Replace a value in a data frame based on a conditional (`if`) statement

In the R data frame coded for below, I would like to replace all of the times that B appears with b.

junk <- data.frame(x <- rep(LETTERS[1:4], 3), y <- letters[1:12])
colnames(junk) <- c("nm", "val")

this provides:

   nm val
1   A   a
2   B   b
3   C   c
4   D   d
5   A   e
6   B   f
7   C   g
8   D   h
9   A   i
10  B   j
11  C   k
12  D   l

My initial attempt was to use a for and if statements like so:

for(i in junk$nm) if(i %in% "B") junk$nm <- "b"

but as I am sure you can see, this replaces ALL of the values of junk$nm with b. I can see why this is doing this but I can't seem to get it to replace only those cases of junk$nm where the original value was B.

NOTE: I managed to solve the problem with gsub but in the interest of learning R I still would like to know how to get my original approach to work (if it is possible)

Easier to convert nm to characters and then make the change:

junk$nm <- as.character(junk$nm)
junk$nm[junk$nm == "B"] <- "b"

EDIT: And if indeed you need to maintain nm as factors, add this in the end:

junk$nm <- as.factor(junk$nm)

another useful way to replace values

library(plyr)
junk$nm <- revalue(junk$nm, c("B"="b"))

Short answer is:

junk$nm[junk$nm %in% "B"] <- "b"

Take a look at Index vectors in R Introduction (if you don't read it yet).

---

EDIT. As noticed in comments this solution works for character vectors so fail on your data.

For factor best way is to change level:

levels(junk$nm)[levels(junk$nm)=="B"] <- "b"

As the data you show are factors, it complicates things a little bit. @diliop's Answer approaches the problem by converting to nm to a character variable. To get back to the original factors a further step is required.

An alternative is to manipulate the levels of the factor in place.

> lev <- with(junk, levels(nm))
> lev[lev == "B"] <- "b"
> junk2 <- within(junk, levels(nm) <- lev)
> junk2
   nm val
1   A   a
2   b   b
3   C   c
4   D   d
5   A   e
6   b   f
7   C   g
8   D   h
9   A   i
10  b   j
11  C   k
12  D   l

That is quite simple and I often forget that there is a replacement function for levels().

Edit: As noted by @Seth in the comments, this can be done in a one-liner, without loss of clarity:

within(junk, levels(nm)[levels(nm) == "B"] <- "b")

The easiest way to do this in one command is to use which command and also need not to change the factors into character by doing this:

junk$nm[which(junk$nm=="B")]<-"b"

You have created a factor variable in nm so you either need to avoid doing so or add an additional level to the factor attributes. You should also avoid using <- in the arguments to data.frame()

Option 1:

junk <- data.frame(x = rep(LETTERS[1:4], 3), y =letters[1:12], stringsAsFactors=FALSE)
junk$nm[junk$nm == "B"] <- "b"

Option 2:

levels(junk$nm) <- c(levels(junk$nm), "b")
junk$nm[junk$nm == "B"] <- "b"
junk

You can use ifelse too, which is very simple to understand

junk$val <- ifelse(junk$nm == "B", "b", junk$val)

If you still want to do it through for loop the correct way of doing it

for(i in 1:nrow(junk)){
  if(junk[i, "nm"] == "B"){
    junk[i, "val"] <- "b"
  }
}

junk
> junk
   nm val
1   A   a
2   B   b
3   C   c
4   D   d
5   A   e
6   B   b
7   C   g
8   D   h
9   A   i
10  B   b
11  C   k
12  D   l

If you are working with character variables (note that stringsAsFactors is false here) you can use replace:

junk <- data.frame(x <- rep(LETTERS[1:4], 3), y <- letters[1:12], stringsAsFactors = FALSE)
colnames(junk) <- c("nm", "val")

junk$nm <- replace(junk$nm, junk$nm == "B", "b")
junk
#    nm val
# 1   A   a
# 2   b   b
# 3   C   c
# 4   D   d
# ...

stata.replace<-function(data,replacevar,replacevalue,ifs) {
  ifs=parse(text=ifs)
  yy=as.numeric(eval(ifs,data,parent.frame()))
  x=sum(yy)
  data=cbind(data,yy)
  data[yy==1,replacevar]=replacevalue
  message=noquote(paste0(x, " replacement are made"))
  print(message)
  return(data[,1:(ncol(data)-1)])
}

Call this function using below line.

d=stata.replace(d,"under20",1,"age<20")

I got the same issue, you can also do the same thing for each columns,

 fix_junk <- function(x){
      #x <- as.character(x)
      x[x == "B"] <- "b"
      x
    }
    junk[] <- lapply(junk, fix_junk); junk # junk[] to get a data frame rather than a list
    junk[1:3] <- lapply(junk[1:3], fix_junk); junk