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Accepted
Álvaro Rodríguez Pardo
Asked: 2021-10-16 10:17:05 +0800 CST 2021-10-16 10:17:05 +0800 CST

How to print in the Linux Shell the data of a CSV file based on the number of digits they have, without using sed or awk?

  • 772

I have a CSV file with two columns, one belonging to MEPs and one to their IDs. These identifiers have between 3 and 6 numbers, and I would like to print on the screen those parliamentarians along with their identifiers, as long as these are 4 digits. I am going to put the first data of the CSV printed on the screen that is obtained by means of the following command that I have used so that it is better understood:

cat europarlamentarios.csv | cut -d "," -f1,2 | head -10

Uma AALTONEN,23752
Damien ABAD,96850
Claudette ABELA BALDACCHINO,118860
Jean-Pierre ABELIN,1829
Victor ABENS,1802
William ABITBOL,4361
Carlos ABOIM INGLEZ,1680
Gérard d'ABOVILLE,2202
Lars ADAKTUSSON,124990
Gordon J. ADAM,1427

As you can see, the second value corresponds to the identifiers, and these have different digits. Well, I would like to print on the screen those names of MEPs with their respective identifiers if they have only 4 digits. I have tried to use the command cutbut I have not known how to apply it in this case since all the lines have different lengths. I have also thought that it would still be necessary to apply control structures such as if\else, but this is still beyond my knowledge since I am starting to program, and I think there must be some way to get the result I want without using this control structure.

If someone has an idea of ​​how it could be done and gives me a cable, I appreciate it!

linux

3 Answers

  1. Best Answer

    The solution is to use egrep to find the ones with that pattern

    cat pepe.csv | cut -d "," -f1,2 | egrep ',.{4}$' | head -10

    Search after a comma, those with 4 characters in the second field. When looking for a pattern, those values ​​in brackets usually indicate the minimum and maximum length (and if it's a single number, it represents both).

    Note the character $at the end, which indicates that it will look for that match at the end of each line. For future cases, if you use strings with something else after the pattern you're checking, you should keep this in mind (say PEPITO,1234,OtraCosa).

    The output is:

    $ cat pepe.csv | cut -d "," -f1,2 | egrep ',.{4}$' | head -10
Jean-Pierre ABELIN,1829
Victor ABENS,1802
William ABITBOL,4361
Carlos ABOIM INGLEZ,1680
Gérard d'ABOVILLE,2202
Gordon J. ADAM,1427
    • 3

  2. Only with bashyou could do something like this:

    while IFS=, read -r col1 col2; 
do
  if [ "${#col2}" -eq 4 ]; then echo "$col1, $col2"; fi
done < europarlamentarios.csv

    Comments:

    • We set the Input Field Separators as the comma ( IFS=,)
    • We do a read cycle of the input file and for each line:
      • We check the length of the second column"${#col2}"
      • If it is equal to 4 ( if [ "${#col2}" -eq 4 ]) we print the line
    • 2

  3. I propose a one-command answer with the extension grepthat allows PERL-like regular expressions.

    $ grep -P '.*?,\d{4}$' europarlamentarios.csv

    Which results:

    Jean-Pierre ABELIN,1829
Victor ABENS,1802
William ABITBOL,4361
Carlos ABOIM INGLEZ,1680
Gérard d'ABOVILLE,2202
Gordon J. ADAM,1427

    In this case, the regular expression '.*?,\d{4}$'indicates:

    • .*?, all characters without being greedy
    • ,, followed by a comma
    • \d{4}, immediately followed by four digits
    • $, then ending with a line ending
    • 0