Efficient Collapse Dummy Variables

Question

What is an efficient way (any solution including non-base packages welcomed) to collapse dummy variables back into a factor.

   race.White race.Hispanic race         


        

Answer 1

Another idea:

ff = function(x)
{
    ans = integer(nrow(x))
    for(i in seq_along(x)) ans[as.logical(x[[i]])] = i
    names(x)[ans]
}                                    
sub("[^.]+\\.", "", ff(dat))
#[1] "White"    "Asian"    "White"    "Black"    "Asian"    "Hispanic" "White"    "White"    "White"    "Black"

And to compare with akrun's alternatives:

akrun1 = function(x) names(x)[max.col(x, "first")]
akrun2 = function(x) names(x)[(as.matrix(x) %*% seq_along(x))[, 1]]
akrun3 = function(x) names(x)[do.call(pmax, x * seq_along(x)[col(x)])]
akrunlike = function(x) names(x)[do.call(pmax, Map("*", x, seq_along(x)))]

DF = setNames(as.data.frame("[<-"(matrix(0L, 1e4, 1e3), 
                                  cbind(seq_len(1e4), sample(1e3, 1e4, TRUE)), 
                                  1L)), 
              paste("fac", 1:1e3, sep = ""))

identical(ff(DF), akrun1(DF))
#[1] TRUE
identical(ff(DF), akrun2(DF))
#[1] TRUE
identical(ff(DF), akrun3(DF))
#[1] TRUE
identical(ff(DF), akrunlike(DF))
#[1] TRUE
microbenchmark::microbenchmark(ff(DF), akrun1(DF), akrun2(DF), 
                               akrun3(DF), akrunlike(DF), 
                               as.matrix(DF), col(DF), times = 30)
#Unit: milliseconds
#          expr        min         lq     median         uq        max neval
#        ff(DF)   61.99124   64.56194   78.62267  102.18424  152.64891    30
#    akrun1(DF)  296.89042  314.28641  327.95059  353.46185  394.46013    30
#    akrun2(DF)  103.76105  114.01497  120.12191  129.86513  166.13266    30
#    akrun3(DF) 1141.46478 1163.96842 1178.92961 1203.83848 1231.70346    30
# akrunlike(DF)  125.47542  130.20826  141.66123  157.92743  203.42331    30
# as.matrix(DF)   19.46940   20.54543   28.22377   35.69575   87.06001    30
#       col(DF)  103.61454  112.75450  116.00120  126.09138  176.97435    30

I included as.matrix() and col() just to show that "list"-y structures can be convenient on efficient looping as is. E.g., in contrast to a by-row looping, a way to use by-column looping doesn't need time to transform the structure of data.

Answer 2

We can use max.col to get the column index, subset the column names based on that and use sub to remove the prefix.

sub('[^.]+\\.', '', names(dat)[max.col(dat)])
#[1] "White"    "Asian"    "White"    "Black"    "Asian"    "Hispanic"
#[7] "White"    "White"    "White"    "Black"  

Here, I assumed that there is a single 1 per each row. If there are multiple 1s, we can use the option ties.method='first' or ties.method='last'.

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Or another option is doing the %*% with the sequence of columns, subset the column names, and remove the prefix with sub.

 sub('[^.]+\\.', '', names(dat)[(as.matrix(dat) %*%seq_along(dat))[,1]])

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Or we can use pmax

sub('[^.]+\\.', '', names(dat)[do.call(pmax,dat*seq_along(dat)[col(dat)])])