Sed/Awk in bash script

Question

I have some data which is in the following format:

bruce   434355/512000 (84.8349609375%)
oliver   217542/512000 (42.488671875%)
kareem   236778/563200 (42.0415482954545%)

The data in the second column is in KB and I want to extract the first part (before the /) and display that only, like so:

 bruce    424.17MB (84.8349609375%)
 oliver   212.44MB (42.488671875%)
 kareem   231.24MB (42.0415482954545%)

Is there a way in sed or awk to do that?

Alternately, what I have done is extract the second column using awk, wrote it into another file and the display one line from each file within a while loop.

Answer 1

awk '$2=$2/1024"MB"' file

如果只想保留两位小数，请执行以下操作：

awk '{printf "%s %.2fMB %s\n", $1, $2/1024, $3 }' file

Answer 2

You can use the following awk script:

function human(x) {
    s="kMGTEPYZ";
    while (x>=1000 && length(s)>1)
        {x/=1024; s=substr(s,2)}
    return int(x+0.5) substr(s,1,1)
}
{
    split($2, a, "/");
    $2 = human(a[1]);
    print
}

save it to a file and execute it this way:

awk -f awk.script your.txt

The output is:

bruce 424M (84.8349609375%)
oliver 212M (42.488671875%)
kareem 231M (42.0415482954545%)

The human function is taken from https://unix.stackexchange.com/a/44087 . Thanks!