Dictionary style replace multiple items

Question

I have a large data.frame of character data that I want to convert based on what is commonly called a dictionary in other languages.

Currently I am going about it like so:

foo <- data.frame(snp1 = c("AA", "AG", "AA", "AA"), snp2 = c("AA", "AT", "AG", "AA"), snp3 = c(NA, "GG", "GG", "GC"), stringsAsFactors=FALSE)
foo <- replace(foo, foo == "AA", "0101")
foo <- replace(foo, foo == "AC", "0102")
foo <- replace(foo, foo == "AG", "0103")

This works fine, but it is obviously not pretty and seems silly to repeat the replace statement each time I want to replace one item in the data.frame.

Is there a better way to do this since I have a dictionary of approximately 25 key/value pairs?

Answer 1

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])}

Answer 2

Note this answer started as an attempt to solve the much simpler problem posted in How to replace all values in data frame with a vector of values?. Unfortunately, this question was closed as duplicate of the actual question. So, I'll try to suggest a solution based on replacing factor levels for both cases, here.

---

In case there is only a vector (or one data frame column) whose values need to be replaced and there are no objections to use factor we can coerce the vector to factor and change the factor levels as required:

x <- c(1, 1, 4, 4, 5, 5, 1, 1, 2)
x <- factor(x)
x
#[1] 1 1 4 4 5 5 1 1 2
#Levels: 1 2 4 5
replacement_vec <- c("A", "T", "C", "G")
levels(x) <- replacement_vec
x
#[1] A A C C G G A A T
#Levels: A T C G

Using the forcatspackage this can be done in a one-liner:

x <- c(1, 1, 4, 4, 5, 5, 1, 1, 2)
forcats::lvls_revalue(factor(x), replacement_vec)
#[1] A A C C G G A A T
#Levels: A T C G

---

In case all values of multiple columns of a data frame need to be replaced, the approach can be extended.

foo <- data.frame(snp1 = c("AA", "AG", "AA", "AA"), 
                  snp2 = c("AA", "AT", "AG", "AA"), 
                  snp3 = c(NA, "GG", "GG", "GC"), 
                  stringsAsFactors=FALSE)

level_vec <- c("AA", "AC", "AG", "AT", "GC", "GG")
replacement_vec <- c("0101", "0102", "0103", "0104", "0302", "0303")
foo[] <- lapply(foo, function(x) forcats::lvls_revalue(factor(x, levels = level_vec), 
                                                       replacement_vec))
foo
#  snp1 snp2 snp3
#1 0101 0101 <NA>
#2 0103 0104 0303
#3 0101 0103 0303
#4 0101 0101 0302

Note that level_vec and replacement_vec must have equal lengths.

More importantly, level_vec should be complete , i.e., include all possible values in the affected columns of the original data frame. (Use unique(sort(unlist(foo))) to verify). Otherwise, any missing values will be coerced to <NA>. Note that this is also a requirement for Martin Morgans's answer.

So, if there are only a few different values to be replaced you will be probably better off with one of the other answers, e.g., Ramnath's.