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awk - operation over time staps

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# 1  
Old 02-03-2009
Error awk - operation over time staps
Hi @ all,
I'm trying to accomplish a simple excel-job with awk, without any result...

What i have is a file with a time stamp for each line like :

21:02:07
21:04:11
21:06:28
21:08:44
21:10:45
21:12:48
21:14:52
21:16:53
21:18:55


What i would, is to calculate the time elapsed since the last time value. In other terms i need a difference between the [line+1] value and the [line] for line=0 to line=EOF.

The output should be :

00:02:04
00:02:17
....and so no


i've done a script that loads these value in a string....but then i can't operate on them by simple numeric aperator probably because i'm trying to subtract two character-strings.

This is what i've done

Code:
awk ' { RS = "" ; FS = "\n" } ; 
{ 
     for (i = 0; i <= NF; i++) 
    string[i]=$i 
} 

END {
    
    for (i = 0; i <= NF; i++) 
    diff[i]=string[i+1] - string[i]
    printf "%s\n", diff[i]


}'

The first part of code (upper the END statement) seems to works fine (do you agree?)...the for cycle correctly populate the string array.
But when i try to assign the [I]string[i+1] - stringvalues to the diff array...doesn't work Smilie

Maybe there is some function for time operations?
Maybe i have to firstly convert the time value to UTC?

Hoping in your help!
# 2  
Old 02-03-2009
who knows?

here's a ksh solution:

Code:
typeset -Z2 hour min sec

prev_time_t=0

sed -e 's/:/ /g' << EOF |
21:02:07
21:04:11
21:06:28
21:08:44
21:10:45
21:12:48
21:14:52
21:16:53
21:18:55
EOF
while read hour min sec ; do

  (( time_t = ( hour * 3600 ) + ( min * 60 ) + sec ))

  (( diff = time_t - prev_time_t ))

  prev_time_t=$time_t

  if [ $diff -eq $time_t ]; then
    continue
  fi

  (( hour = diff / 3600 ))
  (( min  = ( diff % 3600 ) / 60 ))
  (( sec  = diff % 60 ))

  print $hour:$min:$sec

done

Smilie
# 3  
Old 02-03-2009
Quote:
Originally Posted by quirkasaurus
who knows?

here's a ksh solution:

Why write a ksh solution when one that will work in any POSIX shell (which includes ksh) is just as easy?

Code:
prev_time_t=0

while IFS=": " read hour min sec x ; do
  time_t=$(( $hour * 3600 + $min * 60 + sec ))
  diff=$(( time_t - prev_time_t ))
  prev_time_t=$time_t

  if [ $diff -eq $time_t ]; then
    continue
  fi

  hour=$(( diff / 3600 ))
  min=$(( ( diff % 3600 ) / 60 ))
  sec=$(( diff % 60 ))

  printf "%02d:%02d:%02d\n" "$hour" "$min" "$sec"

done < "$FILE"

If the file is large, an awk solution will be faster.
Last edited by cfajohnson; 02-03-2009 at 07:06 PM.. Reason: s/$/%/
# 4  
Old 02-03-2009
Both scripts run perfectly, and this could be sufficient, so thank you all so much!

Therefore the file that i'm being to parse tends to grow up every 30 min, so the fastest solution is still welcome (and i think that awk is the fastest way)

@cfajohnson

there is a little error :
- printf "$02d:%02d:%02d\n" "$hour" "$min" "$sec"
+ printf "%02d:%02d:%02d\n" "$hour" "$min" "$sec"

Smilie
# 5  
Old 02-03-2009
Code:
awk  -F:  'NR>1 { printf "%.02d:%.02d:%.02d\n", $1-s1, $2-s2, $3-s3 } 
                          { s1=$1; s2=$2; s3=$3 }'  file

# 6  
Old 02-03-2009
We forgot to manage these inputs:

23:55:26
23:57:32
00:00:13
00:02:16

@to posix scripters
this is your output

00:02:06
00:-57:-19
00:02:03

@robin
this is yours
00:02:06
-23:-57:-19
00:02:03

thanks @ all!!!
# 7  
Old 02-03-2009
Try...
Code:
awk -F: '{t=$1*60*60+$2*60+$3;d=t-p;if(d<0)d+=24*60*60;if(p)printf "%02d:%02d:%02d\n",d/60/60,d/60%60,d%60;p=t}' file

Tested...
Code:
$ cat file
21:02:07
21:04:11
21:06:28
21:08:44
21:10:45
21:12:48
21:14:52
21:16:53
21:18:55
23:55:26
23:57:32
00:00:13
00:02:16
$ awk -F: '{t=$1*60*60+$2*60+$3;d=t-p;if(d<0)d+=24*60*60;if(p)printf "%02d:%02d:%02d\n",d/60/60,d/60%60,d%60;p=t}' file
00:02:04
00:02:17
00:02:16
00:02:01
00:02:03
00:02:04
00:02:01
00:02:02
02:36:31
00:02:06
00:02:41
00:02:03

Last edited by Ygor; 02-03-2009 at 10:02 PM..
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