使用“登录"重塑 R 中的数据“退出"次 [英] Reshaping data in R with "login" "logout" times
问题描述
我是 R 的新手，正在为自己的目的从事一个副项目.我有这个数据(可重现的 dput 在问题的末尾):
I'm new to R, and am working on a side project for my own purposes. I have this data (reproducable dput of this is at the end of the question):
X datetime user state
1 1 2016-02-19 19:13:26 User1 joined
2 2 2016-02-19 19:21:18 User2 joined
3 3 2016-02-19 19:21:33 User1 joined
4 4 2016-02-19 19:35:38 User1 joined
5 5 2016-02-19 19:44:15 User1 joined
6 6 2016-02-19 19:48:55 User1 joined
7 7 2016-02-19 19:52:40 User1 joined
8 8 2016-02-19 19:53:15 User3 joined
9 9 2016-02-19 20:02:34 User3 joined
10 10 2016-02-19 20:13:48 User3 joined
19 637 2016-02-19 19:13:32 User1 left
20 638 2016-02-19 19:25:26 User1 left
21 639 2016-02-19 19:30:30 User2 left
22 640 2016-02-19 19:42:16 User1 left
23 641 2016-02-19 19:47:59 User1 left
24 642 2016-02-19 19:51:06 User1 left
25 643 2016-02-19 20:02:26 User3 left
我希望它看起来像这样:
I want it to look like this:
user joined left
1 User1 2016-02-19 19:13:26 2016-02-19 19:13:32
2 User2 2016-02-19 19:21:18 2016-02-19 19:30:30
3 User3 2016-02-19 19:53:15 2016-02-19 20:02:26
4 User1 2016-02-19 19:21:33 2016-02-19 19:25:26
.
.
.
我正在研究 tidyr，因为显然涉及一些重塑，但我无法理解到底需要做什么.这甚至可能吗(没有循环/大量的程序代码)?我无法理解如何绕过的问题是，无法知道特定的左"记录应该连接到特定的连接"记录.我可以找到的所有示例都涉及收集其他值的静态月份或日期.我应该补充一点，不一定保证所有记录都具有左"值(用户可能仍会加入").
I'm looking at tidyr as there's some reshaping involved obviously, but I can't wrap my head around what exactly needs to be done. Is this even possible (without looping/massive amounts of procedural code)? The problem I can't grasp how to get around is that there's no way to know that a particular "left" record should be joined to a particular "joined" record. Examples I can find all involve a static month or day over which other values are gathered. I should add that it's not necessarily guaranteed that all records are guaranteed to have a "left" value (a user might still be "joined").
这是数据样本的 dput 输出:
Here's the dput output of a sample of the data:
> dput(samp)
structure(list(X = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 637L, 638L, 639L, 640L,
641L, 642L, 643L, 644L, 645L, 646L, 647L, 648L, 649L, 650L, 651L
), datetime = structure(c(1L, 3L, 4L, 7L, 9L, 11L, 13L, 14L,
16L, 18L, 21L, 22L, 23L, 26L, 27L, 30L, 32L, 33L, 2L, 5L, 6L,
8L, 10L, 12L, 15L, 17L, 19L, 20L, 24L, 25L, 28L, 29L, 31L), .Label = c("2016-02-19 19:13:26",
"2016-02-19 19:13:32", "2016-02-19 19:21:18", "2016-02-19 19:21:33",
"2016-02-19 19:25:26", "2016-02-19 19:30:30", "2016-02-19 19:35:38",
"2016-02-19 19:42:16", "2016-02-19 19:44:15", "2016-02-19 19:47:59",
"2016-02-19 19:48:55", "2016-02-19 19:51:06", "2016-02-19 19:52:40",
"2016-02-19 19:53:15", "2016-02-19 20:02:26", "2016-02-19 20:02:34",
"2016-02-19 20:13:38", "2016-02-19 20:13:48", "2016-02-19 20:42:27",
"2016-02-19 20:48:22", "2016-02-19 20:49:31", "2016-02-19 20:59:58",
"2016-02-19 21:06:20", "2016-02-19 21:10:43", "2016-02-19 21:11:13",
"2016-02-19 21:11:15", "2016-02-19 21:11:22", "2016-02-19 21:17:33",
"2016-02-19 22:02:45", "2016-02-19 22:05:18", "2016-02-19 22:05:37",
"2016-02-19 22:05:47", "2016-02-19 22:30:30"), class = "factor"),
user = structure(c(1L, 2L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L,
3L, 4L, 1L, 1L, 4L, 4L, 4L, 3L, 1L, 1L, 2L, 1L, 1L, 1L, 3L,
3L, 3L, 1L, 4L, 1L, 1L, 4L, 4L), .Label = c("User1", "User2",
"User3", "User4"), class = "factor"), state = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L), .Label = c("joined", "left"), class = "factor")), .Names = c("X",
"datetime", "user", "state"), class = "data.frame", row.names = c(NA,
-33L))
推荐答案
我们可以利用left"和joined"的顺序，每个用户一个跟一个的匹配.
We can make use of the order of "left" and "joined", and match when one follows the other for each user.
为此，我将使用 library(data.table)
library(data.table)
setDT(df)
## order the data by user and datetime
df <- df[order(user, datetime)]
## add an 'order' column, which is a sequence from 1 to lenght()
## for each user
df[, order := seq(1:.N), by=user]
## split the left and joins
dt_left <- df[state == "left"]
dt_joined <- df[state == "joined"]
## assuming 'left' is after 'joined', shift the 'order' back for left
dt_left[, order := order - 1]
## join user an dorder (and subsetting relevant columns)
## keeping when there's a 'joined' but not a 'left'
dt <- dt_left[, .(user, order, datetime)][dt_joined[, .(user, order, datetime)], on=c("user", "order"), nomatch=NA]
## rename columns
setnames(dt, c("datetime", "i.datetime"), c("left", "joined"))
user order left joined
1: User1 1 2016-02-19 19:13:32 2016-02-19 19:13:26
2: User1 3 2016-02-19 19:25:26 2016-02-19 19:21:33
3: User1 5 2016-02-19 19:42:16 2016-02-19 19:35:38
4: User1 7 2016-02-19 19:47:59 2016-02-19 19:44:15
5: User1 9 2016-02-19 19:51:06 2016-02-19 19:48:55
6: User1 11 2016-02-19 20:48:22 2016-02-19 19:52:40
7: User1 13 2016-02-19 21:11:13 2016-02-19 21:06:20
8: User1 15 2016-02-19 21:17:33 2016-02-19 21:11:15
9: User2 1 2016-02-19 19:30:30 2016-02-19 19:21:18
10: User3 1 2016-02-19 20:02:26 2016-02-19 19:53:15
11: User3 3 2016-02-19 20:13:38 2016-02-19 20:02:34
12: User3 5 2016-02-19 20:42:27 2016-02-19 20:13:48
13: User3 7 NA 2016-02-19 20:49:31
14: User3 8 NA 2016-02-19 22:30:30
15: User4 1 2016-02-19 21:10:43 2016-02-19 20:59:58
16: User4 3 2016-02-19 22:02:45 2016-02-19 21:11:22
17: User4 5 2016-02-19 22:05:37 2016-02-19 22:05:18
18: User4 7 NA 2016-02-19 22:05:47
这篇关于使用“登录"重塑 R 中的数据“退出"次的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
