R：將一行拆分為多行，再將列拆分為多列

Question

我陷入了一項看似簡單的任務。 想象以下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"))

^{創建於 2022-11-17，使用reprex v2.0.2}

我的想法是，我想將acceleration_raw列分成 3 個不同的列： acc_x 、 acc_y和acc_z 。 acceleration_raw的每一行都是一串字符，最終導致 120 個數值觀察。 我想將acceleration_raw分開，然后以 3 的步長從第一行來回獲取每個值並將其放入acc_x ，從第二行來回的每個值並將其放入acc_y ，最后是第三行的每個值然后把它放到acc_z 。

我嘗試首先將acceleration_raw與separate_rows從dplyr中分離出來：

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

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

^{創建於 2022-11-17，使用reprex v2.0.2}

在那之后：

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

^{創建於 2022-11-17，使用reprex v2.0.2}

關於如何進行的任何建議？

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"))

這返回

# 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

Answer 2

沒有 NA，我無法使用 pivot_wider，因此解決方案不是最優的：

library(tidyverse)

dt1 %>% 
  as_tibble() %>% 
  separate_rows(acceleration_raw, sep = " ") %>% 
  group_by(group =  as.integer(gl(n(), n()/3, n()))) %>% 
  mutate(id = row_number()) %>% 
  mutate(group = case_when(group == 1 ~ "acc_x",
                           group == 2 ~ "acc_y", 
                           group == 3 ~ "acc_z")) %>% 
  pivot_wider(names_from = group, values_from = acceleration_raw) %>% 
  mutate(acc_y = lead(acc_y,n()/3),
         acc_z = lead(acc_z,n()/3*2)) %>% 
  na.omit()

# A tibble: 120 x 6
   ID    timestamp              id acc_x  acc_y  acc_z 
   <fct> <dttm>              <int> <chr>  <chr>  <chr> 
 1 202E  2017-05-02 00:00:00     1 -0.703 -0.703 -0.750
 2 202E  2017-05-02 00:00:00     2 0.656  0.680  0.586 
 3 202E  2017-05-02 00:00:00     3 0.164  0.117  0.117 
 4 202E  2017-05-02 00:00:00     4 -0.703 -0.680 -0.773
 5 202E  2017-05-02 00:00:00     5 0.656  0.680  0.586 
 6 202E  2017-05-02 00:00:00     6 0.164  0.117  0.117 
 7 202E  2017-05-02 00:00:00     7 -0.703 -0.680 -0.773
 8 202E  2017-05-02 00:00:00     8 0.656  0.680  0.609 
 9 202E  2017-05-02 00:00:00     9 0.164  0.117  0.117 
10 202E  2017-05-02 00:00:00    10 -0.703 -0.680 -0.773
# ... with 110 more rows
# i Use `print(n = ...)` to see more rows