如何打印每个ID的最后一行？

Question

I have a list of IDs (exon) each with multiple suffixes and I want to get the last line of each ID

Input:

NM_203_exon_19
NM_203_exon_20
NM_0217_exon_7
NM_0217_exon_8
NM_0217_exon_9
NM_91_exon_14
NM_91_exon_15
NM_91_exon_16
NM_91_exon_17

Desired output:

NM_203_exon_20
NM_0217_exon_9
NM_91_exon_17

Answer 1

tac INPUTFILE |awk -F'_' '!a[$1FS$2]++' |tac
NM_203_exon_20
NM_0217_exon_9
NM_91_exon_17

Answer 2

As you scan each line, you could check store the previous "ID" and then print it if the new "ID" is different:

$ awk -F'_exon_' '{if($1 != id && last)print last; id=$1; last=$0} END{print last}' file
NM_203_exon_20
NM_0217_exon_9
NM_91_exon_17

Answer 3

As your list is ordered , One idea could be put everything in array and iterate trough it (without using awk), when you change String means the previous was the biggest one and you will print it, it will not get the last one , so you will print the last item of the array in the end .

!/usr/bin/bash
fileString=$(cat  filename |tr "\n" " ")
array=($fileString)
for ((i=0; i < ${#array[@]}-1; i++))
do
  if [${array[$i]} != ${array[$i+1]}]; THEN 
    echo ${array[$i]}
  fi
done
   echo ${array[${#array[@]}-1]}

Answer 4

As the file is already sorted as per suffix for each id, print the last line for each id

awk -F"_" 'NR==1{prev=$2}; $2==prev{a=$0} $2!=prev{print a; prev=$2} END{print $0}' file

Output:

NM_203_exon_20
NM_0217_exon_9
NM_91_exon_17

Answer 5

With GNU sort for -s (stable sort):

$ tac file | sort -t_ -k2,2 -su
NM_0217_exon_9
NM_203_exon_20
NM_91_exon_17