Question

我有一些时间数据

library(data.table); library(lubridate); set.seed(42)
dat <- rbind(data.table(time=as.POSIXct("2019-01-01 08:00:00") + round(runif(10,60,1e4)), val=runif(10),group=1)[order(time), id:=seq_len(.N)],
             data.table(time=as.POSIXct("2019-02-01 18:00:00") + round(runif(10,60,1e4)), val=runif(10),group=2)[order(time), id:=seq_len(.N)])
> dat[order(group,id)]
                   time         val group id
 1: 2019-01-01 08:23:19 0.117487362     1  1
 2: 2019-01-01 08:48:24 0.934672247     1  2
 3: 2019-01-01 09:27:00 0.940014523     1  3
 4: 2019-01-01 09:47:19 0.462292823     1  4
 5: 2019-01-01 09:49:51 0.474997082     1  5
 6: 2019-01-01 09:57:48 0.560332746     1  6
 7: 2019-01-01 10:03:02 0.978226428     1  7
 8: 2019-01-01 10:18:35 0.255428824     1  8
 9: 2019-01-01 10:32:33 0.457741776     1  9
10: 2019-01-01 10:36:15 0.719112252     1 10
11: 2019-02-01 18:14:39 0.003948339     2  1
12: 2019-02-01 18:23:59 0.811055141     2  2
13: 2019-02-01 19:05:39 0.007334147     2  3
14: 2019-02-01 19:15:03 0.906601408     2  4
15: 2019-02-01 19:26:11 0.832916080     2  5
16: 2019-02-01 20:19:30 0.611778643     2  6
17: 2019-02-01 20:30:46 0.737595618     2  7
18: 2019-02-01 20:31:03 0.207658973     2  8
19: 2019-02-01 20:37:50 0.685169729     2  9
20: 2019-02-01 20:44:50 0.388108283     2 10

，我想在接下来的一个小时中为每个val的值计算time的总和。例如，对于ID 1，这将是ID 1和ID 2的val的总和（因为ID 3的时间比ID 1之后的一小时多），对于ID 2，这是{{1} }代表ID 2到4，依此类推。这样会产生所需的输出（仅对于组1）

val

最后可能出现两种行为：

按原样对待序列；
其中仅在直到一个小时后没有> res time val id new1 new2 1: 2019-01-01 08:23:19 0.1174874 1 1.052160 1.052160 2: 2019-01-01 08:48:24 0.9346722 2 2.336979 2.336979 3: 2019-01-01 09:27:00 0.9400145 3 3.671292 3.671292 4: 2019-01-01 09:47:19 0.4622928 4 3.908132 3.908132 5: 2019-01-01 09:49:51 0.4749971 5 3.445839 NA 6: 2019-01-01 09:57:48 0.5603327 6 2.970842 NA 7: 2019-01-01 10:03:02 0.9782264 7 2.410509 NA 8: 2019-01-01 10:18:35 0.2554288 8 1.432283 NA 9: 2019-01-01 10:32:33 0.4577418 9 1.176854 NA 10: 2019-01-01 10:36:15 0.7191123 10 0.719112 NA有ID的id的情况下才计算总和，而所有其他值都设置为NA（首选）。

我怀疑要解决这个问题，我需要在time内子集化，但这是我经常遇到且无法解决的问题。我尚未了解一般的处理方法。

Answer 1

可能是join的循环

dat1 <- dat[order(id)]
out <- rbindlist(lapply(dat1$id, function(i) {
      d1 <- dat1[seq_len(.N) >= match(i, id)]
      d1[d1[, .(time = time %m+% hours(1))], .(time1 = time, val, new1 = sum(val)),
         on = .(time <= time), by = .EACHI][1]
      }))[, time := NULL][]
setnames(out, 1, "time")
out[time < time[2]   %m+% hours(1), new2 := new1]
out
#                   time       val      new1     new2
# 1: 2019-01-01 08:23:19 0.1174874 1.0521596 1.052160
# 2: 2019-01-01 08:48:24 0.9346722 2.3369796 2.336980
# 3: 2019-01-01 09:27:00 0.9400145 3.6712924 3.671292
# 4: 2019-01-01 09:47:19 0.4622928 3.9081319 3.908132
# 5: 2019-01-01 09:49:51 0.4749971 3.4458391       NA
# 6: 2019-01-01 09:57:48 0.5603327 2.9708420       NA
# 7: 2019-01-01 10:03:02 0.9782264 2.4105093       NA
# 8: 2019-01-01 10:18:35 0.2554288 1.4322829       NA
# 9: 2019-01-01 10:32:33 0.4577418 1.1768540       NA
#10: 2019-01-01 10:36:15 0.7191123 0.7191123       NA

更新

对于新数据，我们可以按组split并应用相同的方法

f1 <- function(data) {
              lst1 <- split(data, data[["group"]])
              rbindlist(lapply(lst1, function(.dat) {
                out <- rbindlist(lapply(.dat$id, function(i) {
                      d1 <- .dat[seq_len(.N) >= match(i, id)]
                      d1[d1[, .(time = time %m+% hours(1))], .(time1 = time, val, new1 = sum(val)),
                         on = .(time <= time), by = .EACHI][1]
                      }))[, time := NULL][]
                setnames(out, 1, "time")

                out[time[.N]-time > hours(1), new2 := new1][] 
              })
              )}

 f1(dat1)
 #                  time         val      new1      new2
 #1: 2019-01-01 08:23:19 0.117487362 1.0521596 1.0521596
 #2: 2019-01-01 08:48:24 0.934672247 2.3369796 2.3369796
 #3: 2019-01-01 09:27:00 0.940014523 3.6712924 3.6712924
 #4: 2019-01-01 09:47:19 0.462292823 3.9081319 3.9081319
 #5: 2019-01-01 09:49:51 0.474997082 3.4458391        NA
 #6: 2019-01-01 09:57:48 0.560332746 2.9708420        NA
 #7: 2019-01-01 10:03:02 0.978226428 2.4105093        NA
 #8: 2019-01-01 10:18:35 0.255428824 1.4322829        NA
 #9: 2019-01-01 10:32:33 0.457741776 1.1768540        NA
#10: 2019-01-01 10:36:15 0.719112252 0.7191123        NA
#11: 2019-02-01 18:14:39 0.003948339 0.8223376 0.8223376
#12: 2019-02-01 18:23:59 0.811055141 1.7249907 1.7249907
#13: 2019-02-01 19:05:39 0.007334147 1.7468516 1.7468516
#14: 2019-02-01 19:15:03 0.906601408 1.7395175 1.7395175
#15: 2019-02-01 19:26:11 0.832916080 1.4446947        NA
#16: 2019-02-01 20:19:30 0.611778643 2.6303112        NA
#17: 2019-02-01 20:30:46 0.737595618 2.0185326        NA
#18: 2019-02-01 20:31:03 0.207658973 1.2809370        NA
#19: 2019-02-01 20:37:50 0.685169729 1.0732780        NA
#20: 2019-02-01 20:44:50 0.388108283 0.3881083        NA

在R data.table中计算总和的每个变量的汇总

1 个答案:

更新