如何仅获取 bash 中特定列的重复行

Question

Imagine I have this file in bash:

1 3 6 name1
1 2 7 name2
3 4 2 name1
2 2 2 name3
7 8 2 name2
1 2 9 name4

How could I extract just those lines which present the field "name" repeated and sort them?

My expected output would be:

1 3 6 name1
3 4 2 name1
1 2 7 name2
7 8 2 name2

I was trying to use sort -k4,4 myfile | uniq -D sort -k4,4 myfile | uniq -D , but I don't find how to tell uniq to work with the 4th column. Thanks!

Answer 1

You were close. You need to skip fields preceding the last one.

$ sort -k4 file | uniq -f3 -D
1 3 6 name1
3 4 2 name1
1 2 7 name2
7 8 2 name2

Answer 2

Could you please try following.

awk '
{
  a[$NF]++
  b[$NF]=(b[$NF]?b[$NF] ORS:"")$0
}
END{
  for(i in a){
    if(a[i]>1){
      print b[i]
    }
  }
}
'  Input_file

OR in case you want to sort the output try following then.

awk '
{
  a[$NF]++
  b[$NF]=(b[$NF]?b[$NF] ORS:"")$0
}
END{
  for(i in a){
    if(a[i]>1){
      print b[i]
    }
  }
}
'  Input_file  |  sort -k4

Answer 3

You may use this awk + sort :

awk 'FNR==NR{freq[$NF]++; next} freq[$NF] > 1' file{,} | sort -k4

1 3 6 name1
3 4 2 name1
1 2 7 name2
7 8 2 name2