Convert YYYYMM to MMMYY

Question

Basing on this SO 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

Answer 1

MATL, 18 14 13 bytes

4e!Z{Zc12XOXk

Input is provided as a string (enclosed in single quotes).

This version only runs in MATL on MATLAB since MATLAB is able to automatically parse datestr('2016 04').

Explanation

        % Implicitly grab input as a string
4e!     % Reshape input to be 2 x 4 (puts the year in row 1 and month in row 2)
Z{      % Place each row in a separate cell
Zc      % Join them together using a space to create 'yyyy mm' format
12      % Number literal, pre-defined datestring of 'mmmyy'
XO      % Convert from serial date number to string using this format
Xk      % Convert to uppercase
        % Implicitly display

---

Here is an 18 byte version which works on Octave (and therefore the online interpreter)

'yyyymm'2$YO12XOXk

Try it Online

Modified version for all test cases

Explanation

            % Implicitly grab input as a string
'yyyymm'    % Push the format string as a string literal
2$YO        % Convert to a serial date number
12          % Number literal, pre-defined datestring of 'mmmyy'
XO          % Convert from serial date number to string using this format
Xk          % Convert to uppercase
            % Implicitly display

Answer 2

Python 3, 70 bytes

from time import*
lambda s:strftime("%b%y",strptime(s,"%Y%m")).upper()

This uses the built-in strftime and strptime functions.

For 1 byte more, here's a version which parses the string manually:

lambda s:" JFMAMJJASONDAEAPAUUUECOENBRRYNLGPTVC"[int(s[4:])::12]+s[2:4]

This encodes the month names in an interesting way (thanks to Henry Gomersall for saving a byte).

Answer 3

PowerShell, 49 46 40 bytes

date "$args".insert(4,'-')-U %b%y|% *per

Try it online!

Thanks to @Joey for saving 3 bytes!

Takes input $args as an explicit string (e.g., '201604') via command-line input. Uses the string.Insert() function to put a - in the appropriate space, and that resultant string forms input to the Get-Date cmdlet with the -Uformat parameter specifying the three-month shorthand plus two-digit year. We then tack on a .ToUpper() to make the output string capitalized. That string is left on the pipeline and printing is implicit.

Also, as pointed out, this is locale-sensitive. Here's the locale information that I'm using where this works correctly.

PS C:\Tools\Scripts\golfing> get-culture

LCID             Name             DisplayName
----             ----             -----------
1033             en-US            English (United States)

Answer 4

Bash + coreutils, 18

Requires 64-bit version of date for the given testcases, which recognises dates earlier than 14th December 1901.

date -d$101 +%^b%y

Answer 5

Retina, 71 70 bytes

Thanks to Sp3000 for saving 1 byte.

Byte count assumes ISO 8859-1 encoding. The trailing linefeed is significant.

(..)(..)$
DECNOVOCTSEPAUGJULJUNMAYAPRMARFEBJANXXX$2$*¶$1
+`...¶

R-6`.

Try it online!

Explanation

Taking 201604 as an example:

(..)(..)$
DECNOVOCTSEPAUGJULJUNMAYAPRMARFEBJANXXX$2$*¶$1

This swaps the last two digits of the year with the month while also expanding the month in unary using linefeeds, and prepending the list of months in reverse so we'd get:

20DECNOVOCTSEPAUGJULJUNMAYAPRMARFEBJANXXX¶¶¶¶16

Where the ¶ represent linefeeds (0x0A).

+`...¶

Now we repeatedly remove three non-linefeed characters followed by a linefeed. That is we eat up the list of months from the end for each linefeed representing a month:

20DECNOVOCTSEPAUGJULJUNMAYAPRMARFEBJANXXX¶¶¶¶16
20DECNOVOCTSEPAUGJULJUNMAYAPRMARFEBJAN¶¶¶16
20DECNOVOCTSEPAUGJULJUNMAYAPRMARFEB¶¶16
20DECNOVOCTSEPAUGJULJUNMAYAPRMAR¶16
20DECNOVOCTSEPAUGJULJUNMAYAPR16

This is why we've inserted that XXX: since the months start counting from 1, we'll always remove at least three characters, even for January.

R-6`.

Finally, we remove everything up to the 6th character from the end. In other words we only keep the last five characters.

Answer 6

CJam, 50 46 bytes

q2/1>~i("4H~0ë~³!ò²×¶7Ö"256b25b'Af+3/=\

Try it online. Thanks to Martin Ender for compressing the string to save a few bytes.

Explanation

q2/  e# Get input and divide it into groups of 2, like ["20" "16" "04"]
1>~  e# Discard the first item and dump the remaining array to the stack
i(   e# Convert the top value (month) to an integer and decrement it, because
     e# arrays are zero-indexed
"..."256b25b e# Convert this string from base-256 to base-25
'Af+ e# "Add" a capital A to each number to get the letters
3/   e# Divide into groups of 3 to make an array of month names
=\   e# Get the requested month and swap the elements to put the year on
     e# top, so it is printed last

Answer 7

Bash, 39 28 bytes

date -d$101 +%b%y|tr a-z A-Z

Thanks Digital Trauma!

Answer 8

Java 7, 137 characters (161 bytes)

String d(String i){return Integer.toString("憯䷳烣㘿烪摿摽㛨近筍矯䏔".charAt(Integer.parseInt(i.substring(4))-1),36).toUpperCase()+i.substring(2,4);}

Consider each month name (JAN, FEB etc...) is a number in base 36 and encode it into corresponding Unicode symbol. Then get corresponding symbol from the string encode it back again in base 36 and after that some plain string manipulations.

Slightly ungolfed:

String d(String input){
return 
  Integer.toString("憯䷳烣㘿烪摿摽㛨近筍矯䏔" // encoded month names
  .charAt(Integer.parseInt(input.substring(4))-1),36) // get a symbol from encoded names at position input[4:], decode it to base 36 value
  .toUpperCase()+input.substring(2,4); // get it to upper case and add year
}

You can see it running here: https://ideone.com/IKlnPY

Answer 9

05AB1E, 51 42 41 bytes

2ô¦`ï<•r–ºþ¯Bê€õaPù£—^5AºüLwÇ–è•35B3ôsèsJ

Explanation

                                           # implicit input, 123405
2ô                                         # split input into pieces of 2, ['12','34','05']
  ¦`                                       # push last 2 elements to stack, '05', '34'
    ï<                                     # convert month to its int index, 4
      •r–ºþ¯Bê€õaPù£—^5AºüLwÇ–è•35B        # get compressed string containing 3-letter months, 
                                             JANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC
                                   3ô      # split into pieces of 3
                                             ['JAN', 'FEB', 'MAR', 'APR', 'MAY', 'JUN', 'JUL', 'AUG', 'SEP', 'OCT', 'NOV', 'DEC']
                                     sè    # get month at index retrieved earlier, MAY
                                       sJ  # join with 2-digit year and implicitly print, MAY34

Try it online

9 bytes saved thanks to string compression, courtesy of @Adnan

Answer 10

JavaScript, 87 84 80 79 bytes

x=>(new Date(x.replace(/.{4}/,'$&-'))+'').slice(4,7).toUpperCase()+x.slice(2,4)

To get the month, gets the date (which is formed of "YYYYMM" converted to "YYYY-MM") and retrieves the characters 5 to 8, that are exactly the first three letters of the month. But it costs much to convert it to upper case.

Demo:

function s(x) {
  return (new Date(x.replace(/.{4}/, '$&-')) + '').slice(4,7)
         .toUpperCase() + x.slice(2, 4)
}

console.log(s('201604'));
console.log(s('200001'));
console.log(s('000112'));
console.log(s('123405'));

Answer 11

Python, 83 bytes

from datetime import*
lambda x:datetime.strptime(x,'%Y%m').strftime('%b%y').upper()

Answer 12

Kotlin, 100 bytes

fun f(d:String)=SimpleDateFormat("MMMyy").format(SimpleDateFormat("yyyyMM").parse(d)).toUpperCase()

Pretty straight forward use of Java SimpleDateFormat

Answer 13

MATLAB / Octave, 42 bytes

@(x)upper(datestr(datenum(x,'yyyymm'),12))

Creates an anonymous function named ans that is called with a string representing the date: ans('201604').

Online Demo

This solution uses datenum to convert the input date to a serial date number, and then datestr with the predefined output spec of mmmyy (12) to yield the string representation in the required format. Finally, we use upper to change it to MMMYY since the uppercase month is not an output option.

Answer 14

JavaScript ES6, 77 66 bytes

Saved 11 bytes thanks to @Bálint!

a=>(new Date(0,a[4]+a[5]-1)+"").slice(4,7).toUpperCase()+a[2]+a[3]

Get's the date by extracting the string returned by the Date class. then capitalizes and adds the year.

ES5 version:

var a = prompt("Enter YYYYMM: ");
result = (new Date(0,a[4]+a[5]-1)+"").slice(4,7).toUpperCase()+a[2]+a[3]
alert(result);

Answer 15

Julia, 57 56 53 bytes

s->uppercase(Dates.format(DateTime(s,"yyyym"),"uyy"))

This is an anonymous function that accepts a string and returns a string. To call it, assign it to a variable.

First we construct a DateTime object using the type constructor and a format string. Note that the single m in the format string will get both one- and two-digit months, though the former case is irrelevant here. Since no days are specified, the first of the month is assumed.

We can then format the value as a string using the Dates.format function from the Base.Dates submodule. The string uyy gets the three-letter month name and two-digit year, but the result is in title case, e.g. Apr16 instead of the desired APR16, so we need to uppercase it.

Try it online! (includes all test cases)

Answer 16

C, 147 145 112 bytes

main(m){char a[99]="JANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC";scanf("%4s%d",a+50,&m);printf("%.3s%s",a+--m*3,a+52);}

Online demo

Thanks ugoren!

Answer 17

Japt, 35 34 bytes

ÐUr$/..../$,"$&-")+P s4,7 u +Us2,4

Link.

Uses the same technique as my JavaScript answer.

Answer 18

Java 8, 154 113 bytes

import java.text.*;s->new SimpleDateFormat("MMMyy").format(new SimpleDateFormat("yyyyMM").parse(s)).toUpperCase()

Explanation:

Try it online.

import java.text.*;                 // Required import for SimpleDateFormat
s->                                 // Method with String as both parameter and return-type
  new SimpleDateFormat("MMMyy")     //  Create a formatter with format "MMMyy"
   .format(                         //  Format the following:
     new SimpleDateFormat("yyyyMM") //   Create another formatter with format "yyyyMM"
     .parse(s))                     //   Parse the input with this format
  .toUpperCase()                    //  Convert everything to Uppercase and return

Answer 19

C#, 94 87 bytes

string C(string s)=>System.DateTime.Parse(s.Insert(4,"-")).ToString("MMMyy").ToUpper();

Saved 7 bytes by using C#6 Syntax.

Try Online

Answer 20

Pyth, 45 bytes

+:."AYw2ûDÈëKH§È¼DYÉx\E±oË"J*3sgz5+3J:z2 4

Try it online!

Explanation:

+:."AYw2ûDÈëKH§È¼DYÉx\E±oË"J*3sgz5+3J:z2 4
                                z           Take the input
                               g 5          Slice inclusively from index 5 to the end
                              s             Parse as an int
                            *3              Multiply by 3
                           J                Store in variable J, this also returns J
 :                                          Take a slice
  ."AYw2ûDÈëKH§È¼DYÉx\E±oË"                 Of this packed string
                           J*3sgz5          From the J we defined before
                                  +3J       To J+3
+                                           To this string, append
                                     :z     A slice of the index
                                       2 4  From [2,4).

The packed string contains "JANJANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC". The two JANs are so that I can index it pseudo-one-indexed.

EDIT: Fixed bug that was messing with TIO

Answer 21

R, 65 bytes

function(A)paste0(toupper(month.abb[el(A:1)%%100]),substr(A,3,4))

Try it online!

Takes input as a string, leverages the constant month.abb. Uses modulus and substr to extract relevant values.

Answer 22

Oracle SQL, 49 Bytes

select to_char(to_date(n,'yyyymm'),'MONyy')from t

The data must be inserted in a table called T with a column N of type VARCHAR2(6), CHAR(6) or , only if all the years are > 1000, NUMBER

Usage:

drop table t;
create table t (n VARCHAR2(6));
insert into t values ('201604');
insert into t values ('200001');
insert into t values ('000112');
insert into t values ('123405');    

select to_char(to_date(n,'yyyymm'),'MONyy')from t;

Answer 23

Microsoft SQL Server, 57 Bytes

SELECT UPPER(FORMAT(CAST('201601'+'01' AS DATE),'MMMyy'))

The Upper function is required as format does not produce Upper case months as would be expected with the MMM format pattern.

Usage:

drop table t;
create table t (n VARCHAR(6));
insert into t values ('201604');
insert into t values ('200001');
insert into t values ('000112');
insert into t values ('123405');    

SELECT UPPER(FORMAT(CAST(n+'01' AS DATE),'MMMyy')) FROM t

Answer 24

Microsoft Excel VBA, 48 Bytes

Anonymous VBE immediate window function that takes input from range [A1] and outputs to the VBE immediate window

?UCase(MonthName([Right(A1,2)],1))+[Mid(A1,3,2)]

Current version thanks to input from @Engineer Toast, for pointing out the abbreviate option of the MonthName function.

Previous Version

?UCase(Format(DateSerial([Left(A1,4)],[Right(A1,2)],1),"MMMYY"))

Answer 25

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.

Answer 26

C# - 75 Bytes

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

Answer 27

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!

Answer 28

MediaWiki Template (with ParserFunctions), 29 bytes

{{uc:{{#time:My|{{{1}}}01}}}}

Only works if English locale is chosen.

Try Sandbox

Answer 29

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 :)

Answer 30

Perl 5 -p, 69 bytes

$_=(1,JAN,FEB,MAR,APR,MAY,JUN,JUL,AUG,SEP,OCT,NOV,DEC)[/..$/g].$`%100

Try it online!

Answer 31

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

Answer 32

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.

Answer 33

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.)

• 200001
• 000112

Answer 34

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).

Answer 35

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.

Answer 36

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

Answer 37

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")

Answer 38

dc, 103 bytes

[0n]sP[JAN][FEB][MAR][APR][MAY][JUN][JUL][AUG][SEP][OCT][NOV][DEC]Csi[lid1-si:mz1<M]dsMxA0~;mPA0~dZ2>Pp

Try it online!

[0n]sP is a macro that just prints a 0 w/o a newline for the sake of padding 1-9.

[JAN]...[DEC] places the strings on the stack. Csi[lid1-si:mz1<M]dsMx is a macro that decrements i from 12, popping the strings off of the stack and placing them at index i in the array m. Stops when one item (input) is left on the stack.

100~;mP does quotient/remainder division by 100, and prints the value from m indexed by the remainder. The quotient is left on the stack. 100~dZ2>Pp again gets the remainder after dividing by 100, runs the padding macro P this is one digit long, and then prints.

Without any way of manipulating strings, I don't think that part is golfable. Doing any sort of wizardry with JAN/JUN/JUL would likely take far more bytes than the strings themselves.

Answer 39

Tcl, 98 bytes

proc C d {scan $d %2d%2d%2d d y m
format %s%02d [string tou [clock f [clock sc $m/1/0] -f %b]] $y}

Try it online!

---

Tcl, 115 bytes

proc C d {scan $d %2d%2d%2d d y m
format %s%02d [lindex {. JAN FEV MAR APR MAY JUN JUL AUG SEP OCT NOV DEC} $m] $y}

Try it online!

---

Tcl, 144 bytes

proc C d {scan $d %2d%2d%2d d y m
format %s%02d [string map {10 OCT 11 NOV 12 DEC 1 JAN 2 FEV 3 MAR 4 APR 5 MAY 6 JUN 7 JUL 8 AUG 9 SEP} $m] $y}

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Answer 40

Python, 75 bytes

Here is my obvious Python lambda, takes and returns a string, hope I didn't miss some obvious hole:

lambda x:"JANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC"[3*int(x[4:])-3:][:3]+x[2:4]

Answer 41

PHP, 55 bytes

<?=date(My,strtotime(wordwrap($argn,4,"-",1)))&"___??";

Run as pipe with -n or try it online. Requires PHP 5 or later.

Answer 42

Base R, 78 bytes

paste(casefold(month.abb,upper=T)[strtoi(substr(A,5,6))],substr(A,3,4),sep="")

Not sure if this is required to be formatted as a standalone function, but if so, it is 91 bytes:

function(A){paste(casefold(month.abb,upper=T)[strtoi(substr(A,5,6))],substr(A,3,4),sep="")}

Answer 43

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
}

Answer 44

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]

Answer 45

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);}

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Answer 46

C (gcc), 107 104 bytes

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

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Answer 47

Batch, 148 bytes

@set d=%1
@for %%m in (JAN.01 FEB.02 MAR.03 APR.04 MAY.05 JUN.06 JUL.07 AUG.08 SEP.09 OCT.10 NOV.11 DEC.12)do @if %%~xm==.%d:~4% echo %%~nm%d:~2,2%

No date libraries to speak of so manual parsing ftw. I don't know whether the rules of Code Golf allow creating 12 appropriately-named files in the current directory, resulting in something like this:

@set d=%1
@for %%m in (*.%d:~4%)do @echo %%~nm%d:~2,2%

Answer 48

Ruby 46 + 6 for -rdate = 52

For completeness, if nothing else.

->d{Date.strptime(d,'%Y%m').strftime('%^b%y')}

See the tests on repl.it: https://repl.it/Cj7z

Answer 49

T-SQL, 96 bytes

declare @i varchar(8)
select  @i = '201604'
select upper(left(datename(month,@i+'01'),3)) + substring(@i,5,2)

Answer 50

Java, 142 bytes

s->(new String[]{"JAN","FEB","MAR","APR","MAY","JUN","JUL","AUG","SEP","OCT","NOV","DEC"})[Integer.decode(s.substring(4))-1]+s.substring(2,4);