Dear colleagues,
I am trying to run a PERMANOVA in the vegan package with an appropriate
number of permutations (see example below), ideally . Obviously that
number of permutations does not exists so I would like to use Monte Carlo
permutation tests to derive the probability value, as is do
You might also want to try a different mirror.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.
Jeff et. al (although there are no others so far on this thread).
I finally gave up and decided to use 4.2 below from the "R for Windows FAQ”
_
4.2 I don’t have permission to write to the R-3.2.2\library directory.
You can install packages
The example code works, and reports permutations. Can you provide more
information?
data(dune)
data(dune.env)
adonis(dune ~ Management*A1, data=dune.env, permutations=)
On Tue, Oct 27, 2015 at 3:56 AM, Sean Porter wrote:
> Dear colleagues,
>
>
>
> I am trying to run a PERMANOVA in th
Dear all,
I tried to run the following code on 3 different OS:
```
download.file("
https://raw.githubusercontent.com/wush978/DataScienceAndR/course/RBasic-07-Loading-Dataset/orglist-100.CSV";,
destfile = "orglist-100.CSV")
con <- file("orglist-100.CSV", encoding = "UTF-16LE")
src <- readLines(con
Dear R-team,
I am using HPbayes package, specificaly hp.bm.imis function in it.
When I run the function it sometimes gives this error
Error in eigen(m, only.values = TRUE, symmetric = TRUE) :
infinite or missing values in 'x'
even though in my data there isn't any missing values or infinte value
Hi
I need a data set containing both numerical and categorical variables and a
two-class outcome to be used in examples of my R package (for variable
selection). Do you know any well-known one? I prefer it to be related to
healthcare and to have at least about 15 variables.
Regards
Farideh B
Hi everyone,
I am trying to perform a multiple imputation with mice on a dataset of about
13000 observations and 178 variables. I can start an "empty" imputation
("imp_start <- mice(data, maxit=0)"), but after a few minutes R stops with the
following error message:
Error: cannot allocate vector
Hi Stephen and others,
I am trying to run a one-way permanova where I have only 2 levels in the factor
“time”, and each level contains only 3 replicates. So because I have such few
observations (6 in total) and levels (2) there are not enough possible
permutations to get a reasonable test (i
Dear Jennifer
See inline below
On 27/10/2015 12:20, Lorenz, Jennifer wrote:
Hi everyone,
I am trying to perform a multiple imputation with mice on a dataset of about 13000 observations and
178 variables. I can start an "empty" imputation ("imp_start <- mice(data,
maxit=0)"), but after a few
Dear R list,
I am trying to plot the curve of a function.
Here's the R code:
library(mvtnorm)
p <- function(x, mu){
mu <- c(mu, 0)
dmvnorm(c(x, 1), mu, diag(2))
}
> curve(p(x, 2), from = 0, to =1)
Error in dmvnorm(c(x, 1), mu, diag(2)) :
mean and sigma have non-conforming size
I think
Sean: There are only 20 possible combinations, 6!/(3! x 3!), so you just
need to enumerate them completely (no Monte Carlo approximation required).
I don't know if permanova() can do this but you can do it with the mrpp()
functions and argument (,exact=TRUE) in Blossom package for R.
Brian
Brian
Hi,
I am running through a series of regression in a loop as follows:
results <- vector("list", length(mydata$varnames))
for (i in 1:length(mydata$varnames)) {
results[[i]] <- summary(lm(log(eval(parse(text=varnames[i]))) ~ age + sex +
CMV.status, data=mydata))
}
Now, when I look at the result
Look at the permute package. I believe this functionality is there.
Please excuse my brevity; this message was sent from my telephone.
On Oct 27, 2015 10:55 AM, "Cade, Brian" wrote:
> Sean: There are only 20 possible combinations, 6!/(3! x 3!), so you just
> need to enumerate them completely (n
> On Oct 27, 2015, at 10:19 AM, Ravi Varadhan wrote:
>
> Hi,
>
> I am running through a series of regression in a loop as follows:
>
> results <- vector("list", length(mydata$varnames))
>
> for (i in 1:length(mydata$varnames)) {
> results[[i]] <- summary(lm(log(eval(parse(text=varnames[i])))
Never mind, I figured it out.
You need to use sapply(), for instance, curve(sapply(x, p), from = 0, to
=10)
Thanks all!
On Tue, Oct 27, 2015 at 11:14 AM, C W wrote:
> Dear R list,
>
> I am trying to plot the curve of a function.
>
> Here's the R code:
>
> library(mvtnorm)
>
> p <- function(x,
Hello R,
I'm trying to figure out if it would be possible use a device (maybe simply
a trackball) separate from the mouse that would have the ability to rotate
a 3D plot made with rgl without any buttons. This is to build an
interactive "demo"-like piece for which a mouse or trackpad is an
unsatis
Marc,Ravi:
I may misunderstand, but I think Marc's solution labels the list
components but not necessarily the summary() outputs. This might be
sufficient, as in:
> z <- list(y1=rnorm(10,5),y2 = rnorm(10,8),x=1:10)
>
> ##1
> results1<-lapply(z[-3],function(y)lm(log(y)~x,data=z))
> lapply(results1
On 27/10/2015 1:38 PM, santiago gil wrote:
Hello R,
I'm trying to figure out if it would be possible use a device (maybe simply
a trackball) separate from the mouse that would have the ability to rotate
a 3D plot made with rgl without any buttons. This is to build an
interactive "demo"-like piec
Well, I haven't built the device yet, I'm trying to figure out what my
options will be. In the case of plugging in a simple USB trackball, I
suppose it can work with the HID Manager in Mac to customize the signal
from it.
In general, would it be possible to manipulate the plot, say for example,
wi
Dear R-help,
I am working with a matrix "m" from which I would like to extract some
elements. An toy example is as follows:
## input matrix
m <- structure(c(0L, 0L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 0L, 0L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L,
I'm trying to generate prediction of the column "dubina" using this
algorithm made in R's "neuralnet" package. But I keep getting non-reliable
neural-net output. I have tried changing the number of hidden layers,
normalizing and denormalizing data. Is there a mistake in the algorithm,
maybe because
Thank you very much, Marc & Bert.
Bert - I think you're correct. With Marc's solution, I am not able to get the
response variable name in the call to lm(). But, your solution works well.
Best regards,
Ravi
-Original Message-
From: Bert Gunter [mailto:bgunter.4...@gmail.com]
Sent: Tue
Thanks for helping Boris.
Regards,
Maram Salem
On 25 October 2015 at 23:30, Boris Steipe wrote:
> It may be useful for you to estimate the time complexity of your function:
> try it with smaller input that takes short and noticeable time, see whether
> the time increases linearly, quadratical
If you want to use the numbers you gave a the index into the matrix, then
you can create a matrix with the values and then index into 'm'. I don't
see a '4' in the output example you gave using your index values:
> m <- structure(c(0L, 0L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
+ 1L, 1L, 1L
Dear Jim,
Thank you very much for your quick reply.
I am sorry for the confusion it may have caused, but I messed up the
indexes in my example. I would like, from the following matrix "m"
## input
m <- structure(c(0L, 0L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L,
0L, 0L, 2L, 2L, 2L, 2L, 2L, 2
Sorry to be a late comer to this problem but I'm having a similar
issue. My function is called by ode from deSolve
ADM1_C <- function(t,state,parameters,...){
with(as.list(c(state,parameters)), {
# do some stuff here and return a list containing a vector of
derivatives to ode
...
})
}
in w
On 27/10/2015 2:35 PM, santiago gil wrote:
> Well, I haven't built the device yet, I'm trying to figure out what my
> options will be. In the case of plugging in a simple USB trackball, I
> suppose it can work with the HID Manager in Mac to customize the signal
> from it.
>
> In general, would it
On 27/10/2015 3:52 PM, David Stevens wrote:
> Sorry to be a late comer to this problem but I'm having a similar
> issue. My function is called by ode from deSolve
>
> ADM1_C <- function(t,state,parameters,...){
>
> with(as.list(c(state,parameters)), {
># do some stuff here and return a list
I don't see how you are getting the result you provide.
> m
[,1] [,2] [,3] [,4] [,5]
[1,]00000
[2,]00000
[3,]00000
[4,]12300
[5,]12300
[6,]12300
[7,]123
Maybe it's just me --- I'm notoriously slow --- but I haven't a clue
what you are trying to do. You said:
"... the idea is to extract the diagonal elements until a
zero is found."
I don't see anything that resembles a "diagonal element" (*entry*,
actually; sets have elements, arrays h
Thank you all for your solutions and comments.
As Dr. Carlson mentioned, we leave rows 1 to 3 out as they are all zeroes.
Then, the entries I need to select from m are
entry value
4,1 ---> 1
5,2 ---> 2
6,3 ---> 3
7,1 ---> 1
8,2 ---> 2
9,3 ---> 3
10,4 -
Dear R-help,
In the rms package, when using the ols function with a penalty, the
df.residual appears to always be n-1 (with n being the sample size).
That seems strange to me, but I don't have much knowledge in this
area.
Here's an example:
library(rms)
set.seed(1)
n <- 50
d <- data.frame(x1
Hello all!
I´m fitting a mixed effects cox model with coxme function of coxme package.
I want to konw what is the best way to check the model adequacy, once that
function cox.zph that does not work for coxme objects.
Thanks in advanced,
Raoni
--
Raoni Rosa Rodrigues
Research Associate of Fish
Hi all,
I have this data on "sim" variable
10403.000NANANANA
11178.000NANANANA
11521.000NANANANA
11385.000NANANANA
10102.000NANANANA
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