我有一个大小为 7GB 的文件。现在有两个日期时间,我想使用 awk 来获取这两个日期时间之间的时间差。
下面是我的文件的样子:
A B C D E
18/06/28 09:19:07 295 141536 18-06-28 09:17:47
18/06/28 09:20:07 268 1160 18-06-28 09:18:58
18/06/28 09:21:07 317 1454 18-06-28 09:19:47
18/06/28 09:22:07 275 1491 18-06-28 09:20:59
18/06/28 09:23:07 320 1870 18-06-28 09:21:07
18/06/28 09:24:07 310 1869 18-06-28 09:22:30
18/06/28 09:25:07 150 693 18-06-28 09:23:28
18/06/28 09:26:07 414 2227 18-06-28 09:24:34
我想要 (AB) - (E) 之间的差异。
我试过这个:
cat filename | awk -F " " '{print date -d ($1$2)-($5)}'
输出应该是两个日期时间之间的时间差。就像第一行的差异将是 1 分 20 秒
答案1
使用 GNU awk:
gawk '
function dt2epoch(date, time, timestr) {
timestr = "20" substr(date,1,2) " " substr(date,4,2) " " substr(date,7,2) \
" " substr(time,1,2) " " substr(time,4,2) " " substr(time,7,2)
return mktime(timestr)
}
function epoch2hms(t) {
return strftime("%H:%M:%S", t, 1)
}
function abs(n) {return n<0 ? -1*n : n}
NR == 1 {next}
{ print epoch2hms(abs(dt2epoch($5,$6) - dt2epoch($1,$2))) }
' file
输出
00:01:20
00:01:09
00:01:20
00:01:08
00:02:00
00:01:37
00:01:39
00:01:33
对于perl,我会使用约会时间生态系统:
perl -MDateTime::Format::Strptime -lane '
BEGIN {$f = DateTime::Format::Strptime->new(pattern => "%y-%m-%d %H:%M:%S")}
next if $. == 1;
$F[0] =~ s{/}{-}g;
$t1 = $f->parse_datetime("$F[0] $F[1]");
$t2 = $f->parse_datetime("$F[4] $F[5]");
$d = $t1->subtract_datetime($t2);
printf "%02d:%02d:%02d\n", $d->hours, $d->minutes, $d->seconds;
' file
更快的 perl 版本,不需要任何非核心模块
perl -MTime::Piece -lane '
next if $. == 1;
$t1 = Time::Piece->strptime("$F[0] $F[1]", "%y/%m/%d %H:%M:%S");
$t2 = Time::Piece->strptime("$F[4] $F[5]", "%y-%m-%d %H:%M:%S");
$diff = gmtime(abs($t1->epoch - $t2->epoch));
print $diff->hms;
' file
或者,替代输出:
$ perl -MTime::Piece -lane '
next if $. == 1;
$t1 = Time::Piece->strptime("$F[0] $F[1]", "%y/%m/%d %H:%M:%S");
$t2 = Time::Piece->strptime("$F[4] $F[5]", "%y-%m-%d %H:%M:%S");
print abs($t1 - $t2)->pretty;
' file
1 minutes, 20 seconds
1 minutes, 9 seconds
1 minutes, 20 seconds
1 minutes, 8 seconds
2 minutes, 0 seconds
1 minutes, 37 seconds
1 minutes, 39 seconds
1 minutes, 33 seconds
答案2
使用bash
和awk
组合:
$ awk 'NR>1 {print $1,$2,$5,$6}' input | while read d1 t1 d2 t2; do
i1=$(date -u -d "20$d1 $t1" +%s)
i2=$(date -u -d "20$d1 $t2" +%s)
date -d @"$((i1-i2))" +%M:%S;
done
01:20
01:09
01:20
01:08
02:00
01:37
01:39
01:33
答案3
基准测试:我多次重复了你的样本数据
$ wc -l file
131073 file
现在一些时间安排:
$ time awk 'NR>1 {print $1,$2,$5,$6}' file |
while read d1 t1 d2 t2; do
i1=$(date -u -d "20$d1 $t1" +%s)
i2=$(date -u -d "20$d1 $t2" +%s)
date -d @"$((i1-i2))" +%M:%S
done > /dev/null
real 8m55.533s
user 5m46.956s
sys 1m33.726s
$ time perl -MDateTime::Format::Strptime -lane '
BEGIN {$f = DateTime::Format::Strptime->new(pattern => "%y-%m-%d %H:%M:%S")}
next if $. == 1;
$F[0] =~ s{/}{-}g;
$t1 = $f->parse_datetime("$F[0] $F[1]");
$t2 = $f->parse_datetime("$F[4] $F[5]");
$d = $t1->subtract_datetime($t2);printf "%02d:%02d:%02d\n", $d->hours, $d->minutes, $d->seconds;
' file > /dev/null
real 0m37.684s
user 0m33.168s
sys 0m0.058s
$ time gawk '
function dt2epoch(date, time, timestr) {
timestr = "20" substr(date,1,2) " " substr(date,4,2) " " substr(date,7,2) \
" " substr(time,1,2) " " substr(time,4,2) " " substr(time,7,2)
return mktime(timestr)
}
function epoch2hms(t) {
return strftime("%H:%M:%S", t, 1)
}
function abs(n) {return n<0 ? -1*n : n}
NR == 1 {next}
{ print epoch2hms(abs(dt2epoch($5,$6) - dt2epoch($1,$2))) }
' file > /dev/null
real 0m1.074s
user 0m0.610s
sys 0m0.366s
具有内置时间函数的 GNU awk 显然是赢家,即使进行所有字符串操作也是如此。
更新:新的 perl 实现。仍然比 gawk 慢,但比使用功能丰富但重量级的 DateTime 模块的版本领先很多:
$ time perl -MTime::Piece -lane '
next if $. == 1;
$t1 = Time::Piece->strptime("$F[0] $F[1]", "%y/%m/%d %H:%M:%S");
$t2 = Time::Piece->strptime("$F[4] $F[5]", "%y-%m-%d %H:%M:%S");
$diff = gmtime(abs($t1->epoch - $t2->epoch));
print $diff->hms;
' file > /dev/null
real 0m2.631s
user 0m2.231s
sys 0m0.170s