I've read a bunch of text files to extract some patterns. I need the line number too but the line number must be removed before the final grep (but can be saved for further processing maybe using variables).

I'll explain my procedure splitting the (long oneliner) command for better understanding:

  1. read file with cat, do some cleaning (remove strange characters & line feeds using sed and tr) and such. Here just an example of many piped cleaning tasks:

     cat file | sed 's/,/ /g' | sed '/^$/d'
  2. add line number and tab with nl command & more processing and cleaning

     nl -nrz -w4 -s$'\t' | tr '\n\r' ' '
  3. extract the final desired pattern to CSV file

     grep -Eio '.{0,0}account number.{0,15}' >> account_list.csv

The issue is I need the line number from step 2 to the very same CSV (other column no matter the order) using the SAME ONE LINE COMMAND (no while or loop allowed) but no success so far.

[EDITED for better understanding] Take into account the line number I need is the original one prior to file cleaning. This cleaning process sometimes deletes some paragraphs. Imagine a file with a thousand lines, after processing i got one hundred. New line numbering is wrong. [end edit]

Sample imput after some processing:

0123 the first account number 2345356432 must be used
0345 take it just for billing purposes, not any other.
0657 Meanwhile the second account number 8623525534
0987 user is blocked until the issue is solved with

The desired oputput would be:

 2345356432; 0123
 8623525534; 0657


0123; 2345356432
0657; 8623525534

any hint would be much appreciated

  • Much better if you present sample of input and desired output
    – Costas
    Commented Aug 18, 2016 at 9:08
  • added sample input and desired output Costas
    – jomaweb
    Commented Aug 18, 2016 at 9:15
  • 1
    What is the reason to do pre-cleaning if you need just account number(s)?
    – Costas
    Commented Aug 18, 2016 at 10:17
  • sed -rn 's/ .*account number (\S{,15}) *.*/; \1/p' but if the string account … is splitted in 2 lines? Which number you'l need?
    – Costas
    Commented Aug 18, 2016 at 10:23
  • Hi Costas, I'm trying to simplify a lot to make this issue suitable to be asked here. The program is very big and this is a tiny part (tricky but tiny) from it. The real purpose of the program is preparing text for Machine Learning algorithms and a lot of processing is made prior to getting account numbers. In addition, account numbers is an example (easy one) but the real pattern I need to extract is a very complex one.
    – jomaweb
    Commented Aug 18, 2016 at 10:26

4 Answers 4


Using awk on the original input file, prior to cleaning:

awk '/account number [[:digit:]]+/ { match($0, "account number ([[:digit:]]+)", a); print NR ";" substr($0, a[1, "start"], a[1, "length"]); }' input

This extracts the account number and prints the line number at the start of the line:


If you want to extract the pre-processed number instead from the cleaned-up file:

awk '/account number [[:digit:]]+/ { match($0, "account number ([[:digit:]]+)", a); print $1 ";" substr($0, a[1, "start"], a[1, "length"]); }' input

Splitting this up a little:

  • /account number [[:digit:]]+/ ensures we only process lines matching "account number" followed by a number;
  • match($0, "account number ([[:digit:]]+)", a) looks for the pattern again and stores the positions and lengths of the matched groups (([[:digit:]]+), the number) in array a;
  • print NR ";" substr($0, a[1, "start"], a[1, "length"]) prints the record number (i.e. the line number; use FNR if you want to process multiple files), followed by ;, followed by the substring corresponding to the first group: a[1, "start"] gives its starting index, a[1, "length"] its length (this was filled in by match).

All this assumes there's at most one account number per line.

The second variant prints $1 instead of NR, i.e. the first field in the file, which is the pre-processed line number.

  • Stephen your code did the trick but It's hard for me to understand. Would you be so kind of explaining a little bit the involved parts ?
    – jomaweb
    Commented Aug 18, 2016 at 10:35
  • @jomaweb hopefully that's a little easier to understand... Commented Aug 18, 2016 at 10:40

If your grep version supports Perl regular-expressions, you can use look-behind:

grep -Pnio "(?<=account number.)([0-9]{0,15})" text
  • guido, your command adds NEW line numbers. Take into account the line numbers are in the file prior to all processing. And some of that processing is deleting some lines .What I'm using to feed grep is a subset of the original file. The line number I need is the original, not the added by the final grep. Imagine a file with a thousand lines, after processing i got one hundred. Your grep re-set the numbering from 1 to 100
    – jomaweb
    Commented Aug 18, 2016 at 10:07
  • @jomaweb How we can see the above from your example?
    – Costas
    Commented Aug 18, 2016 at 10:14
  • edited input text to accomodate to better requirements explanation
    – jomaweb
    Commented Aug 18, 2016 at 10:28
  • I'd offer to use -z option + [[:space:]] class symbols instead just <space>
    – Costas
    Commented Aug 18, 2016 at 10:35

Given your input and output, a awk script seems much simpler:

gawk '/account number/ {
    nr=gensub(/.*account\s*number\s*([0-9]+).*/, "\\1", "g")
    print FNR "; " nr

Of course, you may need to adapt the account number extraction and output format to your likes. But you get the idea. (Requires GNU awk due to the use of the gensub function.)

  • 1
    Syntax error. Not working
    – jomaweb
    Commented Aug 18, 2016 at 10:01
  • This requires GNU awk. Edited my answer. Stephen's code should be compatible with other versions. Commented Aug 18, 2016 at 12:44

I would be tempted to use Perl for this, something like this should work:

perl -ne 'print "$1; $2\n" if /^(\d+).*account number (\d+)/' input

On lines that start with some digits (^\d+), and contain the string "account number" followed by some digits, print the 1st and 2nd capture groups (parts in parenthesis, here the numbers). If you want to print Perl's idea of the line number, use $. instead of $1.

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