Renaming levels in multiple factors in a dataframe - r

Question

I have multiple factors in a dataframe that each contain levels with names "Very long text 1" and "Very long text 2". I want to rename the levels to "1" and "2". I can easily do it for each individual factor with
levels(df$factorname1)
[levels(df$factorname1)=="Very long text 1"] <- "1"
But it's cumbersome to repeat it for a few hundred factors. Is there a way to rename the level for a range of factors or simply locate "Very long text 1" anywhere in the dataframe and rename it?
Example data:
structure(list(Q5.2.01 = structure(c(1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("",
"A whole different level\n"), class = "factor"), Q5.2.02 = structure(c(2L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L,
1L), .Label = c("", "Very long text 2\n"), class = "factor"),
Q5.2.03 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA), Q5.2.04 = structure(c(1L, 2L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L
), .Label = c("", "Very long text 2\n"), class = "factor"),
Q5.2.05 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 1L), .Label = c("", "A whole different level\n",
"Very long text 2 blablabla\n"), class = "factor"), Q5.2.06 = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L,
1L, 1L), .Label = c("", "Very long text 2\n"), class = "factor"),
Q5.2.07 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA), Q5.2.08 = c(NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA), Q5.2.09 = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L,
1L, 1L), .Label = c("", "Very long text 1\n"), class = "factor"),
Q5.2.10 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA), Q5.2.11 = structure(c(1L, 1L, 1L,
1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L
), .Label = c("", "Very long text 2\n"), class = "factor"),
Q5.2.12 = structure(c(1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("", "Very long text 2\n"
), class = "factor"), Q5.2.13 = structure(c(1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("",
"Very long text 1\n"), class = "factor"), Q5.2.14 = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 3L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L,
1L, 1L), .Label = c("", "Very long text 1\n", "Very long text 2\n"
), class = "factor"), Q5.2.15 = structure(c(1L, 1L, 1L, 2L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("",
"Very long text 2\n"), class = "factor"), Q5.2.16 = structure(c(1L,
1L, 3L, 1L, 1L, 3L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L,
1L, 2L), .Label = c("", "Very long text 1\n", "Very long text 2\n"
), class = "factor"), respondentID = structure(c(8L, 8L,
8L, 6L, 7L, 7L, 5L, 5L, 5L, 4L, 4L, 4L, 3L, 3L, 3L, 2L, 2L,
1L), .Label = c("EO13", "EO15", "EO17", "EO19", "EO21", "Eo23",
"EO23", "EO24"), class = "factor")), .Names = c("Q5.2.01",
"Q5.2.02", "Q5.2.03", "Q5.2.04", "Q5.2.05", "Q5.2.06", "Q5.2.07",
"Q5.2.08", "Q5.2.09", "Q5.2.10", "Q5.2.11", "Q5.2.12", "Q5.2.13", `"Q5.2.14", "Q5.2.15", "Q5.2.16", "respondentID"), class = "data.frame",` row.names = c(NA,
-18L))

Answer 1

You can use the revalue function from the package plyr. (It can also be done with base R, but I like this solution).
Here's an example
> DF <- data.frame(V1 = factor(c("A", "B", "C", "A", "D", "E")),
V2=factor(c("A", "A", "A", "A", "D", "E")))
> DF
V1 V2
1 A A
2 B A
3 C A
4 A A
5 D D
6 E E
Now let's assume that the factor level D is the one we wish to replace. Then we can use lapply to iterate over columns in the data frame (remember to only select the relevant factors), and revalue to specify the replacement(s). Wrap everything in as.data.frame to convert back to a data frame.
> library("plyr")
> as.data.frame(lapply(DF, function(x) { revalue(x, c("D"="YAY")) }))
V1 V2
1 A A
2 B A
3 C A
4 A A
5 YAY YAY
6 E E
Update
You can restrict attention to factors by adding a subset selection to the data frame
as.data.frame(lapply(DF[,sapply(DF, class) == "factor"], function(x) { revalue(x, c("D"="YAY")) }))