I have a tab file with two columns like below
BB_12 100_AA
BB_13 101_AB
BB_14 102_AD
BB_15 103_AC
I wish to remove the number_ in second column (replace number_ with nothing). For this I tried sed replace in the following ways unsuccessfully.
sed 's/\d+\_//g' infile
sed 's/(\d+\_)//g' infile
But none of the tweaks worked. It looks like it is not searching in 2nd column. How to modify this ? The expected output is
BB_12 AA
BB_13 AB
BB_14 AD
BB_15 AC
Thanks in advance.
You may just process the last column with sed:
sed -E 's/[^ ]*_([^ ]*) *$/\1/' file
The output:
BB_12 AA
BB_13 AB
BB_14 AD
BB_15 AC
Awk alternative:
awk '{ sub(/^[^ ]+_/, "", $2) }1' OFS='\t' file
Following simple sed may help you in same.
sed 's/\([^ ]*\) \([^_]*\)_\(.*\)/\1 \3/g' Input_file
Output will be as follows.
BB_12 AA
BB_13 AB
BB_14 AD
BB_15 AC
Related
To delete/comment 3 lines befor a pattern (including the line with the pattern):
how can i achive it through sed command
Ref:
sed or awk: delete n lines following a pattern
the above ref blog help to achive the this with after a pattern match but i need to know before match
define host{
use xxx;
host_name pattern;
alias yyy;
address zzz;
}
the below sed command will comment the '#' after the pattern match for example
sed -e '/pattern/,+3 s/^/#/' file.cfg
define host{
use xxx;
#host_name pattern;
#alias yyy;
#address zzz;
#}
like this how can i do this for the before pattern?
can any one help me to resolve this
If tac is allowed :
tac|sed -e '/pattern/,+3 s/^/#/'|tac
If tac isn't allowed :
sed -e '1!G;h;$!d'|sed -e '/pattern/,+3 s/^/#/'|sed -e '1!G;h;$!d'
(source : http://sed.sourceforge.net/sed1line.txt)
Reverse the file, comment the 3 lines after, then re-reverse the file.
tac file | sed '/pattern/ {s/^/#/; N; N; s/\n/&#/g;}' | tac
#define host{
#use xxx;
#host_name pattern;
alias yyy;
address zzz;
}
Although I think awk is a little easier to read:
tac file | awk '/pattern/ {c=3} c-- > 0 {$0 = "#" $0} 1' | tac
This might work for you (GNU sed):
sed ':a;N;s/\n/&/3;Ta;/pattern[^\n]*$/s/^/#/mg;P;D' file
Gather up 4 lines in the pattern space and if the last line contains pattern insert # at the beginning of each line in the pattern space.
To delete those 4 lines, use:
sed ':a;N;s/\n/&/3;Ta;/pattern[^\n]*$/d;P;D' file
To delete the 3 lines before pattern but not the line containing pattern use:
sed ':a;N;s/\n/&/3;Ta;/pattern[^\n]*$/s/.*\n//;P;D'
I have lines that start like this: 2141058222 11/22/2017 and I want to append a ; at the end of the ten digit number like this: 2141058222; 11/22/2017.
I've tried sed with sed -i 's/^[0-9]\{10\}\\$/;&/g' which does nothing.
What am I missing?
Try this:
echo "2141058222 11/22/2017" | sed -r 's/^([0-9]{10})/&;/'
echo "2141058222 11/22/2017" | sed 's/ /; /'
Output:
2141058222; 11/22/2017
If the input is always in the format specified, GNU cut works, and might even be more efficient than sed:
cut -c -10,11- --output-delimiter ';' <<< "2141058222 11/22/2017"
Output:
2141058222; 11/22/2017
For an input file that'd be:
cut -c -10,11- --output-delimiter ';' file
My text looks like this:
cat
catch
cat_mouse
catty
I want to replace "cat" with "dog".
When I do
sed "s/cat/dog/"
my result is:
dog
catch
cat_mouse
catty
How do I replace with sed if only part of the word matches?
There's a mistake :
You lack the g modifier
sed 's/cat/dog/g'
g
Apply the replacement to all matches to the regexp, not just the first.
See
http://www.gnu.org/software/sed/manual/html_node/The-_0022s_0022-Command.html
http://sed.sourceforge.net/sedfaq3.html#s3.1.3
If you want to replace only cat by dog only if part of the word matches :
$ perl -pe 's/cat(?=.)/dog/' file.txt
cat
dogch
dog_mouse
dogty
I use Positive Look Around, see http://www.perlmonks.org/?node_id=518444
If you really want sed :
sed '/^cat$/!s/cat/dog/' file.txt
bash-3.00$ cat t
cat
catch
cat_mouse
catty
To replace cat only if it is part of a string
bash-3.00$ sed 's/cat\([^$]\)/dog\1/' t
cat
dogch
dog_mouse
dogty
To replace all occurrences of cat:
bash-3.00$ sed 's/cat/dog/' t
dog
dogch
dog_mouse
dogty
awk solution for this
awk '{gsub("cat","dog",$0); print}' temp.txt
How would like to join two lines usung awk or sed?
For example, I have data like below:
abcd
12:12:12:12:12:12:12:12
efgh001_01
45:45:45:45:45:45:45:45
ijkl7464746
78:78:78:78:78:78:78:78
and I need output like below:
abcd 12:12:12:12:12:12:12:12
efgh001_01 45:45:45:45:45:45:45:45
ijkl7464746 78:78:78:78:78:78:78:78
Running this almost works, but I need the space or tab:
awk '!(NR%2){print$0p}{p=$0}'
You're almost there:
awk '(NR % 2 == 0) {print p, $0} {p = $0}'
With sed you can do that as follows:
sed -n 'N;s/\n/ /p' file
where:
N reads next line
s replaces the new line character with a space to join both lines properly
p prints the result
This might work for you:
sed '$!N;s/\n/ /' file
or this:
paste -sd' \n' file
I want to print next line of matching word with sed.
I tried this command but it gives error :
sed -n '/<!\[CDATA\[\]\]>/ { N p}/' test.xml
what about grep -e -A 1 regex? It will print line below regex.
With sed, looking for pattern "dd", below works fine as you would:
sed -n '/dd/ {n;p}' file
For file content:
dd
aa
ss
aa
It prints:
aa
use awk
awk '/pattern/{getline;print}' file