Bash loop command until file contains n duplicate entries (lines)

Question

Hello, I'm writing a script and I need to create a loop that will execute same commands until file does contain a specified number of duplicate entries. For example, with each loop I will echo random string to file results. And I want loop to stop when there are 10 lines of of the same string.

I thought of something like

while [ `some command here (maybe using uniq)` -lt 10 ]
do
command1
command2
command3
done

Do you have any idea how can this problem be solved? Using grep can't be done since I don't know what string I need to look for.

Thank you for your suggestions.

Answer 1

while [ `sort filename | uniq -c | sort -nr | head -1 | sed -e 's:^ *::' -e 's: .*::'` -lt 10 ]

Not incredibly efficient (the "sort -nr | head -1" bit is particularly sub-optimal), but that's the standard "quick and dirty" solution.

Answer 2

Not the most efficient solution, but this should work:

while [ `sort $file | uniq -c | awk '{print $1}' | sort -nr | head -n1` -lt 10 ]

Answer 3

here's another version, which you do it within one (g)awk process. Also try to use $() syntax as much as possible

while [ "$(gawk '{!_[$0]++}END{b=asort(_,a);print a[b]}' file)" -lt 10 ]

Answer 4

I would use associative arrays in awk or Bash 4 and avoid running sort twice plus uniq and head.

Whenever you write your value to the file, increment that array element.

#!/bin/bash
# Bash 4
while true
do
    one_command
    val=$(command_to_output_val | tee out_file)
    if ((array[$val]++ >= 10))
    then
        break
    fi
done