I want to convert a subset of data.table cols to a new class. There's a popular question here (Convert column classes in data.table) but the answer creates a new object, rather than operating on the starter object.
Take this example:
dat <- data.frame(ID=c(rep("A", 5), rep("B",5)), Quarter=c(1:5, 1:5), value=rnorm(10))
cols <- c('ID', 'Quarter')
How best to convert to just the cols
columns to (e.g.) a factor? In a normal data.frame you could do this:
dat[, cols] <- lapply(dat[, cols], factor)
but that doesn't work for a data.table, and neither does this
dat[, .SD := lapply(.SD, factor), .SDcols = cols]
A comment in the linked question from Matt Dowle (from Dec 2013) suggests the following, which works fine, but seems a bit less elegant.
for (j in cols) set(dat, j = j, value = factor(dat[[j]]))
Is there currently a better data.table answer (i.e. shorter + doesn't generate a counter variable), or should I just use the above + rm(j)
?
You can change data types using as. * where * is the datatype to change to, the other way is using class(). class(df$var) = "Numeric".
Use the lapply() Function to Convert Multiple Columns From Integer to Numeric Type in R. Base R's lapply() function allows us to apply a function to elements of a list.
SD stands for "Subset of Data. table". The dot before SD has no significance but doesn't let it clash with a user-defined column name.
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]
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