如何解决R data.frame中多个变量的小时均值？
我有以下代码，正在尝试查找每个hourly mean中的variables (i.e.,X,Y,and Z)。我的输出应为带有data.frame列的hourlyDate和所有mean hourly data中的variables。任何前进的方式都将不胜感激。
library(lubridate)
set.seed(123)
T <- data.frame(Datetime = seq(ymd_hms("2011-01-01 00:00:00"),to= ymd_hms("2011-12-31 00:00:00"),by = "5 min"),X = runif(104833,5,10),Y = runif(104833,Z = runif(104833,10))
T$Date <- format(T$Datetime,format="%Y-%m-%d")
T$Hour <- format(T$Datetime,format = "%H")
T$Mints <- format(T$Datetime,format = "%M")
解决方法
尝试：
library(lubridate)
library(dplyr)
set.seed(123)
T <- data.frame(Datetime = seq(ymd_hms("2011-01-01 00:00:00"),to= ymd_hms("2011-12-31 00:00:00"),by = "5 min"),X = runif(104833,5,10),Y = runif(104833,Z = runif(104833,10))
T %>% mutate(hourlyDate = floor_date(Datetime,unit='hour')) %>%
select(-Datetime) %>% group_by(hourlyDate) %>%
summarize(across(everything(),mean)) %>%
ungroup()
#> `summarise()` ungrouping output (override with `.groups` argument)
#> # A tibble: 8,737 x 4
#> hourlyDate X Y Z
#> <dttm> <dbl> <dbl> <dbl>
#> 1 2011-01-01 00:00:00 8.00 7.90 6.90
#> 2 2011-01-01 01:00:00 7.93 7.47 7.90
#> 3 2011-01-01 02:00:00 7.83 6.89 7.67
#> 4 2011-01-01 03:00:00 6.61 7.92 7.18
#> 5 2011-01-01 04:00:00 7.27 7.20 6.48
#> 6 2011-01-01 05:00:00 7.88 6.80 7.69
#> 7 2011-01-01 06:00:00 7.07 8.05 7.52
#> 8 2011-01-01 07:00:00 7.40 7.92 6.99
#> 9 2011-01-01 08:00:00 7.97 7.76 7.26
#> 10 2011-01-01 09:00:00 7.57 7.47 6.94
#> # ... with 8,727 more rows
由reprex package（v0.3.0）于2020-08-20创建
,这是一种整洁的方法：
library(dplyr)
group_by(T,Date,Hour) %>%
summarize(X = mean(X),Y = mean(Y),Z = mean(Z)) %>%
transmute(Date = as.POSIXct(paste0(Date," ",Hour,":00:00")),X,Y,Z)
#> # A tibble: 8,737 x 4
#> # Groups: Date [8,714]
#> Date X Y Z
#> <dttm> <dbl> <dbl> <dbl>
#> 1 2011-01-01 00:00:00 8.00 7.90 6.90
#> 2 2011-01-01 01:00:00 7.93 7.47 7.90
#> 3 2011-01-01 02:00:00 7.83 6.89 7.67
#> 4 2011-01-01 03:00:00 6.61 7.92 7.18
#> 5 2011-01-01 04:00:00 7.27 7.20 6.48
#> 6 2011-01-01 05:00:00 7.88 6.80 7.69
#> 7 2011-01-01 06:00:00 7.07 8.05 7.52
#> 8 2011-01-01 07:00:00 7.40 7.92 6.99
#> 9 2011-01-01 08:00:00 7.97 7.76 7.26
#> 10 2011-01-01 09:00:00 7.57 7.47 6.94
#> # ... with 8,727 more rows
库lubridate具有floor_date函数，可将Datetime列修整为指定的单位。
然后按每小时时间戳汇总您想要的变量
library(dplyr)
library(lubridate)
T %>%
group_by(hourlyDate = lubridate::floor_date(Datetime,unit = 'hours')) %>%
summarise(across(.cols = c(X,Z),.fns = ~mean(.x,na.rm=TRUE),.names = "meanHourlyData_{.col}"))
顺便说一句，我建议不要将T用作变量名，因为这也是TRUE的缩写，并且可能会导致一些意外的行为...
三种基本的R解决方案是将split，tapply或rowsum与table结合使用。后者速度特别快（比dplyr答案之一快9倍。）
tl; dr是您获得了以下计算时间
#R> Unit: milliseconds
#R> expr min lq mean median uq max neval
#R> split + sapply 563.9 577.4 636.1 649.8 680.7 697.1 10
#R> tapply + sapply 108.0 117.3 134.0 120.2 124.4 205.1 10
#R> rowsum + table 21.3 21.3 21.5 21.3 21.6 21.9 10
#R> dplyr 172.4 176.6 182.3 180.9 185.9 203.4 10
这是解决方案
# create date-hour column
T$DateH <- format(T$Datetime,format="%Y-%m-%d-%H")
# using split + sapply
options(digits = 3)
out_1 <- sapply(split(T[,c("X","Y","Z")],T$DateH),colMeans)
head(t(out_1),5)
#R> X Y Z
#R> 2011-01-01-00 8.00 7.90 6.90
#R> 2011-01-01-01 7.93 7.47 7.90
#R> 2011-01-01-02 7.83 6.89 7.67
#R> 2011-01-01-03 6.61 7.92 7.18
#R> 2011-01-01-04 7.27 7.20 6.48
# using tapply + sapply
out_2 <- sapply(c("X","Z"),function(var) c(tapply(T[[var]],T$DateH,mean)))
head(out_2)
#R> X Y Z
#R> 2011-01-01-00 8.00 7.90 6.90
#R> 2011-01-01-01 7.93 7.47 7.90
#R> 2011-01-01-02 7.83 6.89 7.67
#R> 2011-01-01-03 6.61 7.92 7.18
#R> 2011-01-01-04 7.27 7.20 6.48
# check that we get the same
all.equal(t(out_1),out_2,check.attributes = FALSE)
#R> [1] TRUE
# with rowsum + table
out_3 <- as.matrix(rowsum(T[,group = T$DateH)) /
rep(table(T$DateH),3)
# check that we get the same
all.equal(out_2,out_3)
#R> [2] TRUE
# compare with dplyr solution
library(dplyr)
out_3 <- group_by(T,Z)
# check that we get the same
all.equal(out_2,as.matrix(out_3[,"Z")]),check.attributes = FALSE)
#R> [1] TRUE
# check computation time
library(microbenchmark)
microbenchmark(
`split + sapply` =
sapply(split(T[,colMeans),`tapply + sapply` =
sapply(c("X",mean))),`rowsum + table` =
as.matrix(rowsum(T[,group = T$DateH)) /
rep(table(T$DateH),3),`dplyr` =
group_by(T,Hour) %>%
summarize(X = mean(X),Z = mean(Z)) %>%
transmute(Date = as.POSIXct(paste0(Date,times = 10)
#R> Unit: milliseconds
#R> expr min lq mean median uq max neval
#R> split + sapply 563.9 577.4 636.1 649.8 680.7 697.1 10
#R> tapply + sapply 108.0 117.3 134.0 120.2 124.4 205.1 10
#R> rowsum + table 21.3 21.3 21.5 21.3 21.6 21.9 10
#R> dplyr 172.4 176.6 182.3 180.9 185.9 203.4 10
我认为使用data.table也可能会很快获得结果。最后，请勿将T用作变量名。 T是TRUE的简写！
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