R:将行拆分为多行，然后将列拆分为多列

Question

我陷入了一项看似简单的任务中。想象一下以下data.table:

dt1 <- data.table(ID = as.factor(c("202E", "202E", "202E")), 
              timestamp = as.POSIXct(c("2017-05-02 00:00:00",
                                       "2017-05-02 00:15:00",
                                       "2017-05-02 00:30:00")), 
              acceleration_raw = c("-0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.727 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.727 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.164 -0.703 0.656 0.141 -0.703 0.656 0.164 -0.703 0.656 0.141 -0.703 0.656 0.141 -0.703 0.656 0.141 -0.703 0.656 0.141",
                                   "-0.703 0.680 0.117 -0.680 0.680 0.117 -0.680 0.680 0.117 -0.680 0.680 0.117 -0.680 0.680 0.117 -0.680 0.680 0.117 -0.680 0.680 0.117 -0.703 0.680 0.117 -0.703 0.680 0.117 -0.703 0.680 0.117 -0.680 0.680 0.117 -0.703 0.680 0.117 -0.680 0.680 0.117 -0.703 0.680 0.117 -0.680 0.680 0.117 -0.703 0.680 0.117 -0.680 0.680 0.117 -0.680 0.680 0.117 -0.680 0.680 0.117 -0.680 0.680 0.117 -0.703 0.680 0.117 -0.703 0.680 0.117 -0.703 0.680 0.117 -0.703 0.680 0.117 -0.703 0.680 0.117 -0.703 0.680 0.117 -0.703 0.680 0.117 -0.680 0.680 0.117 -0.680 0.680 0.117 -0.703 0.680 0.117 -0.703 0.680 0.117 -0.703 0.680 0.117 -0.703 0.680 0.117 -0.703 0.680 0.117 -0.703 0.680 0.117 -0.703 0.680 0.117 -0.703 0.680 0.117 -0.703 0.680 0.117 -0.703 0.680 0.117 -0.703 0.680 0.117", 
                                   "-0.750 0.586 0.117 -0.773 0.586 0.117 -0.773 0.609 0.117 -0.773 0.586 0.117 -0.773 0.586 0.117 -0.773 0.586 0.117 -0.773 0.586 0.117 -0.773 0.586 0.117 -0.773 0.586 0.141 -0.773 0.586 0.141 -0.773 0.586 0.141 -0.773 0.586 0.141 -0.773 0.586 0.141 -0.773 0.586 0.141 -0.773 0.586 0.141 -0.773 0.586 0.141 -0.773 0.586 0.141 -0.773 0.586 0.141 -0.773 0.586 0.141 -0.773 0.586 0.141 -0.773 0.586 0.141 -0.750 0.586 0.141 -0.773 0.586 0.141 -0.773 0.586 0.141 -0.773 0.586 0.141 -0.773 0.586 0.117 -0.773 0.586 0.141 -0.773 0.586 0.117 -0.773 0.586 0.117 -0.773 0.586 0.117 -0.773 0.586 0.117 -0.773 0.586 0.117 -0.773 0.586 0.117 -0.773 0.586 0.141 -0.773 0.586 0.117 -0.773 0.586 0.117 -0.773 0.586 0.117 -0.773 0.586 0.117 -0.773 0.586 0.117 -0.773 0.586 0.117"))

^{创建于 2022 年 11 月 17 日 reprex v2.0.2}

我的想法是，我想将 acceleration_raw 列分成 3 个不同的列:acc_x、acc_y 和 acc_z. acceleration_raw 的每一行都是一个字符串，最终产生 120 个数字观测值。我想分离 Acceleration_raw，然后以 3 的步长获取第一行和第四行的每个值，并将其放入第二行和第四行的每个值，并将其放入 acc_x并将其放入 acc_y，最后将第三行及以后的每个值放入 acc_z。

我尝试首先将 acceleration_raw 与 separate_rows 与 dplyr 分开:

library('tidyverse')
library('data.table')

dt1 <- dt1 %>%
  separate_rows(acceleration_raw, sep = " ", convert = F)

^{创建于 2022 年 11 月 17 日 reprex v2.0.2}

之后:

library('tidyverse')
library('data.table')

dt1 <- dt1 %>%
  separate_rows(acceleration_raw, sep = " ", convert = F) %>%
  mutate(acc_x = seq(acceleration_raw, from = 1, to = length(dt1), by = 3), 
         acc_y = seq(acceleration_raw, from = 2, to = length(dt1), by = 3), 
         acc_z = seq(acceleration_raw, from = 3, to = length(dt1), by = 3))
#> Warning in seq.default(acceleration_raw, from = 1, to = length(dt1), by = 3):
#> first element used of 'length.out' argument
#> Error in `mutate()`:
#> ! Problem while computing `acc_x = seq(acceleration_raw, from = 1, to =
#>   length(dt1), by = 3)`.
#> Caused by error in `ceiling()`:
#> ! non-numeric argument to mathematical function

^{创建于 2022 年 11 月 17 日 reprex v2.0.2}

关于如何进行的任何建议？

Answer 1

您可以使用pivot_wider和unnest:

library(tidyverse)

dt1 %>%
  separate_rows(acceleration_raw, sep = " ", convert = F) %>%
  mutate(id = rep(c("acc_x", "acc_y", "acc_z"), times = nrow(.) / 3)) %>% 
  pivot_wider(names_from = id, values_from = acceleration_raw, values_fn = list) %>% 
  unnest(cols = c("acc_x", "acc_y", "acc_z"))

这会返回

# A tibble: 120 × 5
   ID    timestamp           acc_x  acc_y acc_z
   <fct> <dttm>              <chr>  <chr> <chr>
 1 202E  2017-05-02 00:00:00 -0.703 0.656 0.164
 2 202E  2017-05-02 00:00:00 -0.703 0.656 0.164
 3 202E  2017-05-02 00:00:00 -0.703 0.656 0.164
 4 202E  2017-05-02 00:00:00 -0.703 0.656 0.164
 5 202E  2017-05-02 00:00:00 -0.703 0.656 0.164
 6 202E  2017-05-02 00:00:00 -0.703 0.656 0.164
 7 202E  2017-05-02 00:00:00 -0.703 0.656 0.164
 8 202E  2017-05-02 00:00:00 -0.703 0.656 0.164
 9 202E  2017-05-02 00:00:00 -0.703 0.656 0.164
10 202E  2017-05-02 00:00:00 -0.703 0.656 0.164
# … with 110 more rows