Hello,
The answer is given but there is no need to coerce to matrix first, as
long as the columns are numeric.
From ?exp, right at the beginning of section Details:
Details
All except logb are generic functions: methods can be defined for them
individually or via the Math group generic.
Follow the link Math:
Details
There are four groups for which S3 methods can be written, namely the
"Math", "Ops", "Summary" and "Complex" groups. These are not R objects
in base R, but methods can be supplied for them and base R contains
factor, data.frame and difftime methods for the first three groups.
And exp is the group "Math", 2nd bullet.
class(mtcars)
#[1] "data.frame"
exp(mtcars)
# output omitted
But if a column is not numeric the method Math.data.frame throws an error.
exp(iris)
#Error in Math.data.frame(iris) :
# non-numeric-alike variable(s) in data frame: Species
exp(iris[-5]) # remove the offending column
# output omitted
Hope this helps,
Rui Barradas
Às 18:23 de 14/10/21, Ana Marija escreveu:
Thank you so much!
On Thu, Oct 14, 2021 at 12:17 PM Bert Gunter <bgunter.4...@gmail.com> wrote:
As all of your columns are numeric, you should probably convert your df to
a matrix. Then use exp() on that, of course:
exp(as.matrix(b))
see ?exp
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Thu, Oct 14, 2021 at 10:10 AM Ana Marija <sokovic.anamar...@gmail.com>
wrote:
Hi All,
I have a data frame like this:
head(b)
LRET02 LRET04 LRET06 LRET08 LRET10 LRET12 LRET14
1 0 0.6931472 . 1.0986123 1.0986123 1.0986123 0.6931472
2 2.1972246 2.4849066 2.4849066 . 2.5649494 2.6390573 2.6390573
3 1.6094379 1.7917595 1.6094379 1.7917595 2.0794415 1.9459101 2.0794415
4 0 0 0 0 0 0 0
5 0.6931472 0 1.0986123 1.0986123 0.6931472 0.6931472 0.6931472
6 1.0986123 1.0986123 1.0986123 0.6931472 1.0986123 1.3862944 1.0986123
All values in this data frame are product of natural log. I have to do
inverse of it.
So for example do do inverse of 0.6931472 I would do:
2.718281828^0.6931472
[1] 2
How do I perform this operation for every single value in this data frame?
The original data frame is this dimension:
dim(b)
[1] 1441 18
Thanks
Ana
[[alternative HTML version deleted]]
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
