我有一个包含3个变量的数据框：POSIXct对象-time，数字-RRR和因数-he。其中RRR是液体降水量，he是水文事件编号，这里的时间对应于洪水事件的开始。
df <- structure(list(time = structure(c(1396879200, 1396922400, 1396976400,
1397008800, 1397095200, 1397332800, 1397354400, 1397397600, 1397451600,
1397484000, 1397527200, 1397786400, 1397959200, 1398002400, 1398024000,
1398132000, 1398175200, 1398218400, 1398261600, 1398369600, 1398466800,
1398477600, 1398520800, 1398564000, 1398607200, 1398747600, 1398780000,
1398909600, 1398952800, 1398974400, 1398996000),
class = c("POSIXct", "POSIXt"),
tzone = ""),
RRR = c(NA, 2, NA, 4, NA, NA, 0.9, 3,
NA, 0.4, 11, NA, 0.5, 1, NA, 13, 4, 0.8, 0.3, NA, NA, 8, 4, 11,
1, NA, 7, 1, 0.4, NA, 4),
he = c(1, NA, 2, NA, 3, 4, NA, NA,
5, NA, NA, 6, NA, NA, 7, NA, NA, NA, NA, 8, 9, NA, NA, NA, NA,
10, NA, NA, NA, 11, NA)),
class = "data.frame",
row.names = c(NA, -31L))
我的数据框的标题如下：
> df
time RRR he
1 2014-04-07 18:00:00 NA 1
2 2014-04-08 06:00:00 2.0 NA
3 2014-04-08 21:00:00 NA 2
4 2014-04-09 06:00:00 4.0 NA
5 2014-04-10 06:00:00 NA 3
6 2014-04-13 00:00:00 NA 4
7 2014-04-13 06:00:00 0.9 NA
8 2014-04-13 18:00:00 3.0 NA
9 2014-04-14 09:00:00 NA 5
我需要计算每个he值的时间与最后一个非NA RRR值的时间之间的时间差 强>。例如，对于he = 2，期望的差应为difftime(df$time[3], df$time[2])，而对于he = 4，时间差应为difftime(df$time[6], df$time[4])。因此，最后我想得到一个这样的数据框，其中“ diff”是时差，以小时为单位。
> df
time RRR he diff
1 2014-04-07 18:00:00 NA 1 NA
2 2014-04-08 06:00:00 2.0 NA NA
3 2014-04-08 21:00:00 NA 2 15
4 2014-04-09 06:00:00 4.0 NA NA
5 2014-04-10 06:00:00 NA 3 24
6 2014-04-13 00:00:00 NA 4 90
7 2014-04-13 06:00:00 0.9 NA NA
8 2014-04-13 18:00:00 3.0 NA NA
9 2014-04-14 09:00:00 NA 5 15
答案 0 :(得分：1)
我确信必须有更简单的方法，但是使用tidyverse和data.table可以做到：
df %>%
mutate(time = as.POSIXct(time, format = "%Y-%m-%d %H:%M:%S")) %>% #Transforming "time" into a datetime object
fill(RRR) %>% #Filling the NA values in "RRR" with tha last non-NA value
group_by(temp = rleid(RRR)) %>% #Grouping by run length of "RRR"
mutate(temp2 = seq_along(temp)) %>% #Sequencing around the run length of "RRR"
group_by(RRR, temp) %>% #Group by "RRR" and run length of "RRR"
mutate(diff = ifelse(!is.na(he), difftime(time, time[temp2 == 1], units="hours"), NA)) %>% #Computing the difference in hours between the first occurrence of a non-NA "RRR" value and the non-NA "he" values
ungroup() %>%
select(-temp, -temp2, -RRR) %>% #Removing the redundant variables
rowid_to_column() %>% #Creating unique row IDs
left_join(df %>%
rowid_to_column() %>%
select(RRR, rowid), by = c("rowid" = "rowid")) %>% #Merging with the original df to get the original values of "RRR"
select(-rowid) #Removing the redundant variables
time he diff RRR
<dttm> <dbl> <dbl> <dbl>
1 2014-04-07 16:00:00 1. 0. NA
2 2014-04-08 04:00:00 NA NA 2.00
3 2014-04-08 19:00:00 2. 15. NA
4 2014-04-09 04:00:00 NA NA 4.00
5 2014-04-10 04:00:00 3. 24. NA
6 2014-04-12 22:00:00 4. 90. NA
7 2014-04-13 04:00:00 NA NA 0.900
8 2014-04-13 16:00:00 NA NA 3.00
9 2014-04-14 07:00:00 5. 15. NA
10 2014-04-14 16:00:00 NA NA 0.400
答案 1 :(得分：1)
以下是利用其非等额联接功能的data.table方法：
library(data.table)
setDT(df)
df[df[!is.na(he)][df[!is.na(RRR)], on = .(time>time), rrr_time := i.time],
on = .(time, he), rrr_time := i.rrr_time][, diff := difftime(time, rrr_time)]
结果是：
# time RRR he rrr_time diff
# <POSc> <num> <num> <POSc> <difftime>
# 1: 2014-04-07 16:00:00 NA 1 <NA> NA hours
# 2: 2014-04-08 04:00:00 2.0 NA <NA> NA hours
# 3: 2014-04-08 19:00:00 NA 2 2014-04-08 04:00:00 15 hours
# 4: 2014-04-09 04:00:00 4.0 NA <NA> NA hours
# 5: 2014-04-10 04:00:00 NA 3 2014-04-09 04:00:00 24 hours
# 6: 2014-04-12 22:00:00 NA 4 2014-04-09 04:00:00 90 hours
# 7: 2014-04-13 04:00:00 0.9 NA <NA> NA hours
# 8: 2014-04-13 16:00:00 3.0 NA <NA> NA hours
# 9: 2014-04-14 07:00:00 NA 5 2014-04-13 16:00:00 15 hours
# 10: 2014-04-14 16:00:00 0.4 NA <NA> NA hours
# 11: 2014-04-15 04:00:00 11.0 NA <NA> NA hours
# 12: 2014-04-18 04:00:00 NA 6 2014-04-15 04:00:00 72 hours
# 13: 2014-04-20 04:00:00 0.5 NA <NA> NA hours
# 14: 2014-04-20 16:00:00 1.0 NA <NA> NA hours
# 15: 2014-04-20 22:00:00 NA 7 2014-04-20 16:00:00 6 hours
# 16: 2014-04-22 04:00:00 13.0 NA <NA> NA hours
# 17: 2014-04-22 16:00:00 4.0 NA <NA> NA hours
# 18: 2014-04-23 04:00:00 0.8 NA <NA> NA hours
# 19: 2014-04-23 16:00:00 0.3 NA <NA> NA hours
# 20: 2014-04-24 22:00:00 NA 8 2014-04-23 16:00:00 30 hours
# 21: 2014-04-26 01:00:00 NA 9 2014-04-23 16:00:00 57 hours
# 22: 2014-04-26 04:00:00 8.0 NA <NA> NA hours
# 23: 2014-04-26 16:00:00 4.0 NA <NA> NA hours
# 24: 2014-04-27 04:00:00 11.0 NA <NA> NA hours
# 25: 2014-04-27 16:00:00 1.0 NA <NA> NA hours
# 26: 2014-04-29 07:00:00 NA 10 2014-04-27 16:00:00 39 hours
# 27: 2014-04-29 16:00:00 7.0 NA <NA> NA hours
# 28: 2014-05-01 04:00:00 1.0 NA <NA> NA hours
# 29: 2014-05-01 16:00:00 0.4 NA <NA> NA hours
# 30: 2014-05-01 22:00:00 NA 11 2014-05-01 16:00:00 6 hours
# 31: 2014-05-02 04:00:00 4.0 NA <NA> NA hours
# time RRR he rrr_time diff
答案 2 :(得分：1)
带有base的{{1}}替代项：
findInterval