R data.frame 中多个变量的每小时平均值？

Question

我有以下代码，正在尝试找到每个 variables (i.e., X,Y, and Z) 的 hourly mean。我的输出应该是 data.frame，其中 hourlyDate 列和所有 variables 的 mean hourly data。任何前进的方式将不胜感激。

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, 5,10), Z = runif(104833, 5,10))
T$Date <- format(T$Datetime, format="%Y-%m-%d")
T$Hour <- format(T$Datetime, format = "%H")
T$Mints <- format(T$Datetime, format = "%M")

Answer 1

这里有一个 tidyverse 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

Answer 2

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, 5,10), Z = runif(104833, 5,10))



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

Answer 3

库 lubridate 有一个 floor_date 函数可以将您的日期时间列修剪为指定的单位。

然后只需按您想要的变量的每小时时间戳进行总结

library(dplyr)
library(lubridate)

T %>%
    group_by(hourlyDate = lubridate::floor_date(Datetime, unit = 'hours')) %>%
    summarise(across(.cols = c(X,Y,Z), .fns = ~mean(.x, na.rm=TRUE), .names = "meanHourlyData_{.col}"))

顺便说一句，我建议不要使用 T 作为变量名，因为这也是 TRUE 的简写，可能会导致一些意外行为...

Answer 4

三基 R 解决方案是使用 split、tapply 或 rowsum 结合 table。后者特别快（比 dplyr 答案之一快 9 倍）。

tl;dr 是您得到以下计算时间

#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", "Y", "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[, c("X", "Y", "Z")], 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, Date, Hour) %>% 
  summarize(X = mean(X), Y = mean(Y), Z = mean(Z)) %>%
  transmute(Date = as.POSIXct(paste0(Date, " ", Hour, ":00:00")), X, Y, Z)


# check that we get the same
all.equal(out_2, as.matrix(out_3[, c("X", "Y", "Z")]),
          check.attributes = FALSE)
#R> [1] TRUE

# check computation time
library(microbenchmark)
microbenchmark(
  `split + sapply` = 
    sapply(split(T[, c("X", "Y", "Z")], T$DateH), colMeans), 
  `tapply + sapply` = 
    sapply(c("X", "Y", "Z"), 
           function(var) c(tapply(T[[var]], T$DateH, mean))), 
  `rowsum + table` = 
    as.matrix(rowsum(T[, c("X", "Y", "Z")], group = T$DateH)) / 
    rep(table(T$DateH), 3),
  `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), 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

我假设 data.table 也可能会很快获得结果。最后，不要使用 T 作为变量名。 T 是 TRUE!

的 shorthand

R data.frame 中多个变量的每小时平均值？

Hourly mean of multiple variables in R data.frame?

aggregate

r

mean

dataframe

lubridate