You failed to tell us any of: the actual code you used, the data you were
working with, or the actual error you got. Your "pseudo-code" looks okay to me,
so I suggest you read the Posting Guide and try again. (Anything other than a
complete reproducible example of your problem is unlikely to eli
At 07:41 PM 7/10/2012, you wrote:
Seems to work for me:
> x <- data.frame(old1 = sample(c(1,2,8), 10, TRUE), old2 = 1:10)
> x$new <- ifelse(x$old1 == 8, 1, x$old2)
> x
Thanks Jim and Dan. The problem ended up with a missing values issue
that threw me off. When your post confirmed that my co
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Jeff
> Sent: Tuesday, July 10, 2012 5:26 PM
> To: r-help
> Subject: [R] ifelse help
>
>
> I'm sure this is easy, but I'm new to R and can&
Seems to work for me:
> x <- data.frame(old1 = sample(c(1,2,8), 10, TRUE), old2 = 1:10)
> x$new <- ifelse(x$old1 == 8, 1, x$old2)
> x
old1 old2 new
1 11 1
2 22 2
3 23 3
4 84 1
5 15 5
6 86 1
7 87 1
8 28 8
9 2
I'm sure this is easy, but I'm new to R and can't find any example of
the following.
Here's what I'm trying to do in pseudo-code.
data$newvar <- ifelse(data$oldvar1 == 8, 1,data$oldvar2)
In other words, if the existing variable equals 8, then the new
variable should equal 1, otherwise the n
r-help-boun...@r-project.org napsal dne 19.01.2009 21:08:59:
> Sorry I didn't give the proper initialization of j. But you are right j
should
> also be an array of 5. So x[j + 5] would return 5 values.
>
> So if the array returned from 'ifelse' is the same dimention as test
(h), then
> are all
On Mon, Jan 19, 2009 at 9:08 PM, wrote:
> Sorry I didn't give the proper initialization of j. But you are right j
> should also be an array of 5. So x[j + 5] would return 5 values.
>
> So if the array returned from 'ifelse' is the same dimention as test (h),
> then are all the values of h being
Sorry I didn't give the proper initialization of j. But you are right j should
also be an array of 5. So x[j + 5] would return 5 values.
So if the array returned from 'ifelse' is the same dimention as test (h), then
are all the values of h being tested? So since h as you say has no dimensions
On Mon, 19 Jan 2009, rkevinbur...@charter.net wrote:
I am having a hard time understanding what is happening with ifelse.
Let me illustrate:
h <- numeric(5)
p <- 1:5
j <- floor(j)
And j is 0:4 + epsilon , where 0 <= epsilon < 1, evidently.
x <- 1:1000
ifelse(h == 0, x[j+2], 1:5)
[1] 2 3 4
I am having a hard time understanding what is happening with ifelse.
Let me illustrate:
h <- numeric(5)
p <- 1:5
j <- floor(j)
x <- 1:1000
ifelse(h == 0, x[j+2], 1:5)
[1] 2 3 4 5 6
My question is, "shouldn't this be retruning 25 numbers?" It seems that the
ifelse should check 5 values of h for
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