汇总时间序列的多个组内的数据
[英]Summarize data within multiple groups of a time series
问题描述
I have a series of observations of birds at different locations and times. 
我对不同地点和时间的鸟类进行了一系列观察。 The data frame looks like this: 数据框如下所示：
birdID site ts
1 A 2013-04-15 09:29
1 A 2013-04-19 01:22
1 A 2013-04-20 23:13
1 A 2013-04-22 00:03
1 B 2013-04-22 14:02
1 B 2013-04-22 17:02
1 C 2013-04-22 14:04
1 C 2013-04-22 15:18
1 C 2013-04-23 00:54
1 A 2013-04-23 01:20
1 A 2013-04-24 23:07
1 A 2013-04-30 23:47
1 B 2013-04-30 03:51
1 B 2013-04-30 04:26
2 C 2013-04-30 04:29
2 C 2013-04-30 18:49
2 A 2013-05-01 01:03
2 A 2013-05-01 23:15
2 A 2013-05-02 00:09
2 C 2013-05-03 07:57
2 C 2013-05-04 07:21
2 C 2013-05-05 02:54
2 A 2013-05-05 03:27
2 A 2013-05-14 00:16
2 D 2013-05-14 10:00
2 D 2013-05-14 15:00
I would like to summarize the data in a way that shows the first and last detection of each bird at each site, and the duration at each site, while preserving information about multiple visits to sites (ie if a bird went from site A > B > C > A > B, I would like show each visit to site A and B independently, not lump both visits together). 我想以一种方式总结数据，显示每个站点的每只鸟的第一次和最后一次检测，以及每个站点的持续时间，同时保留有关多次访问站点的信息（即，如果一只鸟从站点A> B出发） > C> A> B，我想独立显示每次访问A站点和B站点，而不是将两次访问放在一起）。
I am hoping to produce output like this, where the start (min_ts), end (max_ts), and duration (days) of each visit are preserved: 我希望产生这样的输出，其中保留每次访问的开始（min_ts），结束（max_ts）和持续时间（天）：
birdID site min_ts max_ts days
1 A 2013-04-15 09:29 2013-04-22 00:03 6.6
1 B 2013-04-22 14:02 2013-04-22 17:02 0.1
1 C 2013-04-22 14:04 2013-04-23 00:54 0.5
1 A 2013-04-23 01:20 2013-04-30 23:47 7.9
1 B 2013-04-30 03:51 2013-04-30 04:26 0.02
2 C 2013-04-30 4:29 2013-04-30 18:49 0.6
2 A 2013-05-01 01:03 2013-05-02 00:09 0.96
2 C 2013-05-03 07:57 2013-05-05 02:54 1.8
2 A 2013-05-05 03:27 2013-05-14 00:16 8.8
2 D 2013-05-14 10:00 2013-05-14 15:00 0.2
I have tried this code, which yields the correct variables but lumps all the information about a single site together, not preserving multiple visits: 我尝试过这段代码，它产生了正确的变量，但是将所有关于单个站点的信息整合在一起，而不是保留多次访问：
df <- df %>%
group_by(birdID, site) %>%
summarise(min_ts = min(ts),
max_ts = max(ts),
days = difftime(max_ts, min_ts, units = "days")) %>%
arrange(birdID, min_ts)
birdID site min_ts max_ts days
1 A 2013-04-15 09:29 2013-04-30 23:47 15.6
1 B 2013-04-22 14:02 2013-04-30 4:26 7.6
1 C 2013-04-22 14:04 2013-04-23 0:54 0.5
2 C 2013-04-30 04:29 2013-05-05 2:54 4.9
2 A 2013-05-01 01:03 2013-05-14 0:16 12.9
2 D 2013-05-14 10:00 2013-05-14 15:00 0.2
I realize grouping by site is a problem, but if I remove that as a grouping variable the data are summarised without site info. 我意识到按站点分组是一个问题，但如果我将其作为分组变量删除，则数据将在没有站点信息的情况下进行汇总。 I have tried this. 我试过这个。 It doesn't run, but I feel it's close to the solution: 它没有运行，但我觉得它接近解决方案：
df <- df %>%
group_by(birdID) %>%
summarize(min_ts = if_else((birdID == lag(birdID) & site != lag(site)), min(ts), NA_real_),
max_ts = if_else((birdID == lag(birdID) & site != lag(site)), max(ts), NA_real_),
min_d = min(yday(ts)),
max_d = max(yday(ts)),
days = max_d - min_d))
2 个解决方案
解决方案1
5 已采纳 2019-06-23 22:21:26
One possibility could be: 一种可能性是：
df %>%
group_by(birdID, site, rleid = with(rle(site), rep(seq_along(lengths), lengths))) %>%
summarise(min_ts = min(ts),
max_ts = max(ts),
days = difftime(max_ts, min_ts, units = "days")) %>%
ungroup() %>%
select(-rleid) %>%
arrange(birdID, min_ts)
birdID site min_ts max_ts days
<int> <chr> <dttm> <dttm> <drtn>
1 1 A 2013-04-15 09:29:00 2013-04-22 00:03:00 6.60694444 days
2 1 B 2013-04-22 14:02:00 2013-04-22 17:02:00 0.12500000 days
3 1 C 2013-04-22 14:04:00 2013-04-23 00:54:00 0.45138889 days
4 1 A 2013-04-23 01:20:00 2013-04-30 23:47:00 7.93541667 days
5 1 B 2013-04-30 03:51:00 2013-04-30 04:26:00 0.02430556 days
6 2 C 2013-04-30 04:29:00 2013-04-30 18:49:00 0.59722222 days
7 2 A 2013-05-01 01:03:00 2013-05-02 00:09:00 0.96250000 days
8 2 C 2013-05-03 07:57:00 2013-05-05 02:54:00 1.78958333 days
9 2 A 2013-05-05 03:27:00 2013-05-14 00:16:00 8.86736111 days
10 2 D 2013-05-14 10:00:00 2013-05-14 15:00:00 0.20833333 days
Here it creates a rleid() -like grouping variable and then calculates the difference. 在这里，它创建一个rleid()的分组变量，然后计算差异。
Or the same using rleid() from data.table explicitly: 或者使用相同rleid()从data.table明确：
df %>%
group_by(birdID, site, rleid = rleid(site)) %>%
summarise(min_ts = min(ts),
max_ts = max(ts),
days = difftime(max_ts, min_ts, units = "days")) %>%
ungroup() %>%
select(-rleid) %>%
arrange(birdID, min_ts)
解决方案2
1 2019-06-23 23:42:55
Another alternative is to use lag and cumsum to create a grouping variable. 另一种方法是使用lag和cumsum来创建分组变量。
library(dplyr)
df %>%
group_by(birdID, group = cumsum(site != lag(site, default = first(site)))) %>%
summarise(min_ts = min(ts),
max_ts = max(ts),
days = difftime(max_ts, min_ts, units = "days")) %>%
ungroup() %>%
select(-group)
# A tibble: 10 x 4
# birdID min_ts max_ts days
# <int> <dttm> <dttm> <drtn>
# 1 1 2013-04-15 09:29:00 2013-04-22 00:03:00 6.60694444 days
# 2 1 2013-04-22 14:02:00 2013-04-22 17:02:00 0.12500000 days
# 3 1 2013-04-22 14:04:00 2013-04-23 00:54:00 0.45138889 days
# 4 1 2013-04-23 01:20:00 2013-04-30 23:47:00 7.93541667 days
# 5 1 2013-04-30 03:51:00 2013-04-30 04:26:00 0.02430556 days
# 6 2 2013-04-30 04:29:00 2013-04-30 18:49:00 0.59722222 days
# 7 2 2013-05-01 01:03:00 2013-05-02 00:09:00 0.96250000 days
# 8 2 2013-05-03 07:57:00 2013-05-05 02:54:00 1.78958333 days
# 9 2 2013-05-05 03:27:00 2013-05-14 00:16:00 8.86736111 days
#10 2 2013-05-14 10:00:00 2013-05-14 15:00:00 0.20833333 days
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