匯總時間序列的多個組內的數據

Question

我對不同地點和時間的鳥類進行了一系列觀察。 數據框如下所示：

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

我想以一種方式總結數據，顯示每個站點的每只鳥的第一次和最后一次檢測，以及每個站點的持續時間，同時保留有關多次訪問站點的信息（即，如果一只鳥從站點A> B出發） > C> A> B，我想獨立顯示每次訪問A站點和B站點，而不是將兩次訪問放在一起）。

我希望產生這樣的輸出，其中保留每次訪問的開始（min_ts），結束（max_ts）和持續時間（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

我嘗試過這段代碼，它產生了正確的變量，但是將所有關於單個站點的信息整合在一起，而不是保留多次訪問：

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

我意識到按站點分組是一個問題，但如果我將其作為分組變量刪除，則數據將在沒有站點信息的情況下進行匯總。 我試過這個。 它沒有運行，但我覺得它接近解決方案：

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)) 

Answer 1

一種可能性是：

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

在這里，它創建一個rleid()的分組變量，然后計算差異。

或者使用相同rleid()從data.table明確：

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)

Answer 2

另一種方法是使用lag和cumsum來創建分組變量。

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