Print min and max with awk-CodePudding

Why wont this awk script work, where I want to print the min and max of an input by just reading the data once. I would also like to initialize the min and max variable with the first value of the data.

Say I have a file

I am using awk to print the min and max with:

cat test.txt | awk 'BEGIN {min=$2; max=$2} {if ($2 < min) {min=$2}; if ($2 > max) {max=$2} } END {print "min: " min "max: " max}'

But I am getting a result of:

min: max: 2345

I dont understand why min is set to blank?

CodePudding user response：

The BEGIN {} block is processed before the first line so $2 is undefined.

$ cat file
a 14
a 34
a 234
a 2345
a 1
a 24
$ awk 'NR == 1  { min = max = $2; next }
       $2 > max { max = $2 }
       $2 < min { min = $2 }
       END      { print "min: " min ", max: " max }' file
min: 1, max: 2345

CodePudding user response：

All numeric values in awk are stored in floating-point format, and there are no equivalents to C INT_MIN, INT_MAX, so you simply need to pick the range of values you wish to handle and initialize your max to the minimum possible value in the range you wish to consider, and min to the maximum possible value in the range you wish to consider.

For signed integer values, simply choosing the min/max values for 32-bit a signed integer is reasonable. For example, you can do:

 awk '
  BEGIN { min=2147483647; max=-2147483648 } 
  $2 > max { max=$2 } 
  $2 < min { min=$2 }
  END { print "max: ", max ORS "min: ", min }
' file

Example Use/Output

With your data in file, you would get:

$ awk '
>   BEGIN { min=2147483647; max=-2147483648 }
>   $2 > max { max=$2 }
>   $2 < min { min=$2 }
>   END { print "max: ", max ORS "min: ", min }
> ' file
max:  2345
min:  1

Which is what you were looking for.