This question already has answers here:
how can I combine these lines
(4 answers)
Closed 8 years ago.
I want convert this text on a given file:
87665
S
3243423
S
334243
N
...
to something like this:
87665,S
3243423,S
334243,N
...
I've been reading some similar questions, but it didn't work... is there a way to do this with a single line command in linux?
Thanks!
Using sed:
sed '$!N;s/\n/,/' filename
Using paste:
paste -d, - - < filename
paste would leave a trailing , in case the input has an odd number of lines.
Something like this might work for you:
$ awk 'NR%2{a=$0;next}{print a","$0}' file
87665,S
3243423,S
334243,N
To handle files with odd lines, you can do:
awk '{printf "%s%s", $0, NR%2?",":ORS}' file
Just for fun, a pure bash solution:
while IFS= read -r l1; do
read -r l2
printf '%s\n' "$l1${l2:+,$l2}"
done < file
If there's an odd number of lines, the last line will not have a trailing comma.
Related
This question already has answers here:
Are shell scripts sensitive to encoding and line endings?
(14 answers)
Closed last year.
I am new on coding in bash Linux and I have the following problem.
I'm trying to concatenate string in loop to create a path. I have a text file in which I stored some strings to use in the loop. I wrote this example just to show you the problem:
for bio in `cat /data/giordano/species_ranges/prova_bio.txt` # list of strings: "bio_01", "bio_02"...
do
echo /data/giordano/species_range/$bio.tif # concatenation
done
The result I expect would be:
/data/giordano/species_range/bio_01.tif
/data/giordano/species_range/bio_02.tif
/data/giordano/species_range/bio_03.tif
But what actually came out was:
.tifa/giordano/species_range/bio_01
.tifa/giordano/species_range/bio_02
.tifa/giordano/species_range/bio_03
/data/giordano/species_range/bio_04.tif
I really don't understand what kind of problem it is...
I suggest that awk would be simpler for this task. We use tr to remove the Cr line endings
~/tests/bash $ tr -d "\r" < data/giordano/species_range/proverbio.txt | awk '{ print "/data/giordano/species_range/" $0 ".tif"
> }'
/data/giordano/species_range/bio_1.tif
/data/giordano/species_range/bio_2.tif
/data/giordano/species_range/bio_3.tif
/data/giordano/species_range/bio_4.tif
Thank you to Charles Duffy for the improvements.
You probably have Windows line endings in your file, which contain an additional carriage return (\r). This makes the cursor go to the beginning of the line. You can remove the \rs from your file by piping to tr. Extend your first line like this:
for bio in `cat /data/giordano/species_ranges/prova_bio.txt | tr -d '\r'`
This question already has answers here:
How to save both matching and non-matching from grep
(3 answers)
Closed 1 year ago.
Currently My shell script iterate the lines in one huge file two times:
(What I want to do is just like the shell script below.)
grep 'some_text' huge_file.txt > lines_contains_a.txt
grep -v 'some_text' huge_file.txt > lines_not_contains_a.txt
but it is slow.
How to do the same thing only iterate the lines once?
Thanks!
With GNU awk:
awk '/some_text/ { print >> "lines_contains_a.txt" }
!/some_text/ { print >> "lines_not_contains_a.txt" }' huge_file.txt
With sed:
sed -n '/some_text/ w lines_contains_a.txt
/some_text/! w lines_not_contains_a.txt' huge_file.txt
This question already has an answer here:
How can I extract the content between two brackets?
(1 answer)
Closed 4 years ago.
I have a large log file I need to sort, I want to extract the text between parentheses. The format is something like this:
<#44541545451865156> (example#6144) has left the server!
How would I go about extracting "example#6144"?
This sed should work here:
sed -E -n 's/.*\((.*)\).*$/\1/p' file_name
There are many ways to skin this cat.
Assuming you always have only one lexeme in parentheses, you can use bash parameter expansion:
while read t; do echo $(t=${t#*(}; echo ${t%)*}); done <logfile
The first substitution: ${t#*(} cuts off everything up and including the left parenthesis, leaving you with example#6144) has left the server!; the second one: ${t%)*} cuts off the right parenthesis and everything after that.
Alternatively, you can also use awk:
awk -F'[)(]' '{print $2}' logfile
-F'[)(]' tells awk to use either parenthesis as the field delimiter, so it splits the input string into three tokens: <#44541545451865156>, example#6144, and has left the server!; then {print $2} instructs it to print the second token.
cut would also do:
cut -d'(' -f 2 logfile | cut -d')' -f 1
Try this:
sed -e 's/^.*(\([^()]*\)).*$/\1/' <logfile
The /^.*(\([^()]*\)).*$/ is a regular expression or regex. Regexes are hard to read until you get used to them, but are most useful for extracting text by pattern, as you are doing here.
This question already has answers here:
How to pass a variable containing slashes to sed
(7 answers)
Combining two sed commands
(2 answers)
Linux, find replace on a folder of files using a list of items for replacement?
(1 answer)
Closed 4 years ago.
I want to read file using a bash script and delete line(s) which are matching with my specific scenario (line(s) starting with 'z').
my code works fine if the 'inputFile' contains only alphabetic characters.
but, if a line with 'specific characters of sed' (line eg : z-2.10.3.2 x/y/z F (&)[]+* ) then i got an error,(error : sed: -e expression #1, char 29: unterminated `y' command).
#!/bin/bash
inputFile="test.txt"
while IFS= read -r line
do
echo "$line"
if [[ $line == z* ]];
then
sed -i "/$line/d" $inputFile
fi
done < "$inputFile"
i want to delete 'z-2.10.3.2 x/y/z F (&)[]+*' kind of lines, how can i do this...?
As you mentioned you don't need line which has z*
Simply use grep -v
grep -vE "^[[:blank:]]*z" file
I have created one scenario where I have a file which contains
root#ubuntu:~/T/e/s/t# cat file
hello world
sample line 1
sample line 2 world
sample line 3
sample line 4
In my case, I want to remove the line contains "world"
root#ubuntu:~/T/e/s/t# grep -v "world" file
sample line 1
sample line 3
sample line 4
If you want you can redirect your output in another file.
This question already has answers here:
Can grep show only words that match search pattern?
(15 answers)
Closed 5 years ago.
I have a requirement of searching a pattern like x=<followed by any values> from a file and displaying the pattern i.e x=<followed by any values>, only in the screen, not the whole line. How can I do it in Linux?
I have 3 answers, from simple (but with caveats) to complex (but foolproof):
1) If your pattern never appears more than once per line, you could do this (assuming your shell is
PATTERN="x="
sed "s/.*\($PATTERN\).*/\1/g" your_file | grep "$PATTERN"
2) If your pattern can appear more than once per line, it's a bit harder. One easy but hacky way to do this is to use a special characters that will not appear on any line that has your pattern, eg, "#":
PATTERN="x="
SPECIAL="#"
grep "$PATTERN" your_file | sed "s/$PATTERN/$SPECIAL/g" \
| sed "s/[^$SPECIAL]//g" | sed "s/$SPECIAL/$PATTERN/g"
(This won't separate the output pattern per line, eg. you'll see x=x=x= if a source line had 3 times "x=", this is easy to fix by adding a space in the last sed)
3) Something that always works no matter what:
PATTERN="x="
awk "NF>1{for(i=1;i<NF;i++) printf FS; print \"\"}" \
FS="$PATTERN" your_file