在 R 中按指定的时间长度扩展时间序列的最快方法



我有来自两种类型的活动记录器的数据。第一个记录器记录记录器处于湿或干状态的秒数(参见act1)。第二个记录器每3秒对湿/干采样进行一次采样,并每10分钟记录一次湿/湿的样品总数。给定 3 秒的采样间隔,每 10 分钟周期结束时记录的值范围从零(始终干燥)到 200(始终潮湿),请参阅act2

我想重塑和重新采样来自第一个记录器的数据,以使用最有效的方法复制第二个记录器的格式。

我在这里提供的示例使用数据的子样本(6 行),但我的实际数据集包含一年多的观察(40,000+ 行),目前 3 天后仍在运行。

act1 <- structure(list(
Valid = c("ok", "ok", "ok", "ok", "ok", "ok"),
Date = structure(c(1425579093, 1425579171, 1425579177, 1425579216, 1425579225, 1425579240),
class = c("POSIXct", "POSIXt"), tzone = ""),
Activity = c(78L, 6L, 39L, 9L, 15L, 9L),
Wet = c("wet", "dry", "wet", "dry", "wet", "dry")),
row.names = c("2", "3", "4", "5", "6", "7"),
class = "data.frame")
act2 <- structure(list(
Valid = c("ok", "ok", "ok", "ok", "ok", "ok"),
Date = structure(c(1425579093, 1425579171, 1425579177, 1425579216, 1425579225, 1425579240),
class = c("POSIXct", "POSIXt"), tzone = ""),
Activity = c(78L, 6L, 39L, 9L, 15L, 9L),
Wet = c("wet", "dry", "wet", "dry", "wet", "dry")), row.names = c("2", "3", "4", "5", "6", "7"),
class = "data.frame")

使用lapply,我根据"活动"列中指定的间隔扩展了act1数据帧中的日期列(POSIXct 格式),并在"湿">列中保留了对相应状态的引用。

act1  <-  lapply(1:nrow(act1),  function(x){
data.frame(
Valid = rep(act1[x, 1], act1[x, 3]), 
Date = strptime(act1[x, 2], format = "%Y-%m-%d%H:%M:%S")+(seq_len(act1[x, 3])-1), 
Activity = rep(1, act1[x, 3]), 
Wet = rep(act1[x, 4], act1[x, 3])
)})
act1 <- as.data.frame(do.call(rbind, act1))

然后,我使用dplyr润滑剂将每个观察结果分组到 3 秒的箱中,并确定每个箱中的最后一个观测值是否湿润。我将剩余的湿观测值分组 10 分钟箱,并总结出湿的样本数量。

library(dplyr)
library(lubridate)
act1 <- act1 %>%
mutate(interval = floor_date(Date, unit="minutes") + seconds(floor(second(Date)/3)*3)) %>% 
group_by(interval) %>%
summarise(Valid = "ok",
Wet = Wet[which(Date==max(Date))]=="wet") %>%
mutate(int10 = floor_date(interval, unit="hour") +
minutes(floor(minute(interval)/10)*10) +
(min(interval) - min(floor_date(interval, unit="hour") + minutes(floor(minute(interval)/10)*10)))) %>% 
group_by(int10) %>%
summarise(Valid = "ok",
Activity = sum(Wet)) %>%
rename(Date = int10) %>%
select(Valid,Date,Activity)

我在这里提供的示例使用了原始数据集的子集(6 行),但是我的实际数据集包含一年多的观察(40,000+ 行),目前3 天后仍在运行!

矢量化、repcutseq应该在你这个任务的工具箱里。

你涉及lapply的第一个主要语句可以缩短 - 你只是想重复这些行。例如,act1[c(1,1), ]将返回第一行act12 次。在您的循环中,您可以访问act1[x, 3]4 次。下面的这一行将复制我们想要的行次数:

act1a <- act1_copy[rep(seq_len(nrow(act1_copy)), act1_copy[['Activity']]), ]
> nrow(act1_copy)
[1] 6
> seq_len(nrow(act1_copy))
[1] 1 2 3 4 5 6
> act1_copy[['Activity']]
[1] 78  6 39  9 15  9
> rep(seq_len(nrow(act1_copy)), act1_copy[['Activity']])
[1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
[60] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
[119] 3 3 3 3 3 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6
# or if you're into external packages, this is a lot nicer looking:
tidyr::uncount(act1_copy, weights = Activity)

下一步是更正秒数并重做Activity1

act1a[['Date']] <- act1a[['Date']] + sequence(act1_copy[['Activity']]) - 1
act1a[['Activity']] <- 1L

现在,使旧日志记录数据与较新的日志记录数据(即每 3 秒一次日志)匹配的最后一步是按 3 秒分组。请务必注意,每 3 秒一次相当于每 3 行一次。因此,根据act$Date完成的信心,我们可以执行以下两种方法之一:

act1b <- act1a[!duplicated(cut(act1a$Date, '3 sec', labels = F)), ]
# or if you're sure there is one reading per second, you can just do once every three rows 
act1b <- act1a[seq(from = 1, to = nrow(act1a), by = 3), ]
# what cut() looks like for reference
cut(act1a$Date, '3 sec', labels = F)
[1]  1  1  1  2  2  2  3  3  3  4  4  4  5  5  5  6  6  6  7  7  7  8  8  8  9  9  9 10 10 10 11 11 11 12 12 12 13 13 13
[40] 14 14 14 15 15 15 16 16 16 17 17 17 18 18 18 19 19 19 20 20 20 21 21 21 22 22 22 23 23 23 24 24 24 25 25 25 26 26 26
[79] 27 27 27 28 28 28 29 29 29 30 30 30 31 31 31 32 32 32 33 33 33 34 34 34 35 35 35 36 36 36 37 37 37 38 38 38 39 39 39
[118] 40 40 40 41 41 41 42 42 42 43 43 43 44 44 44 45 45 45 46 46 46 47 47 47 48 48 48 49 49 49 50 50 50 51 51 51 52 52 52
#or with labels:
cut(act1a$Date, '3 sec')
[1] 2015-03-05 13:11:33 2015-03-05 13:11:33 2015-03-05 13:11:33 2015-03-05 13:11:36 2015-03-05 13:11:36
[6] 2015-03-05 13:11:36 2015-03-05 13:11:39 2015-03-05 13:11:39 2015-03-05 13:11:39 2015-03-05 13:11:42
[11] 2015-03-05 13:11:42 2015-03-05 13:11:42 2015-03-05 13:11:45 2015-03-05 13:11:45 2015-03-05 13:11:45
[16] 2015-03-05 13:11:48 2015-03-05 13:11:48 2015-03-05 13:11:48 2015-03-05 13:11:51 2015-03-05 13:11:51
[21] 2015-03-05 13:11:51 2015-03-05 13:11:54 2015-03-05 13:11:54 2015-03-05 13:11:54 2015-03-05 13:11:57 
# truncated for brevity.

最后一步是聚合数据。就像最后一步一样,我们可以使用cut()来分组使用时间,也可以再次使用rep(seq())来使我们的分组更快一些。

aggregate(act1b$Wet, list(Date = cut(act1b$Date, '10 min')), FUN = function(x) sum(x == 'wet'))
#or if you know there is one reading per second,
aggregate(act1b$Wet,
list(Date = rep(act1b$Date[seq(from = 1, to = nrow(act1b), by = 10 * 60 / 3)]
, each = 10 * 60 / 3
, length.out = nrow(act1b)))
, FUN = function(x) sum(x == 'wet'))

综上所述,您可以得到:

act1a <- act1_copy[rep(seq_len(nrow(act1_copy)), act1_copy[['Activity']]), ]
act1a[['Date']] <- act1a[['Date']] + sequence(act1_copy[['Activity']]) - 1
act1a[['Activity']] <- 1L
act1b <- act1a[seq(from = 1, to = nrow(act1a), by = 3), ]
aggregate(act1b$Wet,
list(Date = rep(act1b$Date[seq(from = 1, to = nrow(act1b), by = 10 * 60 / 3)]
, each = 10 * 60 / 3
, length.out = nrow(act1b)))
, FUN = function(x) sum(x == 'wet'))
library(tidyr)
library(dplyr)
tidyr::uncount(act1_copy, weights = Activity)%>%
mutate(Activity = 1L
, Date = Date + sequence(act1_copy[['Activity']]) - 1)%>%
slice(seq(from = 1, to = nrow(.), by = 3))%>%
group_by(Date =rep(Date[seq(from = 1, to = nrow(.), by = 10 * 60 / 3)]
, each = 10 * 60 / 3
, length.out = nrow(.)))%>%
summarize(Wet = sum(Wet == 'wet'))
# A tibble: 1 x 2
Date                  Wet
<dttm>              <int>
1 2015-03-05 13:11:33    44

是一些性能和代码 - 请注意我在 10 分钟总结中遇到问题的data.table方式,因此它并不完全是苹果对苹果:

Unit: milliseconds
expr       min        lq      mean    median        uq       max neval
cole_base  2.044400  2.174451  2.328061  2.253251  2.340801  6.424400   100
cole_dplyr  3.152901  3.359501  3.502880  3.428101  3.515302  8.248401   100
cole_dt  3.308601  3.541151  3.884475  3.698201  3.796652 13.155701   100
original_all 32.626601 33.061152 34.531462 33.392151 34.237601 50.499501   100
library(microbenchmark)
library(data.table)
library(dplyr)
library(tidyr)
library(lubridate)
act1_copy <- structure(list(
Valid = c("ok", "ok", "ok", "ok", "ok", "ok"),
Date = structure(c(1425579093, 1425579171, 1425579177, 1425579216, 1425579225, 1425579240),
class = c("POSIXct", "POSIXt"), tzone = ""),
Activity = c(78L, 6L, 39L, 9L, 15L, 9L),
Wet = c("wet", "dry", "wet", "dry", "wet", "dry")),
row.names = c("2", "3", "4", "5", "6", "7"),
class = "data.frame")
dt <- as.data.table(act1_copy)
microbenchmark( cole_base = {
act1a <- act1_copy[rep(seq_len(nrow(act1_copy)), act1_copy[['Activity']]), ]
act1a[['Date']] <- act1a[['Date']] + sequence(act1_copy[['Activity']]) - 1
# if you know there is definitly one reading per second
# act1a[['Date']] <- act1a[['Date']] + seq_len(nrow(act1a)) - 1
act1a[['Activity']] <- 1L
# act1b <- act1a[!duplicated(cut(act1a$Date, '3 sec', labels = F)), ]
# or if you're sure there is one reading per second, you can just do once every three rows 
act1b <- act1a[seq(from = 1, to = nrow(act1a), by = 3), ]
# aggregate(act1b$Wet, list(Date = cut(act1b$Date, '10 min')), FUN = function(x) sum(x == 'wet'))
#or if you know there is one reading per second,
aggregate(act1b$Wet,
list(Date = rep(act1b$Date[seq(from = 1, to = nrow(act1b), by = 10 * 60 / 3)]
, each = 10 * 60 / 3
, length.out = nrow(act1b)))
, FUN = function(x) sum(x == 'wet'))
}
, cole_dplyr = {
tidyr::uncount(act1_copy, weights = Activity)%>%
mutate(Activity = 1L
, Date = Date + sequence(act1_copy[['Activity']]) - 1)%>%
# filter(!duplicated(cut(Date, '3 sec', labels = F)))%>%
slice(seq(from = 1, to = nrow(.), by = 3))%>%
# group_by(Date = cut(Date, '10 min'))%>%
group_by(Date =rep(Date[seq(from = 1, to = nrow(.), by = 10 * 60 / 3)]
, each = 10 * 60 / 3
, length.out = nrow(.)))%>%
summarize(Wet = sum(Wet == 'wet'))
}
, cole_dt = {
copy(dt)[rep(seq_len(.N), Activity)
, .(Date = Date + sequence(act1_copy[['Activity']]) - 1
,Wet, Valid, Activity = 1L) 
][seq(from = 1, to = .N, by = 3)
, .(Wet = sum(Wet == 'wet'))
, by = cut(Date, '10 min')]
}
, original_all = {
act1  <-  lapply(1:nrow(act1_copy),  function(x){
data.frame(
Valid = rep(act1_copy[x, 1], act1_copy[x, 3]),
Date = strptime(act1_copy[x, 2], format = "%Y-%m-%d%H:%M:%S")+(seq_len(act1_copy[x, 3])-1),
Activity = rep(1, act1_copy[x, 3]),
Wet = rep(act1_copy[x, 4], act1_copy[x, 3])
)})
act1 <- as.data.frame(do.call(rbind, act1))
act1 <- act1 %>%
mutate(interval = floor_date(Date, unit="minutes") + seconds(floor(second(Date)/3)*3)) %>%
group_by(interval) %>%
summarise(Valid = "ok",
Wet = Wet[which(Date==max(Date))]=="wet") %>%
mutate(int10 = floor_date(interval, unit="hour") +
minutes(floor(minute(interval)/10)*10) +
(min(interval) - min(floor_date(interval, unit="hour") + minutes(floor(minute(interval)/10)*10)))) %>%
group_by(int10) %>%
summarise(Valid = "ok",
Activity = sum(Wet)) %>%
rename(Date = int10) %>%
select(Valid,Date,Activity)
}
)

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