Converting Character Response to "N" over a dataset - r

Question

To start off, and example Dataset :
x <- data.frame(v1=1:5,v2=1:5,v3=1:5,
v4=c("Bob","Green","Curley","Banana","No"),
v5=c("Hello","This question is awful, Mad",NA,"Help","Me"))
I've got a large dataset with a multitude of numeric and character variables (survey data). These responses vary greatly in content and length; the order these variables are in matter, as well. I'm trying to find a way to select all of the character variables in my dataset, and then set any responses to the letter "N"/"Another item" (while leaving the NA values intact).
With the help of other users in the community, I'm able to fill all of these character variables with NA or "N", etc. :
x[,sapply(x, is.character)] <- "N"
But, I would really like to be able to retain those NA values present within the data - Something like this (I'm not very proficient with the apply functions just yet) :
x[ #Contains ANY Text# ,sapply(x, is.character)] <- "NA"
I haven't found anything that will allow me find any and all text within a row/column? It appears something like GREP only works with specific character strings to my knowledge. I'm also unsure of my formatting with the aforementioned function is correct, so please let me know if I'm making an error in placing my #Contains ANY text# argument.
Thanks in advance All!

Answer 1

A data.frame is a list so its columns can be changed using lapply.
Here we can subset x to the character columns, and then lapply over them replacing non-NA values with whatever we want.
x <- data.frame(v1=1:5,v2=1:5,v3=1:5,
v4=c("Bob","Green","Curley","Banana","No"),
v5=c("Hello","This question is awful, Mad",NA,"Help","Me"),
stringsAsFactors = FALSE) # your original data.frame had factors
x
# v1 v2 v3 v4 v5
# 1 1 1 1 Bob Hello
# 2 2 2 2 Green This question is awful, Mad
# 3 3 3 3 Curley <NA>
# 4 4 4 4 Banana Help
# 5 5 5 5 No Me
is_char_col <- sapply(x, is.character)
is_char_col
# v1 v2 v3 v4 v5
# FALSE FALSE FALSE TRUE TRUE
Use replace:
x[is_char_col] <- lapply(x[is_char_col], function(k) replace(k, !is.na(k), "N"))
x
# v1 v2 v3 v4 v5
# 1 1 1 1 N N
# 2 2 2 2 N N
# 3 3 3 3 N <NA>
# 4 4 4 4 N N
# 5 5 5 5 N N
If the replacement logic is actually more complicated, you could modify the anonymous function inside lapply.

Answer 2

Here is a method using a generic function as mentioned by #effel.
x <- data.frame(v1=1:5,v2=1:5,v3=1:5,
v4=c("Bob","Green","Curley","Banana","No"),
v5=c("Hello","This question is awful, Mad",NA,"Help","Me"),
stringsAsFactors = FALSE)
x <- data.frame(lapply(x, function(i) if(is.character(i)) ifelse(!is.na(i), "N", i) else i))