topic Re: Three-Digit Day of Year in Operating System - Linux

Three-Digit Day of Year

Michael Leis — Fri, 02 Sep 2005 11:50:43 GMT

I need help with a unix shell script (prefer K-shell) that will convert a given date in the format yyyymmdd into yyddd. (Input format is negotiable, output format needs to be yyddd or yy seperated from ddd.) For example, I put in 20050831 it would output 05243 because August 31 was the 243 day of 2005. I'm familiar with date '+%y%j' that provides that for todays date, but I need to be able to declare any reasonable given date in the past or future. I don't want the true julian date (number of days since 4713BC), just the number of days a given date is into its year. Any help is appreciated.

Re: Three-Digit Day of Year

Pete Randall — Fri, 02 Sep 2005 11:56:41 GMT

Sounds like a job for Clay's caljd.sh (attached). Invoke as "caljd.sh -u" for usage details.

Pete

Re: Three-Digit Day of Year

Pete Randall — Fri, 02 Sep 2005 11:59:30 GMT

Michael,

I apologise - I assumed that Clay's script would handle such a thing but, going through the usage notes myself, I don't see an option for 3 digit output.

Pete

Re: Three-Digit Day of Year

Pete Randall — Fri, 02 Sep 2005 12:05:46 GMT

Of course you could use caljd to determine the julian day of January 1 and subtract that from the current julian day. That ought to work.

Pete

Re: Three-Digit Day of Year

Pat Lieberg — Fri, 02 Sep 2005 12:38:49 GMT

I whipped this up quick. Its a script that asks for the month and day and then tdells you the day of the year it is. Note that it does not calculate leap year, so Feb was hard-coded to 28 days. That could be programmed in though.

#!/bin/ksh
#
print - n "Enter the month and day: "
read month day

set -A days 31 28 31 30 31 30 31 31 30 31 30 31
((n = 0))
((j_days = 0))

((month = month - 1))

while ((n < $month))
do
((j_days = j_days + ${days[$n]}))
((n = n + 1))
done

((j_days = j_days + $day))
echo $j_days

Re: Three-Digit Day of Year

Pat Lieberg — Fri, 02 Sep 2005 12:45:57 GMT

I forgot your desired result was inthe format dddyy, but you can add this to the end of the script:

year=$(date +%y)
result="$j_days$year"
echo $result

Re: Three-Digit Day of Year

Hein van den Heuvel — Fri, 02 Sep 2005 12:50:45 GMT

In Perl:

---- julian.p ----
use Time::Local;
($y,$m,$d) = unpack "a4 a2 a2", shift @ARGV;
$s1 = timelocal(0,0,1,$d,$m-1,$y-1900);
$s2 = timelocal(0,0,0,1,0,$y-1900);
$u = 1 + ($s1 - $s2)/86400;
printf "%02d%03d\n", $y-2000, $u;

I used the $U = 1 + ... because jan-1 is day 1.
I used the hour 1 in $s1 to deal with integer truncation.

Hein.

Re: Three-Digit Day of Year

James R. Ferguson — Fri, 02 Sep 2005 12:58:48 GMT

Hi Michael:

A small perl script would easily help:

You an embed this in your shell and pass the Month Day Year as agruments. I've let it output a three-digit year and a three-digit day of year. THus 1999 outputs as 099 while 2005 appears as 105.

perl -e 'use Time::Local;$time=timelocal(0,0,0,$ARGV[1],$ARGV[0]-1,$ARGV[2]);($y,$d)=(localtime($time))[5,7];$d++;printf "%03d%03d\n"' 8 31 2005

...returns 105243

Obviously, you can use this in your shell like this:

RESULT=`perl -e `...`` $M $D $Y

Regards!

...JRF...

Re: Three-Digit Day of Year

Hein van den Heuvel — Fri, 02 Sep 2005 13:17:22 GMT

Ah yes, I forgot about the $yday = 8th field in the localtime result. Using that my script could change to:

---- cat julian.p -----
use Time::Local;
($y,$m,$d) = unpack "a4 a2 a2", shift @ARGV;
$u = 1 + (localtime(timelocal(0,0,1,$d,$m-1,$y-1900)))[7];
printf "%02d%03d\n", $y-2000, $u;

There is a minor typo / cut&paste error in the JRF solution.
Try this:

# perl -e 'use Time::Local;$time=timelocal(0,0,0,$ARGV[1],$ARGV[0]-1,$ARGV[2]);($y,$d)=(localtime($time))[5,7];$d++;printf "%03d%03d\n", $y,$d' 8 31 2005

105243

Re: Three-Digit Day of Year

James R. Ferguson — Fri, 02 Sep 2005 13:25:21 GMT

Hi (again):

Yes, Hein, thank you. The paste I did cut off part of the script:

#!/usr/bin/perl
use Time::Local;
$time=timelocal(0,0,0,$ARGV[1],$ARGV[0]-1,$ARGV[2]);
($y,$d) = (localtime($time)) [5,7];
$d++;
printf "%03d%03d\n",$y,$d;

I like your use of 'unpack'. It would allow Michael to pass one argument to the perl script instead of three if that suits his needs better. Thanks for the correction to mine!

Regards!

...JRF...

Re: Three-Digit Day of Year

Michael Leis — Fri, 02 Sep 2005 14:50:32 GMT

Thank you for your help -- sometimes the simple things are the hardest to see. I ended up using Pat's sample script as it worked well for my purposes, however I can see where the PERL code would work well also. Thank you everyone!

Re: Three-Digit Day of Year

Michael Leis — Fri, 02 Sep 2005 14:51:23 GMT