Question

awk a file of dates, timestamps, values

Asked by: mateinone

Okay so I have what i figure should be a fairly straight forward question, but I was wondering what the answer is so thought I would throw it up

I have a file with dates, times, times, bytes

here is an example
2009-08-01,19:00:24,19:31:27,168537840
2009-08-01,19:04:39,19:17:16,321545923
2009-08-01,19:07:00,19:09:53,624757969
2009-08-01,19:07:35,19:09:45,275602323
2009-08-01,19:39:22,20:50:29,5665469256
2009-08-02,19:00:06,19:06:01,1386402864
2009-08-02,19:00:09,20:57:20,98241673
2009-08-02,19:00:10,19:02:30,519669461
2009-08-02,19:00:10,20:00:13,604466725
2009-08-02,19:00:20,19:38:34,189802392
2009-08-02,19:07:11,19:17:16,292963957
2009-08-02,19:15:41,19:16:36,83704532
2009-08-02,19:20:42,19:23:22,578206015
2009-08-03,19:00:09,19:03:20,1283766757
2009-08-03,19:00:10,19:06:39,1095627752
2009-08-03,19:00:12,20:43:39,80770993
2009-08-03,19:00:15,20:06:55,7632777428
2009-08-03,19:00:28,19:42:56,2143945407
2009-08-03,19:12:09,19:26:04,300927680
2009-08-03,19:26:11,19:28:48,560081633
2009-08-03,19:34:32,19:35:23,54324466
2009-08-04,19:00:08,19:06:48,1149345050

I want to get the output so that it grabs
Date, start, end, bytes (as mb)
for each date.
so the date, the earliest start time, the latest end time and the total sum of bytes for the day and I would like it in awk.

I can do it and do it quick with a simple shell
cat $filename |awk -F, '{print $1}'|sort -u |while read date
do
  early=`cat $filename|grep "^$date"|awk -F, '{print $2}'|sort -n|head -n 1`
  late=`cat $filename|grep "^$date"|awk -F, '{print $3}'|sort -n|tail -n 1`
  sum=`cat $filename|awk -F, '/^'"$date"'/ {printf "%-.2f", ($4/1024/1024)}'|awk '{sum = sum + $1} END {print sum"Mb"}'`
  echo date: $date early:$early late:$late sum:$sum
done                                          

for the following output
date: 2009-07-29 early:19:00:06 late:21:14:06 sum:2445.09Mb
date: 2009-07-30 early:19:00:08 late:20:40:34 sum:205.491Mb
date: 2009-07-31 early:19:00:06 late:20:49:37 sum:146.324Mb
date: 2009-08-01 early:19:00:04 late:20:50:29 sum:1172.21Mb
date: 2009-08-02 early:19:00:06 late:20:57:20 sum:1322.19Mb
date: 2009-08-03 early:19:00:09 late:20:43:39 sum:1224.3Mb
date: 2009-08-04 early:19:00:08 late:21:09:29 sum:1096.1Mb

Anyone keen to give it a go?

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Asked On
2009-08-04 at 06:04:00ID24624445
Tags

awk

,

shell

,

unix

Topics

KornShell (ksh)

,

Bourne-Again Shell (bash)

Participating Experts
3
Points
250
Comments
11

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Answers

 

by: CptArAbPosted on 2009-08-04 at 06:15:21ID: 25013325

Ok, but whats wrong with what you got? It seems to be exactly what you need, so what is the question?

 

by: mateinonePosted on 2009-08-04 at 06:27:57ID: 25013464

My question is how do I do the same thing in awk.
I think there will be 1 or 2 liner in awk to grab this data and I would be interested in the construction of that line.

This is purely a theoretical question, I am playing around with awk and I am wondering a way to construct this answer. I guess I am after the answer with somewhat of an explanation.
Yes I can do it my way and in fact I am doing it that way.. just interested in the awk way for the sake of knowing it.

 

by: CptArAbPosted on 2009-08-04 at 06:39:48ID: 25013617

Ahh i see, sorry, should have read the question properly.

I would have to play around with that idea as well...

 

by: mish33Posted on 2009-08-05 at 09:38:38ID: 25025298

Consider Python:

dates = {}
for line in open('filename'):
    date, early, late, bytes = line.split(',')
    if date in dates:
        d = dates[date]
        d[0] = min(d[0], early)
        d[1] = max(d[1], late)
        d[2] += int(bytes)
    else:
        dates[date] = [early, late, int(bytes)]
for date, (early, late, total) in sorted(dates.items()):
    print 'date:%s early:%s late:%s sum:%.2fMb' % (date, early, total/1024./1024)

                                              
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by: mateinonePosted on 2009-08-06 at 00:26:43ID: 25030859

Cheers for the alternate solution mish, I can also do it in perl, but specifically am wondering about the power of awk and would like to find and awk solution (if possible).
I do appreciate the answer though.

 

by: ozoPosted on 2009-08-06 at 02:25:49ID: 25031395

sort filename | awk -F, '{if( $1 != d ){ if( d ){ printf("date: %s early:%s late:%s sum:%.2f\n",d,e,l,s/1024/1024) } d=$1; e=l=$2;  s=0; } if( $3 > l ){ l=$3; } s+=$4 }END{printf("date: %s early:%s late:%s sum:%.2f\n",d,e,l,s/1024/1024)}'

 

by: CptArAbPosted on 2009-08-06 at 05:43:29ID: 25032557

Wow ozo, that just made me realise how much i still need to learn :-(

 

by: mateinonePosted on 2009-08-06 at 09:39:04ID: 25035413

okay ozo that is absolutely perfect..
If I double the points do you want to give a bit of an explanation to each of the elements.. The main exercise is so that I can see how someone would do it and pick up some things.

if not, that is ok the points are of cause yours anyway

 

by: ozoPosted on 2009-08-07 at 01:38:44ID: 25040789

sort ensures that all lines with the same date appear together, and that the earliest start time for a given date appears first.
-F, sets the field separator to , so that $1 will contain the date, $2 the start time, $3 the end time, and $4 the bytes
d stores the previous date, so that we can tell when a new date starts, so we know we are done accumulating the sum for the previous date and output it, unless there was no previous date
l keeps track of the latest end, and s keeps track of the sum
Anything else you want toi know?

 

by: mateinonePosted on 2009-08-17 at 07:12:06ID: 25114597

Points Doubled

Thanks a heap

 

by: mateinonePosted on 2009-08-17 at 07:12:56ID: 31611387

Excellent answer giving me exactly what I required.

20120131-EE-VQP-002

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