我正在尝试计算每个人（trialnumber）的试验之间的时间量（sq_id）。我已经能够弄清楚如何计算试验之间的时差（time_gap），但我的输出中包含了不应该存在的所有重复行。
我的数据的一部分可以找到here。出于可重复性的目的，我在下面包含了数据集（称为export）：
sq_id ageclass sex cohort year age grid trialnumber trialdate trialtime
6244 A F 2000 2005 5 AG 1 05/24/05 0:00
10212 A M 2006 2008 2 KL 1 05/04/08 6:13
10212 A M 2006 2010 4 KL 4 05/20/10 6:12
10212 A M 2006 2009 3 KL 2 06/10/09 6:14
10212 A M 2006 2009 3 KL 3 07/01/09 6:15
23052 J F 2017 2017 0 SU 2 08/02/17 11:00
23052 J F 2017 2017 0 SU 1 07/20/17 10:51
23080 J M 2017 2017 0 KL 2 07/29/17 10:20
23080 J M 2017 2017 0 KL 1 07/07/17 8:35
我做的第一件事是计算试验之间的时间，如下：
#adding time between trials to data
trialdate<-as.POSIXct(data$trialdate,format="%m/%d/%y")
data$datetime=as.POSIXct(paste(trialdate, data$trialtime),format= '%Y-%m-%d',usetz=FALSE)
#calculates time btw first trial and all other trials
timebtw <- data %>% group_by(sq_id) %>%
select(sq_id, trialnumber, datetime) %>%
mutate(time_gap = (datetime - nth(datetime, which.min((datetime)))), time_gap=time_gap/86400) #time_gap units are in seconds, changed to days
然后我将timebtw数据集加入我的原始数据集（称为export）：
new<-dplyr::left_join(export, timebtw, by = "sq_id")
我得到的输出如下：
> export
sq_id ageclass sex cohort year age grid trialnumber.x trialdate trialtime datetime time_gap trialnumber.y
6244 A F 2000 2005 5 AG 1 05/24/05 0:00 2005-05-24 0 secs 1
10212 A M 2006 2008 2 KL 1 05/04/08 6:13 2008-05-04 0 secs 1
10212 A M 2006 2008 2 KL 1 05/04/08 6:13 2008-05-04 746 secs 4
10212 A M 2006 2008 2 KL 1 05/04/08 6:13 2008-05-04 402 secs 2
10212 A M 2006 2008 2 KL 1 05/04/08 6:13 2008-05-04 423 secs 3
10212 A M 2006 2010 4 KL 4 05/20/10 6:12 2010-05-20 0 secs 1
10212 A M 2006 2010 4 KL 4 05/20/10 6:12 2010-05-20 746 secs 4
10212 A M 2006 2010 4 KL 4 05/20/10 6:12 2010-05-20 402 secs 2
10212 A M 2006 2010 4 KL 4 05/20/10 6:12 2010-05-20 423 secs 3
10212 A M 2006 2009 3 KL 2 06/10/09 6:14 2009-06-10 0 secs 1
10212 A M 2006 2009 3 KL 2 06/10/09 6:14 2009-06-10 746 secs 4
10212 A M 2006 2009 3 KL 2 06/10/09 6:14 2009-06-10 402 secs 2
10212 A M 2006 2009 3 KL 2 06/10/09 6:14 2009-06-10 423 secs 3
10212 A M 2006 2009 3 KL 3 07/01/09 6:15 2009-07-01 0 secs 1
10212 A M 2006 2009 3 KL 3 07/01/09 6:15 2009-07-01 746 secs 4
10212 A M 2006 2009 3 KL 3 07/01/09 6:15 2009-07-01 402 secs 2
10212 A M 2006 2009 3 KL 3 07/01/09 6:15 2009-07-01 423 secs 3
23052 J F 2017 2017 0 SU 2 08/02/17 11:00 2017-08-02 13 secs 2
23052 J F 2017 2017 0 SU 2 08/02/17 11:00 2017-08-02 0 secs 1
23052 J F 2017 2017 0 SU 1 07/20/17 10:51 2017-07-20 13 secs 2
23052 J F 2017 2017 0 SU 1 07/20/17 10:51 2017-07-20 0 secs 1
23080 J M 2017 2017 0 KL 2 07/29/17 10:20 2017-07-29 22 secs 2
23080 J M 2017 2017 0 KL 2 07/29/17 10:20 2017-07-29 0 secs 1
23080 J M 2017 2017 0 KL 1 07/07/17 8:35 2017-07-07 22 secs 2
23080 J M 2017 2017 0 KL 1 07/07/17 8:35 2017-07-07 0 secs 1
这是一个问题。每time_gap只应有一个trialnumber值。
因此，例如，对于sq_id 10212，输出应如下所示：
sq_id ageclass sex cohort year age grid trialnumber.x trialdate trialtime datetime time_gap trialnumber.y
10212 A M 2006 2008 2 KL 1 05/04/08 6:13 2008-05-04 0 secs 1
10212 A M 2006 2010 4 KL 4 05/20/10 6:12 2010-05-20 746 secs 4
10212 A M 2006 2009 3 KL 2 06/10/09 6:14 2009-06-10 402 secs 2
10212 A M 2006 2009 3 KL 3 07/01/09 6:15 2009-07-01 423 secs 3
我需要trialnumber.x和trialnumber.y列匹配，这样只有与试验一样多的行（即sq_id 6244将有1行，{{1 10212 4行，sq_id 23052 2行，sq_id 23080 2行）。
有谁知道如何修改我的代码才能获得此输出？
答案 0 :(得分：0)
dat %>%
mutate(trial_dt = lubridate::mdy_hms(paste(trialdate, trialtime))) %>%
group_by(sq_id) %>%
mutate(time_gap = difftime(trial_dt, min(trial_dt), units = "days"))
# # A tibble: 9 x 11
# # Groups: sq_id [4]
# sq_id ageclass sex `cohort year` age grid trialnumber trialdate trialtime trial_dt time_gap
# <int> <chr> <chr> <chr> <int> <chr> <int> <chr> <time> <dttm> <time>
# 1 6244 A F 2000 2005 5 AG 1 05/24/05 00:00 2005-05-24 00:00:00 0
# 2 10212 A M 2006 2008 2 KL 1 05/04/08 06:13 2008-05-04 06:13:00 0
# 3 10212 A M 2006 2010 4 KL 4 05/20/10 06:12 2010-05-20 06:12:00 745.999305555556
# 4 10212 A M 2006 2009 3 KL 2 06/10/09 06:14 2009-06-10 06:14:00 402.000694444444
# 5 10212 A M 2006 2009 3 KL 3 07/01/09 06:15 2009-07-01 06:15:00 423.001388888889
# 6 23052 J F 2017 2017 0 SU 2 08/02/17 11:00 2017-08-02 11:00:00 13.00625
# 7 23052 J F 2017 2017 0 SU 1 07/20/17 10:51 2017-07-20 10:51:00 0
# 8 23080 J M 2017 2017 0 KL 2 07/29/17 10:20 2017-07-29 10:20:00 22.0729166666667
# 9 23080 J M 2017 2017 0 KL 1 07/07/17 08:35 2017-07-07 08:35:00 0
似乎没有必要单独计算时间间隔，因此无需加入：
{{1}}