I have a file with output like this:
MIKE;123456
JOHN-HELPER;654321
SAM.SMITH;182364
I need everything before the delimiter to move to the end of the line, so it'll look like this:
123456;MIKE
654321;JOHN-HELPER
182364;SAM.SMITH
Struggling it work it out with sed... any ideas?
Like this, for example:
$ sed -r 's/([^;]*);(.*)/\2;\1/' a
123456;MIKE
654321;JOHN-HELPER
182364;SAM.SMITH
It "catches" two groups: everything before ; and then the rest. Next step is to print these blocks the other way round: \2;\1.
Or with awk:
$ awk -F";" '{print $2";"$1}' a
123456;MIKE
654321;JOHN-HELPER
182364;SAM.SMITH
It sets ; as field delimiter and then prints the fields the other way round.
Related
i m trying to perform the following substitution on lines of the general format:
BBBBBBB.2018_08,XXXXXXXXXXXXX,01/01/2014,"109,07",DF,CCCCCCCCCCC, .......
as you see the problem is that its a comma separated file, with a specific field containing a comma decimal. I would like to replace that with a dot .
I ve tried this, to replace the first occurence of a pattern after match, but to no avail, could someone help me?
sed -e '/,"/!b' -e "s/,/./"
sed -e '/"/!b' -e ':a' -e "s/,/\./"
Thanks in advance. An awk or perl solution would help me as well. Here's an awk effort:
gawk -F "," 'substr($10, 0, 3)==3 && length($10)==12 { gsub(/,/,".", $10); print}'
That yielded the same file unchanged.
CSV files should be parsed in awk with a proper FPAT variable that defines what constitutes a valid field in such a file. Once you do that, you can just iterate over the fields to do the substitution you need
gawk 'BEGIN { FPAT = "([^,]+)|(\"[^\"]+\")"; OFS="," }
{ for(i=1; i<=NF;i++) if ($i ~ /[,]/) gsub(/[,]/,".",$i);}1' file
See this answer of mine to understand how to define and parse CSV file content with FPAT variable. Also see Save modifications in place with awk to do in-place file modifications like sed -i''.
The following sed will convert all decimal separators in quoted numeric fields:
sed 's/"\([-+]\?[0-9]*\)[,]\?\([0-9]\+\([eE][-+]\?[0-9]+\)\?\)"/"\1.\2"/g'
See: https://www.regular-expressions.info/floatingpoint.html
This might work for you (GNU sed):
sed -E ':a;s/^([^"]*("[^",]*"[^"]*)*"[^",]*),/\1./;ta' file
This regexp matches a , within a pair of "'s and replaces it by a .. The regexp is anchored to the start of the line and thus needs to be repeated until no further matches can be matched, hence the :a and the ta commands which causes the substitution to be iterated over whilst any substitution is successful.
N.B. The solution expects that all double quotes are matched and that no double quotes are quoted i.e. \" does not appear in a line.
If your input always follows that format of only one quoted field containing 1 comma then all you need is:
$ sed 's/\([^"]*"[^"]*\),/\1./' file
BBBBBBB.2018_08,XXXXXXXXXXXXX,01/01/2014,"109.07",DF,CCCCCCCCCCC, .......
If it's more complicated than that then see What's the most robust way to efficiently parse CSV using awk?.
Assuming you have this:
BBBBBBB.2018_08,XXXXXXXXXXXXX,01/01/2014,"109,07",DF,CCCCCCCCCCC
Try this:
awk -F',' '{print $1,$2,$3,$4"."$5,$6,$7}' filename | awk '$1=$1' FS=" " OFS=","
Output will be:
BBBBBBB.2018_08,XXXXXXXXXXXXX,01/01/2014,"109.07",DF,CCCCCCCCCCC
You simply need to know the field numbers for replacing the field separator between them.
In order to use regexp as in perl you have to activate extended regular expression with -r.
So if you want to replace all numbers and omit the " sign, then you can use this:
echo 'BBBBBBB.2018_08,XXXXXXXXXXXXX,01/01/2014,"109,07",DF,CCCCCCCCCCC, .......'|sed -r 's/\"([0-9]+)\,([0-9]+)\"/\1\.\2/g'
If you want to replace first occurrence only you can use that:
echo 'BBBBBBB.2018_08,XXXXXXXXXXXXX,01/01/2014,"109,07",DF,CCCCCCCCCCC, .......'|sed -r 's/\"([0-9]+)\,([0-9]+)\"/\1\.\2/1'
https://www.gnu.org/software/sed/manual/sed.txt
I have a whole bunch of files, and I wish to change something like this:
My line of text
My other line of text
Into
My line of text\\
My other line of text
Seems simple, but somehow it isn't. I have tried sed s,"\n\n","\\\\\n", as well as tr '\n' '\\' and about 20 other incarnations of these commands.
There must be something going on which I don't understand... but I'm completely lost as to why nothing is working. I've had some comical things happening too, like when cat'ing out the file, it doesn't print newlines, only writes over where the rest was written.
Does anyone know how to accomplish this?
sed works on lines. It fetches a line, applies your code to it, fetches the next line, and so forth. Since lines are treated individually, multiline regexes don't work quite so easily.
In order to use multiline regexes with sed, you have to first assemble the file in the pattern space and then work on it:
sed ':a $!{ N; ba }; s/\n\n/\\\\\n/g' filename
The trick here is the
:a $!{ N; ba }
This works as follows:
:a # jump label for looping
$!{ # if the end of the input has not been reached
N # fetch the next line and append it to what we already have
ba # go to :a
}
Once this is over, the whole file is in the pattern space, and multiline regexes can be applied to it. Of course, this requires that the file is small enough to fit into memory.
sed is line-oriented and so is inappropriate to try to use on problems that span lines. You just need to use a record-oriented tool like awk:
$ awk -v RS='^$' -v ORS= '{gsub(/\n\n/,"\\\\\n")}1' file
My line of text\\
My other line of text
The above uses GNU awk for multi-char RS.
Here is an awk that solve this:
If the the blank lines could contains tabs or spaces, user this:
awk '!NF{a=a"//"} b{print a} {a=$0;b=NF} END {print a}' file
My line of text//
My other line of text
If blank line is just blank with nothing, this should do:
awk '!NF{a=a"//"} a!=""{print a} {a=$0} END {print a}' file
This might work for you (GNU sed):
sed 'N;s|\n$|//|;P;D' file
This keeps 2 lines in the pattern space at any point in time and replaces an empty line by a double slash.
I'm very much a junior when it comes to the sed command, and my Bruce Barnett guide sits right next to me, but one thing has been troubling me. With a file, can you filter it using sed to select only specific items? For example, in the following file:
alpha|november
bravo|october
charlie|papa
alpha|quebec
bravo|romeo
charlie|sahara
Would it be possible to set a command to return only the bravos, like:
bravo|october
bravo|romeo
With sed:
sed '/^bravo|/!d' filename
Alternatively, with grep (because it's sort of made for this stuff):
grep '^bravo|' filename
or with awk, which works nicely for tabular data,
awk -F '|' '$1 == "bravo"' filename
The first two use a regular expression, selecting those lines that match it. In ^bravo|, ^ matches the beginning of the line and bravo| the literal string bravo|, so this selects all lines that begin with bravo|.
The awk way splits the line across the field separator | and selects those lines whose first field is bravo.
You could also use a regex with awk:
awk '/^bravo|/' filename
...but I don't think this plays to awk's strengths in this case.
Another solution with sed:
sed -n '/^bravo|/p' filename
-n option => no printing by default.
If line begins with bravo|, print it (p)
2 way (at least) with sed
removing unwanted line
sed '/^bravo\|/ !d' YourFile
Printing only wanted lines
sed -n '/^bravo\|/ p' YourFile
if no other constraint or action occur, both are the same and a grep is better.
If there will be some action after, it could change the performance where a d cycle directly to the next line and a p will print then continue the following action.
Note the escape of pipe is needed for GNU sed, not on posix version
I have a huge file that contains lines that follow this format:
New-England-Center-For-Children-L0000392290
Southboro-Housing-Authority-L0000392464
Crew-Star-Inc-L0000391998
Saxony-Ii-Barber-Shop-L0000392491
Test-L0000392334
What I'm trying to do is narrow it down to just this:
New-England-Center-For-Children
Southboro-Housing-Authority
Crew-Star-Inc
Test
Can anyone help with this?
Using GNU awk:
awk -F\- 'NF--' OFS=\- file
New-England-Center-For-Children
Southboro-Housing-Authority
Crew-Star-Inc
Saxony-Ii-Barber-Shop
Test
Set the input and output field separator to -.
NF contains number of fields. Reduce it by 1 to remove the last field.
Using sed:
sed 's/\(.*\)-.*/\1/' file
New-England-Center-For-Children
Southboro-Housing-Authority
Crew-Star-Inc
Saxony-Ii-Barber-Shop
Test
Simple greedy regex to match up to the last hyphen.
In replacement use the captured group and discard the rest.
Version 1 of the Question
The first version of the input was in the form of HTML and parts had to be removed both before and after the desired text:
$ sed -r 's|.*[A-Z]/([a-zA-Z-]+)-L0.*|\1|' input
Special-Restaurant
Eliot-Cleaning
Kennedy-Plumbing
Version 2 of the Question
In the revised question, it is only necessary to remove the text that starts with -L00:
$ sed 's|-L00.*||' input2
New-England-Center-For-Children
Southboro-Housing-Authority
Crew-Star-Inc
Saxony-Ii-Barber-Shop
Test
Both of these commands use a single "substitute" command. The command has the form s|old|new|.
The perl code for this would be: perl -nle'print $1 if(m{-.*?/(.*?-.*?)-})
We can break the Regex down to matching the following:
- for that's between the city and state
.*? match the smallest set of character(s) that makes the Regex work, i.e. the State
/ matches the slash between the State and the data you want
( starts the capture of the data you are interested in
.*?-.*? will match the data you care about
) will close out the capture
- will match the dash before the L####### to give the regex something to match after your data. This will prevent the minimal Regex from matching 0 characters.
Then the print statement will print out what was captured (your data).
awk likes these things:
$ awk -F[/-] -v OFS="-" '{print $(NF-3), $(NF-2)}' file
Special-Restaurant
Eliot-Cleaning
Kennedy-Plumbing
This sets / and - as possible field separators. Based on them, it prints the last_field-3 and last_field-2 separated by the delimiter -. Note that $NF stands for last parameter, hence $(NF-1) is the penultimate, etc.
This sed is also helpful:
$ sed -r 's#.*/(\w*-\w*)-\w*\.\w*</loc>$#\1#' file
Special-Restaurant
Eliot-Cleaning
Kennedy-Plumbing
It selects the block word-word after a slash / and followed with word.word</loc> + end_of_line. Then, it prints back this block.
Update
Based on your new input, this can make it:
$ sed -r 's/(.*)-L\w*$/\1/' file
New-England-Center-For-Children
Southboro-Housing-Authority
Crew-Star-Inc
Saxony-Ii-Barber-Shop
Test
It selects everything up to the block -L + something + end of line, and prints it back.
You can use also another trick:
rev file | cut -d- -f2- | rev
As what you want is every slice of - separated fields, let's get all of them but last one. How? By reversing the line, getting all of them from the 2nd one and then reversing back.
Here's how I'd do it with Perl:
perl -nle 'm{example[.]com/bp/(.*?)/(.*?)-L\d+[.]htm} && print $2' filename
Note: the original question was matching input lines like this:
<loc>http://www.example.com/bp/Lowell-MA/Special-Restaurant-L0000423916.htm</loc>
<loc>http://www.example.com/bp/Houston-TX/Eliot-Cleaning-L0000422797.htm</loc>
<loc>http://www.example.com/bp/New-Orleans-LA/Kennedy-Plumbing-L0000423121.htm</loc>
The -n option tells Perl to loop over every line of the file (but not print them out).
The -l option adds a newline onto the end of every print
The -e 'perl-code' option executes perl-code for each line of input
The pattern:
/regex/ && print
Will only print if the regex matches. If the regex contains capture parentheses you can refer to the first captured section as $1, the second as $2 etc.
If your regex contains slashes, it may be cleaner to use a different regex delimiter ('m' stands for 'match'):
m{regex} && print
If you have a modern Perl, you can use -E to enable modern feature and use say instead of print to print with a newline appended:
perl -nE 'm{example[.]com/bp/(.*?)/(.*?)-L\d+[.]htm} && say $2' filename
This is very concise in Perl
perl -i.bak -lpe's/-[^-]+$//' myfile
Note that this will modify the input file in-place but will keep a backup of the original data in called myfile.bak
I am very new to sed so please bear with me... I have a file with contents like
a=1
b=2,3,4
c=3
d=8
.
.
I want to append 'x' to a line which starts with 'c=' and does not contain an 'x'. What I am using right now is
sed -i '/^c=/ s/$/x/'
but this does not cover the second part of my explanation, the 'x' should only be appended if the line did not have it already and hence if I run the command twice it makes the line "c=3xx" which I do not want.
Any help here would be highly appreciated and I know there are a lot of sharp heads around here :) I understand that this can be handled pretty easily through bash but using sed here is a hard requirement.
You can do something like this:
sed -i '/^c=/ {/x/b; s/$/x/}'
Curly brackets are used for grouping. The b command branches to the end of the script (stops the processing of the current line).
b label
Branch to label; if label is omitted, branch to end of script.
Edit: as William Pursell suggests in the comment, a shorter version would be
sed -i '/^c=/ { /x/ !s/$/x/ }'
awk is probably a better choice here as you can easily combine regular expression matches with logical operators. Given the input:
$ cat file
a=1
b=2,3,4
c=3
c=x
c=3
d=8
The command would be:
$ awk '/^c=/ && !/x/ {$0=$0"x"; print $0}' file
a=1
b=2,3,4
c=3x
c=x
c=3x
d=8
Where $0 is the awk variable that contains the current line being read.
This might work for you (GNU sed):
sed -i '/^c=[^x]*$/s/$/x/' file
or:
sed -i 's/^c=[^x]*$/&x/' file