I would like to list all files in a linux directory then apply a regular expression on them to format the file name, and print these formatted files names.
Example:
ls -lthrh
.
.
-rwxrwxrwx. 1 root root 633 Oct 31 2016 Oracle_Schedule_ARC-Oracle_ARCH-1477938600005-1002-Oracleorcl-rman1.txt
-rwxrwxrwx. 1 root root 610 Nov 7 2016 MOD-1478512353102-1002-Oracleorcl-rman1.txt
After applying my regex '.+?(?=-)' I would have everything before the first '-' to be:
Oracle_Schedule_ARC
MOD
I've tried using awk, but I couldn't pass a regex to it. I will apply later | sort | uniq to have a unique output of the regex output.
In any POSIX shell (bash, pdksh, ksh93, zsh, dash):
for name in *; do
printf '%s\n' "${name%%-*}"
done
This would go through all the names in the current directory and output the bit before the first - character. It does this by removing the longest suffix string matching -* from the filename using a standard parameter substitution.
Note that -* is a shell globbing pattern, not a regular expression. Regular expressions are useful for working on text, but globbing patterns are fast and efficient for working with filenames and pathnames in general, as you don't have to start another process with a regex engine, such as awk or sed.
In bash, you could also get away from using a loop at all:
set -- *
printf '%s\n' "${#%%-*}"
This first sets the positional parameters to the names in the current directory. printf is then invoked on the set of names, each individually transformed with the same parameter substitution as in the first part of this answer.
The same thing, but using an array variable other than the array of positional parameters:
names=( * )
printf '%s\n' "${names[#]%%-*}"
Related
I have a large binary file. I want to extract certain strings from it and copy them to a new text file.
For example, in:
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZM-^G
I want to take the number '7' (after the #^#^#E) and every character after it stopping at the Z ('ignoring the M-^G).
I want to copy this 7cacscKLrrok9bwC3Z64NTnZ to a new file.
There will be multiple such strings in one file. The end will always be denoted by the M- (which I don't want copied). The start will always be denoted by a 7 (which I do want copied).
Unfortunately, my knowledge of grep, sed, etc, does not extend to this level. Can someone please suggest a viable way to achieve this?
cat -v filename | grep [7][A-Z,a-z] will show all strings with a '7' followed by a letter but that's not much.
Thank you.
I've noticed that my requirements are rather more complicated.
(I've performed the correct - I hope - formatting this time). Thanks to 'tshiono' for his (?) answer to the earlier submission.
I want to check the ending of a string and, if it ends in M-, grep another string that follows it (with junk in between). If the string does not end in M-, then I don't want it copied (let alone any other strings).
So what I would like is:
grep -a -Po "7[[:alnum:]]+(?=M-)" file_name and if the ending is M- then grep -a -Po "5x[[:alnum:]]+(?=\^)" file_name to copy the string that starts with 5x and ends with a ^.
In this example:
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZM-^GwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
The outcome would be:
7cacscKLrrok9bwC3Z64NTnZ
5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk
However, if the ending is not M- (more precisely, if the ending is ^S), then do not try the second grep and do not record anything at all.
In this example:
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZ^SGwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
The outcome would be null (nothing copied) as the 7cacs... string ends in ^S.
Is grep the correct tool? Grep a file and if the condition in the grep command is 'yes' then issue a different grep command but if the condition is 'no' then do nothing.
Thanks again.
I have noticed one addition modification.
Can one add an OR command to the second part? Grep if the second string starts with 5x OR 6x?
In the example below, grep -aPo "7[[:alnum:]]+M-.*?5x[[:alnum:]]+\^" filename | grep -aPo "7[[:alnum:]]+(?=M-)|5x[[:alnum:]]+(?=\^)" will extract the strings starting with 7 and the strings starting with 5x.
How can one change the 5x to 5x or 6x?
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZM-^GwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7AAAAAscKLrrok9bwC3Z64NTnZM-^GwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM6x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
In this example, the desired outcome would be:
7cacscKLrrok9bwC3Z64NTnZ
5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk
7AAAAAscKLrrok9bwC3Z64NTnZ
6x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk
UPDATE MARCH 09:
I need to create a series of complex grep (or perl) commands to extract strings from a series of binary files.
I need two strings from the binary file.
The first string will always start with a 1.
The first string will end with a letter or number. The next letter will always be a lower case k. I do not want this k character.
The difficulty is that the ending k will not always be the first k in the string. It might be the first k but it might not.
After the k, there is a second string. The second string will always start with an A or a B.
The ending of the second string will be in one of two forms:
a) it will end with a space then display the first three characters from the first string in lower case followed by a )
b) it will end with a ^K then display the first three characters from the first string in lower case.
For example:
1pppsx9YPar8Rvs75tJYWZq3eo8PgwbckB4m4zT7Yg042KIDYUE82e893hY ppp)
Should be:
1pppsx9YPar8Rvs75tJYWZq3eo8Pgwbc and B4m4zT7Yg042KIDYUE82e893hY - delete the k and the space then ppp.
For example:
1zzzsx9YPkr8Rvs75tJYWZq3eo8PgwbckA2m4zT7Yg042KIDYUE82e893hY^Kzzz
Should be:
1zzzsx9YPkar8Rvs75tJYWZq3eo8Pgwbc and A4m4zT7Yg042KIDYUE82e893hY - delete the second k and the ^Kzzz.
In the second example, we see that the first k is part of the first string. It is the k before the A that breaks up the first and second strings.
I hope there is a super grep expert who can help! Many thanks!
If your grep supports -P option, would you please try:
grep -a -Po "7[[:alnum:]]+(?=M-)" file
The -a option forces grep to read the input as a text file.
The -P option enables the perl-compatible regex.
The -o option tells grep to print only the matched substring(s).
The pattern (?=M-) is a zero-width lookahead assertion (introduced in
Perl) without including it in the result.
Alternatively you can also say with sed:
sed 's/M-/\n/g' file | sed -n 's/.*\(7[[:alnum:]]\+\).*/\1/p'
The first sed command splits the input file into miltiple lines by
replacing the substring M- with a newline.
It has two benefits: it breaks the lines to allow multiple matches with
sed and excludes the unnecessary portion M- from the input.
The next sed command extracts the desired pattern from the input.
It assumes your sed accepts \n in the replacement, which is
a GNU extension (not POSIX compliant). Otherwise please try (in case you are working on bash):
sed 's/M-/\'$'\n''/g' file | sed -n 's/.*\(7[[:alnum:]]\+\).*/\1/p'
[UPDATE]
(The requirement has been updated by the OP and the followings are solutions according to it.)
Let me assume the string which starts with 7 and ends with M- is always followed
by another (no more and no less than one) string which starts with 5x and ends
with ^ (ascii caret character) with junks in between.
Then would you please try the following:
grep -aPo "7[[:alnum:]]+M-.*?5x[[:alnum:]]+\^" file | grep -aPo "7[[:alnum:]]+(?=M-)|5x[[:alnum:]]+(?=\^)"
It executes the task in two steps (two cascaded greps).
The 1st grep narrows down the input data into the candidate substring
which will include the desired two sequences and junks in between.
The regex .*? in between matches any (ascii or binary) characters
except for a newline character.
The trailing ? enables the shortest match
which avoids the overrun due to the greedy nature of regex. The regex is intended to match junks in between.
The 2nd grep includes two regex's merged with a pipe | meaning logical OR.
Then it extracts two desired sequences.
A potential problem of grep solution is that grep is a line oriented command
and cannot include the newline character in the matched string.
If a newline character is included in the junks in between (I'm not sure about the possibility), the above solution will fail.
As a workaround, perl will provide flexible manipulations with binary data.
perl -0777 -ne '
while (/(7[[:alnum:]]+)M-.*?(5x[[:alnum:]]+)\^/sg) {
printf("%s\n%s\n", $1, $2);
}
' file
The regex is mostly same as that of grep because the -P option of grep means
perl-compatible.
It can capture multiple patterns at once in variables $1 and $2 hence just one regex is enough.
The -0777 option to the perl command tells perl to slurp all data
at once.
The s option at the end the regex makes a dot match a newline character.
The g option enables the global (multiple) match.
[UPDATE2]
In order to make the regex match either 5x or 6x, replace 5x with (5|6)x.
Namely:
grep -aPo "7[[:alnum:]]+M-.*?(5|6)x[[:alnum:]]+\^" file | grep -aPo "7[[:alnum:]]+(?=M-)|(5|6)x[[:alnum:]]+(?=\^)"
As mentioned before, the pipe | means OR. The OR operator has the lowest priority in the evaluation, hence you need to enclose them with parens in this case.
If there is a possibility any other number than 5 or 6 may appear, it will be safer to put [[:digit:]] instead, which matches any one digit betweeen 0 and 9:
grep -aPo "7[[:alnum:]]+M-.*?[[:digit:]]x[[:alnum:]]+\^" file | grep -aPo "7[[:alnum:]]+(?=M-)|[[:digit:]]x[[:alnum:]]+(?=\^)"
[UPDATE3]
(Answering the OP's requirement on March 9th)
Let me start with a perl code which regex will be relatively easier
to explain.
perl -0777 -ne 'while (/(1(.{3}).+)k([AB].*)[\013 ]\2/g){print "$1 $3\n"}' file
Output:
1pppsx9YPar8Rvs75tJYWZq3eo8Pgwbc B4m4zT7Yg042KIDYUE82e893hY
1zzzsx9YPkr8Rvs75tJYWZq3eo8Pgwbc A2m4zT7Yg042KIDYUE82e893hY
[Explanation of regex]
(1(.{3}).+)k([AB].*)[\013 ]\2
( start of the 1st capture group referred by $1 later
1 literal "1"
( start of the 2nd capture group referred by \2 later
.{3} a sequence of the identical three characters such as ppp or zzz
) end of the 2nd capture group
.+ followed by any characters with "greedy" match which may include the 1st "k"
) end of the 1st capture group
k literal "k"
( start of the 3rd capture group referred by $3 later
[AB].* the character "A" or "B" followed by any characters
) end of the 3rd capture group
[\013 ] followed by ^K or a whitespace
\2 followed by the capture group 2 previously assigned
When implementing it with grep, we will encounter a limitation of grep.
Although we want to extract multiple patterns from the input file,
the -e option (which can specify multiple search patterns) does not
work with -P option. Then we need to split the regex into two patterns
such as:
grep -Po "(1(.{3}).+)(?=k([AB].*)[\013 ]\2)" file
grep -Po "(1(.{3}).+)k\K([AB].*)(?=[\013 ]\2)" file
And the result will be:
1pppsx9YPar8Rvs75tJYWZq3eo8Pgwbc
1zzzsx9YPkr8Rvs75tJYWZq3eo8Pgwbc
B4m4zT7Yg042KIDYUE82e893hY
A2m4zT7Yg042KIDYUE82e893hY
Please be noted the order of output is not same as the order of appearance in the original file.
Another option will be to introduce ripgrep or rg which is a fast
and versatile version of grep. You may need to install ripgrep with
sudo apt install ripgrep or using other package handling tool.
An advantage of ripgrep is it supports -r (replace) option in which
you can make use of the backreferences:
rg -N -Po "(1(.{3}).+)k([AB].*)[\013 ]\2" -r '$1 $3' file
The -r '$1 $3' option prints the 1st and the 3rd capture groups and the result will be the same as perl.
In the general case, you can use the strings utility to pluck out ASCII from binary files; then of course you can try to grep that output for patterns that you find interesting.
Many traditional Unix utilities like grep have internal special markers which might get messed up by binary input. For example, the character \xFF was used for internal purposes by some versions of GNU grep so you can't grep for that character even if you can figure out a way to represent it in the shell (Bash supports $'\xff' for example).
A traditional approach would be to run hexdump or a similar utility, and then grep that for patterns. However, more modern scripting languages like Perl and Python make it easy to manipulate arbitrary binary data.
perl -ne 'print if m/\xff\xff/' </dev/urandom
This might work for you (GNU sed):
sed -En '/\n/!{s/M-\^G/\n/;s/7[^\n]*\n/\n&/};/^7[^\n]*/P;D' file
Split each line into zero or more lines that begin with 7 and end just before M-^G and only print such lines.
I want to use grep/awk/sed to extract matched strings for each line of a log file. Then place it into csv file.
Highlighted strings (1432,53,http://www.espn.com/)
If the input is:
2018-10-31
18:48:01.717,INFO,15592.15627,PfbProxy::handlePfbFetchDone(0x1d69850,
pfbId=561, pid=15912, state=4, fd=78, timer=61), FETCH DONE: len=45,
PFBId=561, pid=0, loadTime=1434 ms, objects=53, fetchReqEpoch=0.0,
fetchDoneEpoch:0.0, fetchId=26, URL=http://www.espn.com/
2018-10-31
18:48:01.806,DEBUG,15592.15621,FETCH DONE: len=45, PFBId=82, pid=0,
loadTime=1301 ms, objects=54, fetchReqEpoch=0.0, fetchDoneEpoch:0.0,
fetchId=28, URL=http://www.diply.com/
Expected output for the above log lines:
URL,LoadTime,Objects
http://www.espn.com/,1434,53
http://www.diply.com/,1301,54
This is an example, and the actual Log File will have much more data.
--My-Solution-So-far-
For now I used grep to get all lines containing keyword 'FETCH DONE' (these lines contain strings I am looking for).
I did come up with regular expression that matches the data I need, but when I grep it and put it in the file it prints each string on the new line which is not quite what I am looking for.
The grep and regular expression I use (online regex tool: https://regexr.com/42cah):
echo -en 'url,loadtime,object\n'>test1.csv #add header
grep -Po '(?<=loadTime=).{1,5}(?= )|((?<=URL=).*|\/(?=.))|((?<=objects=).{1,5}(?=\,))'>>test1.csv #get matching strings
Actual output:
URL,LoadTime,Objects
http://www.espn.com
1434
53
http://www.diply.com
1301
54
Expected output:
URL,LoadTime,Objects
http://www.espn.com/,1434,53
http://www.diply.com/,1301,54
I was trying using awk to match multiple regex and print comma in between. I couldn't get it to work at all for some reason, even though my regex matches correct strings.
Another idea I have is to use sed to replace some '\n' for ',':
for(i=1;i<=n;i++)
if(i % 3 != 0){
sed REPLACE "\n" with "," on i-th line
}
Im pretty sure there is a more efficient way of doing it
Using sed:
sed -n 's/.*loadTime=\([0-9]*\)[^,]*, objects=\([0-9]*\).* URL=\(.*\)/\3,\1,\2/p' input | \
sed 1i'URL,LoadTime,Objects'
I'm trying to remove whitespace in file names and replace them.
Input:
echo "File Name1.xml File Name3 report.xml" | sed 's/[[:space:]]/__/g'
However the output
File__Name1.xml__File__Name3__report.xml
Desired output
File__Name1.xml File__Name3__report.xml
You named awk in the title of the question, didn't you?
$ echo "File Name1.xml File Name3 report.xml" | \
> awk -F'.xml *' '{for(i=1;i<=NF;i++){gsub(" ","_",$i); printf i<NF?$i ".xml ":"\n" }}'
File_Name1.xml File_Name3_report.xml
$
-F'.xml *' instructs awk to split on a regex, the requested extension plus 0 or more spaces
the loop {for(i=1;i<=NF;i++) is executed for all the fields in which the input line(s) is(are) splitted — note that the last field is void (it is what follows the last extension), but we are going to take that into account...
the body of the loop
gsub(" ","_", $i) substitutes all the occurrences of space to underscores in the current field, as indexed by the loop variable i
printf i<NF?$i ".xml ":"\n" output different things, if i<NF it's a regular field, so we append the extension and a space, otherwise i equals NF, we just want to terminate the output line with a newline.
It's not perfect, it appends a space after the last filename. I hope that's good enough...
▶ A D D E N D U M ◀
I'd like to address:
the little buglet of the last space...
some of the issues reported by Ed Morton
generalize the extension provided to awk
To reach these goals, I've decided to wrap the scriptlet in a shell function, that changing spaces into underscores is named s2u
$ s2u () { awk -F'\.'$1' *' -v ext=".$1" '{
> NF--;for(i=1;i<=NF;i++){gsub(" ","_",$i);printf "%s",$i ext (i<NF?" ":"\n")}}'
> }
$ echo "File Name1.xml File Name3 report.xml" | s2u xml
File_Name1.xml File_Name3_report.xml
$
It's a bit different (better?) 'cs it does not special print the last field but instead special-cases the delimiter appended to each field, but the idea of splitting on the extension remains.
This seems a good start if the filenames aren't delineated:
((?:\S.*?)?\.\w{1,})\b
( // start of captured group
(?: // non-captured group
\S.*? // a non-white-space character, then 0 or more any character
)? // 0 or 1 times
\. // a dot
\w{1,} // 1 or more word characters
) // end of captured group
\b // a word boundary
You'll have to look-up how a PCRE pattern converts to a shell pattern. Alternatively it can be run from a Python/Perl/PHP script.
Demo
Assuming you are asking how to rename file names, and not remove spaces in a list of file names that are being used for some other reason, this is the long and short way. The long way uses sed. The short way uses rename. If you are not trying to rename files, your question is quite unclear and should be revised.
If the goal is to simply get a list of xml file names and change them with sed, the bottom example is how to do that.
directory contents:
ls -w 2
bob is over there.xml
fred is here.xml
greg is there.xml
cd [directory with files]
shopt -s nullglob
a_glob=(*.xml);
for ((i=0;i< ${#a_glob[#]}; i++));do
echo "${a_glob[i]}";
done
shopt -u nullglob
# output
bob is over there.xml
fred is here.xml
greg is there.xml
# then rename them
cd [directory with files]
shopt -s nullglob
a_glob=(*.xml);
for ((i=0;i< ${#a_glob[#]}; i++));do
# I prefer 'rename' for such things
# rename 's/[[:space:]]/_/g' "${a_glob[i]}";
# but sed works, can't see any reason to use it for this purpose though
mv "${a_glob[i]}" $(sed 's/[[:space:]]/_/g' <<< "${a_glob[i]}");
done
shopt -u nullglob
result:
ls -w 2
bob_is_over_there.xml
fred_is_here.xml
greg_is_there.xml
globbing is what you want here because of the spaces in the names.
However, this is really a complicated solution, when actually all you need to do is:
cd [your space containing directory]
rename 's/[[:space:]]/_/g' *.xml
and that's it, you're done.
If on the other hand you are trying to create a list of file names, you'd certainly want the globbing method, which if you just modify the statement, will do what you want there too, that is, just use sed to change the output file name.
If your goal is to change the filenames for output purposes, and not rename the actual files:
cd [directory with files]
shopt -s nullglob
a_glob=(*.xml);
for ((i=0;i< ${#a_glob[#]}; i++));do
echo "${a_glob[i]}" | sed 's/[[:space:]]/_/g';
done
shopt -u nullglob
# output:
bob_is_over_there.xml
fred_is_here.xml
greg_is_there.xml
You could use rename:
rename --nows *.xml
This will replace all the spaces of the xml files in the current folder with _.
Sometimes it comes without the --nows option, so you can then use a search and replace:
rename 's/[[:space:]]/__/g' *.xml
Eventually you can use --dry-run if you want to just print filenames without editing the names.
I have spent some time trying to find the answer to this question but with no luck...
I am writing a script which takes a string as input, then searches through a cvs log for a match. The output should be the filename prior to the matching string. I can write a regex to match the filename without any trouble, however I am unsure how to look back for the last occurrence of a regex after a match is made...
There are obviously many files in the cvs log that I will be parsing and the number of lines between the filename and the input string is unknown so cannot do anything like 'grep -B4' etc...
EXAMPLE BELOW (cvs log with irrelevant bits removed) - Need to match a string (which will be given as input to script (e.g. 120233)) and retrieve the filename associated with it, which here will be (Aliases.xml). As I said there are many files so although the regex will match many filenames, i am only interested in the one prior to my matched string.
This is my first time posting here and I am new to programming so I hope this makes sense.
----------
RCS file: /user1/cvs/Aliases.xml
revision 1.18
date: 2015/03/13 16:21:07;
FIX - 217427 - fixed error....
revision 1.17
date: 2013/03/27 08:03:36;
IMPROVEMENT - 120233 - some improvement
You can use awk command for this:
cvs log | awk -F '[:, ]+' -v s='120233' '/RCS file:/{f=$3} f && $0 ~ s{print f; f=""}'
Explanation:
-F '[:, ]+' # make one or more of colon, comma or <space> a field saprator
-v s='120233' # pass search string to awk in variable s
/RCS file:/ # search for string "RCS file:"
{f=$3} # store filename in variable f
f && $0 ~ s # if f is NOT_EMPTY and line matches variable s
print f; f="" # print filename from variable f and set f to EMPTY
im documenting a Shell Script of the server of my job which takes a series of files that starts with the word "dat" and performs a particular task with all those files. The problem is that the script is filtering files using a regular expression with sed command as follows:
namecmp=`grep -l $name dat*.p |sed -e "s/^\(......\)\(..\)\(..\)\(....\)\(.*\)/\1\4\3\2\5/g"| sort -t '.' -k 1.7,1.14 |sed -e "s/^\(......\)\(....\)\(..\)\(..\)\(.*\)/\1\4\3\2\5/g" | tail -1 `
I don't understand how exactly is doing this regular expression to filter out files. It would be helpful to know any expected output or examples files filtered by that expression.
Is there a way to find possible expressions that are accepted by that expression?
grep -l searches in a list of files (dat*.p) for a regular expression ($name in your case, or better: whatever $name evaluates to) and then prints only the files' names in which this was found.
These file names are then passed through the sed command which replaces (s for substitute) something, namely ^\(......\)\(..\)\(..\)\(....\)\(.*\) by \1\4\3\2\5 (so it just regroups parts of the file names). The transformed file names are then passed to sort, and then to sed again which just seems to undo the regrouping of the file name.
Finally, just the last file name is taken (tail -1) and all the rest is thrown away. This can be achieved a lot cheaper than by sorting all the file names, but who cares ;-)
Effectively, this line finds the name of the "last" file matching the regexp in $name. The meaning of "last" is determined by the sorting of the file names after regrouping; assuming from the size of the groups, I think a time stamp is modified so that it is changed from DDMMYYYY to YYYYMMDD which makes sense in a way.
There are libraries designed to do that (e.g. Xeger) but for this I can just provide you with an example:
abcdef02122014foobarfoobarfoobar
^ ^ ^ ^ ^
| | | | |
1 2 3 4 5
becomes
abcdef20140212foobarfoobarfoobar
^ ^ ^ ^ ^
| | | | |
1 4 3 2 5
and then I don't know what the sort does but the next sed simply puts all of the above back in order.
So it seems the regular expressions are used to temporarily change the format of lines for sorting, before restoring the original format.
echo "1111112233444456789" | sed -e "s/^\(......\)\(..\)\(..\)\(....\)\(.*\)/\1\4\3\2\5/g"
-> 1111114444332256789
explain:
Begin 111111 22 33 4444 56789
^ \(......\)\(..\)\(..\)\(....\)\(.*\)
\1 \2 \3 \4 \5
optimization:
The last \(.*\) is not needed and thus the corresponding \5 must be removed
the last g is also not needed (there is only 1 substitution possible dur to ^ meaning start of string)