r的时间秒

Question

有人知道如何按“时间”秒进行分组？时间变量每秒有30个观测值。

   structure(list(time = structure(c(1475049600, 1475049600.03333, 
1475049600.06667, 1475049600.1, 1475049600.13333, 1475049600.16667, 
1475049600.2, 1475049600.23333, 1475049600.26667, 1475049600.3, 
1475049600.33333, 1475049600.36667, 1475049600.4, 1475049600.43333, 
1475049600.46667, 1475049600.5, 1475049600.53333, 1475049600.56667, 
1475049600.6, 1475049600.63333, 1475049600.66667, 1475049600.7, 
1475049600.73333, 1475049600.76667, 1475049600.8, 1475049600.83333, 
1475049600.86667, 1475049600.9, 1475049600.93333, 1475049600.96667, 
1475049601, 1475049601.03333, 1475049601.06667, 1475049601.1, 
1475049601.13333, 1475049601.16667, 1475049601.2, 1475049601.23333, 
1475049601.26667, 1475049601.3), class = c("POSIXct", "POSIXt"
), tzone = "GMT"), X = c(-0.059, -0.023, -0.016, -0.016, -0.016, 
-0.012, -0.016, -0.012, -0.016, -0.012, -0.016, -0.016, -0.016, 
-0.02, -0.02, -0.027, -0.016, -0.02, -0.016, -0.016, -0.02, -0.023, 
-0.016, -0.02, -0.012, -0.016, -0.012, -0.016, -0.016, -0.02, 
-0.02, -0.02, -0.02, -0.008, -0.02, -0.016, -0.016, -0.012, -0.016, 
-0.016), Y = c(-0.777, -0.777, -0.777, -0.777, -0.777, -0.777, 
-0.777, -0.781, -0.773, -0.777, -0.777, -0.777, -0.777, -0.773, 
-0.777, -0.777, -0.777, -0.777, -0.773, -0.777, -0.777, -0.777, 
-0.77, -0.777, -0.777, -0.777, -0.781, -0.777, -0.777, -0.773, 
-0.777, -0.777, -0.777, -0.773, -0.777, -0.777, -0.777, -0.777, 
-0.777, -0.777), Z = c(0.688, 0.688, 0.691, 0.691, 0.688, 0.688, 
0.688, 0.691, 0.691, 0.695, 0.688, 0.691, 0.691, 0.695, 0.691, 
0.688, 0.688, 0.688, 0.691, 0.691, 0.691, 0.691, 0.695, 0.691, 
0.691, 0.688, 0.695, 0.688, 0.691, 0.691, 0.688, 0.688, 0.688, 
0.688, 0.688, 0.691, 0.691, 0.695, 0.695, 0.691)), subject_name = "1", time_zone = "02:00:00", missingness = structure(list(
    time = structure(c(1475747248, 1475747249, 1475747250, 1475747292, 
    1475747293, 1475747294), class = c("POSIXct", "POSIXt"), tzone = "GMT"), 
    n_missing = c(30L, 30L, 1230L, 30L, 30L, 32490L)), class = "data.frame", row.names = c("1475747248", 
"1475747249", "1475747250", "1475747292", "1475747293", "1475747294"
)), old_version = FALSE, firmware = "1.5.0", last_sample_time = structure(1475748377, tzone = "GMT", class = c("POSIXct", 
"POSIXt")), serial_prefix = "TAS", sample_rate = 30L, acceleration_min = "-8.0", acceleration_max = "8.0", header = structure(list(
    `Serial Number` = "TAS1E44150325", `Device Type` = "Link", 
    Firmware = "1.5.0", `Battery Voltage` = "3,88", `Sample Rate` = 30, 
    `Start Date` = structure(1475049600, tzone = "GMT", class = c("POSIXct", 
    "POSIXt")), `Stop Date` = structure(1475859600, tzone = "GMT", class = c("POSIXct", 
    "POSIXt")), `Last Sample Time` = structure(1475748377, tzone = "GMT", class = c("POSIXct", 
    "POSIXt")), TimeZone = "02:00:00", `Download Date` = structure(1475748377, tzone = "GMT", class = c("POSIXct", 
    "POSIXt")), `Board Revision` = "5", `Unexpected Resets` = "0", 
    `Acceleration Scale` = 256L, `Acceleration Min` = "-8.0", 
    `Acceleration Max` = "8.0", Mass = "77,1107028999572", Age = "22", 
    Limb = "Waist", DateOfBirth = "628890912000000000", `Subject Name` = "1", 
    `Serial Prefix` = "TAS"), class = c("gt3x_info", "list")), start_time = structure(1475049600, tzone = "GMT", class = c("POSIXct", 
"POSIXt")), stop_time = structure(1475859600, tzone = "GMT", class = c("POSIXct", 
"POSIXt")), total_records = 20929470L, bad_samples = FALSE, n_head = 40, row.names = c(NA, 
20929470L), class = c("activity_df", "activity_df", "activity_df", 
"data.frame"))

Answer 1

一种简单的方法是从{lubritation} package使用第二个函数。这是使用简单数据的示例：

example <- structure(list(time = structure(c(1569632400, 1569632400, 1569632400, 
1569632401, 1569632401, 1569632401), class = c("POSIXct", "POSIXt"
), tzone = ""), X = c(0, 3, 3, 2, 2, 0), Y = c(1, 2, 1, 1, 1, 
0), Z = c(1, 1, 1, 0, 0, 1)), class = "data.frame", row.names = c(NA, 
-6L))

example
                 time X Y Z
1 2019-09-28 08:00:00 0 1 1
2 2019-09-28 08:00:00 3 2 1
3 2019-09-28 08:00:00 3 1 1
4 2019-09-28 08:00:01 2 1 0
5 2019-09-28 08:00:01 2 1 0
6 2019-09-28 08:00:01 0 0 1

library(lubridate)
library(dplyr)

example |> group_by(second(time)) |> summarise(mean(X), mean(Y), mean(Z))

# A tibble: 2 × 4
  `second(time)` `mean(X)` `mean(Y)` `mean(Z)`
           <dbl>     <dbl>     <dbl>     <dbl>
1              0      2        1.33      1    
2              1      1.33     0.667     0.333