Head command queries

Shell Programming and Scripting
1

we have a file as below

AREA,COUNTRY,RANK
A,MX,1
A,MX,2
A,MX,5
A,MX,8
A,IN,7
A,IN,5
A,IN,21
B,CN,6
B,CN,2
B,CN,8
B,CN,0

we need the TOP 2 RANK records for the combination of Area, Country as below. i know

head -2

, which gives top 2 records from file but not sure it lists based on specific fields. plz help me out.

A,MX,8
A,MX,5
A,IN,21
A,IN,7
B,CN,8
B,CN,6
2

Any attempt from your side?

3

below is my code

sort -r file.txt > sort.txt
PRE_SIZE=""
content="middle.txt"
rm -f $content
rm -f final.txt
while read line
do
 dt=`echo $line | awk -F, '{print $1,$2}'`
 if [ "$PRE_SIZE" = "" ] || [ "$PRE_SIZE" = "$dt" ]
 then

 echo "$line" >> "${content}"
 PRE_SIZE="$dt"

 else

 cat $content | head -2 >> final.txt
 echo "$line" > "${content}"
 PRE_SIZE="$dt"

 fi

done < sort.txt
cat $content | head -2 >> final.txt

my code is creating temporary files, i feel they may degrade performance in case of huge file.
Any suggestions ???

4

Try

{ head -1 file; tail -n+2 file | sort -t, -k1,2 -k3nr; } | awk -F, '!T[$1,$2]++ {P=NR+1} NR<=P'
AREA,COUNTRY,RANK
A,IN,21
A,IN,7
A,MX,8
A,MX,5
B,CN,8
B,CN,6

---------- Post updated at 11:19 ---------- Previous update was at 11:16 ----------

You may save a process and a pipe by taking advantage of the fact that once the header is found an printed, any later occurrence will be filtered out by awk:

{ head -1 file; sort -t, -k1,2 -k3nr file; } | awk -F, '!T[$1,$2]++ {P=NR+1} NR<=P'
1 Like
5

Thanks for the suggestion.
i will research on the awk code you provided :slight_smile:

6

You can achieve the same result with (recent) bash only:

head -1 file
tail -n+2 file | sort -t, -k1,2 -k3nr |
  while IFS=, read A C R X
    do [ "$TMP" != "$A,$C" ] && CNT=2
       [ $(( CNT-- )) -gt 0 ] && { echo $A,$C,$R      
                                   TMP="$A,$C"
                                 } 
    done
7

Here is another awk approach:-

awk -F, '
        NR == 1 {
                print
                next
        }
        {
                idx = $1 FS $2
                if ( idx in A )
                {
                        if( A[idx] < $3 )
                        {
                                P[idx] = A[idx]
                                A[idx] = $3
                        }
                }
                else
                        A[idx] = $3
        }
        END {
                for ( k in A )
                        print k, A[k] RS k, P[k]
        }
' OFS=, file
8

@Yoda: you should provide for $3 less than A[idx] but greater than P[idx]...