Ответ 1
С melt
от data.table
(см. ?patterns
):
library(data.table)
melt(setDT(df), measure = patterns("^qID", "^time_taken"),
value.name = c("qID", "time_taken"))
Результат:
age gender education previous_comp_exp tutorial_time variable qID time_taken
1: 18 Male Undergraduate casual_gamer 62.17926 1 sor9 39.61206
2: 24 Male Undergraduate casual_gamer 85.01288 1 sor9 50.92343
3: 18 Male Undergraduate casual_gamer 62.17926 2 sor8 19.48920
4: 24 Male Undergraduate casual_gamer 85.01288 2 sor8 16.15616
или с помощью tidyr
:
library(dplyr)
library(tidyr)
df %>%
gather(variable, value, qID.1:time_taken.2) %>%
mutate(variable = sub("\\.\\d$", "", variable)) %>%
group_by(variable) %>%
mutate(ID = row_number()) %>%
spread(variable, value, convert = TRUE) %>%
select(-ID)
Результат:
# A tibble: 4 x 7
age gender education previous_comp_exp tutorial_time qID time_taken
<int> <fctr> <fctr> <fctr> <dbl> <chr> <dbl>
1 18 Male Undergraduate casual_gamer 62.17926 sor9 39.61206
2 18 Male Undergraduate casual_gamer 62.17926 sor8 19.48920
3 24 Male Undergraduate casual_gamer 85.01288 sor9 50.92343
4 24 Male Undergraduate casual_gamer 85.01288 sor8 16.15616
Примечание:
Для метода tidyr
convert=TRUE
используется для преобразования time_taken
обратно в numeric
, поскольку он был принудительно принят, когда gather
ed с столбцами qID
.
Данные:
df = structure(list(age = c(18L, 24L), gender = structure(c(1L, 1L
), .Label = "Male", class = "factor"), education = structure(c(1L,
1L), .Label = "Undergraduate", class = "factor"), previous_comp_exp = structure(c(1L,
1L), .Label = "casual_gamer", class = "factor"), tutorial_time = c(62.17926,
85.01288), qID.1 = structure(c(1L, 1L), .Label = "sor9", class = "factor"),
time_taken.1 = c(39.61206, 50.92343), qID.2 = structure(c(1L,
1L), .Label = "sor8", class = "factor"), time_taken.2 = c(19.4892,
16.15616)), .Names = c("age", "gender", "education", "previous_comp_exp",
"tutorial_time", "qID.1", "time_taken.1", "qID.2", "time_taken.2"
), class = "data.frame", row.names = c(NA, -2L))