I want that rows to rearrangement to columns with spaces between the strings.
q9I6OEFg003411
q9I5IHv5006818
q9I6Gi6P024439
q9I5RoA0019541
Expected view:
q9I6OEFg003411 q9I5IHv5006818 q9I6Gi6P024439 q9I5RoA0019541
You can use tr to translate <newline> to <space>:
tr '\n' ' ' < file
Also in sed:
sed -n '1h;1!H;${g;s/\n/ /g;p}' file
In awk:
awk -vORS=' ' 1 file
If file is small, you can use cat:
echo `cat file`
If you know vim:
:%s/\n/ /
maybe more ...
Using paste command:
paste -sd" " file
d option to set delimiter.
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 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
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 have a tab delimited file like this (without the headers and in the example I use the pipe character as delimiter for clarity)
ID1|ID2|VAL1|
1|1|3
1|1|4
1|2|3
1|2|5
2|2|6
I want add a new field to this file that changes whenever ID1 or ID2 change. Like this:
1|1|3|1
1|1|4|1
1|2|3|2
1|2|5|2
2|2|6|3
Is this possible with an one liner in sed,awk, perl etc... or should I use a standard programming language (Java) for this task. Thanks in advance for your time.
Here is an awk
awk -F\| '$1$2!=a {f++} {print $0,f;a=$1$2}' OFS=\| file
1|1|3|1
1|1|4|1
1|2|3|2
1|2|5|2
2|2|6|3
Simple enough with bash, though I'm sure you could figure out a 1-line awk
#!/bin/bash
count=1
while IFS='|' read -r id1 id2 val1; do
#Can remove next 3 lines if you're sure you won't have extraneous whitespace
id1="${id1//[[:space:]]/}"
id2="${id2//[[:space:]]/}"
val1="${val1//[[:space:]]/}"
[[ ( -n $old1 && $old1 -ne $id1 ) || ( -n $old2 && $old2 -ne $id2 ) ]] && ((count+=1))
echo "$id1|$id2|$val1|$count"
old1="$id1" && old2="$id2"
done < file
For example
> cat file
1|1|3
1|1|4
1|2|3
1|2|5
2|2|6
> ./abovescript
1|1|3|1
1|1|4|1
1|2|3|2
1|2|5|2
2|2|6|3
Replace IFS='|' with IFS=$'\t' for tab delimited
Using awk
awk 'FNR>1{print $0 FS (++a[$1$2]=="1"?++i:i)}' FS=\| file
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