如何在R中将行拆分为3个较短的行

Question

So I have a table in R like this:

id    col1    col2    col3    col4    col5     col6     col7     col8   col9 
101   1       1111    202     2       1120     5512     3        1221   900
102   1       2999    1110    2       2000     5000     3        80     200
103   1       1121    333     2       111      222      3        101    1000
. 
.

I'm trying to bring the long row for each subject into multiple rows like this:

id   trial   col1   col2   
101  1       1111   202
101  2       1120   5512
101  3       1221   900
102  1       2999   1110
102  2       2000   5000
102  3       80     200
103  1       1121   333
103  2       111    222
103  3       101    1000

I'd appreciate any help as I'm a R neophyte here. I imagine I'd want to read the col as a triplet and then compile them up but have no idea how to go about to do it.

Answer 1

The problem with your data is the it is stored in a very unconventional way. Generally when data is transformed from a wide format to a long format the variable names in the wide data become data points within the long data, hence the name pivot table. To overcome this problem I suggest you transform the data as follows:

d <- d[, !grepl("col[147]", names(d))] 

names(d)[-1] <- paste(sort(rep(1:3, 2)), paste0("col", 1:2))

Once you have done this it is relatively straightforward to reshape the data with the tidyr package.

d %>%
  gather(key, value, -id) %>%
  separate(key, c("trial", "new"), sep = "\\s") %>%
  spread(new, value)

Answer 2

(A bit manual) What about this?

res <- cbind(rep(df[,1], each = nrow(df)), matrix(c(t(df[-1])), ncol = 3, byrow = TRUE))
colnames(res) <- c("id", "trial", "col1", "col2")
res
       id trial col1 col2
 [1,] 101     1 1111  202
 [2,] 101     2 1120 5512
 [3,] 101     3 1221  900
 [4,] 102     1 2999 1110
 [5,] 102     2 2000 5000
 [6,] 102     3   80  200
 [7,] 103     1 1121  333
 [8,] 103     2  111  222
 [9,] 103     3  101 1000

Answer 3

Here is another option with array

cbind(rep(df1$id, 
  each=nrow(df1)),apply(aperm(array(unlist(df1[-1]), 
           dim=c(3,3,3)), c(3,2,1)), 2, c))
#      [,1] [,2] [,3] [,4]
# [1,]  101    1 1111  202
# [2,]  101    2 1120 5512
# [3,]  101    3 1221  900
# [4,]  102    1 2999 1110
# [5,]  102    2 2000 5000
# [6,]  102    3   80  200
# [7,]  103    1 1121  333
# [8,]  103    2  111  222
# [9,]  103    3  101 1000

data

df1 <- structure(list(id = 101:103, col1 = c(1L, 1L, 1L), 
 col2 = c(1111L, 
 2999L, 1121L), col3 = c(202L, 1110L, 333L), col4 = c(2L, 2L, 
2L), col5 = c(1120L, 2000L, 111L), col6 = c(5512L, 5000L, 222L
 ), col7 = c(3L, 3L, 3L), col8 = c(1221L, 80L, 101L), 
col9 = c(900L, 
200L, 1000L)), .Names = c("id", "col1", "col2", "col3", 
"col4", 
"col5", "col6", "col7", "col8", "col9"), class = "data.frame", 
 row.names = c(NA, -3L))

Answer 4

text1 = "
id    col1    col2    col3    col4    col5     col6     col7     col8   col9 
101   1       1111    202     2       1120     5512     3        1221   900
102   1       2999    1110    2       2000     5000     3        80     200
103   1       1121    333     2       111      222      3        101    1000
"

df1 <- read.table(text=text1, head=T, as.is=T)

library(plyr)

ddply(df1, .(id), function(df){
  df1 <- df[, 2:4]
  df2 <- df[, 5:7]
  df3 <- df[, 8:10]
  names(df1) <- c("trial", "col1", "col2")
  names(df2) <- c("trial", "col1", "col2")
  names(df3) <- c("trial", "col1", "col2")
  df.n <- do.call(rbind, list(df1, df2, df3))
  return(df.n)
})
#    id trial col1 col2
# 1 101     1 1111  202
# 2 101     2 1120 5512
# 3 101     3 1221  900
# 4 102     1 2999 1110
# 5 102     2 2000 5000
# 6 102     3   80  200
# 7 103     1 1121  333
# 8 103     2  111  222
# 9 103     3  101 1000