Unix中的时间戳数据操作

Question

I have a csv data file that has two timestamp fields - start_time and end_time. They are strings in the form of "2014-02-01 00:06:22" . Each line of the data file is a record with multiple fields. The file is pretty small.

I want to calculate the average duration among all records. Other than using shell scripts, is there any one-liner command that I could use for this kind of simple calculation, possibly using awk?

I'm very new to awk. Here's what I have but does not work. $6 and $7 are fields for start_time and end_time.

awk -F, 'BEGIN { count=0 total=0 }
    { sec1=date +%s -d $6 sec2=date +%s -d $7
    total+=sec2-sec1 count++} 
    END {print "avg trip time: ", total/count}' dataset.csv

Sample of the csv file:

"start_time","stop_time","start station name","end station name","bike_id"
"2014-02-01 00:00:00","2014-02-01 00:06:22","Washington Square E","Stanton St & Chrystie St","21101"

Answer 1

Using GNU awk for mktime() and gensub():

$ cat tst.awk
BEGIN { FS="^\"|\",\"" }
function t2s(time) { return mktime(gensub(/[-:]/," ","g",time)) }
NR>1 { totDurs += (t2s($3) - t2s($2)) }
END { print totDurs / (NR-1) }

$ gawk -f tst.awk file
382

with other awks you need to call the shell date function:

$ cat tst2.awk
BEGIN { FS="^\"|\",\"" }
function t2s(time,      cmd,secs) {
    cmd = "date +%s -d \"" time "\""
    if ( (cmd | getline secs) <= 0 ) {
        secs = -1
    }
    close(cmd)
    return secs
}
NR>1 { totDurs += (t2s($3) - t2s($2)) }
END { print totDurs / (NR-1) }

$ awk -f tst2.awk file                               
382