convert a string into value in R - r

Question

Take for example:
s <- c("1y1","2y1","1.5y2","1.8y2")
y1 and y2 means to multiply 2 and 1.8.
1y1 = 1*2=2
1.5y2 = 1.5*1.8 = 2.7
So, the result would be c(2,4,2.7,3.24).
How can I simply get it?
I can do it by c(1,2.5,3,2.7)*c(2,2,1.8,1.8), but it is not elegant, if the string is very long.

Answer 1

Using gsub to replace y1 and y2 by their respective numeric values then one can use the classical eval(parse(..)) to evaluate the resulted numeric expressions.
s <- c("1y1","2y1","1.5y2","1.8y2")
s <- gsub('y1','*2',s)
s <-gsub('y2','*1.8',s)
sapply(s,function(x)eval(parse(text=x)))
1*2 2*2 1.5*1.8 1.8*1.8
2.00 4.00 2.70 3.24
EDIT more elegant option using gsubfn, and its flexible function substitution version.
> library(gsubfn)
> sub.f <- function(x,y) as.numeric(x) * ifelse(y == 1, 2, 1. 8)
> as.numeric(gsubfn('(.*)y(.*)', sub.f, s))
[1] 2.00 4.00 2.70 3.24
we split, each expression into 2 components x and y and use them (once coerced to numeric) to compute manually the expression.
Alternately, this one liner:
> strapply(s, '(.*)y(.*)', ~ as.numeric(x) * ifelse(y == 1, 2, 1.8), simplify = TRUE)
[1] 2.00 4.00 2.70 3.24

Answer 2

You can try this:
s = sub(pattern="y", replacement="*y", x=s)
sub will replace all the ocurrences of y with *y, so the expression is syntactically correct to R. However, it's not really an expression but a character vector. Take notice sub will replace only the first ocurrence of y. If there are more than one "y" per string, you have to use gsub instead. For more information see ?sub.
y1 = 2
y2 = 1.8
s = parse(text=s) # parse will convert s to a R expression
s.num = sapply(s, eval) # and this will evaluate the expression to give the results
[1] 2.00 4.00 2.70 3.24

Answer 3

Here is another idea:
s <- c("1y1", "2y1", "1.5y2", "1.8y2")
s2 <- as.numeric(sub("y[0-9]", "", s))
mult <- rep(NA, len = length(s))
mult[grep("y1", s)] <- 2
mult[grep("y2", s)] <- 1.8
s2 * mult
#[1] 2.00 4.00 2.70 3.24
On a larger scale:
S = sample(s, 1e4, T)
f1 = function() {
S <- gsub('y1','*2',S)
S <-gsub('y2','*1.8',S)
sapply(S,function(x)eval(parse(text=x)))
}
f2 = function() {
S = sub(pattern="y", replacement="*y", x=S)
S = parse(text=S)
S.num = sapply(S, eval)
S.num
} ; y1 = 2 ; y2 = 1.8
f3 = function() {
s2 <- as.numeric(sub("y[0-9]", "", S))
mult <- rep(NA, len = length(S))
mult[grep("y1", S)] <- 2
mult[grep("y2", S)] <- 1.8
s2 * mult
}
library(microbenchmark)
microbenchmark(f1(), f2(), f3(), times = 10)
#Unit: milliseconds
# expr min lq median uq max neval
# f1() 1940.63876 2037.03500 2064.4605 2072.98875 2101.73358 10
# f2() 93.43298 98.69724 115.8716 125.07774 153.43385 10
# f3() 12.91724 13.01781 13.1805 13.30283 17.94973 10
sum(f1() != f2())
#[1] 0
sum(f2() != f3())
#[1] 0