我已经搜索了SO，以便在没有运气的情况下实现我的需要，所以在这里。 前段时间我发现了包装dplyr及其潜力。我认为这个包可以做我想要的，我只是不知道如何。这是我数据的一小部分，但应该代表我的问题。
dummy<-structure(list(time = structure(1:20, .Label = c("2015-03-25 12:24:00",
"2015-03-25 21:08:00", "2015-03-25 21:13:00", "2015-03-25 21:47:00",
"2015-03-26 03:08:00", "2015-04-01 20:30:00", "2015-04-01 20:34:00",
"2015-04-01 20:42:00", "2015-04-01 20:45:00", "2015-09-29 18:26:00",
"2015-09-29 19:11:00", "2015-09-29 21:21:00", "2015-09-29 22:03:00",
"2015-09-29 22:38:00", "2015-09-30 00:48:00", "2015-09-30 01:38:00",
"2015-09-30 01:41:00", "2015-09-30 01:45:00", "2015-09-30 01:47:00",
"2015-09-30 01:49:00"), class = "factor"), ID = c(1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L), station = c(1L, 1L, 1L, 2L, 3,
4L, 4L, 4L, 4L, 5L, 5L, 6L,
6L, 5, 5, 5L, 7, 7, 7L,
7)), .Names = c("time", "ID", "station"), class = "data.frame", row.names = c(NA,
-20L))
我希望根据ID和station列来评估时间列中的行。具体来说，我希望函数（dplyr？）评估每个时间行，并将时间与前一个时间（第一行）和下一个时间（行+ 1）进行比较。如果当前行的时间在前一行和/或下一行的1小时之内，并且当前行的ID和站与上一行和/或下一行的ID和站匹配，那么我想在新行中添加一行1 ，否则为0。
我如何使用dplyr实现这一目标？
预期结果应该是这样的：
time ID station new.value
1 2015-03-25 12:24:00 1 1 0
2 2015-03-25 21:08:00 1 1 1
3 2015-03-25 21:13:00 1 1 1
4 2015-03-25 21:47:00 1 2 0
5 2015-03-26 03:08:00 1 3 0
6 2015-04-01 20:30:00 1 4 1
7 2015-04-01 20:34:00 1 4 1
8 2015-04-01 20:42:00 1 4 1
9 2015-04-01 20:45:00 1 4 1
10 2015-09-29 18:26:00 2 5 1
11 2015-09-29 19:11:00 2 5 1
12 2015-09-29 21:21:00 2 6 1
13 2015-09-29 22:03:00 2 6 1
14 2015-09-29 22:38:00 2 5 0
15 2015-09-30 00:48:00 2 5 1
16 2015-09-30 01:38:00 2 5 1
17 2015-09-30 01:41:00 2 7 1
18 2015-09-30 01:45:00 2 7 1
19 2015-09-30 01:47:00 2 7 1
20 2015-09-30 01:49:00 2 7 1
答案 0 :(得分：5)
以下是使用difftime和dplyr mutate函数的选项。首先，我们使用group_by操作来确保比较在ID和Station的每个唯一组合中。 difftime可用于计算差异时间，为方便起见，此处单位将设置为hours。 lag和lead函数也来自dplyr包，它会向后或向前移动所选列。结合difftime的矢量化操作，您可以计算当前行与上一行/下一行之间的时差。我们使用abs来确保结果是绝对值。 <1的条件确保差异在一小时内。 as.integer相应地将逻辑值（T或F）转换为（1或0）。
library(dplyr)
dummy %>% group_by(ID, station) %>%
mutate(new.value = as.integer(
abs(difftime(time, lag(time, default = Inf), units = "hours")) < 1 |
abs(difftime(time, lead(time, default = Inf), units = "hours")) < 1))
Source: local data frame [20 x 4]
Groups: ID, station [7]
time ID station new.value
(time) (int) (dbl) (int)
1 2015-03-25 12:24:00 1 1 0
2 2015-03-25 21:08:00 1 1 1
3 2015-03-25 21:13:00 1 1 1
4 2015-03-25 21:47:00 1 2 0
5 2015-03-26 03:08:00 1 3 0
6 2015-04-01 20:30:00 1 4 1
7 2015-04-01 20:34:00 1 4 1
8 2015-04-01 20:42:00 1 4 1
9 2015-04-01 20:45:00 1 4 1
10 2015-09-29 18:26:00 2 5 1
11 2015-09-29 19:11:00 2 5 1
12 2015-09-29 21:21:00 2 6 1
13 2015-09-29 22:03:00 2 6 1
14 2015-09-29 22:38:00 2 5 0
15 2015-09-30 00:48:00 2 5 1
16 2015-09-30 01:38:00 2 5 1
17 2015-09-30 01:41:00 2 7 1
18 2015-09-30 01:45:00 2 7 1
19 2015-09-30 01:47:00 2 7 1
20 2015-09-30 01:49:00 2 7 1
答案 1 :(得分：5)
Psidom的答案很棒 - 这是data.table方法。
library(data.table)
setDT(dummy)
# you do NOT want a factor for your time variable
dummy[, time := as.POSIXct(time) ]
dummy[, `:=`(lag_diff = c(Inf, diff(as.numeric(time))),
lead_diff = c(diff(as.numeric(time)), Inf)),
by = .(ID, station) ]
dummy[, new.value := as.integer(lag_diff < 3600 | lead_diff < 3600) ]
dummy
答案 2 :(得分：3)
使用R基函数（sapply和difftime）的另一种解决方案：
n=nrow(dummy)
dummy$new.value=
as.numeric(sapply(1:n, function(i)
(i<n && (dummy[i,"ID"]==dummy[i+1,"ID"] && dummy[i,"station"]==dummy[i+1,"station"])
&& abs(as.numeric(difftime(dummy[i,"time"], dummy[i+1,"time"]), "hours"))<=1)
||
(i>1 && (dummy[i,"ID"]==dummy[i-1,"ID"] && dummy[i,"station"]==dummy[i-1,"station"])
&& abs(as.numeric(difftime(dummy[i,"time"], dummy[i-1,"time"]), "hours"))<=1)
))
# > dummy
# time ID station new.value
# 1 2015-03-25 12:24:00 1 1 0
# 2 2015-03-25 21:08:00 1 1 1
# 3 2015-03-25 21:13:00 1 1 1
# 4 2015-03-25 21:47:00 1 2 0
# 5 2015-03-26 03:08:00 1 3 0
# 6 2015-04-01 20:30:00 1 4 1
# 7 2015-04-01 20:34:00 1 4 1
# 8 2015-04-01 20:42:00 1 4 1
# 9 2015-04-01 20:45:00 1 4 1
# 10 2015-09-29 18:26:00 2 5 1
# 11 2015-09-29 19:11:00 2 5 1
# 12 2015-09-29 21:21:00 2 6 1
# 13 2015-09-29 22:03:00 2 6 1
# 14 2015-09-29 22:38:00 2 5 0
# 15 2015-09-30 00:48:00 2 5 1
# 16 2015-09-30 01:38:00 2 5 1
# 17 2015-09-30 01:41:00 2 7 1
# 18 2015-09-30 01:45:00 2 7 1
# 19 2015-09-30 01:47:00 2 7 1
# 20 2015-09-30 01:49:00 2 7 1