How do I replace G with white space? - regex

Question

I have to check the free space on a disk volume and am using the command df -h, which produces this output:
Filesystem Size Used Avail Use% Mounted on
/dev/md0 27G 24G 2.1G 92% /
udev 506M 112K 506M 1% /dev
/dev/sda1 102M 40M 63M 39% /boot
This piece of script
my (#space, #freesp);
#space = grep /\/dev\/md0/,`df -h`;
print "#space\n";
is giving me
/dev/md0 27G 24G 2.1G 92% /
Which I can split on whitespace giving the values in $freesp[0], $freesp[1] etc.
But I want to remove or replace that G with white space, so that I can compare $freesp[3] with some value and can proceed with my script.
So what is the regex I have to use for this to happen, or is there any other better way to do this?
This code is working for me, but I'm looking in a direction I have mentioned.
#!/usr/bin/perl
use strict;
use warnings;
my (#space, #freesp);
#space = grep /\/dev\/md0/, `df`;
print "#space\n";
for (#space) {
chomp;
#freesp = split /\s+/, $_;
}
if ($freesp[3]/1024/1024 < 2.0) {
print "Space is less\n";
}
else {
print "Space is OK\n";
}
#print ($freesp[3]/1024/1024);

Answer 1

The real solution is not to use the -h option to df if you intend to parse the output with a script. That way, the output won't contain the "human-readable" K / M / G suffixes in the first place.
You may wish to instead use the -B1 option, which causes all sizes to be reported in bytes.
Edit: To answer the literal question, you could remove the suffixes and rescale the values appropriately like this:
my %units = (K => 2**10, M => 2**20, G => 2**30, T => 2**40,
P => 2**50, E => 2**60, Z => 2**70, Y => 2**80);
s/^(\d+(?:\.\d+)?)([KMGTPEZY])$/$1 * $units{$2}/e for #freesp[1 .. 3];
However, using the -B1 switch instead of -h would give the same result more easily, not to mention more accurately and reliably.