删除数据行:重叠的时间区间?
编辑:我现在也在寻找其他编程语言的解决方案。
根据我之前提到的另一个问题,我有一个数据集,下面是给R用户的dput格式,这个数据集表示用户的电脑使用记录:
username machine start end
1 user1 D5599.domain.com 2011-01-03 09:44:18 2011-01-03 09:47:27
2 user1 D5599.domain.com 2011-01-03 09:46:29 2011-01-03 10:09:16
3 user1 D5599.domain.com 2011-01-03 14:07:36 2011-01-03 14:56:17
4 user1 D5599.domain.com 2011-01-05 15:03:17 2011-01-05 15:23:15
5 user1 D5599.domain.com 2011-02-14 14:33:39 2011-02-14 14:40:16
6 user1 D5599.domain.com 2011-02-23 13:54:30 2011-02-23 13:58:23
7 user1 D5599.domain.com 2011-03-21 10:10:18 2011-03-21 10:32:22
8 user1 D5645.domain.com 2011-06-09 10:12:41 2011-06-09 10:58:59
9 user1 D5682.domain.com 2011-01-03 12:03:45 2011-01-03 12:29:43
10 USER2 D5682.domain.com 2011-01-12 14:26:05 2011-01-12 14:32:53
11 USER2 D5682.domain.com 2011-01-17 15:06:19 2011-01-17 15:44:22
12 USER2 D5682.domain.com 2011-01-18 15:07:30 2011-01-18 15:42:43
13 USER2 D5682.domain.com 2011-01-25 15:20:55 2011-01-25 15:24:38
14 USER2 D5682.domain.com 2011-02-14 15:03:00 2011-02-14 15:07:43
15 USER2 D5682.domain.com 2011-02-14 14:59:23 2011-02-14 15:14:47
>
同一个用户名在同一台电脑上可能会有多个同时进行的(时间上重叠的)会话。我该如何去掉这些重复的记录,只保留一个会话呢?原始数据集大约有50万行。
期望的输出是(去掉第2行和第15行)
username machine start end
1 user1 D5599.domain.com 2011-01-03 09:44:18 2011-01-03 09:47:27
3 user1 D5599.domain.com 2011-01-03 14:07:36 2011-01-03 14:56:17
4 user1 D5599.domain.com 2011-01-05 15:03:17 2011-01-05 15:23:15
5 user1 D5599.domain.com 2011-02-14 14:33:39 2011-02-14 14:40:16
6 user1 D5599.domain.com 2011-02-23 13:54:30 2011-02-23 13:58:23
7 user1 D5599.domain.com 2011-03-21 10:10:18 2011-03-21 10:32:22
8 user1 D5645.domain.com 2011-06-09 10:12:41 2011-06-09 10:58:59
9 user1 D5682.domain.com 2011-01-03 12:03:45 2011-01-03 12:29:43
10 USER2 D5682.domain.com 2011-01-12 14:26:05 2011-01-12 14:32:53
11 USER2 D5682.domain.com 2011-01-17 15:06:19 2011-01-17 15:44:22
12 USER2 D5682.domain.com 2011-01-18 15:07:30 2011-01-18 15:42:43
13 USER2 D5682.domain.com 2011-01-25 15:20:55 2011-01-25 15:24:38
14 USER2 D5682.domain.com 2011-02-14 15:03:00 2011-02-14 15:07:43
>
这是数据集:
structure(list(username = c("user1", "user1", "user1",
"user1", "user1", "user1", "user1", "user1",
"user1", "USER2", "USER2", "USER2", "USER2", "USER2", "USER2"
), machine = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 3L,
3L, 3L, 3L, 3L, 3L, 3L), .Label = c("D5599.domain.com", "D5645.domain.com",
"D5682.domain.com", "D5686.domain.com", "D5694.domain.com", "D5696.domain.com",
"D5772.domain.com", "D5772.domain.com", "D5847.domain.com", "D5855.domain.com",
"D5871.domain.com", "D5927.domain.com", "D5927.domain.com", "D5952.domain.com",
"D5993.domain.com", "D6012.domain.com", "D6048.domain.com", "D6077.domain.com",
"D5688.domain.com", "D5815.domain.com", "D6106.domain.com", "D6128.domain.com"
), class = "factor"), start = structure(c(1294040658, 1294040789,
1294056456, 1294232597, 1297686819, 1298462070, 1300695018, 1307603561,
1294049025, 1294835165, 1295269579, 1295356050, 1295961655, 1297688580,
1297688363), class = c("POSIXct", "POSIXt"), tzone = ""), end =
structure(c(1294040847,
1294042156, 1294059377, 1294233795, 1297687216, 1298462303, 1300696342,
1307606339, 1294050583, 1294835573, 1295271862, 1295358163, 1295961878,
1297688863, 1297689287), class = c("POSIXct", "POSIXt"), tzone = "")),
.Names = c("username",
"machine", "start", "end"), row.names = c(NA, 15L), class = "data.frame")
5 个回答
1
我不知道这是不是你想要的,或者它是否比你现在的方案更好。这是一个使用PowerShell的解决方案,它用一个哈希表来存储数据,哈希表的键是用户名和计算机名的组合。值则是开始和结束时间的哈希。
如果一个键(会话)已经存在,它就会更新结束时间。如果不存在,它会创建一个新的键,并设置开始时间和初始结束时间。当它在日志中遇到该用户/计算机的新会话记录时,它会更新这个会话键的结束时间。
$ht = @{}
import-csv <logfile> |
foreach{
$key = $_.username + $_.computername
if ($ht.ContainsKey($key)){$ht.$key.end = $_.end}
else{$ht.add("$key",@{start=$_.start;end=$_.end}}
}
完成后,你需要从键中分离出用户和计算机名。
1
这里介绍了一种使用 lubridate
里的 interval
类的替代方案。
library(lubridate)
int <- with(d, new_interval(start, end))
现在我们需要一个函数来检查时间段是否重叠。可以参考这个链接:判断两个日期范围是否重叠。
int_overlaps <- function(int1, int2)
{
(int_start(int1) <= int_end(int2)) &
(int_start(int2) <= int_end(int1))
}
接下来,对所有的时间段组合进行这个检查。
index <- combn(seq_along(int), 2)
overlaps <- int_overlaps(int[index[1, ]], int[index[2, ]])
重叠的行:
int[index[1, overlaps]]
int[index[2, overlaps]]
需要删除的行就是 index[2, overlaps]
。
4
试试这个intervals包:
library(intervals)
f <- function(dd) with(dd, {
r <- reduce(Intervals(cbind(start, end)))
data.frame(username = username[1],
machine = machine[1],
start = structure(r[, 1], class = class(start)),
end = structure(r[, 2], class = class(end)))
})
do.call("rbind", by(d, d[1:2], f))
用这个示例数据,这个操作把原本的15行数据减少到了13行(通过把原数据框中的第1行和第2行,以及第12行和第13行合并在一起):
username machine start end
1 user1 D5599.domain.com 2011-01-03 02:44:18 2011-01-03 03:09:16
2 user1 D5599.domain.com 2011-01-03 07:07:36 2011-01-03 07:56:17
3 user1 D5599.domain.com 2011-01-05 08:03:17 2011-01-05 08:23:15
4 user1 D5599.domain.com 2011-02-14 07:33:39 2011-02-14 07:40:16
5 user1 D5599.domain.com 2011-02-23 06:54:30 2011-02-23 06:58:23
6 user1 D5599.domain.com 2011-03-21 04:10:18 2011-03-21 04:32:22
7 user1 D5645.domain.com 2011-06-09 03:12:41 2011-06-09 03:58:59
8 user1 D5682.domain.com 2011-01-03 05:03:45 2011-01-03 05:29:43
9 USER2 D5682.domain.com 2011-01-12 07:26:05 2011-01-12 07:32:53
10 USER2 D5682.domain.com 2011-01-17 08:06:19 2011-01-17 08:44:22
11 USER2 D5682.domain.com 2011-01-18 08:07:30 2011-01-18 08:42:43
12 USER2 D5682.domain.com 2011-01-25 08:20:55 2011-01-25 08:24:38
13 USER2 D5682.domain.com 2011-02-14 07:59:23 2011-02-14 08:14:47