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James Palmer
Asked: 2020-07-15 14:59:27 +0800 CST

Get human readable cpu frequency

  • 772

I am trying to get a simplified output of the cpu frequency to 3 digits, such that if the frequency is less than 1000 it will output xxx Mhz and if it's above 1000 it will outpute x.xx Ghz. I can get only the frequency with lscpu | sed -n 's/CPU MHz:[ \t]*//p', and the first 4 digits with lscpu | sed -n 's/CPU MHz:[ \t]*//p' | cut -c1-4, however I'm not sure how to parse this to achieve the desired result.

bash sed awk

1 Answers

  1. Best Answer

    Try:

    lscpu | awk '/CPU MHz/{if($NF+0>1000)printf "%.3f GHz\n",$NF/1000; else printf "%.3f MHz\n",$NF}'

    How it works

    • /CPU MHz/{...}

      This selects the line we want and executes the commands in curly braces only for that line.

    • if($NF+0>1000)printf "%.3f GHz\n",$NF/1000; else printf "%.3f MHz\n",$NF

      If the final field, $NF, is greater than 1000, then we divide it by 1000 and print it as GHz. Otherwise, we print it as MHz. The format %.3f determines how many significant digits are printed. Adjust this to suit your tastes.

    Multi-line version

    For those who prefer their commands spread over multiple lines:

    lscpu | awk '
    /CPU MHz/{
        if($NF+0>1000)
            printf "%.3f GHz\n",$NF/1000
        else
            printf "%.3f MHz\n",$NF
    }'
    • 3