Hi i have log like this :
Actually i will process the data become
Anybody can help me ?
Try this:
awk -F"[ :]" ' BEGIN {
printf "Date Hour average_RSS average_PCPU\n";
}
NR>1{
if(arr[$1" "$2]=="") { rss="";cpu="";cnt=0;}
rss+=$(NF-2); cpu+=$NF; cnt++;
arr[$1" "$2]=rss" "cpu" "cnt; }
END {
for (i in arr) { split(arr,a," ");
printf "%s %.1f %.1f\n", i,a[1]/a[3],a[2]/a[3]; }
}' filename
Output:
Hi Dennis,
I think the average_PCPU for the date 20091116 at hours 08 should be 22.1818
Change the below line in the code according to your required precision. ie, %.4f here
printf "%s %.1f %.1f\n", i,a[1]/a[3],a[2]/a[3]; }
awk -F[\ \:] 'BEGIN {print "Date Hour average_RSS average_PCPU"}
{if (NR>1) rss[$2]+=$(NF-2); cpu[$2]+=$NF; count[$2]++}
END { for (i in rss) {printf "11-16-2009 %s %.4f %.4f\n",i,rss/count,cpu/count} }' urfile
Date Hour average_RSS average_PCPU
11-16-2009 07 1342177.0000 19.7727
11-16-2009 08 1380012.0909 22.1818
11-16-2009 09 1403367.0000 26.7500
Python, if you have it
#!/usr/bin/env python
d={}
e={}
f=open("file")
line=f.readline()
for li in f:
s = li.split()
year,mth,day=s[0][:4],s[0][4:6],s[0][6:8]
ti = s[1][:2]
STR= day+"-"+mth+"-"+year+" "+ti
d.setdefault(STR,[])
e.setdefault(STR,[])
rss,pcpu=s[7],s[8]
d[STR].append(rss)
e[STR].append(pcpu)
f.close()
for date,values in d.iteritems():
print date, sum(map(int,values))/len(values), sum(map(int,e[date]))/len(e[date])
output
# python script.py
16-11-2009 09 1682302 1403367
16-11-2009 08 1667231 1380012
16-11-2009 07 1646110 1342177
# cat awk.script
BEGIN {
FS="[ :]"
}
NR == 1 {
print "Date Time average_RSS average_PCPU"
}
a && a != $1 OFS $2 {
printf "%s %.4f %.4f\n",a,(b[a]/d),(c[a]/d)
b[$1 OFS $2] = c[$1 OFS $2] = ""
d = 0
}
NR > 1 {
a = $1 OFS $2
b[$1 OFS $2] += $(NF-2)
c[$1 OFS $2] += $NF
d++
next
}
END {
printf "%s %.4f %.4f\n",a,(b[a]/d),(c[a]/d)
}
# awk -f awk.script LOG.FILE
Date Time average_RSS average_PCPU
20091116 07 1342177.0000 19.7727
20091116 08 1380012.0909 22.1818
20091116 09 1403367.0000 26.7500