R: Converting multiple binary columns into one factor variable whose factors are binary column names

Question

I am a new R user. Currently I am working on a dataset wherein I have to transform the multiple binary columns into single factor column

Here is the example:

Answer 1

Melt is certainly a solution. I'd suggest using the reshape2 melt as follows:

library(reshape2)

df=data.frame(Property.RealEstate=c(0,0,1,0,0,0),
              Property.Insurance=c(0,1,0,1,0,0),
              Property.CarOther=c(0,0,0,0,1,0),
              Property.Unknown=c(0,0,0,0,0,1))

#add id column (presumably you have ids more meaningful than row numbers)
df$row=1:nrow(df)

#melt to "long" format
long=melt(df,id="row")

#only keep 1's
long=long[which(long$value==1),]

#merge in ids for NA entries
long=merge(df[,"row",drop=F],long,all.x=T)

#clean up to match example output
long=long[order(long$row),"variable",drop=F]
names(long)="Property"
long$Property=gsub("Property.","",long$Property,fixed=T)

#results
long

Answer 2

Alternately, you can just do it in the naïve way. I think it's more transparent than any of the other suggestions (including my other suggestion).

df=data.frame(Property.RealEstate=c(0,0,1,0,0,0),
              Property.Insurance=c(0,1,0,1,0,0),
              Property.CarOther=c(0,0,0,0,1,0),
              Property.Unknown=c(0,0,0,0,0,1))

propcols=c("Property.RealEstate", "Property.Insurance", "Property.CarOther", "Property.Unknown")

df$Property=NA

for(colname in propcols)({
  coldata=df[,colname]
  df$Property[which(coldata==1)]=colname
})

df$Property=gsub("Property.","",df$Property,fixed=T)

Answer 3

> mat <- matrix(c(0,1,0,0,0,
+                 1,0,0,0,0,
+                 0,0,0,1,0,
+                 0,0,1,0,0,
+                 0,0,0,0,1), ncol = 5, byrow = TRUE)
> colnames(mat) <- c("Level1","Level2","Level3","Level4","Level5")
> mat
     Level1 Level2 Level3 Level4 Level5
[1,]      0      1      0      0      0
[2,]      1      0      0      0      0
[3,]      0      0      0      1      0
[4,]      0      0      1      0      0
[5,]      0      0      0      0      1

Create a new factor based upon the index of each 1 in each row Use the matrix column names as the labels for each level

NewFactor <- factor(apply(mat, 1, function(x) which(x == 1)), 
                    labels = colnames(mat)) 

> NewFactor 
[1] Level2 Level1 Level4 Level3 Level5 
Levels: Level1 Level2 Level3 Level4 Level5 

also you can try:

factor(mat%*%(1:ncol(mat)), labels = colnames(mat)) 

also use Tomas solution - ifounf somewhere in SO

as.factor(colnames(mat)[mat %*% 1:ncol(mat)])

Answer 4

Something different:

Get the data:

dat <- data.frame(Property.RealEstate=c(1,0,1,0,0,0),Property.Insurance=c(0,1,0,1,0,0),Property.CarOther=c(0,0,0,0,1,0),Property.Unknown=c(0,0,0,0,0,1))

Reshape it:

names(dat)[row(t(dat))[t(dat)==1]]
#[1] "Property.RealEstate" "Property.Insurance"  "Property.RealEstate"
#[4] "Property.Insurance"  "Property.CarOther"   "Property.Unknown" 

If you want it cleaned up, do:

gsub("Property\\.","",names(dat)[row(t(dat))[t(dat)==1]])
#[1] "RealEstate" "Insurance"  "RealEstate" "Insurance"  "CarOther"   "Unknown" 

If you prefer a factor output:

factor(row(t(dat))[t(dat)==1],labels=names(dat))

...and cleaned up:

factor(row(t(dat))[t(dat)==1],labels=gsub("Property\\.","",names(dat)) )