How do I evaluate columns inside data.table with different conditions

data.table, r

Solution

The approach I came up with looks like the following:

dt[, c("a", "b", "c") := lapply(.SD, `==`, 3), 
   .SDcols = c("a", "b", "c")][, d := (d == 6)][]
#        a     b     c           x          y     d
# 1: FALSE FALSE FALSE -0.50219235  0.1169713 FALSE
# 2: FALSE FALSE FALSE  0.13153117  0.3186301  TRUE
# 3:  TRUE FALSE FALSE -0.07891709 -0.5817907  TRUE
# 4: FALSE  TRUE  TRUE  0.88678481  0.7145327 FALSE

It doesn't win any points in terms of readability, but seems to be OK in terms of performance.

Here's some sample data to test:

library(data.table)
set.seed(100)
Nrow = 3000000
dt <- data.table(a = sample(10, Nrow, TRUE), 
                 b = sample(10, Nrow, TRUE), 
                 c = sample(10, Nrow, TRUE), 
                 x = rnorm(Nrow), 
                 y = rnorm(Nrow),
                 d = sample(10, Nrow, TRUE)) 

... some functions to test...

fun1 <- function(indt) {
  indt[, c("a", "b", "c") := lapply(.SD, `==`, 3), 
     .SDcols = c("a", "b", "c")][, d := (d == 6)][]
}

fun2 <- function(indt) {
  for (i in c("a","b","c")) indt[, (i):=get(i)==3]
  for (i in c("d"))         indt[, (i):=get(i)==6]
  indt
}

fun3 <- function(indt) {
  f <- function(col,x) indt[,(col):=(.SD==x),.SDcols=col]
  lapply(list("a","b","c"), f, 3)
  lapply(list("d"), f, 6)
  indt
}

... and some timings...

microbenchmark(fun1(copy(dt)), fun2(copy(dt)), fun3(copy(dt)), times = 10)
# Unit: milliseconds
#            expr      min        lq    median        uq       max neval
#  fun1(copy(dt)) 518.6034  535.0848  550.3178  643.2968  695.5819    10
#  fun2(copy(dt)) 830.5808 1037.8790 1172.6684 1272.6236 1608.9753    10
#  fun3(copy(dt)) 922.6474 1029.8510 1097.7520 1145.1848 1340.2009    10

identical(fun1(copy(dt)), fun2(copy(dt)))
# [1] TRUE
identical(fun2(copy(dt)), fun3(copy(dt)))
# [1] TRUE

At this scale, I would go for whatever is most readable to you (unless those milliseconds really count), but if your data are larger, you might want to experiment a little more with the different options.

Addition from Matt

Agreed. To follow up comment, here's `fun4` but it's only a smidgen fastest on this size (3e6 rows, 90MB)

fun4 <- function(indt) {
  for (i in c("a","b","c")) set(indt,NULL,i,indt[[i]]==3)
  for (i in c("d"))         set(indt,NULL,i,indt[[i]]==6)
  indt
}

microbenchmark(copy(dt), fun1(copy(dt)), fun2(copy(dt)), fun3(copy(dt)), 
               fun4(copy(dt)), times = 10)
# Unit: milliseconds
#            expr        min         lq     median         uq       max neval
#        copy(dt)   64.13398   65.94222   68.32217   82.39942  110.3293    10
#  fun1(copy(dt))  601.84611  618.69288  690.47179  713.56760  766.1534    10
#  fun2(copy(dt))  887.99727  950.33821  978.98988 1071.31253 1180.1281    10
#  fun3(copy(dt)) 1566.90858 1574.30635 1603.55467 1673.38625 1771.4054    10
#  fun4(copy(dt))  566.43528  568.91103  575.06881  672.44021  692.9839    10

> identical(fun1(copy(dt)), fun4(copy(dt)))
[1] TRUE

Next I increased the data size by 10 times to 30 million rows, 915MB.

Note these timings are now in seconds, and on my slow netbook.

set.seed(100)
Nrow = 30000000
dt <- data.table(a = sample(10, Nrow, TRUE), 
              b = sample(10, Nrow, TRUE), 
              c = sample(10, Nrow, TRUE), 
              x = rnorm(Nrow), 
              y = rnorm(Nrow),
              d = sample(10, Nrow, TRUE)) 
object.size(dt)/1024^2
# 915 MB
microbenchmark(copy(dt),fun1(copy(dt)), fun2(copy(dt)), fun3(copy(dt)), 
                 fun4(copy(dt)), times = 3)
# Unit: seconds
#            expr       min        lq    median       uq      max neval
#        copy(dt)   8.04262  53.68556  99.32849 269.4414 439.5544     3
#  fun1(copy(dt)) 207.70646 260.16710 312.62775 317.8966 323.1654     3
#  fun2(copy(dt)) 421.78934 502.03503 582.28073 658.0680 733.8553     3
#  fun3(copy(dt)) 104.30914 187.49875 270.68836 384.7804 498.8724     3
#  fun4(copy(dt)) 158.17239 165.35898 172.54557 183.4851 194.4246     3

Here, `fun4` is on average fastest by quite a bit, I guess, due to the memory efficiency of a `for` loop one column at a time. In `fun1` and `fun3`, the RHS of `:=` is three columns wide before that's then assigned to the three target columns. Having said that, why is my previous `fun2` slowest then? It goes column by column after all. Maybe `get()` copies the column before going into `==`.

There was one run where `fun3` was fastest (104 vs 158). I'm not sure I trust `microbenchmark` on that. I seem to remember some criticism by Radford Neal of `microbenchmark`, but don't recall the outcome.

Those timings were on my really slow netbook :

$ lscpu
Architecture:          x86_64
CPU op-mode(s):        32-bit, 64-bit
Byte Order:            Little Endian
CPU(s):                2
On-line CPU(s) list:   0,1
Thread(s) per core:    1
Core(s) per socket:    2
Socket(s):             1
NUMA node(s):          1
Vendor ID:             AuthenticAMD
CPU family:            20
Model:                 2
Stepping:              0
CPU MHz:               800.000
BogoMIPS:              1995.06
Virtualisation:        AMD-V
L1d cache:             32K
L1i cache:             32K
L2 cache:              512K
NUMA node0 CPU(s):     0,1

> sessionInfo()
R version 3.1.0 (2014-04-10)
Platform: x86_64-pc-linux-gnu (64-bit)   

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] microbenchmark_1.3-0 data.table_1.9.2     bit64_0.9-3          bit_1.1-11

Problem

Given the data.table as follows: ``` library(data.table) set.seed(100) dt <- data.table(a=c(1:3, 1), b = c(1,0,1, 3), c = c(1,2,1,3), x = rnorm(4), y = rnorm(4), d = c(4, 6, 6, 7)) ``` `dt` return, ``` a b c x y d 1: 1 1 1 -0.50219235 0.1169713 4 2: 2 0 2 0.13153117 0.3186301 6 3: 3 1 1 -0.07891709 -0.5817907 6 4: 1 3 3 0.88678481 0.7145327 7 ``` Any numbers in column "a", "b", and "c" that are equal to 3 will be TRUE Also, any numbers in column "d" that are equal to 6 will be TRUE How do I evaluate inside `dt` by using column's name ("a", "b","c", and "d") so that my return would be: ``` a b c x y d 1: FALSE FALSE FALSE -0.50219235 0.1169713 FALSE 2: FALSE FALSE FALSE 0.13153117 0.3186301 TRUE 3: TRUE FALSE FALSE -0.07891709 -0.5817907 TRUE 4: FALSE TRUE TRUE 0.88678481 0.7145327 FALSE ``` Thank you

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