使用data.table计算新列

Question

I have the following data

set.seed(5)
dt <- data.table(ID=letters, x = rnorm(26), y = rnorm(26), z = c(rep(15, 13), rep(20,13)))

return,

    ID           x           y  z
 1:  a -0.84085548  1.41858907 15
 2:  b  1.38435934  1.49877383 15
 3:  c -1.25549186 -0.65708209 15
 4:  d  0.07014277 -0.85279544 15
 5:  e  1.71144087  0.31591504 15
 6:  f -0.60290798  1.10969417 15
 7:  g -0.47216639  2.21546057 15
 8:  h -0.63537131  1.21710364 15
 9:  i -0.28577363  1.47922179 15
10:  j  0.13810822  0.95157383 15
11:  k  1.22763034 -1.00953265 15
12:  l -0.80177945 -2.00047274 15
13:  m -1.08039260 -1.76218587 15
14:  n -0.15753436 -0.14260813 20
15:  o -1.07176004  1.55006037 20
16:  p -0.13898614 -0.80242318 20
17:  q -0.59731309 -0.07457892 20
18:  r -2.18396676  1.89566795 20
19:  s  0.24081726 -0.45656894 20
20:  t -0.25935541  0.56222336 20
21:  u  0.90051195 -0.88700851 20
22:  v  0.94186939 -0.46024458 20
23:  w  1.46796190 -0.72432849 20
24:  x  0.70676109 -0.06921116 20
25:  y  0.81900893  1.46324856 20
26:  z -0.29348185  0.18772610 20

I am trying to update columns x and y by dividing both with z , at the same time, keep the column ID . That is, the final output should contain columns ID , x/z , and y/z

I have try the following code, but it returns me the error

dt[,c('x', 'y'):=lapply(.SD, function(x) x/z), .SDcols = names(dt)]

FYI, there are over 100 columns in the actual data that have to be divided by column z .

Could you please give me suggestions?

Answer 1

Update: Issue #495 is solved now with this recent commit , we can now do this just fine:

require(data.table) # v1.9.7+
nam <- setdiff(names(dt), c("ID", "z"))    
dt[, (nam) := lapply(.SD, `/`, z), .SDcols = nam]

---

nam <- setdiff(names(dt), c("ID", "z"))    
dt[, (nam) := lapply(.SD, `/`, dt[,z]), .SDcols = nam]

Note that i use dt[, z] inside lapply due this data.table bug #495 .
If you use .SDcols you can not use other columns within your function calls.

As a workaround, until #495 is done, you can use mget() as follows:

dt[, (nam) := lapply(mget(nam), `/`, z)]

Answer 2

How about

dt[, `:=`(x=x/z, y=y/z, z=NULL)]

EDIT: After the addition to the original question that there are more than the two columns in the data table I'd go with Floo0's answer

Answer 3

Doesn't this work?

dt$x <- dt$x / dt$z
dt$y <- dt$y / dt$z

dt <- dt[ , seq(1, 3)]

EDIT: If you have many columns you have to divide with z, you can try this instead:

dt[, seq(2, 101)] <- sapply(dt[, seq(2, 101)], '/', dt$z)
dt <- dt[, seq(1, 101)] #replace with boundaries of your choosing