transform all columns that are NAs from numeric to factors - r

Question

I have a data.table called td.br.2, in which some columns are completely NAs. These columns are of type numeric. What I would like to do, is only for these columns to transform them to factors.
I have tried the following, but it does not work ( I do not get an error but it does not do the job either)
td.br.2[] <- td.br.2[,lapply(.SD, function(x) {ifelse(sum(is.na(x)==nrow(td.br.2)),as.factor(x),x)})]

Answer 1

n=10#nr of rows
m=10#nr of cols
N<-n*m
m1<-matrix(runif(N),nrow=n,ncol = m)
dt<-data.table(m1)
names(dt)<-letters[1:m]
dt<-cbind(dt,xxx=rep(NA,nrow(dt)))#adding NA column
At this point
str(dt)
Classes ‘data.table’ and 'data.frame': 10 obs. of 11 variables:
$ a : num 0.661 0.864 0.152 0.342 0.989 ...
$ b : num 0.06036 0.67587 0.00847 0.37674 0.30417 ...
$ c : num 0.3938 0.6274 0.0514 0.882 0.1568 ...
$ d : num 0.777 0.233 0.619 0.117 0.132 ...
$ e : num 0.655 0.926 0.277 0.598 0.237 ...
$ f : num 0.649 0.197 0.547 0.585 0.685 ...
$ g : num 0.6877 0.3676 0.009 0.6975 0.0327 ...
$ h : num 0.519 0.705 0.457 0.465 0.966 ...
$ i : num 0.43777 0.00961 0.30224 0.58172 0.37621 ...
$ j : num 0.44 0.481 0.485 0.125 0.263 ...
$ xxx: logi NA NA NA NA NA NA ...
So by executing:
dt<-dt[, lapply(.SD, function(x){ if(all(is.na(x)))as.factor(as.character(x)) else x}),]
yields:
str(dt)
Classes ‘data.table’ and 'data.frame': 10 obs. of 11 variables:
$ a : num 0.0903 0.0448 0.5956 0.418 0.1316 ...
$ b : num 0.672 0.582 0.687 0.113 0.371 ...
$ c : num 0.404 0.16 0.848 0.863 0.737 ...
$ d : num 0.073 0.129 0.243 0.334 0.285 ...
$ e : num 0.485 0.186 0.539 0.486 0.784 ...
$ f : num 0.4685 0.4815 0.585 0.3596 0.0764 ...
$ g : num 0.958 0.194 0.549 0.71 0.737 ...
$ h : num 0.168 0.355 0.552 0.765 0.605 ...
$ i : num 0.665 0.88 0.23 0.575 0.413 ...
$ j : num 0.1113 0.8797 0.1244 0.0741 0.8724 ...
$ xxx: Factor w/ 0 levels: NA NA NA NA NA NA NA NA NA NA

Answer 2

I am not sure why you would want to do that, but here you are:
naColumns <- sapply(td.br.2, function(x) { all(is.na(x)) })
for (col in which(naColumns))
set(td.br.2, j=col, value=as.factor(x[[col]]))
The factors will have no levels, but you can deal with that as necessary.
(The for loop is partly based on this.)