另一种类似于Sotos的方法，但是1）使用 data.table ，2）不使用 factor 和3）将表替换为 Rfast :: rowTabulate ：

Another option similar to Sotos' approach but 1) using data.table, 2) not using factor and 3) replacing table with Rfast::rowTabulate:

v <- c('Hard', 'Match', 'Easy')
vv <- do.call(paste, expand.grid(v, v))
DT[, (vv) := {
        mat <- mapply(paste, .SD[, -ncol(.SD), with=FALSE], .SD[, -1L])
        as.data.table(Rfast::rowTabulate(matrix(match(mat, vv, 0L), nrow=.N)))
    }, .SDcols=Task_Alpha:Task_Delta]

输出：

   userID Score Task_Alpha Task_Beta Task_Charlie Task_Delta Hard Hard Match Hard Easy Hard Hard Match Match Match Easy Match Hard Easy Match Easy Easy Easy
1:   3108 -8.00       Easy      Easy         Easy       Easy         0          0         0          0           0          0         0          0         3
2:   3207  3.00       Hard      Easy        Match      Match         0          0         0          0           1          1         1          0         0
3:   3350  5.78       Hard      Easy         Hard       Hard         1          0         1          0           0          0         1          0         0
4:   3961 10.00       Easy      <NA>         Hard       Hard         1          0         0          0           0          0         0          0         0
5:   4021 10.00       Easy      Easy         <NA>       Hard         0          0         0          0           0          0         0          0         1

数据：

library(data.table)
library(Rfast)
DT <- structure(list(
    userID = c(3108L, 3207L, 3350L, 3961L, 4021L), 
    Score = c(-8, 3, 5.78, 10, 10), 
    Task_Alpha = structure(c(1L, 2L, 2L, 1L, 1L), .Label = c("Easy", "Hard"), class = "factor"), 
    Task_Beta = structure(c(1L, 1L, 1L, NA, 1L), .Label = "Easy", class = "factor"), 
    Task_Charlie = structure(c(1L, 3L, 2L, 2L, NA), .Label = c("Easy", "Hard", "Match"), class = "factor"), 
    Task_Delta = structure(c(1L, 3L, 2L, 2L, 2L), .Label = c("Easy", "Hard", "Match"), class = "factor")), 
    class = "data.frame", row.names = c(NA, -5L))
setDT(DT)

了解这种方法在实际数据集上的运行速度以及实际数据集是否很大会很有趣。

Would be interesting to know how fast this approach works on actual dataset and if actual dataset is large.

编辑：添加了一些时间

library(data.table)
nr <- 1e6
vec <- c('Hard', 'Match', 'Easy', NA)
DT <- data.table(userID=1:nr, Task_Alpha=sample(vec, nr, TRUE), Task_Beta=sample(vec, nr, TRUE),
    Task_Charlie=sample(vec, nr, TRUE), Task_Delta=sample(vec, nr, TRUE))
df <- as.data.frame(DT)
DT0 <- copy(DT)
DT1 <- copy(DT)
DT2 <- copy(DT)

mtd0 <- function() {
    t(apply(df[-1L], 1, function(i) {
        i1 <- paste(i[-length(i)], i[-1L]);
        i1 <- factor(i1, levels = do.call(paste, expand.grid(c('Easy', 'Match', 'Hard'),
            c('Easy', 'Match', 'Hard'))));
        table(i1)
    }))
}

mtd1 <- function() {
    f_cols <- names(DT0)[ sapply( DT0, is.factor ) ]
    DT0[, (f_cols) := lapply(.SD, as.character), .SDcols = f_cols ]
    #melt to long format
    DT.melt <- melt( DT0, id.vars = "userID", measure.vars = patterns( task = "^Task_"))
    #set order of Aplha-Beta-etc...
    DT.melt[ grepl( "Alpha",   variable ), order := 1 ]
    DT.melt[ grepl( "Beta",    variable ), order := 2 ]
    DT.melt[ grepl( "Charlie", variable ), order := 3 ]
    DT.melt[ grepl( "Delta",   variable ), order := 4 ]
    #order DT.melt
    setorder( DT.melt, userID, order )
    #fill in codes EE, etc...
    DT.melt[, `:=`( code1 = gsub( "(^.).*", "\\1", value ),
        code2 = gsub( "(^.).*", "\\1", shift( value, type = "lead" ) ) ),
        by = userID ]
    #filter only rows without NA
    DT.melt <- DT.melt[ complete.cases( DT.melt ) ]
    #cast to wide output
    dcast( DT.melt, userID ~ paste0( code2, code1 ), fun.aggregate = length )
}

mtd2 <- function() {
    v <- c('Hard', 'Match', 'Easy')
    vv <- do.call(paste, expand.grid(v, v))
    DT2[, (vv) := {
        mat <- mapply(paste, .SD[, -ncol(.SD), with=FALSE], .SD[, -1L])
        as.data.table(Rfast::rowTabulate(matrix(match(mat, vv, 0L), nrow=.N)))
    }, .SDcols=Task_Alpha:Task_Delta]
}

bench::mark(mtd0(), mtd1(), mtd2(), check=FALSE)

时间：

# A tibble: 3 x 13
  expression      min   median `itr/sec` mem_alloc `gc/sec` n_itr  n_gc total_time result                     memory                 time     gc              
  <bch:expr> <bch:tm> <bch:tm>     <dbl> <bch:byt>    <dbl> <int> <dbl>   <bch:tm> <list>                     <list>                 <list>   <list>          
1 mtd0()        2.19m    2.19m   0.00760     252MB    2.26      1   297      2.19m <int[,9] [1,000,000 x 9]>  <df[,3] [171,481 x 3]> <bch:tm> <tibble [1 x 3]>
2 mtd1()       33.16s   33.16s   0.0302      856MB    0.754     1    25     33.16s <df[,10] [843,688 x 10]>   <df[,3] [8,454 x 3]>   <bch:tm> <tibble [1 x 3]>
3 mtd2()     844.95ms 844.95ms   1.18        298MB    1.18      1     1   844.95ms <df[,14] [1,000,000 x 14]> <df[,3] [8,912 x 3]>   <bch:tm> <tibble [1 x 3]>