Andrew -
   I believe

character.data$z = ifelse(is.na(character.data$x),
                          character.data$y,character.data$x)

should do what you want.
                                        - Phil Spector
                                         Statistical Computing Facility
                                         Department of Statistics
                                         UC Berkeley
                                         spec...@stat.berkeley.edu


On Thu, 24 Feb 2011, Andrew Anglemyer wrote:

Thanks!  Unfortunately, in my effort to simply the question, I didn't really
adequately describe the problem.  This solution is perfect in the numeric
case I presented, but what about in the case of character classes!  Let me
try again:

I have
character.data
          id     x        y
      1   1    "a"    NA
      2   2    "b"     "b"
      3   3    "c"      "c"
      4   4   NA      "d"


And I want first
new.character.data
          id     x        y        z
      1   1    "a"    NA      "a"
      2   2    "b"     "b"     "b"
      3   3    "c"      "c"    "c"
      4   4   NA      "d"     "d"

Thanks again!




On Thu, Feb 24, 2011 at 4:27 PM, Ista Zahn <iz...@psych.rochester.edu>wrote:

I think the easiest way is probably

data$z <- rowMeans(data[, c("x", "y")], na.rm=TRUE)

Best,
Ista

On Fri, Feb 25, 2011 at 12:12 AM, Andrew Anglemyer
<andrew.anglem...@gmail.com> wrote:
I am trying to combine two columns in a data frame into one column.  Some
values in either column are missing, but not in the same row for the two
different columns.  Additionally, when both columns in a row contain
data,
the data are identical.  I want a new column with the identical data or
the
data from the column with observed data.  For example:

I have
data
          id   x    y
      1   a   1   NA
      2   b   2    2
      3   c   3    3
      4   d  NA  4

And I want
new.data
          id   x    y     z
      1   a   1   NA   1
      2   b   2    2     2
      3   c   3    3     3
      4   d  NA  4     4

I've looked through the help and there are column combining solutions,
but
they don't seem to work well for this solution.
Thanks for any help!
Andy

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--
Ista Zahn
Graduate student
University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.org


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