计算给定数据集的时间差

Question

I have a list of down/up times and need to calculate the total down time, per day

Here is a snippet of the original dataset:

Jan  1 00:09:56 changed state to down
Jan  1 00:10:02 changed state to up
Jan  1 00:10:45 changed state to down
Jan  1 00:10:52 changed state to up
Jan  1 00:11:45 changed state to down
Jan  1 00:11:52 changed state to up
Jan  2 00:14:55 changed state to down
Jan  2 00:15:05 changed state to up
Jan  2 00:15:35 changed state to down
Jan  2 00:16:41 changed state to up
Jan  3 00:05:22 changed state to down
Jan  3 00:05:28 changed state to up
Jan  3 23:59:58 changed state to down
Jan  4 00:00:19 changed state to up
Jan  4 00:49:28 changed state to down
Jan  4 00:49:34 changed state to up

What I'm asking for is an output like this:

Jan 1: 20 seconds
Jan 2: 76 seconds
Jan 3: 27 seconds
Jan 4: 6 seconds

---

I've written this script to get the timestamps into epoch time:

while read i; do
  datetime=$(echo $i | awk '{print $1" "$2" "$3}')
  epochtime=$(date -d "$datetime" +\%s)
  echo $epochtime
done < ./file

which will give me the data like this:

1640995796
1640995802
1640995845
1640995852
...

I can't seem to find a way to combine them to a single line so I can use this code (or similar):

for i in $epochtime; do
  starttime=$(awk '{print $2}')
  endtime=$(awk '{print $1}')
  delta=$(( endtime - starttime ))
  echo $delta
done

Thanks in advance!

Answer 1

This doesn't produce exactly the expected output and I cant be bothered counting the seconds and up/down indicators to try to figure out the difference between what I thought the requirements are vs what they actually are but it's the right idea for what you're trying to do:

$ cat tst.awk
{
    prevEt = epochtime
    date = $1 " " $2
    $1 = (index("JanFebMarAprMayJunJulAugSepOctNovDec",$1) + 2) / 3
    epochtime = mktime("2022 " gensub(/:/," ","g"), 1)
}
NR%2 == 0 {
    delta += (epochtime - prevEt)
    if ( date != prevDate ) {
        printf "%s: %d seconds\n", prevDate, delta
        delta = 0
        prevDate = date
    }
}
END {
    printf "%s: %d seconds\n", prevDate, delta
}

$ awk -f tst.awk file
: 6 seconds
Jan 1: 24 seconds
Jan 2: 72 seconds
Jan 3: 21 seconds
Jan 4: 6 seconds

---

Original answer:

This may be what you're trying to do, using GNU awk for time functions:

$ cat tst.awk
{
    prevEt = epochtime
    $1 = (index("JanFebMarAprMayJunJulAugSepOctNovDec",$1) + 2) / 3
    epochtime = mktime("2022 " gensub(/:/," ","g"), 1)
}
NR%2 == 0 {
    print prevEt, epochtime, epochtime - prevEt
}

$ awk -f tst.awk file
1640995796 1640995802 6
1640995845 1640995852 7