Convert column classes in data.table
I have a problem using data.table: How do I convert column classes? Here is a simple example: With data.frame I don't have a problem converting it, with data.table I just don't know how:
df <- data.frame(ID=c(rep("A", 5), rep("B",5)), Quarter=c(1:5, 1:5), value=rnorm(10))
#One way: http://stackoverflow.com/questions/2851015/r-convert-data-frame-columns-from-factors-to-characters
df <- data.frame(lapply(df, as.character), stringsAsFactors=FALSE)
#Another way
df[, "value"] <- as.numeric(df[, "value"])
library(data.table)
dt <- data.table(ID=c(rep("A", 5), rep("B",5)), Quarter=c(1:5, 1:5), value=rnorm(10))
dt <- data.table(lapply(dt, as.character), stringsAsFactors=FALSE)
#Error in rep("", ncol(xi)) : invalid 'times' argument
#Produces error, does data.table not have the option stringsAsFactors?
dt[, "ID", with=FALSE] <- as.character(dt[, "ID", with=FALSE])
#Produces error: Error in `[<-.data.table`(`*tmp*`, , "ID", with = FALSE, value = "c(1, 1, 1, 1, 1, 2, 2, 2, 2, 2)") :
#unused argument(s) (with = FALSE)
Do I miss something obvious here?
Update due to Matthew's post: I used an older version before, but even after updating to 1.6.6 (the version I use now) I still get an error.
Update 2: Let's say I want to convert every column of class "factor" to a "character" column, but don't know in advance which column is of which class. With a data.frame, I can do the following:
classes <- as.character(sapply(df, class))
colClasses <- which(classes=="factor")
df[, colClasses] <- sapply(df[, colClasses], as.character)
Can I do something similar with data.table?
Update 3:
sessionInfo() R version 2.13.1 (2011-07-08) Platform: x86_64-pc-mingw32/x64 (64-bit)
locale:
[1] C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] data.table_1.6.6
loaded via a namespace (and not attached):
[1] tools_2.13.1
For a single column:
dtnew <- dt[, Quarter:=as.character(Quarter)]
str(dtnew)
Classes ‘data.table’ and 'data.frame': 10 obs. of 3 variables:
$ ID : Factor w/ 2 levels "A","B": 1 1 1 1 1 2 2 2 2 2
$ Quarter: chr "1" "2" "3" "4" ...
$ value : num -0.838 0.146 -1.059 -1.197 0.282 ...
Using lapply and as.character :
dtnew <- dt[, lapply(.SD, as.character), by=ID]
str(dtnew)
Classes ‘data.table’ and 'data.frame': 10 obs. of 3 variables:
$ ID : Factor w/ 2 levels "A","B": 1 1 1 1 1 2 2 2 2 2
$ Quarter: chr "1" "2" "3" "4" ...
$ value : chr "1.487145280568" "-0.827845218358881" "0.028977182770002" "1.35392750102305" ...
尝试这个
DT <- data.table(X1 = c("a", "b"), X2 = c(1,2), X3 = c("hello", "you"))
changeCols <- colnames(DT)[which(as.vector(DT[,lapply(.SD, class)]) == "character")]
DT[,(changeCols):= lapply(.SD, as.factor), .SDcols = changeCols]
This is a BAD way to do it! I'm only leaving this answer in case it solves other weird problems. These better methods are the probably partly the result of newer data.table versions... so it's worth while to document this hard way. Plus, this is a nice syntax example for eval substitute syntax.
library(data.table)
dt <- data.table(ID = c(rep("A", 5), rep("B",5)),
fac1 = c(1:5, 1:5),
fac2 = c(1:5, 1:5) * 2,
val1 = rnorm(10),
val2 = rnorm(10))
names_factors = c('fac1', 'fac2')
names_values = c('val1', 'val2')
for (col in names_factors){
e = substitute(X := as.factor(X), list(X = as.symbol(col)))
dt[ , eval(e)]
}
for (col in names_values){
e = substitute(X := as.numeric(X), list(X = as.symbol(col)))
dt[ , eval(e)]
}
str(dt)
which gives you
Classes ‘data.table’ and 'data.frame': 10 obs. of 5 variables:
$ ID : chr "A" "A" "A" "A" ...
$ fac1: Factor w/ 5 levels "1","2","3","4",..: 1 2 3 4 5 1 2 3 4 5
$ fac2: Factor w/ 5 levels "2","4","6","8",..: 1 2 3 4 5 1 2 3 4 5
$ val1: num 0.0459 2.0113 0.5186 -0.8348 -0.2185 ...
$ val2: num -0.0688 0.6544 0.267 -0.1322 -0.4893 ...
- attr(*, ".internal.selfref")=<externalptr>
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