I have a data.table with which I\'d like to perform the same operation on certain columns. The names of these columns are given in a character vector. In this particular ex
UPDATE: Following is a neat way to do it without for loop
dt[,(cols):= - dt[,..cols]]
It is a neat way for easy code readability. But as for performance it stays behind Frank's solution according to below microbenchmark result
mbm = microbenchmark(
base = for (col in 1:length(cols)) {
dt[ , eval(parse(text = paste0(cols[col], ":=-1*", cols[col])))]
},
franks_solution1 = dt[ , (cols) := lapply(.SD, "*", -1), .SDcols = cols],
franks_solution2 = for (j in cols) set(dt, j = j, value = -dt[[j]]),
hannes_solution = dt[, c(out_cols) := lapply(.SD, function(x){log(x = x, base = exp(1))}), .SDcols = cols],
orhans_solution = for (j in cols) dt[,(j):= -1 * dt[, ..j]],
orhans_solution2 = dt[,(cols):= - dt[,..cols]],
times=1000
)
mbm
Unit: microseconds
expr min lq mean median uq max neval
base_solution 3874.048 4184.4070 5205.8782 4452.5090 5127.586 69641.789 1000
franks_solution1 313.846 349.1285 448.4770 379.8970 447.384 5654.149 1000
franks_solution2 1500.306 1667.6910 2041.6134 1774.3580 1961.229 9723.070 1000
hannes_solution 326.154 405.5385 561.8263 495.1795 576.000 12432.400 1000
orhans_solution 3747.690 4008.8175 5029.8333 4299.4840 4933.739 35025.202 1000
orhans_solution2 752.000 831.5900 1061.6974 897.6405 1026.872 9913.018 1000
as shown in below chart
My Previous Answer: The following also works
for (j in cols)
dt[,(j):= -1 * dt[, ..j]]