Please don't post in HTML since your code was all messed up. You did
not mention what problems you were having with your code. Now a
couple of things to check is to look at what the structure of "r" that
you are trying to add to "sum" (which should have been "Sum" according
to your assignment ear
CRAN (and crantastic) updates this week
New packages
* activity (1.0)
Maintainer: Marcus Rowcliffe
Author(s): Marcus Rowcliffe
License: GPL-3
http://crantastic.org/packages/activity
Provides functions to fit kernel density functions to animal activity
time data; plot ac
Hello ,I am trying to write a loop for sum of integrals .
the integral
is:integrand4<-function(x,a=1.5,n=3,k=0){(((a+1)*x)^k)*((2-x)^n)*(exp(-a*x-2))/(factorial(k)*factorial(n))}
integrate(integrand4,0,2).
I need a loop to give me the sum of integrals over k = 0,.n , for every
positive inte
Thanks a lot.
At this point then I wonder: seen that my response consists of 5
outcomes for each set of features, should I then train 5 different
models (one for each of them)?
Cheers
Lorenzo
On Sun, Oct 05, 2014 at 11:04:01AM -0700, Jia Xu wrote:
Hi, Lorenzo:
For 1) I think the formula is not
Hi:
I am trying to compute n-grams using package tm and tau with following code:
tokenize_ngrams <- function(x, n=3)
return(rownames(as.data.frame(unclass(textcnt(x,method="string",n=n)
texts <- c("This is the first document.", "This is the second file.",
"This is the third text.")
corpus <-
Hi, Lorenzo:
For 1) I think the formula is not correct. The formula should be outcome
~ features, and that's why you have weird result in 3)
2) predict in caret will automatically find the best result one if
there is one(sometimes it fails). You can print the model to see the cross
validation
... yes.
... And do note that in sampling, truncated != censored.
(They are often confused)
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
Clifford Stoll
On Sun, Oct 5
Bob O'Hara gmail.com> writes:
>
> This isn't an R question at all, so I don't know why it's on this list. But
> the best answer I've got is "a truncated t-distribution with an infinite
> number of degrees of freedom".
>
> Bob
Or, perhaps more productively: "since your question is a general
Dear All,
I am learning the ropes of CARET for automatic model training, more or
less following the steps of the tutorial at
http://bit.ly/ZJQINa
However, there are a few things about which I would like a piece of
advice.
Consider for instance the following model
##
This isn't an R question at all, so I don't know why it's on this list. But
the best answer I've got is "a truncated t-distribution with an infinite
number of degrees of freedom".
Bob
On 5 October 2014 17:18, thanoon younis wrote:
> Dear all R-users
> I have a question regarding truncated norma
Dear all R-users
I have a question regarding truncated normal distribution
: which type of probability distribution has same properties of truncated
normal distribution?
Many thanks in advance
--
Thanoon Y. Thanoon
PhD Candidate
Department of Mathematical Sciences
Faculty of Science
University T
Thats it! Most splendid! Thanks! The final result, just the value alone ,
for future reference.
> x <- c(1,5,3,1)> y <- c(5,8,2,3)> lm(x~y)$coefficient[1](Intercept)
0.5714286 > cat(lm(x~y)$coefficient[1])0.5714286
>
Thanks everyone!
Regards
Billy
On Sun, Oct 5, 2014 at 10:37 PM, peter
On 05 Oct 2014, at 16:06 , billy am wrote:
> Thank you both very much. It is the unname that is what I am looking for.
> Thanks!
>
> Btw , must the [1] be there? I am writing a Shiny web app hence I would
> like to display the value alone.
>
It's part of standard printing of unnamed vectors.
On 05/10/2014, 10:06 AM, billy am wrote:
> Thank you both very much. It is the unname that is what I am looking
> for. Thanks!
>
> Btw , must the [1] be there? I am writing a Shiny web app hence I would
> like to display the value alone.
Try leaving it out, and you'll see what it's for. (It is g
Thank you both very much. It is the unname that is what I am looking for.
Thanks!
Btw , must the [1] be there? I am writing a Shiny web app hence I would
like to display the value alone.
Thanks!
> unname(lm(x~y)$coefficient[1])[1] 0.5714286>
> coef(lm(x~y))["(Intercept)"](Intercept)
0.5714286
Arne Henningsen writes:
> Dear Rainer
>
> On 3 October 2014 14:51, Rainer M Krug wrote:
>> I am using the function frontier::sfa (from the package frontier) to
>> estimate several "half-normal production" stochastic frontier functions.
>>
>> Now I want to compare the coefficients of the linear
On 05.10.2014 15:02, Duncan Murdoch wrote:
On 05/10/2014, 7:21 AM, billy am wrote:
Hi ,
When I run the following code , I get both the description and the value ,
eg : Intercept and 0.5714286.
Is there a way to extract just the value 0.5714286? Thanks!
x <- c(1,5,3,1)> y <- c(5,8,2,3)> lm
On 05/10/2014, 7:21 AM, billy am wrote:
> Hi ,
>
> When I run the following code , I get both the description and the value ,
> eg : Intercept and 0.5714286.
>
> Is there a way to extract just the value 0.5714286? Thanks!
>
>
>> x <- c(1,5,3,1)> y <- c(5,8,2,3)> lm(x~y)
> Call:
> lm(formula = x
Hi ,
When I run the following code , I get both the description and the value ,
eg : Intercept and 0.5714286.
Is there a way to extract just the value 0.5714286? Thanks!
> x <- c(1,5,3,1)> y <- c(5,8,2,3)> lm(x~y)
Call:
lm(formula = x ~ y)
Coefficients:
(Intercept)y
0.5714
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