I want to add a new line to the top of a data file with sed, and write something to that line.
I tried this as suggested in How to add a blank line before the first line in a text file with awk :
sed '1i\
\' ./filename.txt
but it printed a backslash at the beginning of the first line of the file instead of creating a new line. The terminal also throws an error if I try to put it all on the same line ("1i\": extra characters after \ at the end of i command).
Input :
1 2 3 4
1 2 3 4
1 2 3 4
Expected output
14
1 2 3 4
1 2 3 4
1 2 3 4
$ sed '1i\14' file
14
1 2 3 4
1 2 3 4
1 2 3 4
but just use awk for clarity, simplicity, extensibility, robustness, portability, and every other desirable attribute of software:
$ awk 'NR==1{print "14"} {print}' file
14
1 2 3 4
1 2 3 4
1 2 3 4
Basially you are concatenating two files. A file containing one line and the original file. By it's name this is a task for cat:
cat - file <<< 'new line'
# or
echo 'new line' | cat - file
while - stands for stdin.
You can also use cat together with command substitution if your shell supports this:
cat <(echo 'new line') file
Btw, with sed it should be simply:
sed '1i\new line' file
Related
I would like to extract n-th line from file and save it to a new file. For example I have index.txt :
cat index.txt
1 AAAGCGT
2 ACGAAGT
3 ACCTTGT
4 ATAATGT
5 AGGGTGT
6 AGCCAGT
7 AGTTCGT
8 AATGCAG
9 AAAGCGT
10 ACGAAGT
and output should be
cat index.1.txt:
1 AAAGCGT
2 ACGAAGT
cat index.2.txt:
3 ACCTTGT
4 ATAATGT
cat index.3.txt:
5 AGGGTGT
6 AGCCAGT
And so on.. So I would like to extract form input file first 2 rows in cycle and save to new file.
It doesn't give you exactly the names you want, but:
split -l 2 index.txt index.
seems like the easiest solution. It will create files with names beginning with the final argument, so will get names like 'index.aa' and 'index.bb'
This will work for any number of grouped lines just by changing the 2 to a 3 or whatever number you like:
$ awk 'NR%2==1{++i} {print > ("index." i ".txt")}' index.txt
$ ls index.?.txt
index.1.txt index.2.txt index.3.txt index.4.txt index.5.txt
$ tail index.?.txt
==> index.1.txt <==
1 AAAGCGT
2 ACGAAGT
==> index.2.txt <==
3 ACCTTGT
4 ATAATGT
==> index.3.txt <==
5 AGGGTGT
6 AGCCAGT
==> index.4.txt <==
7 AGTTCGT
8 AATGCAG
==> index.5.txt <==
9 AAAGCGT
10 ACGAAGT
awk '{print >"index."(x+=NR%2)".txt"}' file
This increments x every two lines starting from 1 and then prints the line into a file with that name
cat index.1.txt:
1 AAAGCGT
2 ACGAAGT
cat index.2.txt:
3 ACCTTGT
4 ATAATGT
cat index.3.txt:
5 AGGGTGT
6 AGCCAGT
In some awks, extra parens may be required as shown below (As commented by Ed Morton)
awk '{print >("index."(x+=NR%2)".txt")}' file
I would say:
awk '{file=int((NR+1)/2)".txt"; print > file}' file
int((NR+1)/2 maps every line number:
1 --> 1
2 --> 1
3 --> 2
x --> (x+1) / 2
So you get these files:
$ cat 1.txt
1 AAAGCGT
2 ACGAAGT
or
$ cat 3.txt
5 AGGGTGT
6 AGCCAGT
How to add the number of line at the end of the corresponding line into a file text in linux ?
I know this command, but it add the number of line at the beginning and not at the end !
nl -ba -s ', ' fileName.txt > fileName2.txt
e.g.
5,10,33
7,17,77
have to be
5,10,33,1
7,17,77,2
in awk:
awk '{print $0","NR;}' file
on my system:
$ cat file
1 2
3 4
5 6
7 8
$ awk '{print $0","NR;}' file
1 2,1
3 4,2
5 6,3
7 8,4
I need to move the last 4 lines of a text file and move them to the second row in the text file.
I'm assuming that tail and sed are used but, I haven't much luck so far.
Here is a head and tail solution. Let us start with the same sample file as Glenn Jackman:
$ seq 10 >file
Apply these commands:
$ head -n1 file ; tail -n4 file; tail -n+2 file | head -n-4
1
7
8
9
10
2
3
4
5
6
Explanation:
head -n1 file
Print first line
tail -n4 file
Print last four lines
tail -n+2 file | head -n-4
Print the lines starting with line 2 and ending before the fourth-to-last line.
If I'm assuming correctly, ed can handle your task:
seq 10 > file
ed file <<'COMMANDS'
$-3,$m1
w
q
COMMANDS
cat file
1
7
8
9
10
2
3
4
5
6
lines 7,8,9,10 have been moved to the 2nd line
$-3,$m1 means, for the range of lines from "$-3" (3 lines before the last line) to "$" (the last line, move them ("m") below the first line ("1")
Note that the heredoc has been quoted so the shell does not try to interpret the strings $- and $m1 as variables
If you don't want to actually modify the file, but instead print to stdout:
ed -s file <<'COMMANDS'
$-3,$m1
%p
Q
COMMANDS
Here is an awk solution:
seq 10 > file
awk '{a[NR]=$0} END {for (i=1;i<=NR-4;i++) if (i==2) {for (j=NR-3;j<=NR;j++) print a[j];print a[i]} else print a[i]}' file
1
7
8
9
10
2
3
4
5
6
I have a file with many lines of tab separated data in the following format:
1 1 2 2
3 3 4 4
5 5 6 6
...
and I would like to change the format to:
1 1
2 2
3 3
4 4
5 5
6 6
Is there a not too complicated way to do this? I don't have any experience with using awk, sed, etc.
Thanks
If you just want to group your file in blocks of X columns, you can make use of xargs -nX:
$ xargs -n2 < file
1 1
2 2
3 3
4 4
5 5
6 6
To have more control and print an empty line after 4th field, you can also use this awk:
$ awk 'BEGIN{FS=OFS="\t"} {for (i=1;i<=NF;i++) printf "%s%s", $i, (i%2?OFS:RS); print ""}' file
1 1
2 2
3 3
4 4
5 5
6 6
# <-- note there is an empty line here
Explanation
On odd fields, it print FS after it.
On even fields, print RS.
Note FS stands for field separator, which defaults to space, and RS stands for record separator, which defaults to new line. As you have tab as field separator, we redefine it in the BEGIN block.
This is probably the simplest way which allows for customisation
awk '{print $1,$2"\n"$3,$4}' file
For a line between
awk '{print $1,$2"\n"$3,$4"\n"}' file
although fedorquis answer with xargs is probably the simplest if this isn't needed
As Ed pointed out this wouldn't work if there were blanks in the fields, this could be resolved using
awk 'BEGIN{FS=OFS="\t"} {print $1,$2 ORS $3,$4 ORS}' file
Through perl,
perl -pe 's/\t(\d\t\d)$/\n$1\n/g' file
Fed the above command's output to the sed command to delete the last blank line.
perl -pe 's/\t(\d\t\d)$/\n$1\n/g' file | sed '$d'
How do I findout missing number from two sequence using bash script
from example I have file which contain following data
1 1
1 2
1 3
1 5
2 1
2 3
2 5
output : missing numbers are
1 4
2 2
2 4
This awk one-liner gives the requested output for the specified input:
$ awk '$2!=l2+1&&$1==l1{for(i=l2+1;i<$2;i++)print l1,i}{l1=$1;l2=$2}' file
1 4
2 2
2 4
a solution using grep:
printf "%s\n" {1..2}" "{1..5} | grep -vf file