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Something like:
awk '
v[$6] < $12 { v[$6] = $12;l[$6] = $0 }
END { for(i in l) print l[i] }' file
seems to do what you want as long as the output order doesn't matter.

If you want to try this...

Another way:

sort -k6,6 -k12,12nr filename | awk '!line[$6]++'

You may try awk solution its easy, Others might have sed solution

$ awk -F'[|]' '$5==100 && !A[$4]++ {print $4}' fileOR
$ grep -E '(100)' file | cut -d '|' -f4 | sort | uniqResulting
...

That is because of history expansion in bash, which interprets the exclamation mark, even within double quotes. You need to use single quotes like Akshay showed, not double quotes.
You can turn off...

awk 'NR==FNR {A[$1]=$2; next} ($1 in A) {$1=$1 FS A[$1]} 1' FILE1 FILE2

Using awk:
awk 'NR==FNR{A[$1]=$2;next}A[$1]{$2=A[$1]}1' FILE1 FILE2

awk '{split($2,a,".");print $1" fx"a[1]}' file

aran24352 fx14
aran08740 fx07
aran76153 fx17
aran76513 fx01
aran54894 fx04

awk -F"\t" '$4<=1.0e-02' tmp

awk -F= '$2 <= 1.0e-02' infile