【请教】日志提取指定字段

[d4shman]

背景：每天要处理10G的log文件，格式如下：
name=xxx age=yy date=yyyymmdd Pc=xxxx （可能还有其他字段，awk NF会变）
现在要提取age=yy Pc=xxxx这样的内容，我用grep 写了一个脚本，很慢！
#! /bin/bash
#set -x

# the default delimiter
fs=" "
ret=""

# get options
while getopts F:k: opt
do
case "$opt" in
F)
fs=$OPTARG;;
k)
keys=$OPTARG;;
esac
done
# split string into an array
keys=(${keys//,/ })
num=${#keys[@]}
if [ $num -eq 0 ];
then
echo "not enough parameter";
exit;
fi

while read line
do
retstr=""
found=0
for key in ${keys[@]}
do
# \bxxx : word start by xxx
tmpstr=`echo ${line} | grep -oE "\b${key}=[^${fs}]*" `
# do only when string is not empty
[ -n "${tmpstr}" ] && retstr=${retstr}${tmpstr}${fs} && let found+=1
done
# check num
[ -n "${retstr}" -a $found -eq $num ] && echo -e ${retstr}"\n" | sed '/^$/d'
done

该怎样做呢？求大牛们指教！

[d4shman]

awk -F " " '{for(i=1;i<=NF;i++)s=($i~/^name=|^Pc=/)?s"\t"$i:s;$0=s;s=""}NF+=0'