Question

我有时间序列数据，我希望按小时和计数折叠。

library(tidyverse)
library(lubridate)
library(tibbletime)

df.have <- data.frame(v1=ymd_hms("2018-01-01 01:00:00",
                                 "2018-01-01 01:24:20",
                                 "2018-01-01 02:00:00",
                                 "2018-01-01 03:00:00",
                                 "2018-01-01 03:21:20",
                                 "2018-01-01 03:22:20",
                                 #"2018-01-01 04:00:00",
                                 "2018-01-01 05:00:00",
                                 "2018-01-01 06:00:00",
                                 "2018-01-01 06:10:00",
                                 #"2018-01-01 07:00:00",
                                 #"2018-01-01 08:00:00",
                                 "2018-01-01 09:00:00",
                                 "2018-01-02 01:00:00",
                                 "2018-01-02 01:24:20",
                                 "2018-01-02 02:00:00",
                                 "2018-01-02 03:00:00",
                                 "2018-01-02 03:21:20",
                                 "2018-01-02 03:22:20",
                                 #"2018-01-02 04:00:00",
                                 "2018-01-02 05:00:00",
                                 "2018-01-02 06:00:00",
                                 "2018-01-02 06:10:00",
                                 #"2018-01-02 07:00:00",
                                 #"2018-01-02 08:00:00",
                                 "2018-01-02 09:00:00"),
                      v2=c(1, 0, 1, 1, NA, 1, 1, 1, 1, 1,
                           1, 0, 1, 1, NA, 1, 1, 1, 1, 1))

df.have %>%
  as_tbl_time(index = v1) %>%
  collapse_by("hourly", side="start", clean=TRUE) %>%
  group_by(v1) %>%
  mutate(sumv2_byhour = sum(v2, na.rm=TRUE),
         countv1_byhour = n()) %>%
  distinct(v1, .keep_all = TRUE)

# A time tibble: 12 x 4
# Index: v1
# Groups: v1 [12]
   v1                     v2 sumv2_byhour countv1_byhour
   <dttm>              <dbl>        <dbl>          <int>
 1 2018-01-01 01:00:00  1.00         1.00              2
 2 2018-01-01 02:00:00  2.00         2.00              1
 3 2018-01-01 03:00:00  1.00         2.00              3
 4 2018-01-01 05:00:00  3.00         3.00              1
 5 2018-01-01 06:00:00  1.00         2.00              2
 6 2018-01-01 09:00:00  1.00         1.00              1
 7 2018-01-02 01:00:00  1.00         1.00              2
 8 2018-01-02 02:00:00  2.00         2.00              1
 9 2018-01-02 03:00:00  1.00         2.00              3
10 2018-01-02 05:00:00  1.00         1.00              1
11 2018-01-02 06:00:00  4.00         5.00              2
12 2018-01-02 09:00:00  1.00         1.00              1

到目前为止一直很好，但我想对失踪时间零计数。有没有其他方法可以解决这个问题，让我使用dplyr::complete？

Answer 1

我们可以在ungroup步骤之后distinct然后执行complete

df.have %>%
  as_tbl_time(index = v1) %>%
  collapse_by("hourly", side="start", clean=TRUE) %>%
  group_by(v1) %>%
  mutate(sumv2_byhour = sum(v2, na.rm=TRUE),
         countv1_byhour = n()) %>%
  distinct(v1, .keep_all = TRUE) %>%
  ungroup %>%
  complete(v1 = seq(v1[1], v1[length(v1)], by = "1 hour"), 
               fill = list(v2 = 0, sumv2_byhour = 0, countv1_byhour = 0))
# A tibble: 33 x 4
#   v1                     v2 sumv2_byhour countv1_byhour
#   <dttm>              <dbl>        <dbl>          <dbl>
# 1 2018-01-01 01:00:00  1.00         1.00           2.00
# 2 2018-01-01 02:00:00  1.00         1.00           1.00
# 3 2018-01-01 03:00:00  1.00         2.00           3.00
# 4 2018-01-01 04:00:00  0            0              0   
# 5 2018-01-01 05:00:00  1.00         1.00           1.00
# 6 2018-01-01 06:00:00  1.00         2.00           2.00
# 7 2018-01-01 07:00:00  0            0              0   
# 8 2018-01-01 08:00:00  0            0              0   
# 9 2018-01-01 09:00:00  1.00         1.00           1.00
#10 2018-01-01 10:00:00  0            0              0   
# ... with 23 more rows

Answer 2

另一种避免使用c并使用dplyr::complete()包的解决方案。我发现这更能描述你想要做什么。

padr

按小时完成缺失时间序列

2 个答案: