问题:

How to find maximum and minimum value from the below table using awk command.

20 90 60 30
55 75 80 85
10 15 99 95
55 95 70 20
9  35 85 75

I want the output like max value=99 and min=9

回答1:

with gnu awk:

awk '{for(x=1;x<=NF;x++)a[++y]=$x}END{c=asort(a);print "min:",a[1];print "max:",a[c]}'

output:

min: 9
max: 99

without awk:

xargs -n1|sort -n|head or tail -1

e.g.

min:

kent$  echo "20 90 60 30
55 75 80 85
10 15 99 95
55 95 70 20
9  35 85 75"|xargs -n1|sort -n|head -1
9

max:

kent$  echo "20 90 60 30
55 75 80 85
10 15 99 95
55 95 70 20
9  35 85 75"|xargs -n1|sort -n|tail -1
99

you can of course xargs -n1|sort -n then pipe to awk to pick first and last and print in one shot.

回答2:

If you have GNU awk:

# using array
awk '{x[NR]=$1}END{asort(x);print "max="x[NR],"min="x[1]}' RS=' +|\n' file
max=99 min=9

# No array
awk 'NR==1{m=n=$1}{$1>m?m=$1:m;$1<n?n=$1:n}END{print "max="m,"min="n}' RS=' +|\n' file
max=99 min=9

回答3:

awk '
NR == 1 { min=max=$1 }
{
    for (i=1;i<=NF;i++) {
        min = (min < $i ? min : $i)
        max = (max > $i ? max : $i)
    }
}
END {
    printf "min value = %s\n", (min == "" ? "NaN" : min)
    printf "max value = %s\n", (max == "" ? "NaN" : max)
}
' file

The test resulting in "NaN" is to accommodate empty input files.