使用AWK在大于x的第二列中找到最小数字

Question

I have a file with two columns,

sdfsd 1.3
sdfds 3
sdfsdf 2.1
dsfsdf -1 

if x is 2

I want to print sdfsdf 2.1

How to express it in awk (bash or sed is fine too)

Answer 1

It's awfully tempting to do this:

sort -k 2 -g  | awk '$2 >= 2 { print; exit }'

Tested and works on your example. If no second column is at least 2, it prints nothing.

Answer 2

awk:

BEGIN {
  min=0
  mint=""
  threshold=2
}
{
  if($2 > threshold && ($2 < min || min == 0)) {
    min = $2
    mint = $1
  }
}
END
{
  print mint, min
}