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
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.
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 text file that consists of 45999 lines. Each line has a word (unigram). I want to create two-sequential words (bigrams). For example:
apple
pie
red
vine
I want 'apple pie', 'pie red', 'red vine'. I tried with sed 'N;s/\n/ /' but it creates just 'apple pie' and 'red vine'. How can I solve this problem? Thank you..
Could you please try following if you are ok with awk.
awk -v RS="" '
BEGIN{
OFS=","
s1="\047"
}
{
for(i=2;i<=NF;i++){
print s1 $(i-1) s1, s1 $i s1
}
}' Input_file
Output will be as follows.
'apple','pie'
'pie','red'
'red','vine'
2nd solution: since output of OP is not clear so adding this one too.
awk -v RS="" '
BEGIN{
OFS=","
s1="\047"
}
{
for(i=2;i<=NF;i++){
val=(val?val OFS:"")s1 $(i-1) s1 OFS s1 $i s1
}
}
END{
print val
}' Input_file
Output will be as follows.
'apple','pie','pie','red','red','vine'
This might work for you (GNU sed):
sed -nE 'N;s/\n(.*)/ \1&/;P;D' file
Append the next line to the current line, then replace the newline by a space and append the second line again. Print/delete the first line and repeat.
N.B. This does not print the last line as it is not a pair, if the last line is needed use:
sed -E 'N;s/\n(.*)/ \1&/;P;D' file
If the output is to be printed as a single line with each pair surrounded by single quotes and separated by a comma, use:
sed -E ':a;$!N;s/(\S+)\n(.*)/'\''\1 \2'\'', \2/;ta;s/ (\S+)$/ '\''\1'\''/' file
Or:
sed -E ':a;$!N;s/(\S+)\n(.*)/'\''\1 \2'\'', \2/;ta;s/, \S+$/' file
I have a file split in blocks like the following:
AGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTGGGG
AGGTAGTTATTATTTTTTTGGTTTTTAGTATTTAATTGAGTGTTT
ATGTAGGTGTTTATGTATTAGTTTTTTTTAGGTTTAGGGTGTTGT
ATTTAGGTTTTGTGTTTTGTGTATTATTGAATTTAATTAAAGTTA
AGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTT
AGTTTTTTTTTATTTGTCGGGATATTTTAGTTGATTTTAGATTGC
TATATTTTTAGTTTCGATTCGTCGTAAGTTTTATTTTTTTTTAAT
GGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTTT
I've truncated/wrapped the lines for clarity's sake, but imagine very long lines. The point of my question is that I want a final file that looks like this:
AGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTGGGGAGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTT
AGGTAGTTATTATTTTTTTGGTTTTTAGTATTTAATTGAGTGTTTAGTTTTTTTTTATTTGTCGGGATATTTTAGTTGATTTTAGATTGC
ATGTAGGTGTTTATGTATTAGTTTTTTTTAGGTTTAGGGTGTTGTTATATTTTTAGTTTCGATTCGTCGTAAGTTTTATTTTTTTTTAAT
ATTTAGGTTTTGTGTTTTGTGTATTATTGAATTTAATTAAAGTTAGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTTT
Where this new block has:
the same number of lines as the initial blocks,
each of the lines of the resulting block is a concatenation of the lines with the same line-number in the initial blocks.
this concatenation should be in-order (i.e. "1st line of 1st block" + "1st line of 2nd block", etc
Is it possible to achieve this final block using sed and/or awk, could you show me how it could be done?
In bash with paste:
$ paste <(head -4 file) <(tail -4 file) | tr -d '\t'
AGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTGGGGAGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTT
AGGTAGTTATTATTTTTTTGGTTTTTAGTATTTAATTGAGTGTTTAGTTTTTTTTTATTTGTCGGGATATTTTAGTTGATTTTAGATTGC
ATGTAGGTGTTTATGTATTAGTTTTTTTTAGGTTTAGGGTGTTGTTATATTTTTAGTTTCGATTCGTCGTAAGTTTTATTTTTTTTTAAT
ATTTAGGTTTTGTGTTTTGTGTATTATTGAATTTAATTAAAGTTAGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTTT
try this:
awk -vOFS="" '$0{a[NR]=$0}END{for(i=1;i<=NR/2;i++)print a[i],a[i+5]}' file
test with your example:
kent$ cat tmp.txt
AGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTGGGG
AGGTAGTTATTATTTTTTTGGTTTTTAGTATTTAATTGAGTGTTT
ATGTAGGTGTTTATGTATTAGTTTTTTTTAGGTTTAGGGTGTTGT
ATTTAGGTTTTGTGTTTTGTGTATTATTGAATTTAATTAAAGTTA
AGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTT
AGTTTTTTTTTATTTGTCGGGATATTTTAGTTGATTTTAGATTGC
TATATTTTTAGTTTCGATTCGTCGTAAGTTTTATTTTTTTTTAAT
GGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTTT
kent$ awk -vOFS="" '$0{a[NR]=$0}END{for(i=1;i<=NR/2;i++)print a[i],a[i+5]}' tmp.txt
AGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTGGGGAGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTT
AGGTAGTTATTATTTTTTTGGTTTTTAGTATTTAATTGAGTGTTTAGTTTTTTTTTATTTGTCGGGATATTTTAGTTGATTTTAGATTGC
ATGTAGGTGTTTATGTATTAGTTTTTTTTAGGTTTAGGGTGTTGTTATATTTTTAGTTTCGATTCGTCGTAAGTTTTATTTTTTTTTAAT
ATTTAGGTTTTGTGTTTTGTGTATTATTGAATTTAATTAAAGTTAGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTTT
awk -F'\n' -v RS= '{for (i=1;i<=NF;i++) {str[i] = str[i] $i} END {for (i=1;i<=NF;i++) print str[i]}' file
I need help with using sed to comment a matching lines and 4 lines which follows it.
in a text file.
my text file is like this:
[myprocess-a]
property1=1
property2=2
property3=3
property4=4
[anotherprocess-b]
property1=gffgg
property3=gjdl
property2=red
property4=djfjf
[myprocess-b]
property1=1
property4=4
property2=2
property3=3
I want to prefix # to all the lines having text '[myprocess' and 4 lines that follows it
expected output:
#[myprocess-a]
#property1=1
#property2=2
#property3=3
#property4=4
[anotherprocess-b]
property1=gffgg
property3=gjdl
property2=red
property4=djfjf
#[myprocess-b]
#property1=1
#property4=4
#property2=2
#property3=3
Greatly appreciate your help on this.
You can do this by applying a regular expression to a set of lines:
sed -e '/myprocess/,+4 s/^/#/'
This matches lines with 'myprocess' and the 4 lines after them. For those 4 lines it then inserts a '#' at the beginning of the line.
(I think this might be a GNU extension - it's not in any of the "sed one liner" cheatsheets I know)
sed '/\[myprocess/ { N;N;N;N; s/^/#/gm }' input_file
Using string concatenation and default action in awk.
http://www.gnu.org/software/gawk/manual/html_node/Concatenation.html
awk '/myprocess/{f=1} f>5{f=0} f{f++; $0="#" $0} 1' foo.txt
or if the block always ends with empty line
awk '/myprocess/{f=1} !NF{f=0} f{$0="#" $0} 1' foo.txt