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.我在想这个包可以做我想做的事,我只是不知道怎么做.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和站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
这是一个使用difftimewith dplyrmutate函数的选项.首先,我们使用一个group_by操作来确保比较在ID和Station的每个唯一组合中.该difftime可以用来计算时间差,这里的单位将被设置为hours提供方便.的lag和lead从功能也dplyr包,其中所选择的列向后或向前移位.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abs用来确保结果是绝对值.的条件<1,确保不同的是在一个小时内.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
Psidom的答案很棒 - 这是一种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