Issue
I've got a line with an identifier, then a pattern (in my case, a semicolon) and then a list of numbers:
echo "sp16;111111111111111111111111111111211"
I'd like to insert a semicolon between all numbers as in (desired output):
echo "sp16;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;2;1;1"
So far, I found how to insert a semicolon between all characters using sed:
sed 's/.\{1\}/&;/g'
But, then it also inserts semicolons before matching the first semicolon and also it adds a semicolon ad the end of the line.
Solution
With a Perl's one-liner:
perl -pe 's/^([^;]+;)([0-9]+)/$1 . join ";", split "", $2/e' file
sp16;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;1;2;1;1
The regular expression matches as follows:
| Node | Explanation |
|---|---|
^ |
the beginning of the string anchor |
( |
group and capture to \1: |
[^;]+ |
any character except: ; (1 or more times (matching the most amount possible)) |
; |
; |
) |
end of \1 |
( |
group and capture to \2: |
[0-9]+ |
any character of: '0' to '9' (1 or more times (matching the most amount possible)) |
) |
end of \2 |
Perl's code explanations
| Operator | Meaning |
|---|---|
$1 |
captured group 1 like \1 in sed |
. |
concatenation |
join ";", |
join with the character ';' the following list... |
split "", $2 |
split captured group $2 or \2 in a list of digits |
e |
modifier that allow Perl's code/expression in the right part of substitution s/// |
Answered By - Gilles Quénot Answer Checked By - Clifford M. (WPSolving Volunteer)