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How do I reference the entire row when creating a new column in a data.table?

Tags:

r

data.table

I have a data.table with more than 200 variables which are all binary. I want to create a new column in it that counts the difference between each row and a reference vector:

#Example
dt = data.table(
"V1" = c(1,1,0,1,0,0,0,1,0,1,0,1,1,0,1,0),
"V2" = c(0,1,0,1,0,1,0,0,0,0,1,1,0,0,1,0),
"V3" = c(0,0,0,1,1,1,1,0,1,0,1,0,1,0,1,0),
"V4" = c(1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0),
"V5" = c(1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0)  
)

reference = c(1,1,0,1,0)

I can do that with a small for loop, such as

distance = NULL
for(i in 1:nrow(dt)){      
  distance[i] = sum(reference != dt[i,])  
}

But it's kind of slow and surely not the best way to do this. I tried:

dt[,"distance":= sum(reference != c(V1,V2,V3,V4,V5))]
dt[,"distance":= sum(reference != .SD)]

But neither works, as they return the same value for all rows. Also, a solution where I don't have to type all the variable names would be much better, as the real data.table has over 200 columns

like image 894
Fino Avatar asked Jan 24 '19 12:01

Fino


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2 Answers

You can use sweep() with rowSums, i.e.

rowSums(sweep(dt, 2, reference) != 0)
 #[1] 2 2 2 2 4 4 3 2 4 3 2 1 3 4 1 3

BENCHMARK

HUGH <- function(dt) {
    dt[, I := .I] 
    distance_by_I <- melt(dt, id.vars = "I")[, .(distance = sum(reference != value)), keyby = "I"]
    return(dt[distance_by_I, on = "I"])
}

Sotos <- function(dt) {
    return(rowSums(sweep(dt, 2, reference) != 0))
}

dt1 <- as.data.table(replicate(5, sample(c(0, 1), 100000, replace = TRUE)))
microbenchmark(HUGH(dt1), Sotos(dt1))

#Unit: milliseconds
#       expr       min        lq      mean   median        uq       max neval cld
#  HUGH(dt1) 112.71936 117.03380 124.05758 121.6537 128.09904 155.68470   100   b
# Sotos(dt1)  23.66799  31.11618  33.84753  32.8598  34.02818  68.75044   100  a 
like image 175
Sotos Avatar answered Sep 21 '22 00:09

Sotos


Another:

ref = as.list(reference)
dt[, Reduce(`+`, Map(`!=`, .SD, ref))]

How it works. So we're taking each vector column in .SD and comparing it to the single corresponding value in ref. The != function is vectorized, so each element of ref is recycled out to match the length of each vector.

This Map call returns a list of TRUE/FALSE vectors, one for each column. When we add up TRUE/FALSE values, they are treated as 1/0, so we just need to add these columns up. This can be achieved by passing the pairwise operator + between the first column and the second; and then again between the result of that computation and the third column; and so on. This is how Reduce works. It might be more readable as

x = dt[, Map(`!=`, .SD, ref)]
Reduce(`+`, x, init = 0L)

which can be read as

  • v = 0
  • for each xi in x, update v = v + xi

See also ?Map and ?Reduce.


Timings. I'm modifying the benchmark data, since using integers seems a lot saner if the OP really has 0-1 data. Also, adding more columns since the OP says they have a lot. Finally, editing Hugh's answer to be comparable to the others:

HUGH <- function(dt, r) {
  dt[, I := .I] 
  res <- melt(dt, id.vars = "I")[, .(distance = sum(r != value)), keyby = "I"]$distance
  dt[, I := NULL]
  res
}

Sotos <- function(dt, r) {
  return(rowSums(sweep(dt, 2, r) != 0))
}

mm <- function(dt, r){
  colSums(t(dt) != r)
}

ff <- function(DT, r){
  DT[, Reduce(`+`, Map(`!=`, .SD, r))]
}

nr = 20000
nc = 500
dt1 <- as.data.table(replicate(nc, sample(0:1, nr, replace = TRUE)))
ref <- rep(as.integer(reference), length.out=nc)
lref = as.list(ref)

identical(HUGH(dt1, ref), ff(dt1, lref)) # integer output
identical(mm(dt1, ref), Sotos(dt1, ref)) # numeric output
all.equal(HUGH(dt1, ref), mm(dt1, ref))  # but they match
# all TRUE

microbenchmark::microbenchmark(times = 3, 
 HUGH(dt1, ref), 
 Sotos(dt1, ref), 
 mm(dt1, ref), 
 ff(dt1, lref)
)

Result:

Unit: milliseconds
            expr      min        lq     mean   median         uq       max neval
  HUGH(dt1, ref) 365.0529 370.05233 378.8826 375.0517  385.79737  396.5430     3
 Sotos(dt1, ref) 871.5693 926.50462 961.5527 981.4400 1006.54437 1031.6488     3
    mm(dt1, ref) 104.5631 121.74086 131.7157 138.9186  145.29197  151.6653     3
   ff(dt1, lref)  87.0800  87.48975  93.1361  87.8995   96.16415  104.4288     3
like image 38
Frank Avatar answered Sep 22 '22 00:09

Frank