Clever Geek Handbook

Add multiple lines at the beginning of the file

I have many files in the same subdirectory that looks like this:

10 A
11 T
12 A
13 C
15 A

i.e. two columns, where the first row always starts with 10.

I would just like to add lines 1-9 to the beginning of each ie file

1
2
3
4
5
6
7
8
9 
10 A
11 T
12 A
13 C
15 A

I don't need to add a second column.

I know what I could do

sed -i '1i1' FILE

to add the first line, but do I need to enter this command for every additional line I want to add?

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8 answers


Try

seq 1 9 | cat - file.in > file.out

Where file.in

is the input file you specified and file.out

is the name of the desired output file.



Or, even shorter (thanks @ william-threll):

{ seq 1 9; cat file.in; } > file.out
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If you have gnu awk, you can use inplace

awk editing and avoid the final hardcoding number by reading the existing value from the first entry:

awk -i inplace 'FNR==1{for (i=1; i<$1; i++) print i} 1' *.txt


Change *.txt

to a different globe that matches your files.

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This is probably the shortest slot you can use.

sed -i -e1i{1..9} *.txt

This is equivalent to

sed -i -e1i1 -e1i2 -e1i3 -e1i4 -e1i5 -e1i6 -e1i7 -e1i8 -e1i9 *.txt

It uses the extension bash

.

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You can use Unix / POSIX / Linux utility cat

to concatenate and print files.

You can use it <(seq 5)

as a replacement file to create an initial sequence. So you can do:

$ cat <(seq 5) file
1
2
3
4
5
10 A
11 T
12 A
13 C
15 A

Then, if you want the edit effect to be set, you can do:

$ cat <(seq 5) file > tmp && mv tmp file

Then:

$ cat file
1
2
3
4
5
10 A
11 T
12 A
13 C
15 A

If you haven't seq

, you can do:

$ cat <(echo {1..5} | tr ' ' '\n') file

instead of this. (Using 5 lines versus 9 in your example. Obviously how to change that ...)

If you have many files or want to create a second column, you can create a header row:

$ header=$(for ((x=1; x<=5; x++)); do printf "%s #\n" $x; done)
$ echo "$header"
1 #
2 #
3 #
4 #
5 #

Then you can use this line instead of running seq

or something:

$ cat <(echo "$header") file
1 #
2 #
3 #
4 #
5 #
10 A
11 T
12 A
13 C
15 A
+3


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sed -i -e '1!b;'"$(printf 'i%d\n' {1..9})"

This is equivalent to running

sed -i -e '1!b;i1
i2
i3
i4
i5
i6
i7
i8
i9
'

but easier to write. 1!b

skips the rest of the commands on lines other than the first, but you can also use the address 1

.

sed -i -e "$(printf '1i%d\n' {1..9})"
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Alternatively, you might want a pure bash:

$ cat file4
10 A
11 T
12 A
13 C
15 A
$ printf '%s\n%s\n' {1..9} "$(<file4)"
1
2
3
4
5
6
7
8
9
10 A
11 T
12 A
13 C
15 A

To replace the original content of a file, you can do it like this:

printf '%s\n%s\n' {1..9} "$(<file4)" >file44 && mv -f file44 file4

Also sed -i

this is what it does in the background (check sed manpages)

For multiple files, you can use a loop:

for f in *;do printf '%s\n%s\n' {1..9} $(<"$f") >tmp"$f" && mv tmp"$f" "$f";done
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You can insert multiple lines with by sed

placing a backslash after the command i

.

sed -i '1i\
1\
2\
3\
4\
5\
6\
7\
8\
9\
' file
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This might work for you (GNU sed):

sed -i '1e seq 9' file1 file2 file3 ...

The first line of each file evaluates the command seq 9

.

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