Generate a dummy matrix from multiple factor columns - r

Question

I already searched the web and found no answer. I have a big data.frame that contains multiple columns. Each column is a factor variable.
I want to transform the data.frame such that each possible value of the factor variables is a variable that either contains a "1" if the variable is present in the factor column or "0" otherwise.
Here is an example of what I mean.
labels <- c("1", "2", "3", "4", "5", "6", "7")
#create data frame (note, not all factor levels have to be in the columns,
#NA values are possible)
input <- data.frame(ID = c(1, 2, 3),
Cat1 = factor(c( 4, 1, 1), levels = labels),
Cat2 = factor(c(2, NA, 4), levels = labels),
Cat3 = factor(c(7, NA, NA), levels = labels))
#the seven factor levels now are the variables of the data.frame
desired_output <- data.frame(ID = c(1, 2, 3),
Dummy1 = c(0, 1, 1),
Dummy2 = c(1, 0, 0),
Dummy3 = c(0, 0, 0),
Dummy4 = c(1, 0, 1),
Dummy5 = c(0, 0, 0),
Dummy6 = c(0, 0, 0),
Dummy7 = c(1, 0, 0))
input
ID Cat1 Cat2 Cat3
1 4 2 7
2 1 <NA> <NA>
3 1 4 <NA>
desired_output
ID Dummy1 Dummy2 Dummy3 Dummy4 Dummy5 Dummy6 Dummy7
1 0 1 0 1 0 0 1
2 1 0 0 0 0 0 0
3 1 0 0 1 0 0 0
My actual data.frame has over 3000 rows and factors with more than 100 levels.
I hope you can help me converting the input to the desired output.
Greetings
sush

Answer 1

A couple of methods, that riff off of Gregor's and Aaron's answers.
From Aaron's. factorsAsStrings=FALSE keeps the factor variables hence all labes when using dcast
library(reshape2)
dcast(melt(input, id="ID", factorsAsStrings=FALSE), ID ~ value, drop=FALSE)
ID 1 2 3 4 5 6 7 NA
1 1 0 1 0 1 0 0 1 0
2 2 1 0 0 0 0 0 0 2
3 3 1 0 0 1 0 0 0 1
Then you just need to remove the last column.
From Gregor's
na.replace <- function(x) replace(x, is.na(x), 0)
options(na.action='na.pass') # this keeps the NA's which are then converted to zero
Reduce("+", lapply(input[-1], function(x) na.replace(model.matrix(~ 0 + x))))
x1 x2 x3 x4 x5 x6 x7
1 0 1 0 1 0 0 1
2 1 0 0 0 0 0 0
3 1 0 0 1 0 0 0
Then you just need to cbind the ID column

Answer 2

One way to do this is with matrix indexing. You have data specifying which locations in your output matrix should be 1 (the rest should be zero), so we'll make a matrix of zeros and then fill in the 1's based on your data. To do that, your data needs to be in a two column matrix, with the first column being the row (ID) of the output and the second column being the columns.
Put input data in long format, remove missings, convert values to integers matching the labels, then make a matrix as needed.
in2 <- reshape2::melt(input, id.vars="ID")
in2 <- subset(in2, !is.na(value))
in2$value <- match(in2$value, labels)
in2$variable <- NULL
in2 <- as.matrix(in2)
Then make the new output matrix with all zeros, and fill in the ones using that matrix.
out <- matrix(0, nrow=nrow(input), ncol=length(labels))
colnames(out) <- labels
rownames(out) <- input$ID
out[in2] <- 1
out
## 1 2 3 4 5 6 7
## 1 0 1 0 1 0 0 1
## 2 1 0 0 0 0 0 0
## 3 1 0 0 1 0 0 0

Answer 3

Here's a way using model.matrix. We convert the missing values to 0s, and specify 0 as the reference level for the factor contrasts. Then we just add the individual model matrices together and stick on the IDs:
new_lab = as.character(0:7)
for (i in 2:4) {
temp = as.character(input[[i]])
temp[is.na(temp)] = "0"
input[[i]] = factor(temp, levels = new_lab)
}
mm =
model.matrix(~ Cat1, data = input) +
model.matrix(~ Cat2, data = input) +
model.matrix(~ Cat3, data = input)
mm[, 1] = input$ID
colnames(mm) = c("ID", paste0("Dummy", 1:(ncol(mm) - 1)))
mm
# ID Dummy1 Dummy2 Dummy3 Dummy4 Dummy5 Dummy6 Dummy7
# 1 1 0 1 0 1 0 0 1
# 2 2 1 0 0 0 0 0 0
# 3 3 1 0 0 1 0 0 0
# attr(,"assign")
# [1] 0 1 1 1 1 1 1 1
# attr(,"contrasts")
# attr(,"contrasts")$Cat1
# [1] "contr.treatment"
You can leave the result as a model matrix, change it back to a data frame, or whatever else.

Answer 4

This should work on your data frame. I converted the values to numeric before running the ifelse statement. Hope it works:
# Make dummy df
Cat1 = factor(c( 4, 1, 1))
Cat2 = factor(c(2, NA, 4))
Cat3 = factor(c(7, NA, NA))
df <- data.frame(Cat1,Cat2,Cat3)
# Specify columns
cols <- c(1:length(df))
# Convert Values To Numeric
df[,cols] %<>% lapply(function(x) as.numeric(as.character(x)))
# Perform ifelse. If its NA print 0, else print 1
df[,cols] %<>% lapply(function(x) ifelse(x == is.na(x) | (x) %in% NA, 0, 1))
Based on input:
Cat1 Cat2 Cat3
1 4 2 7
2 1 <NA> <NA>
3 1 4 <NA>
Output looks like this:
Cat1 Cat2 Cat3
1 1 1 1
2 1 0 0
3 1 1 0