grep - 找到一个子串，替换并乘以1024

Question

我有一个包含如下内容的文件:

03:14.27,"31K" 
03:13.59,"50M" 
04:11.51,"435K" 

问题是如何获取以字节为单位的数字并替换为旧值，以便我可以得到(也去掉引号很有用):

03:14.27,"31744"
...... 

用什么比较好？ grep 还是 awk？谢谢!

Answer 1

perl!

fg@erwin $ cat t.pl
#!/usr/bin/perl -W

use strict;

my %suffixes = (
        "K" => 10,
        "M" => 20,
        "G" => 30
);

while (my $line = <STDIN>) {
    $line =~ s/"(\d+)(\w)"/ '"' . ($1 << $suffixes{$2}) . '"'/ge;
    print $line;
}
fge@erwin ~ $ cat <<EOF | perl t.pl
> 03:14.27,"31K" 
> 03:13.59,"50M" 
> 04:11.51,"435K"
> EOF
03:14.27,"31744" 
03:13.59,"52428800" 
04:11.51,"445440"

(编辑:新输入)