How to programmatically create binary columns based on a categorical variable in data.table?

Question

I have a big (12 million rows) data.table which looks like this:

library(data.table)
set.seed(123)
dt <         


        

Answer 1

data.table has its own dcast implementation using data.table's internals and should be fast. Give this a try:

dcast(dt, id ~ y, fun.aggregate = function(x) 1L, fill=0L)
#    id a b c d e
# 1:  1 0 1 1 1 1
# 2:  2 1 0 1 0 1
# 3:  3 1 0 1 1 1

---

Just thought of another way to handle this by preallocating and updating by reference (perhaps dcast's logic should be done like this to avoid intermediates).

ans = data.table(id = unique(dt$id))[, unique(dt$y) := 0L][]

All that's left is to fill existing combinations with 1L.

dt[, {set(ans, i=.GRP, j=unique(y), value=1L); NULL}, by=id]
ans
#    id b d c e a
# 1:  1 1 1 1 1 0
# 2:  2 0 0 1 1 1
# 3:  3 0 1 1 1 1

---

Okay, I've gone ahead on benchmarked on OP's data dimensions with ~10 million rows and 10 columns.

require(data.table)
set.seed(45L)
y = apply(matrix(sample(letters, 10L*20L, TRUE), ncol=20L), 1L, paste, collapse="")
dt = data.table(id=sample(1e5,1e7,TRUE), y=sample(y,1e7,TRUE))

system.time(ans1 <- AnsFunction())   # 2.3s
system.time(ans2 <- dcastFunction()) # 2.2s
system.time(ans3 <- TableFunction()) # 6.2s

setcolorder(ans1, names(ans2))
setcolorder(ans3, names(ans2))
setorder(ans1, id)
setkey(ans2, NULL)
setorder(ans3, id)

identical(ans1, ans2) # TRUE
identical(ans1, ans3) # TRUE

where,

AnsFunction <- function() {
    ans = data.table(id = unique(dt$id))[, unique(dt$y) := 0L][]
    dt[, {set(ans, i=.GRP, j=unique(y), value=1L); NULL}, by=id]
    ans
    # reorder columns outside
}

dcastFunction <- function() {
    # no need to load reshape2. data.table has its own dcast as well
    # no need for setDT
    df <- dcast(dt, id ~ y, fun.aggregate = function(x) 1L, fill=0L,value.var = "y")
}

TableFunction <- function() {
    # need to return integer results for identical results
    # fixed 1 -> 1L; as.numeric -> as.integer
    df <- as.data.frame.matrix(table(dt$id, dt$y))
    df[df > 1L] <- 1L
    df <- cbind(id = as.integer(row.names(df)), df)
    setDT(df)
}

Answer 2

For small data sets the table function seems to be more efficient, but on large datasets dcast seems to be the most efficient and convenient option.

TableFunction <- function(){
    df <- as.data.frame.matrix(table(dt$id, dt$y))
    df[df > 1] <- 1
    df <- cbind(id = as.numeric(row.names(df)), df)
    setDT(df)
}


AnsFunction <- function(){
    ans = data.table(id = unique(dt$id))[, unique(dt$y) := 0L][]
    dt[, {set(ans, i=id, j=unique(y), value=1L); NULL}, by=id]
}

dcastFunction <- function(){
    df <-dcast.data.table(dt, id ~ y, fun.aggregate = function(x) 1L, fill=0L,value.var = "y")

}

library(data.table)
library(microbenchmark)
set.seed(123)
N = 10000
dt <- data.table(id=rep(1:N, each=5),y=sample(letters[1 : 5], N*5, replace = T)) 


microbenchmark(
    "dcast" = dcastFunction(),
    "Table" = TableFunction(),
    "Ans"   = AnsFunction()
    )


 Unit: milliseconds
  expr       min        lq      mean    median        uq       max neval cld
 dcast  42.48367  45.39793  47.56898  46.83755  49.33388  60.72327   100  b 
 Table  28.32704  28.74579  29.14043  29.00010  29.23320  35.16723   100 a  
   Ans 120.80609 123.95895 127.35880 126.85018 130.12491 156.53289   100   c

> all(test1 == test2)
[1] TRUE
> all(test1 == test3)
[1] TRUE

y = apply(matrix(sample(letters, 10L*20L, TRUE), ncol=20L), 1L, paste, collapse="")
dt = data.table(id=sample(1e5,1e7,TRUE), y=sample(y,1e7,TRUE))

microbenchmark(
    "dcast" = dcastFunction(),
    "Table" = TableFunction(),
    "Ans"   = AnsFunction()
)
Unit: seconds
  expr      min       lq     mean   median       uq      max neval cld
 dcast 1.985969 2.064964 2.189764 2.216138 2.266959 2.643231   100 a  
 Table 5.022388 5.403263 5.605012 5.580228 5.830414 6.318729   100   c
   Ans 2.234636 2.414224 2.586727 2.599156 2.645717 2.982311   100  b 

Answer 3

If you already know the range of the rows (as in you know that there are no more than 3 rows in your example) and you know the columns you can start with an array of zeros and use the apply function to update values in that secondary table.

My R is a little rust but i think that should work. Additionally the function you pass to the apply method could contain conditions to add necessary rows and columns as is needed.

My R is a little rust so I'm a bit tentative to write it up right now, but I think that's the way to do it.

If you are looking for something a little more plug and play I found this little blerb:

There are two sets of methods that are explained below:

gather() and spread() from the tidyr package. This is a newer interface to the reshape2 package.

melt() and dcast() from the reshape2 package.

There are a number of other methods which aren’t covered here, since they are not as easy to use:

The reshape() function, which is confusingly not part of the reshape2 package; it is part of the base install of R.

stack() and unstack()

from here :: http://www.cookbook-r.com/Manipulating_data/Converting_data_between_wide_and_long_format/

If I was better versed in R I would tell you how those various methods handle collisions going from long lists to wide on. I was googling up "Make a table from flat data in R" to come up with this...

Also Check out this It's that same website as above with my personal comment wrapper : p