awk - 列的一半大於x的所有行

Question

正如標題所示，我試圖在大型tsv文件中查找所有行，其中至少有50％的列的值大於值x AWK。

E.g對於x = 5：

9 6 7 2  3 
0 1 2 7  6 
1 3 8 9 10 

應該返回

9 6 7 2  3 
1 3 8 9 10 

Answer 1

awk來救援！

$ awk -v t=5 '{c=0; for(i=1;i<=NF;i++) c+=($i>t)} c/NF>0.5' file 

9 6 7 2  3 
1 3 8 9 10 

Answer 2

使用Perl：

perl -ane '$x = 5; print if @F/2 <= grep $_ > $x, @F' -- file.tsv 

Answer 3

使用輸入.tsv文件看起來像這樣：

Num1 Num2 Num3 Num4 Num5 
9  6  7  2  3 
0  1  2  7  6 
1  3  8  9  10 

此代碼將做一個awk腳本。我已經留下了意見，看到 腳本的形式，所以你可以相應地調整。

#!/usr/bin/awk -f 

# reads from stdin. 
# Usage: $ ./bigcols.awk < input1.tsv 


# Run at start. 
BEGIN { 
#  print "Start" 
#  print "TSV setting. Field seperator set to tab." 
     FS = "\t" 
     # He wants to find lines with avg greater than var x 
     x=5 
} 

# main. Run for each record. This code uses newlines to denote records. 
{ 
     # Find lines which are of this form: (skip header) 
     # #+, 
     # ie. start with one or more numbers in column 1. 
     if ($1 ~ /^[0-9]+/) { 
       the_avg = ($1 + $2 + $3 + $4 + $5)/5 
       if (the_avg > x) { 
        print $1, $2, $3, $4, $5 
       } 
     } 
} 

# run at end 
#END { print "Stop" }