Dictionary style replace multiple items

If you're open to using packages, plyr is a very popular one and has this handy mapvalues() function that will do just what you're looking for:

foo <- mapvalues(foo, from=c("AA", "AC", "AG"), to=c("0101", "0102", "0103"))

Note that it works for data types of all kinds, not just strings.

map = setNames(c("0101", "0102", "0103"), c("AA", "AC", "AG"))
foo[] <- map[unlist(foo)]

assuming that map covers all the cases in foo. This would feel less like a 'hack' and be more efficient in both space and time if foo were a matrix (of character()), then

matrix(map[foo], nrow=nrow(foo), dimnames=dimnames(foo))

Both matrix and data frame variants run afoul of R's 2^31-1 limit on vector size when there are millions of SNPs and thousands of samples.

Here is a quick solution

dict = list(AA = '0101', AC = '0102', AG = '0103')
foo2 = foo
for (i in 1:3){foo2 <- replace(foo2, foo2 == names(dict[i]), dict[i])}