12 hour time conversion

Question

Challenge

Convert and print out a time in a 12-hour format. HH:MM AM/PM

Examples

Input:

• 'Fri Jun 30 2017 21:14:20 GMT-0700 (PDT)'
• 'Fri Jun 30 2017 00:10:23 GMT-0700 (PDT)'
• 'Fri Jun 30 2017 12:10:23 GMT-0700 (PDT)'
• 'Sat Jun 31 2018 8:06:20 GMT-0700 (PDT)'
• 'Fri Jul 01 2017 01:14:20 GMT-0700 (PDT)'
• 'Sat Apr 10 2020 09:06:20 GMT-0700 (PDT)'

Ouput:

• 9:14 PM
• 12:10 AM
• 12:10 PM
• 08:06 AM
• 1:14 AM
• 09:06 AM

Fine Points

• A zero before a one digit number is okay, no zero is also allowed. The following examples are both allowed:

  • 9:06 AM

  • 09:06 AM

• All tested years will be after 999 (each year will be exactly 4 digits)

Rules

• This is code-golf so the shortest solution in bytes wins
• Standard Loopholes Apply
• You may use functions as well as programs

• Arnauld: Suggested test cases: 00:10:23 --> 12:10 AM and 12:10:23 --> 12:10 PM.
• zoecarver: there we go. Feel free to edit if you see anything else I should change. thanks for the suggestion!
• Not a tree: Related
• ETHproductions: May we assume that the time is given in the timezone the program is being run in? (e.g. 'Fri Jun 30 2017 21:14:20 GMT-0400 (EDT)' for me)
• Eric Lagergren: erm, June 31 doesn't exist. Is that accurate?

Answer 1

JavaScript (ES6), 69 bytes

d=>new Date(d).toLocaleString(0,{hour:n='numeric',minute:n,hour12:1})

f=
d=>new Date(d).toLocaleString(0,{hour:n='numeric',minute:n,hour12:1})
console.log(
 f('Fri Jun 30 2017 21:14:20 GMT-0700 (PDT)'),
 f('Fri Jun 30 2017 00:10:23 GMT-0700 (PDT)'),
 f('Fri Jun 30 2017 12:10:23 GMT-0700 (PDT)'),
 f('Sat Jun 31 2018 8:06:20 GMT-0700 (PDT)'),
 f('Fri Jul 01 2017 01:14:20 GMT-0700 (PDT)'),
 f('Sat Apr 10 2020 09:06:20 GMT-0700 (PDT)'),
)

---

JavaScript (ES6), (削除) 58 (削除ここまで) 55 bytes

Assumes you are in the United States.

d=>new Date(d).toLocaleTimeString().replace(/:.. /,' ')

f=
d=>new Date(d).toLocaleTimeString().replace(/:.. /,' ')
console.log(
 f('Fri Jun 30 2017 21:14:20 GMT-0700 (PDT)'),
 f('Fri Jun 30 2017 00:10:23 GMT-0700 (PDT)'),
 f('Fri Jun 30 2017 12:10:23 GMT-0700 (PDT)'),
 f('Sat Jun 31 2018 8:06:20 GMT-0700 (PDT)'),
 f('Fri Jul 01 2017 01:14:20 GMT-0700 (PDT)'),
 f('Sat Apr 10 2020 09:06:20 GMT-0700 (PDT)')
)

---

JavaScript (ES6), (削除) 81 (削除ここまで) 78 bytes

Answer before outputting a leading 0 in single-digit hours was made optional and test cases without leading 0s were added.

d=>([m,s]=d.slice(16).split`:`,`0${m%12||12}:${s} ${m<12?'A':'P'}M`.slice(-8))

f=
d=>([m,s]=d.slice(16).split`:`,`0${m%12||12}:${s} ${m<12?'A':'P'}M`.slice(-8))
console.log(
 f('Fri Jun 30 2017 21:14:20 GMT-0700 (PDT)'),
 f('Sat Jun 31 2018 08:06:20 GMT-0700 (PDT)'),
 f('Fri Jul 01 2017 01:14:20 GMT-0700 (PDT)'),
 f('Sat Apr 10 2020 09:06:20 GMT-0700 (PDT)')
)

• kamoroso94: I was thinking about changing ${m>12?'P':'A'}M to ${"AP"[m>12]}M, but it doesn't work without casting the index to an integer (like with |0), which makes it the same length. Nicely golfed.
• Rick Hitchcock: You can omit ,21 in the slice to save 3 bytes.
• Rick Hitchcock: Returns 12:10 AM for Fri Jun 30 2017 12:10:23 GMT-0700 (PDT). Should be PM. Changing m>12 to m>11 should fix it.
• zoecarver: to save you 4 bytes: d=>(new Date(d).toLocaleString(0,{hour:'numeric',minute:'numeric',hour12:1}))
• zoecarver: In fact, you could actually save 25 bytes with the following: d=>(new Date(d).toLocaleTimeString().replace(/:\d+/,''))

Answer 2

Python 2, 66 bytes

lambda s:`int(s[15:18])%12`+s[18:21]+' APMM'[int(s[15:18])>11::2]

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• notjagan: This fails for years that have less than 4 digits (although I'm unsure whether it has to work with those).
• zoecarver: I think that is alright.
• Fedone: @notjagan it doesn't have to, it says so in the question.
• Fedone: @totallyhuman Whoops, sorry.
• ETHproductions: Can you change int(s[15:18])>11 to s[15:18]>"11"?

Answer 3

sh + coreutils, 22 bytes

date +%I:%M\ %p -d"1ドル"

(If seconds are allowed, then date +%r -d"1ドル" suffices.)

• zoecarver: impressively few bytes! also well done

Answer 4

JavaScript (ES6), 77 bytes

Assumes that the year has 4 digits.

s=>`${([,,,h,m]=s.match(/\d./g),x=h%12||12)>9?x:'0'+x}:${m} ${'AP'[h/12|0]}M`

Test cases

let f =
s=>`${([,,,h,m]=s.match(/\d./g),x=h%12||12)>9?x:'0'+x}:${m} ${'AP'[h/12|0]}M`
console.log(f('Fri Jun 30 2017 21:14:20 GMT-0700 (PDT)')) // 09:14 PM
console.log(f('Fri Jun 30 2017 00:10:23 GMT-0700 (PDT)')) // 12:10 AM
console.log(f('Fri Jun 30 2017 12:10:23 GMT-0700 (PDT)')) // 12:10 PM
console.log(f('Sat Jun 31 2018 08:06:20 GMT-0700 (PDT)')) // 08:06 AM
console.log(f('Fri Jul 01 2017 01:14:20 GMT-0700 (PDT)')) // 01:14 AM
console.log(f('Sat Apr 10 2020 09:06:20 GMT-0700 (PDT)')) // 09:06 AM

• zoecarver: Like above, I have found a smaller way to create and answer. Feel free to use my code, I don't think that it is really fare to answer my own question. d=>(new Date(d).toLocaleTimeString().replace(/:\d+/,''))
• Arnauld: @pudility This would only work if your Locale is en-US and your timezone is GMT-0700 (PDT). For instance, none of these assumptions are true for me.

Answer 5

Japt, 15 bytes

ÐU ̄24)s8 r.3+SS

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12 bytes if we can assume that the time will be given in the computer's local time:

ÐU s8 r.3+SS

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Mathy approach, 40 bytes

tG5 r"^.."_<C?+Z+B:°TnZ)%CÄÃ+" {"AP"gT}M

Test it online!

Answer 6

V, 36 bytes

16x3wC AMÇ^0ü^1[0-2]/12WrP
ç^ä:/é0

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Hexdump:

00000000: 3136 7833 7743 2041 4d1b c75e 30fc 5e31 16x3wC AM..^0.^1
00000010: 5b30 2d32 5d2f 3132 1857 7250 0ae7 5ee4 [0-2]/12.WrP..^.
00000020: 3a2f e930 :/.0

Answer 7

PHP, 45 bytes

Answer improved thanks to manatwork

<?=(new DateTime($argv[1]))->format('h:i A');

First attempt:

<? $d=new DateTime($argv[1]);echo$d->format('h:i A');

Example usage through php CLI:

php d.php "Sat Apr 10 2020 09:06:20 GMT-0700 (PDT)"

This is my first golf try.

• manatwork: No need for variable $d and that way you can get rid of the explicit echo: <?=(new DateTime($argv[1]))->format('h:i A');.

Answer 8

Jelly, 43 bytes

Ḳ5ịṣ":Ṗṁ3μV’%12‘Dμ1¦μV>11ị)PAμ3¦"0: M"żFṫ-7

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This is superfluously too long! That is, Jelly sucks at time manipulation.

EDIT: I'm even outgolfed by PHP!

Answer 9

Go, 103 bytes

func f(s*string){t,_:=time.Parse("Mon Jan 02 2006 15:04:05 MST-0700 (MST)",*s)
*s=t.Format("03:04 PM")}

Test here: https://play.golang.org/p/P1zRWGske-

Answer 10

05AB1E, 39 bytes

#4è':¡ ̈`sD11›„APès<12%>0ìR2£R)Á... :M‚øJJ

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Answer 11

PHP, 42 bytes

<?=date_create($argv[1])->format('h:i A');

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Answer 12

C#, 145 bytes

namespace System.Linq{s=>{var d=DateTime.Parse(string.Join(" ",s.Split(' ').Skip(1).Take(4)));return d.ToString("h:mm ")+(d.Hour>11?"PM":"AM");}}

Full/Formatted version:

namespace System.Linq
{
 class P
 {
 static void Main()
 {
 Func<string, string> f = s =>
 {
 var d = DateTime.Parse(string.Join(" ", s.Split(' ').Skip(1).Take(4)));
 return d.ToString("h:mm ") + (d.Hour > 11 ? "PM" : "AM");
 };
 Console.WriteLine(f("Fri Jun 30 2017 21:14:20 GMT-0700 (PDT)"));
 Console.WriteLine(f("Fri Jun 30 2017 00:10:23 GMT-0700 (PDT)"));
 Console.WriteLine(f("Fri Jun 30 2017 12:10:23 GMT-0700 (PDT)"));
 Console.WriteLine(f("Fri Jul 01 2017 01:14:20 GMT-0700 (PDT)"));
 Console.WriteLine(f("Sat Apr 10 2020 09:06:20 GMT-0700 (PDT)"));
 Console.ReadLine();
 }
 }
}

Answer 13

,,,, 41 bytes

::18⊢3⊣⇆15⊢3⊣i11>" APMM"⇆⊢2⟛↔15⊢3⊣i12%s#

Explanation

WIP

Answer 14

MATL, 9 bytes

5:24)16XO

Try it at MATL online! Or verify all test cases.

Explanation

5:24 % Push array [5 6 ... 24]
) % Implicit input. Get characters at those positions. This
 % removes the first four characters with the day of the week
16 % Push 16
XO % Convert to date string format 16, which is 'HH:MM PM'
 % Implicitly display

• totallyhuman: Can we have an explanation?
• Luis Mendo: @totallyhuman Sure, thanks for reminding me. Edited. Basically the builtin XO does most of the work

Answer 15

AWK, 66 bytes

split(5,ドルa,":")&&0ドル=((b=a[1]%12)?b:12)":"a[2](a[1]>11?" PM":" AM")

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