Convert *some* column classes in data.table

Besides using the option as suggested by Matt Dowle, another way of changing the column classes is as follows:

dat[, (cols) := lapply(.SD, factor), .SDcols = cols]

By using the := operator you update the datatable by reference. A check whether this worked:

> sapply(dat,class)
       ID   Quarter     value 
 "factor"  "factor" "numeric" 

As suggeted by @MattDowle in the comments, you can also use a combination of for(...) set(...) as follows:

for (col in cols) set(dat, j = col, value = factor(dat[[col]]))

which will give the same result. A third alternative is:

for (col in cols) dat[, (col) := factor(dat[[col]])]

On a smaller datasets, the for(...) set(...) option is about three times faster than the lapply option (but that doesn't really matter, because it is a small dataset). On larger datasets (e.g. 2 million rows), each of these approaches takes about the same amount of time. For testing on a larger dataset, I used:

dat <- data.table(ID=c(rep("A", 1e6), rep("B",1e6)),
                  Quarter=c(1:1e6, 1:1e6),
                  value=rnorm(10))

Sometimes, you will have to do it a bit differently (for example when numeric values are stored as a factor). Then you have to use something like this:

dat[, (cols) := lapply(.SD, function(x) as.integer(as.character(x))), .SDcols = cols]

WARNING: The following explanation is not the data.table-way of doing things. The datatable is not updated by reference because a copy is made and stored in memory (as pointed out by @Frank), which increases memory usage. It is more an addition in order to explain the working of with = FALSE.

When you want to change the column classes the same way as you would do with a dataframe, you have to add with = FALSE as follows:

dat[, cols] <- lapply(dat[, cols, with = FALSE], factor)

A check whether this worked:

> sapply(dat,class)
       ID   Quarter     value 
 "factor"  "factor" "numeric" 

If you don't add with = FALSE, datatable will evaluate dat[, cols] as a vector. Check the difference in output between dat[, cols] and dat[, cols, with = FALSE]:

> dat[, cols]
[1] "ID"      "Quarter"

> dat[, cols, with = FALSE]
    ID Quarter
 1:  A       1
 2:  A       2
 3:  A       3
 4:  A       4
 5:  A       5
 6:  B       1
 7:  B       2
 8:  B       3
 9:  B       4
10:  B       5

You can use .SDcols:

dat[, cols] <- dat[, lapply(.SD, factor), .SDcols=cols]