Awk question: Sum dates

BufferExploder · #1 (**permalink**) 09-01-2008

Hi there,

I have a logfile which has the following layout:

20080812 0 20
20080812 12 10
20080812 12 10
20080812 12 10

I want to sum the "12" on the last 3 lines and save the "20" on the first line. The final output should be

20080812 36 20

I think that should me more easier with awk ? Many thanks in advice.

dennis.jacob · #2 (**permalink**) 09-01-2008

Am not sure whether I get the issue properly, but here is the code: I just assume that you will be taking the first and third fields from first line.

Code:

awk 'BEGIN {sum=0; } {if (NR==1){ var1=$1; var3=$3;}  sum += $2; } END { print var1" "sum" "var3 }' filename

cfajohnson · #3 (**permalink**) 09-02-2008

Code:
awk '
 NR==1 { var1=$1; var3=$3; next }
 { var2 += $2 }
END { print var1" "var2" "var3 }' filename
I suspect that what you really want is more complex than this, If so, you'll probably have to describe your input file more completely.

BufferExploder · #4 (**permalink**) 09-02-2008

Sorry thanks for your replies but i didn't put the entire file so here it is:

200808260640 0 11383
200808210640 0
200808300640 0
200808300640 0
200808300640 0
200808260640 336528522 8844
200808260640 724271039 8080
200808260640 583502861 8077
200808210640 0
200808210640 0
200808210640 0
200808290640 0
200808290640 0
200808290640 0
200808290640 0
200808150640 0 7667
200808160640 0 3285
200808310640 0
200808150640 634799861 4703
200808150640 329658775 4704
200808150640 588901581 4875
200808160640 201718658 1424

What i want to do is sum all the $2 for the same date and save the $3 for that date. E.g like a posted in the first post.

200808150640 0 7667

200808150640 634799861 4703
200808150640 329658775 4704
200808150640 588901581 4875

Expected result:

200808150640 (634799861+329658775+588901581) 7667

i want to do this for the entirely file . Many thanks in advice.

radoulov · #5 (**permalink**) 09-02-2008

Just guessing:
(use nawk or /usr/xpg4/bin/awk on Solaris)

Code:

awk 'END \
{ for (dt in third) print dt, second[dt], third[dt] }
{ if (!_[$1]++) third[$1] = $3; second[$1] += $2 }
' filename

If you need the output sorted, pipe it to sort (or use asorti if you have GNU Awk). Or just use Perl

Code:

perl -ane'
  $third{$F[0]} = $F[2] unless $x{$F[0]}++;
  $second{$F[0]} += $F[1];
  print map "$_ $second{$_} $third{$_}\n", sort keys %x
    if eof' filename

dennis.jacob · #6 (**permalink**) 09-02-2008

Try this:

Code:

for each in $(cut -c1-8 filename | sort -u)
        do
        awk '/^'$each'/{ if ($2==0){ var1=$1;var3=$3;} sum+=$2; }END { print var1" "sum" "var3;}' filename
        done

BufferExploder · #7 (**permalink**) 09-02-2008

Thank you very much my friend works like a charm.

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