Creating Dummy Variables from List

问题

So I'm trying to create dummy variables to attach to a data frame based on whether or not a specific column of the frame has specific words in it. The column would look something like this:

 dumcol = c("good night moon", "good night room", "good morning room", "hello moon")

and I'd be creating dummy variables based on which words are contained in each row, e.g. for the first one, it contains "good", "night", and "moon", but not "room", "morning" or "hello".

The way I've been going about it so far, in a hugely primitive way, is creating a 0-valued matrix of appropriate size, and then using a for loop like so:

result=matrix(ncol=6,nrow=4)
wordlist=unique(unlist(strsplit(dumcal, " ")))
for (i in 1:6)
{ result[grep(wordlist[i], dumcol),i] = 1 }

or something similar. I'm guessing there's a faster/more resource efficient way to do it. Any advice?

回答1:

You could try:

library(tm)
myCorpus <- Corpus(VectorSource(dumcol))
myTDM <- TermDocumentMatrix(myCorpus, control = list(minWordLength = 1))
as.matrix(myTDM)

Which gives:

#         Docs
#Terms     1 2 3 4
#  good    1 1 1 0
#  hello   0 0 0 1
#  moon    1 0 0 1
#  morning 0 0 1 0
#  night   1 1 0 0
#  room    0 1 1 0

If you want the dummy variables in columns, you could use DocumentTermMatrix instead:

#    Terms
#Docs good hello moon morning night room
#   1    1     0    1       0     1    0
#   2    1     0    0       0     1    1
#   3    1     0    0       1     0    1
#   4    0     1    1       0     0    0

回答2:

Try

 library(qdapTools)
 mtabulate(strsplit(dumcol, ' '))
 #    good hello moon morning night room
 #1    1     0    1       0     1    0
 #2    1     0    0       0     1    1
 #3    1     0    0       1     0    1
 #4    0     1    1       0     0    0

 library(splitstackshape)
 cSplit_e(as.data.frame(dumcol), 'dumcol', sep=' ', 
                      type='character', fill=0, drop=TRUE)
 #  dumcol_good dumcol_hello dumcol_moon dumcol_morning dumcol_night dumcol_room
 #1           1            0           1              0            1           0
 #2           1            0           0              0            1           1
 #3           1            0           0              1            0           1
 #4           0            1           1              0            0           0

回答3:

I would do

sdum <- strsplit(dumcol," ")
us   <- unique(unlist(sdum))
res  <- sapply(sdum,function(x)table(factor(x,levels=us)))
#         [,1] [,2] [,3] [,4]
# good       1    1    1    0
# night      1    1    0    0
# moon       1    0    0    1
# room       0    1    1    0
# morning    0    0    1    0
# hello      0    0    0    1

The result can be transposed with t(res) for the dummy variables in columns (the R convention).

来源：https://stackoverflow.com/questions/30513029/creating-dummy-variables-from-list

标签

dummy-data