
                                #strtr allows u to set an array to replace the ocurrences
echo strtr(ltrim(substr($d,-2),'0'),        #Substring of the last two numbers of the input, 
                                           #removing leading zero
    array_combine(range(1,12),      
    str_split(JANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC,3)))
                                           #create an array of ["1"=>JAN, "2"=>FEB...12=>DEC], 
                                           #split the string by 3 characters
    .substr($d,2,-2);
                                           #Concatenate the 2 middle characters
}

l4m2 · Accepted Answer · 2018-04-11 00:17:06Z

1

JavaScript ES6, 61 bytes

@Downgoat seems not there so I repost to optimize

a=>(new Date(0,a%100-1)+0).slice(4,7).toUpperCase()+a[2]+a[3]

answered Apr 11, 2018 at 0:17

l4m2

28.4k2 gold badges25 silver badges103 bronze badges

Add a comment |

foemre · Accepted Answer · 2018-07-20 14:51:41Z

1

C (tcc), 155 bytes

Longest answer here.

main(i){char*k,s[6],*a="JANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC";gets(s);strcpy(k,s+2);memset(k+2,0,2);i=atoi(s+4)*3;memset(a+i,0,i);strcat(k,a+i-3);puts(k);}

Try it online!

edited Jul 20, 2018 at 14:51

answered Jul 20, 2018 at 14:46

foemre

515 bronze badges

\$\begingroup\$ Welcome to PPCG! :) \$\endgroup\$
– Shaggy
Commented Jul 20, 2018 at 14:51
\$\begingroup\$ 136 bytes \$\endgroup\$
– ceilingcat
Commented Aug 20, 2019 at 0:32

Add a comment |

ceilingcat · Accepted Answer · 2018-08-23 20:36:26Z

1

C (gcc), 107 104 bytes

main(){char s[7];gets(s+1);write(1,memcpy(s,"JANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC"+~-atoi(s+5)*3,3),5);}

Try it online!

edited Aug 23, 2018 at 20:36

answered Aug 23, 2018 at 5:38

ceilingcat

7,6652 gold badges30 silver badges30 bronze badges

Add a comment |

Scott · Accepted Answer · 2020-11-03 03:33:16Z

Pyth, 42 bytes

Ac2>Q2+@c12."AY4H~0ë™~³!ò²×Œ¶7Ö"tiHTG

Try it online!

Ac2>Q2+@c12."AY4H~0ë™~³!ò²×Œ¶7Ö"tiHTG   // With Input '123405'

   >Q2                                       // Get input from index 2 (3405)
Ac2 ^                                        // Split it in two and assign to G and H (G='34', H='05'])
           ."AY4H~0ë™~³!ò²×Œ¶7Ö"             // Packed string representing: "JANFEBMAR..."
        c12."    JANFEBMAR...     "        // Chop it into 12 pieces (["JAN", "FEB", "MAR"...])
                                     tiHT    // Convert "05" to 5, subtract 1.
       @  ["JAN", "FEB", "MAR"...]   tiHT    // Get element at that index (MAY)
      +           "MAY"                  G   // "MAY" + 34

M. Justin · Accepted Answer · 2020-11-04 17:26:17Z

1

Groovy, 69 55 51 bytes

f={sprintf('%Tb',(it[4..5]as int)*22**7L)+it[2..3]}

Try it online!

This formats a millisecond value as an uppercase month name: sprintf('%Tb',value). The magic value of 22^7 is a value such that n * 22^7 milliseconds after epoch is in the nth month of the year of 1970. This value was derived by testing all possible 1 to 2 digit values. Therefore month * 22^7 will yield a value within that month.

1970-01-29
1970-02-27
1970-03-28
1970-04-26
1970-05-25
1970-06-23
1970-07-22
1970-08-19
1970-09-17
1970-10-16
1970-11-14
1970-12-13

edited Nov 4, 2020 at 17:26

answered Nov 3, 2020 at 8:39

M. Justin

2911 silver badge7 bronze badges

Add a comment |

lynn · Accepted Answer · 2020-11-04 17:54:44Z

1

Haskell, 85 bytes

f s=3`take`drop(read(drop 4s)*3-3)"JANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC"++[s!!2,s!!3]

Try it online!

answered Nov 4, 2020 at 17:54

lynn

69.2k11 gold badges133 silver badges283 bronze badges

Add a comment |

emanresu A · Accepted Answer · 2021-07-06 20:42:10Z

1

Vyxal `o`, 37 bytes

4ȯI«×g~Þ¯p-Ḃ:u‹Ẇ¹²Ṗ¨€r;↓§%«3ẇ$‹i⇧₴4Ẏ∷

Try it Online!

4ȯI                # Month as int
   «...«3ẇ         # Compressed string of month names
          $‹i      # Get the correct month
             ⇧₴    # Uppercase and print
               4Ẏ∷ # Get the last two digits of year

answered Jul 6, 2021 at 20:42

emanresu A

44.2k5 gold badges100 silver badges240 bronze badges

Add a comment |

devRicher · Accepted Answer · 2021-07-07 00:52:59Z

1

Lua 5.3, 82 bytes

i=io.read()print(os.date("%b%y",os.time({year=i//100,month=i%100,day=1})):upper())

Needs to be run on an english machine, as the month name is language specific.

Version that works in any locale 91 bytes

i=io.read()j=i%100*3 print(("JANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC"):sub(j-2,j)..i:sub(3,4))

answered Jul 7, 2021 at 0:52

devRicher

1,65913 silver badges35 bronze badges

\$\begingroup\$ %100 can be %20 \$\endgroup\$
– Jo King ♦
Commented Jun 22, 2023 at 4:27

Add a comment |

cubic lettuce · Accepted Answer · 2022-02-06 09:58:19Z

1

Scala, 85 bytes

s=>"JANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC".drop(30*s(4)+3*s(5)-1587).take(3)+s(2)+s(3)

The index lookup 30*s(4)+3*s(5)-1587 is a golfed version (via ASCII arithmetics) of the more readable s.slice(4,6).toInt*3-3.

Try it online!

I also tried the nice trick by @cliffroot to represent the names of the months via Chinese characters. This gives the following variant that is 23 characters shorter (62 instead of 85), but at the end 1 byte longer (86 instead of 85):

s=>""+BigInt(""+"憯䷳烣㘿烪摿摽㛨近筍矯䏔"(10*s(4)+s(5)-528),36)+s(2)+s(3)

edited Feb 6, 2022 at 9:58

answered Feb 6, 2022 at 9:47

cubic lettuce

7693 silver badges6 bronze badges

Add a comment |

chunes · Accepted Answer · 2022-02-06 10:31:36Z

1

Factor, 56 bytes

[ 2 tail 2 cut dec> months "%b"strftime >upper prepend ]

Try it online!

Explanation

          ! "201604"
2         ! "201604" 2
tail      ! "1604"
2         ! "1604" 2
cut       ! "16" "04"
dec>      ! "16" 4
months    ! "16" T{ duration f 0 4 0 0 0 0 }
"%b"      ! "16" T{ duration f 0 4 0 0 0 0 } "%b"
strftime  ! "16" "Apr"
>upper    ! "16" "APR"
prepend   ! "APR16"

answered Feb 6, 2022 at 10:31

chunes

26.8k3 gold badges24 silver badges53 bronze badges

Add a comment |

Oliver · Accepted Answer · 2022-02-06 17:38:18Z

1

Python, 115 bytes

m='JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC'.split();d=input();print("%s%s"%(m[int(d[4]+d[5])-1],d[2]+d[3]))

Try it online

answered Feb 6, 2022 at 17:38

Oliver

1756 bronze badges

\$\begingroup\$ I suggest that you look into lambda functions and generally try to avoid use of split() - That said, I think that JQBLZ's manual string parsing answer may already be optimal for Py3 - you should take a look at that one \$\endgroup\$
– Taylor Raine
Commented Feb 6, 2022 at 21:13

Add a comment |

The Thonnu · Accepted Answer · 2023-06-17 13:39:11Z

Thunno 2, 32 bytes

²ẸN“¤ɼḋġŀ§P¡^Ȥḥ)ñt:²æṆµṖƒr“R³İs+

Attempt This Online!

Explanation

²ẸN“...“R³İs+    # Implicit input        ->  "123405"
²                # Split into pairs      ->  ["12","34","05"]
 Ẹ               # Dump onto stack       ->  "12", "34", "05"
  N              # Cast to integer       ->  "12", "34", 5
   “...“         # Compressed string     ->  "12", "34", 5, "janfeb...novdec"
        R        # Uppercase it          ->  "12", "34", 5, "JANFEB...NOVDEC"
         ³       # Split into threes     ->  "12", "34", 5, ["JAN",...,"DEC"]
          İ      # One-based indexing    ->  "12", "34", "MAY"
           s+    # Swap and concatenate  ->  "12", "MAY34"
                 # Implicit output       ->  "MAY34"

bigyihsuan · Accepted Answer · 2023-06-21 14:05:13Z

1

Go, 107 bytes

import(."time";."strings")
func f(s string)string{t,_:=Parse("200601",s)
return ToUpper(t.Format("Jan06"))}

Attempt This Online!

I wish Go's time formatting strings allowed all caps months like JAN06; that would save 22 bytes.

answered Jun 21, 2023 at 14:05

bigyihsuan

10.8k1 gold badge23 silver badges66 bronze badges

\$\begingroup\$ 1 year later, I thought I hadn't answered this question but it seems I have and come up with the exact same solution 1 year later \$\endgroup\$
– bigyihsuan
Commented Jun 11 at 4:08

Add a comment |

int 21h Glory to Ukraine · Accepted Answer · 2024-06-11 07:19:13Z

R, 61 bytes

toupper(format(as.Date(paste0(scan(,""),1),"%Y%m%d"),"%b%y"))

Try it online!

R, 62 bytes

(a former version featuring an addition of 10^7 to the date)

Despite that this question has already 3 answers in R, mine is slightly golfier than the shortest one and it's the only one which does not rely on the predefined month.abb vector.

To play fair, I have put my code onto TIO, which also meant I could not use the new shortcut function definition (=\(...))

If I understand correctly, apparently, there is a bug in the string parser and therefore the as.Date cannot read directly a string in "%Y%m" format.

My shortest solution (and this is code golf) was simply to paste a day number "1" at the end of the YYYYMM date, parse and output the desired format in uppercase letters.

noodle person · Accepted Answer · 2024-06-11 22:42:55Z

1

Uiua, 54 23 bytes

⊂⊡:⌵≡◇↙3Months-1⋕⊃↘↙⟜↘2

Try it online

TIL Uiua has a Months constant, thanks chunes

edited Jun 11 at 22:42

answered Jun 11 at 16:35

noodle person

10.3k1 gold badge22 silver badges76 bronze badges

\$\begingroup\$ 25 bytes \$\endgroup\$
– chunes
Commented Jun 11 at 22:14
\$\begingroup\$ TIL about Months, also the parentheses aren't needed because of how rows works with a scalar argument \$\endgroup\$
– noodle person
Commented Jun 11 at 22:40

Add a comment |

noodle person · Accepted Answer · 2024-06-11 23:19:12Z

TypeScript's type system, 231 bytes

type F<S>=S extends`${any}${any}${infer A}${infer B}${infer C}${infer D}`?`${[0,"JAN","FEB","MAR","APR","MAY","JUN","JUL","AUG","SEP","OCT","NOV","DEC"][(C extends"0"?D:`${C}${D}`)extends`${infer N extends number}`?N:0]}${A}${B}`:S

Try it at the TS Playground

This should be 172 bytes:

type F<S>=S extends`${any}${any}${infer A}${infer B}${infer R extends number}`?`${[0,"JAN","FEB","MAR","APR","MAY","JUN","JUL","AUG","SEP","OCT","NOV","DEC"][R]}${A}${B}`:S

but annoyingly TS doesn't let numeric types start with a 0 because that could be confused with legacy JavaScript octal literals, so this version just bugs out and we spend 59 bytes getting around that.

Convert YYYYMM to MMMYY

Rules:

Examples:

70 Answers 70

Zsh + BSD/MacOS date, 40 bytes

J types/datetime, 52 bytes

J, 70 bytes

Usage

Explanation

Pyth, 39 bytes

jq, 35 characters

Factor, 82 78 bytes

PHP, 78 bytes

Swift 2.2, 149 bytes

Swift 3, 136 bytes

R, 154 150 114 112 bytes

SmileBASIC, 85 bytes

C# - 75 Bytes

K4, 65 59 bytes

Caché ObjectScript - 65 bytes

PHP, 146 Bytes.

JavaScript ES6, 61 bytes

C (tcc), 155 bytes

C (gcc), 107 104 bytes

Pyth, 42 bytes

Groovy, 69 55 51 bytes

Haskell, 85 bytes

Vyxal o, 37 bytes

Lua 5.3, 82 bytes

Scala, 85 bytes

Factor, 56 bytes

Explanation

Python, 115 bytes

Thunno 2, 32 bytes

Explanation

Go, 107 bytes

R, 61 bytes

R, 62 bytes

Uiua, 54 23 bytes

TypeScript's type system, 231 bytes

Your Answer

Sign up or log in

Post as a guest

Not the answer you're looking for? Browse other questions tagged code-golfstringdate or ask your own question.

Linked

Related

Hot Network Questions

Zsh + BSD/MacOS `date`, 40 bytes

J `types/datetime`, 52 bytes

Vyxal `o`, 37 bytes

Not the answer you're looking for? Browse other questions tagged
code-golf
string
date
or ask your own question.