I am trying to Replace a character with #(hash symble) only in 5th & 6th field.
eg. I have to replace 'Z' with '#' only in 5th & 6th field (using perl or AWK script). And remaining fields containng 'Z' symbol should not be affected.
(just I'm updating the post to replace double quote(") instead of Z by #. Can I achive this? thanks for precious help)
eg: i/p file:
aa",bb,ccc,ddd,eee",ddd",fff
aa1",ba1,ccc1,"ddd1,eee"1,ddd1,fff1
z,aa2,bb2",ccc2,ddd2","eee2",ddd2,fff2"
Expected O/p file:
aa",bb,ccc,ddd,eee#,ddd#,fff
aa1",ba1,ccc1,#ddd1,eee#1,ddd1,fff1
aa2,bb2",ccc2,ddd2#,#eee2#,ddd2,fff2"
Thanks.
$ awk 'BEGIN{FS=OFS=","} {for (i=5;i<=6;i++) gsub(/Z/,"#",$i)} 1' file
x,aaZ,bb,ccc,ddd,eee#,dddZ,fff
y,aa1Z,ba1,ccc1,#ddd1,eee#1,ddd1,fff1
z,aa2,bb2Z,ccc2,ddd2#,#eee2,ddd2,fff2Z
Since its only two filed, loop can be omitted.
awk -F, -v OFS=, '{gsub(/Z/,"#",$5);gsub(/Z/,"#",$6)} 1' file
x,aaZ,bb,ccc,ddd,eee#,dddZ,fff
y,aa1Z,ba1,ccc1,#ddd1,eee#1,ddd1,fff1
z,aa2,bb2Z,ccc2,ddd2#,#eee2,ddd2,fff2Z
To replace " in fifth and sixth field:
awk -F, -v OFS=, '{gsub(/\"/,"#",$5);gsub(/\"/,"#",$6)} 1' file
aa",bb,ccc,ddd,eee#,ddd#,fff
aa1",ba1,ccc1,"ddd1,eee#1,ddd1,fff1
z,aa2,bb2",ccc2,ddd2#,#eee2#,ddd2,fff2"
Here is a Perl way to do the job:
perl -anF, -e '$"=","; s/Z/#/ for (#F)[4,5];print"#F";' < in1.txt
If you have mutiple Z in a field, you could use:
perl -anF, -e '$"=","; s/Z/#/g for (#F)[4,5];print"#F";' < in1.txt
Output:
aaZ,bb,ccc,ddd,eee#,ddd#,fff
aa1Z,ba1,ccc1,Zddd1,eee#1,ddd1,fff1
aa2,bb2Z,ccc2,ddd2Z,#eee2,ddd2,fff2Z
Edit according to comment:
in1.txt
aa",bb,ccc,ddd,eee",ddd",fff
aa1",ba1,ccc1,"ddd1,eee"1,ddd1,fff1
aa2,bb2",ccc2,ddd2","eee2,ddd2,fff2"
Command:
perl -anF'','' -e '$"=",";s/"/#/ for (#F)[4,5];print"#F";' < in1.txt
result:
aa",bb,ccc,ddd,eee#,ddd#,fff
aa1",ba1,ccc1,"ddd1,eee#1,ddd1,fff1
aa2,bb2",ccc2,ddd2",#eee2,ddd2,fff2"
Related
I have a file with text as follows:
###interest1 moreinterest1### sometext ###interest2###
not-interesting-line
sometext ###interest3###
sometext ###interest4### sometext othertext ###interest5### sometext ###interest6###
I want to extract all strings between ### .
My desired output would be something like this:
interest1 moreinterest1
interest2
interest3
interest4
interest5
interest6
I have tried the following:
grep '###' file.txt | sed -e 's/.*###\(.*\)###.*/\1/g'
This almost works but only seems to grab the first instance per line, so the first line in my output only grabs
interest1 moreinterest1
rather than
interest1 moreinterest1
interest2
Here is a single awk command to achieve this that makes ### field separator and prints each even numbered field:
awk -F '###' '{for (i=2; i<NF; i+=2) print $i}' file
interest1 moreinterest1
interest2
interest3
interest4
interest5
interest6
Here is an alternative grep + sed solution:
grep -oE '###[^#]*###' file | sed -E 's/^###|###$//g'
This assumes there are no # characters in between ### markers.
With GNU awk for multi-char RS:
$ awk -v RS='###' '!(NR%2)' file
interest1 moreinterest1
interest2
interest3
interest4
interest5
interest6
You can use pcregrep:
pcregrep -o1 '###(.*?)###' file
The regex - ###(.*?)### - matches ###, then captures into Group 1 any zero o more chars other than line break chars, as few as possible, and ### then matches ###.
o1 option will output Group 1 value only.
See the regex demo online.
sed 't x
s/###/\
/;D; :x
s//\
/;t y
D;:y
P;D' file
Replacing "###" with newline, D, then conditionally branching to P if a second replacement of "###" is successful.
This might work for you (GNU sed):
sed -n 's/###/\n/g;/[^\n]*\n/{s///;P;D}' file
Replace all occurrences of ###'s by newlines.
If a line contains a newline, remove any characters before and including the first newline, print the details up to and including the following newline, delete those details and repeat.
I need to replace all occurrences of a string after nth occurrence in every line of a Unix file.
My file data:
:account_id:12345:6789:Melbourne:Aus
:account_id:98765:43210:Adelaide:Aus
My output data:
:account_id:123456789MelbourneAus
:account_id:9876543210AdelaideAus
tried using sed: sed 's/://3g' test.txt
Unfortunately, the g option with the occurrence is not working as expected. instead, it is replacing all the occurrences.
Another approach using awk
awk -v c=':' -v n=2 'BEGIN{
FS=OFS=""
}
{
j=0;
for(i=0; ++i<=NF;)
if($i==c && j++>=n)$i=""
}1' file
$ cat file
:account_id:12345:6789:Melbourne:Aus
:account_id:98765:43210:Adelaide:Aus
$ awk -v c=':' -v n=2 'BEGIN{FS=OFS=""}{j=0;for(i=0; ++i<=NF;)if($i==c && j++>=n)$i=""}1' file
:account_id:123456789MelbourneAus
:account_id:9876543210AdelaideAus
With GNU awk, using gensub please try following. This is completely based on your shown samples, where OP wants to remove : from 3rd occurrence onwards. Using gensub to segregate parts of matched values and removing all colons from 2nd part(from 3rd colon onwards) in it as per OP's requirement.
awk -v regex="^([^:]*:)([^:]*:)(.*)" '
{
firstPart=restPart=""
firstPart=gensub(regex, "\\1 \\2", "1", $0)
restPart=gensub(regex,"\\3","1",$0)
gsub(/:/,"",restPart)
print firstPart restPart
}
' Input_file
I have inferred based on the limited data you've given us, so it's possible this won't work. But I wouldn't use regex for this job. What you have there is colon delimited fields.
So I'd approach it using split to extract the data, and then some form of string formatting to reassemble exactly what you like:
#!/usr/bin/perl
use strict;
use warnings;
while (<DATA>) {
chomp;
my ( undef, $first, #rest ) = split /:/;
print ":$first:", join ( "", #rest ),"\n";
}
__DATA__
:account_id:12345:6789:Melbourne:Aus
:account_id:98765:43210:Adelaide:Aus
This gives you the desired result, whilst IMO being considerably clearer for the next reader than a complicated regex.
You can use the perl solution like
perl -pe 's~^(?:[^:]*:){2}(*SKIP)(?!)|:~~g if /^:account_id:/' test.txt
See the online demo and the regex demo.
The ^(?:[^:]*:){2}(*SKIP)(?!)|: regex means:
^(?:[^:]*:){2}(*SKIP)(?!) - match
^ - start of string (here, a line)
(?:[^:]*:){2} - two occurrences of any zero or more chars other than a : and then a : char
(*SKIP)(?!) - skip the match and go on to search for the next match from the failure position
| - or
: - match a : char.
And only run the replacement if the current line starts with :account_id: (see if /^:account_id:/').
Or an awk solution like
awk 'BEGIN{OFS=FS=":"} /^:account_id:/ {result="";for (i=1; i<=NF; ++i) { result = result (i > 2 ? $i : $i OFS)}; print result}' test.txt
See this online demo. Details:
BEGIN{OFS=FS=":"} - sets the input/output field separator to :
/^:account_id:/ - line must start with :account_id:
result="" - sets result variable to an empty string
for (i=1; i<=NF; ++i) { result = result (i > 2 ? $i : $i OFS)}; print result} - iterates over the fields and if the field number is greater than 2, just append the current field value to result, else, append the value + output field separator; then print the result.
I would use GNU AWK following way if n fixed and equal 2 following way, let file.txt content be
:account_id:12345:6789:Melbourne:Aus
:account_id:98765:43210:Adelaide:Aus
then
awk 'BEGIN{FS=":";OFS=""}{$2=FS $2 FS;print}' file.txt
output
:account_id:123456789MelbourneAus
:account_id:9876543210AdelaideAus
Explanation: use : as field separator and nothing as output field separator, this itself does remove all : so I add : which have to be preserved: 1st (before second column) and 2nd (after second column). Beware that I tested it solely for this data, so if you would want to use it you should firstly test it with more possible inputs.
(tested in gawk 4.2.1)
This might work for you (GNU sed):
sed 's/:/\n/3;h;s/://g;H;g;s/\n.*\n//' file
Replace the third occurrence of : by a newline.
Make a copy of the line.
Delete all occurrences of :'s.
Append the amended line to the copy.
Join the two lines by removing everything from third occurrence of the copy to the third occurrence of the amended line.
N.B. The use of the newline is the best delimiter to use in the case of sed, as the line presented to seds commands are initially devoid of newlines. However the important property of the delimiter is that it is unique and therefore can be any such character as long as it is not found anywhere in the data set.
An alternative solution uses a loop to remove all :'s after the first two:
sed -E ':a;s/^(([^:]*:){2}[^:]*):/\1/;ta' file
With GNU awk for the 3rd arg to match() and gensub():
$ awk 'match($0,/(:[^:]+:)(.*)/,a){ $0=a[1] gensub(/:/,"","g",a[2]) } 1' file
:account_id:123456789MelbourneAus
:account_id:9876543210AdelaideAus
and with any awk in any shell on every Unix box:
$ awk 'match($0,/:[^:]+:/){ tgt=substr($0,1+RLENGTH); gsub(/:/,"",tgt); $0=substr($0,1,RLENGTH) tgt } 1' file
:account_id:123456789MelbourneAus
:account_id:9876543210AdelaideAus
Working with the example log file below:
1;000117;20190529;055529;9521;0988388019
1;000015;20190529;071944;2222;2231
1;000012;20190529;072734;4258;4252
1;000006;20190529;073336;2226;1000
3;000005;20190529;073715;1000;037760967
3;000004;20190529;073751;1000;037760967
I need to normalize the last column filling with spaces until they has the lenght = 25
Tryed with unsuccessful perl code:
perl -F';' -lane '$F[5] = $F[5], sprintf "% 25d"; $" = ";"; print "#F"'
I need the output below:
1;000117;20190529;055529;9521;0988388019
1;000015;20190529;071944;2222;2231
1;000012;20190529;072734;4258;4252
1;000006;20190529;073336;2226;1000
3;000005;20190529;073715;1000;037760967
3;000004;20190529;073751;1000;037760967
$ awk 'BEGIN{FS=OFS=";"} {$NF=sprintf("%-25s",$NF)}1' file
1;000117;20190529;055529;9521;0988388019
1;000015;20190529;071944;2222;2231
1;000012;20190529;072734;4258;4252
1;000006;20190529;073336;2226;1000
3;000005;20190529;073715;1000;037760967
3;000004;20190529;073751;1000;037760967
So you can see the blanks:
$ awk 'BEGIN{FS=OFS=";"} {$NF=sprintf("%-25s",$NF)}1' file | tr ' ' '#'
1;000117;20190529;055529;9521;0988388019###############
1;000015;20190529;071944;2222;2231#####################
1;000012;20190529;072734;4258;4252#####################
1;000006;20190529;073336;2226;1000#####################
3;000005;20190529;073715;1000;037760967################
3;000004;20190529;073751;1000;037760967################
You were on the right track. More successful Perl codes:
perl -F';' -lane '$F[5]=sprintf("%-25s",$F[5]);print join ";",#F'
perl -F';' -pane '$F[5]=sprintf("%-25s",$F[5]);$_=join ";",#F'
This might work for you (GNU sed):
sed -i ':a;/;[^;]\{25\}$/!s/$/ /;ta' file
If the last field is not 25 characters long, add a space until it is.
I am trying to remove all but the first character of a specific field in a .tab file. I want to keep only first character in fields 10 and 11.
Normally the fields have 35 characters in them, so I used:
awk '{gsub ("..................................$","",$10;print} file
however, there are some fields which have less than 35, and were ignored by this replace function. I tired using substring, but I cannot figure out how to make it field specific. I believe there is a way to use perl inside awk so that I can use the function
perl -pe 's/(.).*/$1/g'
but I am not sure how to do that and use the field as the input value, so the file comes out identical except for the altered field.
is there a way to do the perl equivalent with gsub, or the awk equivalent with perl?
help is appreciated!
One way using awk:
awk '{ for (i=10;i<=11;i++) { $i = substr( $i, 1, 1) } } { print }' infile
Another way using gensub function of gawk
gawk '{ for (i=10;i<=11;i++) { $i = gensub(/(.).*/ , "\\1", G , $i) } }1' infile
A shortest awk version, I could figure out:
awk '($10=substr($10,1,1))&&$11=substr($11,1,1)' infile
If the 10th and/or 11th field is not existing then the line is not printed.
Similar version in perl
perl -ane '$F[9]=~s/(.).*/$1/;$F[10]=~s/(.).*/$1/;print "#F\n"' infile
This prints the line even if 10th and/or 11th field is not defined.
Another way with perl:
perl -pe '$c=0; s/(\S+)/(++$c < 10 || $c > 11) ? $1 : substr($1,0,1)/eg' filename
I have a results.txt file that is structured in this format:
Uncharted 3: Javithaxx l Rampant l Graveyard l Team Deathmatch HD (D1VpWBaxR8c)
Matt Darey feat. Kate Louise Smith - See The Sun (Toby Hedges Remix) (EQHdC_gGnA0)
The Matrix State (SXP06Oax70o)
Above & Beyond - Group Therapy Radio 014 (guest Lange) (2013-02-08) (8aOdRACuXiU)
I want to create a new file extracting the youtube URL ID specified in the last characters in each line line "8aOdRACuXiU"
I'm trying to build a URL like this in a new file:
http://www.youtube.com/watch?v=8aOdRACuXiU&hd=1
Note, I appended the &hd=1 to the string that I am trying to be replaced. I have tried using Linux reverse and cut but reverse or rev munges my data. The hard part here is that each line in my text file will have entries with parentheses and I only care about getting the data between the last set of parentheses. Each line has a variable length so that isn't helpful either. What about using grep and .$ for the end of the line?
In summary, I want to extract the youtube ID from results.txt and export it to a new file in the following format: http://www.youtube.com/watch?v=8aOdRACuXiU&hd=1
Using awk:
awk '{
v = substr( $NF, 2, length( $NF ) - 2 )
printf "%s%s%s\n", "http://www.youtube.com/watch?v=", v, "&hd=1"
}' infile
It yields:
http://www.youtube.com/watch?v=D1VpWBaxR8c&hd=1
http://www.youtube.com/watch?v=EQHdC_gGnA0&hd=1
http://www.youtube.com/watch?v=SXP06Oax70o&hd=1
http://www.youtube.com/watch?v=8aOdRACuXiU&hd=1
$ sed 's!.*(\(.*\))!http://www.youtube.com/watch?v=\1\&hd=1!' results.txt
http://www.youtube.com/watch?v=D1VpWBaxR8c&hd=1
http://www.youtube.com/watch?v=EQHdC_gGnA0&hd=1
http://www.youtube.com/watch?v=SXP06Oax70o&hd=1
http://www.youtube.com/watch?v=8aOdRACuXiU&hd=1
Here, .*(\(.*\)) looks for the last occurrence of a pair of parentheses, and captures the characters inside those parentheses. The captured group is then inserted into the URL using \1.
Using a perl one-liner :
perl -lne 'printf "http://www.youtube.com/watch?v=%s&hd=1\n", $& if /[^\(]+(?=\)$)/' file.txt
Or multi-line version :
perl -lne '
printf(
"http://www.youtube.com/watch?v=%s&hd=1\n",
$&
) if /[^\(]+(?=\)$)/
' file.txt