Prepend a letter to each line of a text, in ascending order

Question

A bash script writes its output, which is a list of files, into a file.

file.txt:

/home/user/dir1/dir2/foo00
/home/user/dir1/dir2/foo01
/home/user/dir1/dir2/foo02

I want to prepend a letter to each line, starting with a, and after having reached z, going on with aa, ab…

in the end, the output should look like

file.txt:

a /home/user/dir1/dir2/foo00
b /home/user/dir1/dir2/foo01
c /home/user/dir1/dir2/foo02
...
z /home/user/dir1/dir2/foo26
aa /home/user/dir1/dir2/foo27

Being a newbie in shell scripting, I have no clue, which tool may be appropriate. I So my question keeps surely somewhat imprecise.

I'd prefer bash built ins, if possible.

How can I do this operation?

duplicate : http://stackoverflow.com/questions/2099471/add-a-prefix-string-to-beginning-of-each-line — Amir Afghani, Feb 02 '15 at 21:27
@AmirAfghani, I don't think it is a duplicate, because the question asks for a specific prefix. — Marat Talipov, Feb 02 '15 at 21:34

score 3 · Accepted Answer · answered Feb 02 '15 at 21:42

3

Using only bash builtins, without subshells:

prefixes=({a..z}  {a..z}{a..z}  {a..z}{a..z}{a..z})
i=0
while IFS= read -r line
do
  printf "%s %s\n" "${prefixes[i++]}" "$line"
done < file.txt

answered Feb 02 '15 at 21:42

that other guy

score 1 · Answer 2 · answered Feb 02 '15 at 21:37

1

This isn't bash, but perl's built-in incrementing can do what you want:

#!/usr/bin/env perl
my $pfx = 'a';
print $pfx++, ' ', $_ while (<>);

This script reads from stdin and produces the output you seek.

answered Feb 02 '15 at 21:37

Steve Vinoski

2 Answers2