How to add a new line after an expression

Question

I have a file that reads similar to this:

random_string
83: some words 45: large error report 326: send emails to certain peple
random_string
34: some words 143: job success

I want to target the pattern "#:" (a number followed by a colon) and add a new line after the text that follows it; so it reads like this:

random_string
83: some words
45: large error report
326: email certain people

random_string
34: some words
143: job success

I've tried a sed commands of:

sed "s#'([0-9]*[0-9]:)'#a '/n'#" file.txt
sed "s#'([0-9]*[0-9]:)'#\n#g" file.txt

(I don't like using slashes as delimiters, the fence posts make it hard to read)

and an awk command of:

awk '/[0-9]*[0-9]:/ {printf "%s\n",$0}' file.txt

And neither have worked. I've looked at similar problems posted here and tried their solutions but nothing has worked. I know the answer is most likely very similar and it may even have something to do with the syntax in my number expression but I can't figure it out myself. I have no preference to either awk or sed but I think they would be the best tools at my disposal.

Help?

Answer 1

Here's a Perl solution:

$ perl -pe 's/(\d+:.*?)(?=\d+:|$)/$1\n/g' file
random_string
83: some words 
45: large error report 
326: send emails to certain peple

random_string
34: some words 
143: job success

Explanation

• (\d+:.*?) : match one or more numbers (\d+) followed by a :, and then the smallest string (the ? in .*? makes it non-greedy, it will stop as soon as the first match is found) matching the rest of the regex. Here, it will continue until the part explained below.
• (?=\d+:|$) : The (?=foo) is called a positive lookahead. It will match but what is matched will not be included in the actual result. So, bar(?=foo) will match all cases of bar that are followed by foo. Here, we are looking for either a string of numbers followed by an : (\d+:) or the end of the line ($).

Now, the substitution operator will replace all occurrences of the first pattern with itself and a newline which should give you the desired output.

Answer 2

$ cat file
random_string
83: some words 45: large error report 326: send emails to certain peple
random_string
34: some words 143: job success

You can use sed:

$ sed 's/[0-9]*: [a-z ]*/&\n/g' file

Output:

random_string
83: some words 
45: large error report 
326: send emails to certain peple

random_string
34: some words 
143: job success

Answer 3

awk seems to do the trick:

$ awk '{ for( i=1; i<=NF; i++ ) { if( match( $i, ":" ) ) { printf "\n" } printf( "%s ", $i ) } }' /path/to/file
random_string
83: some words
45: large error report
326: email certain people random_string
34: some words
143: job success

Answer 4

Comparing your input to your output, it seems as if your description of what you want is incorrect. You say "I want to target the pattern #: (a number followed by a colon) and add a new line after" when a more accurate description would be:

• replace the space before any number of digits followed by a colon with a newline.
• insert a newline before every non-empty line not beginning with a digit.
• skip the first line of input because there's nothing that needs to be changed on it.

This sed script implements that. It uses extended regular expressions (-E) instead of sed's default basic regex to minimise the number of backslash escapes required and improve readability.

$ sed -E -e '2,$ {s/ ([0-9]+:)/\n\1/g; s/^[^0-9]/\n&/}' file.txt
random_string
83: some words
45: large error report
326: send emails to certain peple

random_string
34: some words
143: job success

BTW, if there's a possibility of tabs rather than spaces before the [0-9]+:, or multiple whitespace characters then use [[:space:]]+ instead of just a space. e.g.

sed -E -e '2,$ {s/[[:space:]]+([0-9]+:)/\n\1/g; s/^[^0-9]/\n&/}' file.txt