How to add some symbol (or just add newline) if the numbers in the text are not continuous

Question

Ex:

Input file

A<0>
A<1>
A_D2<2>
A_D2<3>
A<4>
A_D2<6>
A<9>
A_D2<10>
A<13>

Desired Output:

A<0>
A<1>
A_D2<2>
A_D2<3>
A<4>
-----
A_D2<6>
-----
-----
A<9>
A_D2<10>
-----
-----
A<13>

Just care about the number in the angle bracket.

If the number is not continuous ,then add some symbol (or just add newline) until the number continue agian.

In this case, number 5,7,8,11 and 12 are missing.

Can anyone solve this problem by using awk or sed (even grep) command?

I am a beginner in Linux. Please explain the details of the whole command line.

Answer 1

Using grep or sed for doing this would not be recommended as grep can't count and sed is really difficult to do any kind of arithmetics in (it would have to be regular expression-based counting, a non-starter for most people except for the dedicated).

$ awk -F '[<>]' '{ while ($2 >= ++nr) print "---"; print }' file
A<0>
A<1>
A_D2<2>
A_D2<3>
A<4>
---
A_D2<6>
---
---
A<9>
A_D2<10>
---
---
A<13>

The awk code assumes that 0 should be the first number, and then maintains the wanted line number for the current line in the variable nr. If a number is read from the input that requires one or several lines to be inserted, this is done by the while loop (which also increments the nr variable).

The number in <...> is parsed out by specifying that < and > should be used as field delimiters. The number is then in $2 (the 2nd field).

Answer 2

This is probably far from efficient...

$ tr '<' '\t' < testfile | tr '>' ' ' \
  | awk '{ while (NR + shift <= $2) { print "-----"; shift++ }; print }' \
  | tr '\t' '<' \
  | tr ' ' '>'
A<0>
A<1>
A_D2<2>
A_D2<3>
A<4>
-----
A_D2<6>
-----
-----
A<9>
A_D2<10>
-----
-----
A<13>

First, I use tr to get two tab-separated fields from the file.

Second, I use tr again to replace '>' with a space, because otherwise my awk command will fail :-/

The awk-professionals around here will likely laugh now :-)

Third, the awk-command will compare the number of rows processed to the second field. If the number of rows is smaller, it will print the marker and increase shift which is added to the number of rows in the previous comparison.

Fourth and fifth: I'm undoing the changes I previously made with tr.

I got some inspiration from https://unix.stackexchange.com/a/190707/364705

Answer 3

I'm not an awk guy, but this also seems to do it. I'm always open to improvements:

awk -F '[<>]' -v num=0 '
{
  while(num < $2) {
    print "----";
    num++
  }
  print $1"<"$2">"
  num++
}' file

At first we set the field separator to match the characters < and >, so each line is split at these characters. For example the first line would be assigned to $1=A and $2=0.

Then we set variable num=0. We use it as the line counter: If the number of the current line $2 is greater than the line counter, print ----, increment the counter repeat until both values are equal. Then print $1<$2> and increment the counter.

Answer 4

To not print dashed lines before the first line even when the number on the first line is not zero:

$ cat tst.awk
BEGIN { FS="[<>]+" }
{
    curr = $(NF-1)
    if ( NR > 1 ) {
        for (i=prev+1; i<curr; i++) {
            print "-----"
        }
    }
    print
    prev = curr
}

$ awk -f tst.awk file
A<0>
A<1>
A_D2<2>
A_D2<3>
A<4>
-----
A_D2<6>
-----
-----
A<9>
A_D2<10>
-----
-----
A<13>

Answer 5

We can approach this problem via regexes using lookahead and lookbehind and only adding the dashes:

$ perl -0777 -pe 's/^.*<(\d+)>.*\n\K(?=.*<(\d+)>.*$)/qq[-----\n] x ($2-$1-1)/gem' file

Results:

A<0>
A<1>
A_D2<2>
A_D2<3>
A<4>
-----
A_D2<6>
-----
-----
A<9>
A_D2<10>
-----
-----
A<13>