Fractional part of epoch - perl

I have an epoch value of 1372252864.901871 which needs to be converted to localtime in perl. I know that localtime($val) where $val is set to 1372252864.901871 converts the value to local time. But I also need to add values of 100 msecs or say 86.7 msecs to this and determine the computed local time. How do I do this? Does the function localtime() take care of fractional epoch value that I have given?

If you just want to add to an epoch time and get another epoch time, just do:
1372252864.901871 + .0867
If you are doing something else, no, localtime does not preserve fractional seconds. You probably want to use DateTime to deal with fractional seconds:
use DateTime;
my $time = DateTime->from_epoch(epoch=>1372252864.901871, time_zone=>'local');
$time->add(nanoseconds=>86.7e6);
print $time->epoch + $time->nanosecond / 1e9;
(time_zone isn't needed for this example, but sounds like what you might want for other things you might be doing with this time)

You might want Time::HiRes:
The Time::HiRes module implements a Perl interface to the usleep,
nanosleep, ualarm, gettimeofday, and setitimer/getitimer system calls,
in other words, high resolution time and timers

The localtime function breaks down a raw time like 1372252864 into year, month, day, etc. Any fractional part of the raw time is quietly ignored.
I'm not aware of any function, either built-in or in a module, that supports broken-down times with a resolution finer than 1 second.
But since you already have the fractional seconds, I'd say you can already do what you need to do. Just use localtime (or gmtime) to break down the integer part of your raw timestamp, and then take the fractional part:
$timestamp = 1372252864.901871;
#parts = localtime($timestamp);
$fraction = $timestamp - int($timestamp);
$milliseconds = int($fraction * 1000.0);
It should be easy enough to write your own high-resolution replacement for localtime that does this for you.
If you need to add 86.7 milliseconds to a timestamp, you'll need to use the raw form anyway:
$timestamp = 1372252864.901871;
$timestamp += 86.7/1000.0;

The simplest way that occurs to me is to use Time::HiRes::Value to handle arithmetic on times to 1μs accuracy, and Time::Piece to format the result as a date/time string.
Time::HiRes::Value holds the time as a two-element array; the first field has the whole seconds while the second has the number of microseconds. That's the same format that Time::HiRes functions work with, but there's nothing that I know of that will natively convert to a date/time string so I've used Time::Piece to convert the whole seconds and just appended the fractional seconds.
It looks like this
use strict;
use warnings;
use Time::HiRes::Value;
use Time::Piece;
my $dt = Time::HiRes::Value->new(1372252864.901871);
my $delta = Time::HiRes::Value->new(86.7E-3); # 86.7ms
$dt += $delta;
print dt_to_string($dt), "\n";
sub dt_to_string {
my ($dt) = #_;
Time::Piece->new($dt->[0])->datetime =~ s{(\d+)\z}{$1 + $dt->[1] / 1E6}er;
}
output
2013-06-26T14:21:4.98857

Related

Subtract two date strings in Perl with conversion to unix time and reverting back

I want to subtract two timestamps in Perl. I converted them to unix-time via the function below and convert the unix timestamp back to how it was. In the example below the result is 01:20:00 instead of 00:20:00
(I think it has sth to do with the start of the unix timestamp 1.1.1970 01:00:00 but not sure how to resolve it)
Any idea? Many thanks for your help in advance.
use POSIX qw( strftime );
use Time::Local qw( timelocal );
sub to_epoch {
$_ = shift;
my #a = split /\W+/, $_;
my $b = timelocal($a[5],$a[4],$a[3],$a[2],$a[1],$a[0]);
return $b;
}
my $h_end = "2018.11.12 00:50:00";
my $h_start = "2018.11.12 00:30:00";
my $duration = to_epoch($h_end) - to_epoch($h_start);
my $convert_back = POSIX::strftime("%H:%M:%S", localtime($duration));
print $convert_back , "\n";
Ouptut: 01:20:00
It works for me. But I think that's because I'm in GMT and you're in CET (GMT+1).
The flaw is in your final step. You are confusing two concepts - a point in time and a duration.
You correctly convert your two points in time to Unix epoch numbers and then you subtract those numbers to get the number of seconds between them. That number is a duration. And you want to convert that duration into a human-readable format. Using localtime() and POSIX::strtime() is not the way to do that. POSIX::strftime() and localtime() deal with points in time, not durations.
The number you get is 1,200. By passing that to localtime() you are saying "what is the epoch number 1,200 when converted to a date and time in my local timezone?" 1,200 is 20 past midnight on Jan 1st 1970 GMT. But in your local, Frankfurt, timezone, it's 20 past 1am. Which is why you're getting 1:20 and I'm getting 0:20.
There are a couple of ways to fix this. You can do the conversion manually.
my $duration = 1_200;
my $mins = int($duration/60);
my $secs = $duration % 60;
Or you can use a proper date/time handling module like DateTime (along with its associated module DateTime::Duration).
It might work if you use timegm() and gmtime() in place of timelocal() and localtime() - but I really don't recommend this approach as it perpetuates the confusion between points in time and durations.
Update: A version using DateTime.
#/usr/bin/perl
use strict;
use warnings;
use DateTime;
use DateTime::Format::Strptime;
my $h_end = '2018.11.12 00:50:00';
my $h_start = '2018.11.12 00:30:00';
my $date_p = DateTime::Format::Strptime->new(
pattern => '%Y.%m.%d %H:%M:%S'
);
my $duration = $date_p->parse_datetime($h_end)
- $date_p->parse_datetime($h_start);
printf '%02d:%02d:%02d', $duration->in_units('hours', 'minutes', 'seconds');
1200, the value of $duration, signifies the following when treated as a epoch timestamp
1970-01-01T01:20:00+01:00
^^^^^^^^
The solution is to replace
strftime("%H:%M:%S", localtime($duration));
with
strftime("%H:%M:%S", gmtime($duration));
This gives
1970-01-01T00:20:00Z
^^^^^^^^
Of course, this is still a hack. You're not suppose to be passing a duration to gmtime. Use an appropriate module instead.
use DateTime::Format::Strptime qw( );
my $format = DateTime::Format::Strptime->new(
pattern => '%Y.%m.%d %H:%M:%S',
on_error => 'croak',
);
my $h_end = $format->parse_datetime('2018.11.12 00:50:00');
my $h_start = $format->parse_datetime('2018.11.12 00:30:00');
my $dur = $h_end - $h_start;
printf "%02d:%02d:%02d\n", $dur->in_units(qw( hours minutes seconds ));
By the way,
timelocal($a[5],$a[4],$a[3],$a[2],$a[1],$a[0])
should be
timelocal($a[5],$a[4],$a[3],$a[2],$a[1]-1,$a[0])

Adding MM::SS times, dividing them and printing the average

Let's say I just ran for 10 km and I now have 10 1km split times in the form of MM::SS. I want a simple way to add up an arbitrary number of the split times and average them. For instance, maybe I want to see how much faster (or slower) the last 5 km were when compared with the first 5 km.
I can do this myself by parsing the times, dividing them into seconds and then converting them back to MM::SS. The math isn't hard, but I was wondering if something on CPAN already does this in a simple way. My first attempt was using DateTime, but it doesn't convert from seconds to minutes, because of leap seconds.
To be clear, I don't care about leap seconds in this context, but I'm curious as to whether a library exists. As an example, here is what I have tried.
#!/usr/bin/env perl
use strict;
use warnings;
use feature qw( say );
use DateTime::Format::Duration;
my $formatter = DateTime::Format::Duration->new( pattern => '%M:%S' );
my #splits = #ARGV;
my $split_number = #splits;
my $total = $formatter->parse_duration( shift #splits );
foreach my $split (#splits) {
$total->add_duration(
$formatter->parse_duration($split) );
}
say 'Total time: ' . join ':', $total->minutes, $total->seconds;
$total->multiply( 1 / $split_number );
say 'Average time: ' . join ':', $total->minutes, $total->seconds;
say 'Using DateTime::Format::Duration: ' . $formatter->format_duration( $total );
And a sample run:
$ perl script/add-splits.pl 1:30 2:30
Total time: 3:60
Average time: 1.5:30
Using DateTime::Format::Duration: 01:30
So, you can see that the duration object itself, gives me a correct answer, but not something that a human wants to decipher.
DateTime::Format::Duration tries to be helpful, but tosses out 30 seconds in the process.
I'm not looking for the raw code to do this. I am interested in whether this already exists on CPAN.
The problem with finding this exact functionality on CPAN is that you want to manipulate strings that are time intervals. Most modules are concerned with the context of such strings, working with them as date and time. So it's hard to find something that simply adds mm:ss format. Since this quest is rather specific, and simple to write, why not wrap it in your own package?
Having said that, see whether the snippet below fits what you are looking for.
This is a simple solution with the core module Time::Piece. It does go to seconds to do the math, but it can be wrapped in a few subs that are then also easily extended for other calculations.
use warnings 'all';
use strict;
use feature 'say';
use Time::Piece;
use POSIX 'strftime';
use List::Util 'sum';
my #times = #ARGV;
my $fmt = '%M:%S';
my $tot_sec = sum map { Time::Piece->strptime($_, $fmt)->epoch } #times;
my $ave_sec = sprintf("%.0f", $tot_sec/#times); # round the average
my ($tot, $ave) = map { strftime $fmt, gmtime $_ } ($tot_sec, $ave_sec);
say "Total time: $tot";
say "Average time: $ave";
For manip_times.pl 2:30 1:30 this prints
Total time: 04:00
Average time: 02:00
We use strptime from Time::Piece to get the object, and then its epoch method returns seconds, which are added and averaged. This is converted back to mm:ss using strftime from POSIX. The Time::Piece also has strftime but to use it we'd have to have an object.
Note that Time::Piece does subtract its objects directly, $t1 - $t2, but it cannot add them. One can add an integer (seconds) to an object though, $t1 + 120. Also see the core Time::Seconds, a part of the distribution.
Comments on the method used in the question
The DateTime::Duration objects that are used cannot convert between different units
See the How DateTime Math Works section of the DateTime.pm documentation for more details. The short course: One cannot in general convert between seconds, minutes, days, and months, so this class will never do so.
From a bit further down the page, we see what conversions can be done and the related ones are only "hours <=> minutes" and "seconds <=> nanoseconds". The reasons have to do with leap seconds, DST, and such. Thus the calculation has to produce results such as 1.5 minutes.
The Class::Date also looks suitable for these particular requirements.
Here is a version using the DateTime object. Since it doesn't handle date parsing, we have to split the strings ourselves...
add_splits.pl
#!/usr/bin/env perl
use strict;
use warnings;
use feature qw( say );
use DateTime;
my #splits = #ARGV;
my $split_number = #splits;
my $fmt = "%M:%S";
my $total = DateTime->from_epoch( epoch => 0 );
foreach my $split (#splits) {
my ($split_min, $split_sec) = split /:/, $split;
$total->add( minutes => $split_min, seconds => $split_sec );
}
say 'Total time: ' . $total->strftime($fmt);
my $avg_seconds = $total->epoch / $split_number;
my $avg = DateTime->from_epoch( epoch => 0 )->add( seconds => $avg_seconds );
say 'Average time: ' . $avg->strftime($fmt);
Output
$ perl add_splits.pl 1:30 2:30
Total time: 04:00
Average time: 02:00

Create time variable in perl

I want to create a variable that will store hardcoded hours and minutes value. Rest stuff like min, day, date, year should be current date and time. How can I do this? I tried timelocal, localtime and many others, but they didn't work.
If you are interested in a complete date time object (like it seems according your description), use DateTime. Example:
use DateTime;
my $dt = DateTime->now; # initialize with current date time
$dt->set_hour(20); #set your own hour
$dt->set_minute(20); #set your own minute
Unless I'm missing something, you indeed use localtime, as in perldoc:
http://perldoc.perl.org/functions/localtime.html
($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst) = localtime(time);
Then, if you want to stick to using the broken-down components, you could overwrite whichever variables you want to "fix":
$hour = $my_hardcoded_hour;
$min = $my_hardcoded_minute;
If instead you need to translate this back to "seconds since the epoch" (Jan 1, 1970, 12:00AM GMT) format, you need the Time::Local package:
http://perldoc.perl.org/Time/Local.html
use Time::Local;
$epoch_current_time_with_fixed_hour_min =
timelocal($sec, $my_hardcoded_minutes, $my_hardcoded_hour, $mday, $mon, $year);

Parse date String to milliseconds and vice versa

I have following date in string format and I need to convert it to milliseconds using a Perl script. I have tried to convert it using DateTime::Format::Strptime and it returns 0 after convert to millisecond.
Date String : 01-13-15:14.16
#!/usr/bin/perl
use DateTime::Format::Strptime;
my $strp = DateTime::Format::Strptime->new(
pattern => '%m-%d-%y:%H.%M',
on_error => 'croak',
);
my $dt = $strp->parse_datetime("01-13-15:14.16");
print $dt->millisecond."\n";
How can I convert datetime to milliseconds
I think it's likely that what you want is not milliseconds but epoch, which is the number of seconds since January 1 1970. Your date-time format has neither seconds nor milliseconds, so both of these fields will be reported as zero, and if you really want milliseconds then you can simply multiply the epoch by 1000.
It looks like this
use strict;
use warnings;
use 5.010;
use DateTime::Format::Strptime;
my $strp = DateTime::Format::Strptime->new(
pattern => '%m-%d-%y:%H.%M',
on_error => 'croak',
);
my $dt = $strp->parse_datetime('01-13-15:14.16');
say $dt->epoch;
output
1421158560
However, the DateTime module is enormous and slow, and there is a better solution in the core module Time::Piece, which would look like this
use strict;
use warnings;
use 5.010;
use Time::Piece;
my $dt = Time::Piece->strptime('01-13-15:14.16', '%m-%d-%y:%H.%M');
say $dt->epoch;
output
1421158560
It is impossible to convert a date to milliseconds. A date is a point in time. Milliseconds measure a duration.
The only way your question makes sense is if we assume you mean "how many milliseconds are there between my given date/time and some other fixed date/time". Because of the way that Unix measures time, let's assume that the fixed date/time you're measuring from is the Unix Epoch (1970-01-01 00:00:00).
You get the number of seconds since the epoch from a DateTime object with the epoch() method.
say $dt->epoch;
To get the number of milliseconds, just multiply that by a thousand.
say 1000 * $dt->epoch;
Your current timestamp doesn't include milliseconds. If you change to parsing strings that include them, then you can add that value on too.
say $dt->milliseconds + (1000 * $dt->epoch);
Adding my comment as an answer:
If you want to convert the datetime value into seconds since the epoch, try one of these:
print sprintf("%f", $dt->epoch + ($dt->millisecond/1000));
print sprintf("%s.%s", $dt->epoch, $dt->millisecond);
(My original comment did not account for millisecond being an integer)
Similar for the values in milliseconds, rather than seconds.millis:
print sprintf("%d", ($dt->epoch * 1000) + $dt->millisecond);
print sprintf("%d%04d", $dt->epoch, $dt->millisecond);
Remember, that if you don't supply DateTime with a value containing milliseconds (or higher resolution), the value will simply be zero.

How can I convert a log4j timestamp to milliseconds in Perl?

The log4j logs I have contain timestamps in the following format:
2009-05-10 00:48:41,905
I need to convert it in perl to millseconds since epoch, which in this case would be 124189673005, using the following gawk function. How do I do it in perl?
I have little or no experience in perl, so appreciate if someone can post an entire script that does this
function log4jTimeStampToMillis(log4jts) {
# log4jts is of the form 2009-03-02 20:04:13,474
# extract milliseconds that is after the command
split(log4jts, tsparts, ",");
millis = tsparts[2];
# remove - : from tsstr
tsstr = tsparts[1];
gsub("[-:]", " ", tsstr);
seconds = mktime(tsstr);
print log4jts;
return seconds * 1000 + millis;
}
Though I almost always tell people to go use one of the many excellent modules from the CPAN for this, most of them do have one major drawback - speed. If you're parsing a large number of log files in real-time, that can sometimes be an issue. In those cases, rolling your own can often be a more suitable solution, but there are many pitfalls and nuances that must be considered and handled properly. Hence the preference for using a known-correct, proven, reliable module written by somebody else. :)
However, before I even considered my advice above, I looked at your code and had converted it to perl in my head... therefore, here is a more-or-less direct conversion of your gawk code into perl. I've tried to write it as simply as possible, so as to highlight some of the more delicate parts of dealing with dates and times in perl by hand.
# import the mktime function from the (standard) POSIX module
use POSIX qw( mktime );
sub log4jTimeStampToMillis {
my ($log4jts, $dst) = #_;
# extract the millisecond field
my ($tsstr, $millis) = split( ',', $log4jts );
# extract values to pass to mktime()
my #mktime_args = reverse split( '[-: ]', $tsstr );
# munge values for posix compatibility (ugh)
$mktime_args[3] -= 1;
$mktime_args[4] -= 1;
$mktime_args[5] -= 1900;
# print Dumper \#mktime_args; ## DEBUG
# convert, make sure to account for daylight savings
my $seconds = mktime( #mktime_args, 0, 0, $dst );
# return that time as milliseconds since the epoch
return $seconds * 1000 + $millis;
}
One important difference between my code and yours - my log4jTimeStampToMillis subroutine takes two parameters:
the log timestamp string
whether or not that timestamp is using daylight savings time ( 1 for true, 0 for false )
Of course, you could just add code to detect if that time falls in DST or not and adjust automatically, but I was trying to keep it simple. :)
NOTE: If you uncomment the line marked DEBUG, make sure to add "use Data::Dumper;" before that line in your program so it will work.
Here's an example of how you could test that subroutine:
my $milliseconds = log4jTimeStampToMillis( "2009-05-10 00:48:41,905", 1 );
my $seconds = int( $milliseconds / 1000 );
my $local = scalar localtime( $seconds );
print "ms: $milliseconds\n"; # ms: 1241844521905
print "sec: $seconds\n"; # sec: 1241844521
print "local: $local\n"; # local: Sat May 9 00:48:41 2009
You should take advantage of the great DateTime package, specifically use DateTime::Format::Strptime:
use DateTime;
use DateTime::Format::Strptime;
sub log4jTimeStampToMillis {
my $log4jts=shift(#_);
#see package docs for how the pattern parameter works
my $formatter= new DateTime::Format::Strptime(pattern => '%Y-%m-%d %T,%3N');
my $dayObj = $formatter->parse_datetime($log4jts);
return $dayObj->epoch()*1000+$dayObj->millisecond();
}
print log4jTimeStampToMillis('2009-05-10 10:48:41,905')."\n";
#prints my local version of the TS: 1241952521905
This saves you the pain of figuring out DST yourself (although you'll have to pass your server's TZ to Strptime via the time_zone parameter). It also saves you from dealing with leap everything if it becomes relevant (and I'm sure it will).
Haven't used it, but you might want to check out Time::ParseDate.
SimpleDateFormat dateFormat = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss,SSS");
Date time = dateFormat.parse(log4jts);
long millis = time.getTime();