[R] replace values in data frame

[Marc Schwartz (via MN)]

On Fri, 2006-07-07 at 11:20 -0400, Wade Wall wrote:
> Hi all,
> 
> I have a three columned list that I have imported into R.  The first column
> is a plot (ex. Plot1), the second is a species name (ex ACERRUB) and the
> third a numeric value.  I want to replace some of the second column names
> with other names (for example replace ACERRUB with ACERDRU).  The original
> and replacement values are in separate lists (not vectors), but I can't seem
> to find the right function to perform this.  The replace function seems to
> only want to work with numbers.
> 
> Any clues?
> 
> Wade

Without seeing the code you are using, we can only guess a syntax error
of some sort.  It works fine using the iris dataset:

> head(iris)
   Sepal.Length Sepal.Width Petal.Length Petal.Width Species
1           5.1         3.5          1.4         0.2  setosa
2           4.9         3.0          1.4         0.2  setosa
3           4.7         3.2          1.3         0.2  setosa
4           4.6         3.1          1.5         0.2  setosa
5           5.0         3.6          1.4         0.2  setosa
6           5.4         3.9          1.7         0.4  setosa
7           4.6         3.4          1.4         0.3  setosa
8           5.0         3.4          1.5         0.2  setosa
9           4.4         2.9          1.4         0.2  setosa
10          4.9         3.1          1.5         0.1  setosa

> iris$Species <- replace(iris$Species, iris$Species == "setosa",
                          "NewValue")

> head(iris)
   Sepal.Length Sepal.Width Petal.Length Petal.Width  Species
1           5.1         3.5          1.4         0.2 NewValue
2           4.9         3.0          1.4         0.2 NewValue
3           4.7         3.2          1.3         0.2 NewValue
4           4.6         3.1          1.5         0.2 NewValue
5           5.0         3.6          1.4         0.2 NewValue
6           5.4         3.9          1.7         0.4 NewValue
7           4.6         3.4          1.4         0.3 NewValue
8           5.0         3.4          1.5         0.2 NewValue
9           4.4         2.9          1.4         0.2 NewValue
10          4.9         3.1          1.5         0.1 NewValue


HTH,

Marc Schwartz