Unlikely that someone will be interested in guessing. You are much more likely
to get a constructive response if you provide the minimal reproducible example
requested in the footer.
Of course, it is possible that you just might find the answer if you carefully
read the help page for nlm, since
Hi R community
I have written a loglikelihood function which I am minimizing using
nlm().
nlm() is giving me no results...I mean, I am getting initial values as
estimates. No iteration.
I have tried many initials value close to true values and far away from
tru values. But every time I
am gett
Hi Dr Singh,
The object mtcars is a data frame and the mean is not defined for a
data frame. If you try it on a component of the data frame for which
mean is defined:
by(mtcars$mpg,mtcars$am,mean)
mtcars$am: 0
[1] 17.14737
mtcars$am: 1
Hi Tom,
What you want is a list rather than a data frame. So:
df<-read.table(text=" Dat1 Dat2 Dat3
1154
2779
3335
42 NA 5
59 NA NA",
header=TRUE)
dflist<-as.list(df)
na.remove<-function(x) return(x[!is.na(x)])
sapply(dflist,na.remove)
Jim
O
> On Apr 14, 2016, at 2:33 PM, Tom Mosca wrote:
>
> Hello,
>
> I�ve tried several times to learn R, but have never gotten past a particular
> gate. My data are organized by column in Excel, with column headers in the
> first row. The columns are of unequal lengths. I export them as CSV, th
You're right, but I think this fails to pinpoint the error. The
problem is that FUN's argument is "applied to (usually data-frame)
subsets of data," and the OP has used FUN = mean, which takes a
vector (+ a few other classes), not a data frame, as argument. See
?mean
Morals:
1. It is rather pr
Hello,
I�ve tried several times to learn R, but have never gotten past a particular
gate. My data are organized by column in Excel, with column headers in the
first row. The columns are of unequal lengths. I export them as CSV, then
import the CSV file into R. I wish to summarize the data b
A mess!
Please follow the posting guide: post in *plain text*, not HTML.
Cheers.
Bert
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, Apr 14, 2016 at 1:1
I think you are not using the best function for what your intentions are.
Try:
> by(data=mtcars, INDICES=list(as.factor(mtcars$am)), FUN=colMeans)
: 0
mpg cyldisp hpdrat wt
qsec vs
17.1473684 6.9473684 290.3789474 160.2631579 3.286
Hi,I have a data set (mydata), which a part of this is like the following:
'data.frame': 36190 obs. of 16 variables:$ RE : int 38 41
11 67 30 18 38 41 41 30 ...$ LU : int 4200 3330 530 4500
3000 1790 4700 3400 3640 4000 ...$ COUNTRY : Factor w/
Hello,
I have some trouble with DataControl and DCARContControl functions. I can use
the following codes for multivariate normal distribution.
sigma <- matrix(c(-1, 0, 0, 1), 2, 2)
dc <- DataControl(size = 100, distribution = rmvnorm, dots = list(sigma =
sigma))
cc <- DCARContControl(epsilon
Dear Sirs,
I am Professor at Indira Gandhi Krishi Vishwavidyalaya, Raipur,
Chhattisgarh, India.
While taking classes, I found the *by() *function producing following error
when I use FUN=mean or median and some other functions, however,
FUN=summary works.
Given below is the output of the example
Howdy everyone
I’m trying to get Odds ratio and OR confidence intervals using a probit model,
but I'm not getting.
Do you think you can help me?
I’m new with R L
naive =
summary(glm(pcr.data[,7]~boldBeta_individual+pcr.data$age,family=binomial(link=probit)))
naive_a
Howdy everyone
I’m trying to get Odds ratio and OR confidence intervals using a probit model,
but I'm not getting.
Do you think you can help me?
I’m new with R L
naive =
summary(glm(pcr.data[,7]~boldBeta_individual+pcr.data$age,family=binomial(link=probit)))
naiv
Thanks Bill, it's very useful to know how parsing and evaluation works.
It seems that quoting is the least complicated solution which is guaranteed
to work.
Best,
Adrian
On 13 Apr 2016 6:04 p.m., "William Dunlap" wrote:
> %=>% would have precendence ('order of operations') problems also.
>
>
On 14 Apr 2016, at 09:47 , Ole C. Brudvik wrote:
> Hi!
> I just started this course and just installed R and RStudio but I got this
> warning/error messages when I open R and RStudio:
>
This is not a help list for any course...
However, that particular issue is just that the FAQ was edited a
1. These are warnings, not error. So you could just ignore them and proceed.
2. However, you may wish to go to the tools -->General menu in RStudio
and set the default text encoding to UTF-8. I think that will get rid
of the warnings.
Happy R-ing. Please make use of R's docs and web resources (es
The 3.2.4-revised version turned out to give trouble for some of CRAN's
subsystems.
Accordingly, a rebadged version 3.2.5 is now released; it only differs in the
version number and a few clean-up items. If you have a working install of
3.2.4-revised there should be no reason to upgrade it.
You
This is a good place to start ...
https://cran.r-project.org/doc/manuals/R-intro.pdf
Jean
On Thu, Apr 14, 2016 at 5:39 AM, kipkorirfrankli...@gmail.com <
kipkorirfrankli...@gmail.com> wrote:
>
> Hello. I am Franklin from University of Eldoret. I really want to study
> the R package. What should
Hello. I am Franklin from University of Eldoret. I really want to study the R
package. What should I first of all do?
[[alternative HTML version deleted]]
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https://stat.ethz.ch
Hi!
I just started this course and just installed R and RStudio but I got this
warning/error messages when I open R and RStudio:
During startup - Warning messages:
1: Setting LC_CTYPE failed, using "C"
2: Setting LC_COLLATE failed, using "C"
3: Setting LC_TIME failed, using "C"
4: Setting LC_
John,
as I know Microsoft has owned the Revolution Analytics and is continuing
the development of the Revolurion R with a new name Microsoft R. Note that
Microsoft R has an open source stable version. I'm not quite sure which
modifications they are doing on it. I suppose that this modified version
Yes. Yes. That info is on their site. That info is on their site. They
have paid support for their customers and
non-Microsoft-R-platform-dependent packages will (most likely) still
be answered by the community.
This is just a re-branding and expansion of what was Revolution R
which has been aroun
Has anyone ever heard of or used Microsoft R server? Does the product work?
What are requirements for running it? How much does it cost and is it supported
by the R community?
Thank you,
John
Confidentiality Statement:
This email message, including any attachments, is for th...{{dropped:6}}
Hello Jim,
You're amazing. This is what finally worked:
arrows(-1,19,35.6,19,code=3, xpd=T).
Don't know the coordinates were giving so much trouble. Maybe something to
do with maps in rworldmap. Thanks again!
Sincerely,
Milu
On Wed, Apr 13, 2016 at 6:51 AM, Jim Lemon wrote:
> Hi Milu,
> My
On 14/04/16 19:28, Marna Wagley wrote:
Hi R user,
Can we calculate sample size when only mean and SE are given?
Let say one example, I have mean with SE is 0.54+-0.0517 (mean+-SE). Is
there any way to find the samples (sample size n) in that condition in R?
i think this question is not related
You are right, your question is not related to R, so not appropraie here. I
pasted your question into google instead, and got all sorts of suggestions,
ncluding
https://www.isixsigma.com/tools-templates/sampling-data/how-determine-sample-size-determining-sample-size/
Sarah
On Thursday, April 14,
So. Given that the second and third panels of the first figure in the first
link I gave show a decision tree with decision rules at each split and the
number of samples at each direction, what _exactly_ is your problem?
On Wednesday, April 13, 2016, Michael Eugene wrote:
> I still need the out
Hi R user,
Can we calculate sample size when only mean and SE are given?
Let say one example, I have mean with SE is 0.54+-0.0517 (mean+-SE). Is
there any way to find the samples (sample size n) in that condition in R?
i think this question is not related to R, I hope you won't mind.
Thanks
On Thu, 14 Apr 2016, Michael Artz wrote:
Ah yes I will have to use the predict function. But the predict function
will not get me there really. If I can take the example that I have a
model predicting whether or not I will play golf (this is the dependent
value), and there are three independen
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