如何用 tidyr 替换数组的 reshape2::melt？

Question

I would like to convert a matrix/array (with dimnames) into a data frame. This can be done very easily using reshape2::melt but seems harder with tidyr , and in fact not really possible in the case of an array. Am I missing something? (In particular since reshape2 describes itself as being retired; see https://github.com/hadley/reshape ).

For example, given the following matrix

MyScores <- matrix(runif(2*3), nrow = 2, ncol = 3, 
                   dimnames = list(Month = month.name[1:2], Class = LETTERS[1:3]))

we can turn it into a data frame as follows

reshape2::melt(MyScores, value.name = 'Score') # perfect

or, using tidyr as follows:

as_tibble(MyScores, rownames = 'Month') %>% 
  gather(Class, Score, -Month)

In this case reshape2 and tidyr seem similar (although reshape2 is shorter if you are looking for a long-format data frame).

However for arrays, it seems harder. Given

EverybodyScores <- array(runif(2*3*5), dim = c(2,3,5), 
                         dimnames = list(Month = month.name[1:2], Class = LETTERS[1:3], StudentID = 1:5))

we can turn it into a data frame as follows:

reshape2::melt(EverybodyScores, value.name = 'Score') # perfect

but using tidyr it's not clear how to do it:

as_tibble(EverybodyScores, rownames = 'Month') # looses month information and need to distange Class and StudentID

Is this a situation where the right solution is to stick to using reshape2 ?

Answer 1

One way I just found by playing around is to coerce via tbl_cube . I have never really used the class but it seems to do the trick in this instance.

EverybodyScores <- array(
  runif(2 * 3 * 5),
  dim = c(2, 3, 5),
  dimnames = list(Month = month.name[1:2], Class = LETTERS[1:3], StudentID = 1:5)
)
library(tidyverse)
library(cubelyr)
EverybodyScores %>%
  as.tbl_cube(met_name = "Score") %>%
  as_tibble
#> # A tibble: 30 x 4
#>    Month    Class StudentID Score
#>    <chr>    <chr>     <int> <dbl>
#>  1 January  A             1 0.366
#>  2 February A             1 0.254
#>  3 January  B             1 0.441
#>  4 February B             1 0.562
#>  5 January  C             1 0.313
#>  6 February C             1 0.192
#>  7 January  A             2 0.799
#>  8 February A             2 0.277
#>  9 January  B             2 0.631
#> 10 February B             2 0.101
#> # ... with 20 more rows

Created on 2018-08-15 by the reprex package (v0.2.0).

Answer 2

Making a tibble drops the row names, but instead of going straight into a tibble, you can make the array into a base R data.frame , then use tidyr::rownames_to_column to make a column for months. Notice that converting to a data frame creates columns with names like A.1 , sticking the class and ID together; you can separate these again with tidyr::separate . Calling as_tibble is optional, just for if you care about it being a tibble in the end, and also can come at any point in the workflow once you've made a column from the row names.

library(tidyverse)

EverybodyScores <- array(runif(2*3*5), dim = c(2,3,5), 
                         dimnames = list(Month = month.name[1:2], Class = LETTERS[1:3], StudentID = 1:5))

EverybodyScores %>%
  as.data.frame() %>%
  rownames_to_column("Month") %>%
  gather(key = class_id, value = value, -Month) %>%
  separate(class_id, into = c("Class", "StudentID"), sep = "\\.") %>%
  as_tibble()
#> # A tibble: 30 x 4
#>    Month    Class StudentID value
#>    <chr>    <chr> <chr>     <dbl>
#>  1 January  A     1         0.576
#>  2 February A     1         0.229
#>  3 January  B     1         0.930
#>  4 February B     1         0.547
#>  5 January  C     1         0.761
#>  6 February C     1         0.468
#>  7 January  A     2         0.631
#>  8 February A     2         0.893
#>  9 January  B     2         0.638
#> 10 February B     2         0.735
#> # ... with 20 more rows

Created on 2018-08-15 by the reprex package (v0.2.0).

Answer 3

Here is the new tidyr way to do the same:

library(tidyr)

EverybodyScores <- array(
  runif(2 * 3 * 5),
  dim = c(2, 3, 5),
  dimnames = list(Month = month.name[1:2], Class = LETTERS[1:3], StudentID = 1:5)
)

as_tibble(EverybodyScores, rownames = "Month") %>%
  pivot_longer(
    cols = matches("^A|^B|^C"),
    names_sep = "\\.",
    names_to = c("Class", "StudentID")
  )
#> # A tibble: 30 x 4
#>    Month   Class StudentID  value
#>    <chr>   <chr> <chr>      <dbl>
#>  1 January A     1         0.0325
#>  2 January B     1         0.959 
#>  3 January C     1         0.593 
#>  4 January A     2         0.0702
#>  5 January B     2         0.882 
#>  6 January C     2         0.918 
#>  7 January A     3         0.459 
#>  8 January B     3         0.849 
#>  9 January C     3         0.901 
#> 10 January A     4         0.328 
#> # … with 20 more rows

^{Created on 2021-02-23 by the reprex package (v1.0.0)}