How to apply a function over two series of sequentially labelled variables without using column numbers?

Question

I need to apply a function using two sets of sequentially labelled variables and attach the new set of variables to the data frame. I need to do this without referring to the column numbers in the code.

More specifically, here is the simple task I am trying to do:

dat <- data.frame(sec1 = sample(c(0:3),10,replace=T) , sec2 = sample(c(0:4),replace=T) , sec3 = sample(c(0:4),replace=T),pri1 = sample(c(0:3),10,replace=T) , pri2 = sample(c(0:4),replace=T) , pri3 = sample(c(0:4),replace=T) )
dat$rel1 <- ifelse(dat$pri1>0,dat$sec1/dat$pri1,NA)
dat

I want to repeat the "ifelse" function shown above without typing it repeatedly for each set of variables.

I must say, I asked similar questions and received helpful answers ( eg1 and eg2 ) previously but in those case the responses either used the column number in the code, or the example was on a single set of sequentially labelled variable. I could not manage to revise the suggested code to solve this particular problem.

Any suggestion is very much appreciated.

Answer 1

dat_n <- cbind(dat, mapply(function(x, y) ifelse(y>0,x/y,NA) ,dat[grepl("sec",names(dat))], dat[grepl("pri",names(dat))]))
> dat_n
   sec1 sec2 sec3 pri1 pri2 pri3      rel1      sec1      sec2 sec3
1     2    1    2    3    3    0 0.6666667 0.6666667 0.3333333   NA
2     3    3    4    0    2    4        NA        NA 1.5000000 1.00
3     1    0    3    1    4    4 1.0000000 1.0000000 0.0000000 0.75
4     2    4    1    3    3    2 0.6666667 0.6666667 1.3333333 0.50
5     2    0    2    3    4    1 0.6666667 0.6666667 0.0000000 2.00
6     1    1    2    1    3    0 1.0000000 1.0000000 0.3333333   NA
7     1    3    4    0    2    4        NA        NA 1.5000000 1.00
8     1    0    3    1    4    4 1.0000000 1.0000000 0.0000000 0.75
9     3    4    1    2    3    2 1.5000000 1.5000000 1.3333333 0.50
10    1    0    2    2    4    1 0.5000000 0.5000000 0.0000000 2.00

Answer 2

You could use Vectorize on an ifelse and clean this up a lot

set.seed(1)
dat <- data.frame(sec1 = sample(c(0:3),10,replace=T) , sec2 = sample(c(0:4),replace=T) , sec3 = sample(c(0:4),replace=T),pri1 = sample(c(0:3),10,replace=T) , pri2 = sample(c(0:4),replace=T) , pri3 = sample(c(0:4),replace=T) )
dat$rel1 <- ifelse(dat$pri1>0,dat$sec1/dat$pri1,NA)
dat

f <- Vectorize(function(x, y) ifelse(y > 0, x / y, NA))


f(dat[1:3], dat[4:6])

#           sec1 sec2      sec3
# [1,]  0.3333333 0.50 0.6666667
# [2,]         NA 0.00 1.0000000
# [3,]  1.0000000 1.50        NA
# [4,]         NA   NA 0.3333333
# [5,]  0.0000000 0.75 1.5000000
# [6,]  3.0000000 0.50 0.6666667
# [7,]         NA 0.00 1.0000000
# [8,]  2.0000000 1.50        NA
# [9,]  0.6666667   NA 0.3333333
# [10,] 0.0000000 0.75 1.5000000

v <- lapply(c('sec','pri'), function(x) grep(x, names(dat)))

cbind(dat, `colnames<-`(f(dat[v[[1]]], dat[v[[2]]]), paste0('rel',1:3)))

#    sec1 sec2 sec3 pri1 pri2 pri3      rel1      rel1 rel2      rel3
# 1     1    1    2    3    2    3 0.3333333 0.3333333 0.50 0.6666667
# 2     1    0    3    0    2    3        NA        NA 0.00 1.0000000
# 3     2    3    4    2    2    0 1.0000000 1.0000000 1.50        NA
# 4     3    1    1    0    0    3        NA        NA   NA 0.3333333
# 5     0    3    3    1    4    2 0.0000000 0.0000000 0.75 1.5000000
# 6     3    1    2    1    2    3 3.0000000 3.0000000 0.50 0.6666667
# 7     3    0    3    0    2    3        NA        NA 0.00 1.0000000
# 8     2    3    4    1    2    0 2.0000000 2.0000000 1.50        NA
# 9     2    1    1    3    0    3 0.6666667 0.6666667   NA 0.3333333
# 10    0    3    3    1    4    2 0.0000000 0.0000000 0.75 1.5000000

(yo dawg I heard you like to vectorize so we put some Vectorize in your vectorized so you can Vectorize while you vectorize)