mjhjmtu

How to sort a file containing below? (s=second, h=hour, d=day m=minute)

1s
2s
1h
2h
1m
2m
2s
1d
1m

awk ' unitvalue=$1; ; 
 /s/ m=1 ; /m/ m=60 ; /h/ m=3600 ; /d/ m=86400 ; 
 sub("[smhd]","",unitvalue); unitvalue=unitvalue*m; 
 print unitvalue " " $1; ' input |
 sort -n | awk ' print $2 '
1s
2s
2s
1m
1m
2m
1h
2h
1d

First version - FPAT is used

gawk '
BEGIN [smhd]";

/s/ factor = 1 
/m/ factor = 60 
/h/ factor = 3600 
/d/ factor = 86400 

 print $1 * factor, $0;
' input.txt | sort -n | awk 'print $2'

FPAT - A regular expression describing the contents of the fields
in a record. When set, gawk
parses the input into fields, where the fields match the regular expression, instead of
using the value of the FS variable as the field separator.

Second version

I was surprised to discover, that without FPAT it also works.
It is caused the number conversion mechanism of awk - How awk Converts Between Strings and Numbers, namely:

A string is converted to a number by interpreting any numeric prefix of the string as numerals: "2.5" converts to 2.5, "1e3" converts to 1,000, and "25fix" has a numeric value of 25. Strings that can’t be interpreted as valid numbers convert to zero.

gawk '
/s/ factor = 1 
/m/ factor = 60 
/h/ factor = 3600 
/d/ factor = 86400 

 print $0 * factor, $0;
' input.txt | sort -n | awk 'print $2'

Input (changed a little bit)

1s
122s
1h
2h
1m
2m
2s
1d
1m

Output

Note: 122 seconds more than 2 minutes, so it sorted after 2m.

1s
2s
1m
1m
2m
122s
1h
2h
1d

If you only have times in the format of your question:

sort -k 1.2,1.2 -k 1.1,1.1 <file>

Where <file> is the file your data resides in. This command sorts on the second letter (ascending) and then sorts on the first letter (ascending). This works because it just so happes that the ordering of the letters for the time units (d > h > m > s) is exactly the order we want (day > hours > minutes > seconds).

This an extension of MiniMax’ answer that can handle a broader range of duration value like 1d3h10m40s.

GNU Awk program (stored in parse-times.awk for the sake of this answer):

#!/usr/bin/gawk -f
BEGIN
 FPAT = "[0-9]+[dhms]";
 duration["s"] = 1;
 duration["m"] = 60;
 duration["h"] = duration["m"] * 60;
 duration["d"] = duration["h"] * 24;



 t=0;
 for (i=1; i<=NF; i++)
 t += $i * duration[substr($i, length($i))];
 print(t, $0);

Invocation:

gawk -f parse-times.awk input.txt | sort -n -k 1,1 | cut -d ' ' -f 2

Solution in Python 3:

#!/usr/bin/python3
import re, fileinput

class RegexMatchIterator:
 def __init__(self, regex, string, error_on_incomplete=False):
 self.regex = regex
 self.string = string
 self.error_on_incomplete = error_on_incomplete
 self.pos = 0

 def __iter__(self):
 return self

 def __next__(self):
 match = self.regex.match(self.string, self.pos)
 if match is not None:
 if match.end() > self.pos:
 self.pos = match.end()
 return match
 else:
 fmt = '0!s returns an empty match at position 1:d for "3!r"'

 elif self.error_on_incomplete and self.pos < len(self.string):
 if isinstance(self.error_on_incomplete, str):
 fmt = self.error_on_incomplete
 else:
 fmt = '0!s didn't match the suffix 3!r at position 1:d of 2!r'

 else:
 raise StopIteration(self.pos)

 raise ValueError(fmt.format(
 self.regex, self.pos, self.string, self.string[self.pos:]))


DURATION_SUFFIXES = 's': 1, 'm': 60, 'h': 3600, 'd': 24*3600 
DURATION_PATTERN = re.compile(
 '(\d+)(' + '|'.join(map(re.escape, DURATION_SUFFIXES.keys())) + ')')

def parse_duration(s):
 return sum(
 int(m.group(1)) * DURATION_SUFFIXES[m.group(2)]
 for m in RegexMatchIterator(DURATION_PATTERN, s,
 'Illegal duration string 3!r at position 1:d'))


if __name__ == '__main__':
 with fileinput.input() as f:
 result = sorted((l.rstrip('n') for l in f), key=parse_duration)
 for item in result:
 print(item)

As you can see I spent about ⅔ of the line count towards a useful iterator over regex.match() results because regex.finditer() doesn't tie matches to the beginning of the current region and there are no other suitable ways to iterate over match results. *grrr*