Hi all,
While using the following package and code:
library(rms)
library(Hmisc)
test.trans-aregImpute(
~ NearestWeekGestation + MaternalBMI + MomAge_Years +
WtGain + ethnicity , n.impute=10
,data=data.1)
test.mod-fit.mult.impute(BirthWeight_g ~
Thanks in advance!
2013/5/18 Ben Bolker bbol...@gmail.com
Ross Marriott Ross.Marriott at fish.wa.gov.au writes:
Dear R Help,
I would like to use the glmer.nb function for mixed modelling using
negative binomial distribution please.
On the CRAN website apparently this function is
On 05/19/2013 09:19 AM, Xianwen Chen wrote:
Hi,
I want to plot grouped bars to compare 95% confidence interval estimates
from two models. Each bar represents a 95% confidence interval estimate
of a coefficient from one of the two models. Each group represents
confidence interval estimates of
Dear all,
I would like to compute index of quantile variation (iqv) for each
observation in my survey data according to their responses on three
categorical variable.
Do you know iqv function for this purpose in a package?
All the best,
Niklas
[[alternative HTML version deleted]]
You should be able to pass a weights argument through fit.mult.impute to ols.
I assume that weights are with respect to variables that you, for some
reason, do not want to have as predictors. Otherwise you are already
conditioning on the weighting factors and no weights are needed.
Frank
Mullah
Dear Simon,
I respectfully disagree with you, when you say that it is NOT a bug. Empty
clusters can only happen if one provides a set of centers; if one enters
something like k=4, it cannot happen, so it must be a bug. Which one is
your case?
From the help:
Except for the Lloyd-Forgy method, k
Hi there,
I have the following code:
z - matrix(c(A, A, B, B, C, C, A, B, C), ncol = 3)
apply(z, 2, table, c(A, B, C))
which give correct results.
However, the following code:
apply(z[,1,drop=FALSE], 2, table, c(A, B, C))
which does not give what I expect. I have been thought it should give
Hi gurus,
I have a big data frame df, with columns named as :
age, income, country
what I want to do is very simpe actually, do
fitFunc-function(thisCountry){
subframe-df[which(country==thisCountry),];
fit-lm(income~0+age, data=subframe);
return(coef(fit));}
for each individual
Dear list,
I am very new to R and have been struggling with extracting data from a netcdf
file. Basically I have read in a file containing vegetation height data
organised in 0.5 degree grid cells spanning the whole globe. Each cell contains
a histogram representing the height distribution and
Hi, I have a question for my simulation problem:
I would like to generate a positive (or semi def positive) covariance
matrix, non singular, in wich the spectral decomposition returns me the same
values for all dimensions but differs only in eigenvectors.
Ex.
sigma
[,1] [,2]
[1,]
Hi: I am using termstrc package and not sure if TODAY field (govbonds
dataset) is today's date or the settlement date for the bonds. Although the
name suggests that it should be today's date, but how the formula is setup
indicates that it should be settlement date. Please help and thanks in
Sorry in the previuos message I've inverted: I would like generate ...
same eigenvectors and different eigenvalues
--
View this message in context:
http://r.789695.n4.nabble.com/Generate-positive-definite-matrix-with-constraints-tp4667449p4667450.html
Sent from the R help mailing list
On May 19, 2013, at 4:20 AM, Jess Baker wrote:
Dear list,
I am very new to R and have been struggling with extracting data from a
netcdf file. Basically I have read in a file containing vegetation height
data organised in 0.5 degree grid cells spanning the whole globe. Each cell
Dear all,
I encountered this strange behaviour with loops and lists. Consider this:
xl - list()
for(i in 5:7){##loop over numeric vector
xl[[i]] - rnorm(i)
}
xl
[[1]]
NULL
[[2]]
[1] -0.4448192 -1.3395014
[[3]]
[1] 1.3214195 -1.2968560 -0.6327795
The above lists contained a NULL element
HI,
May be this helps:
set.seed(24)
dat1-
data.frame(age=sample(30:70,120,replace=TRUE),income=sample(4:8,120,replace=FALSE),country=rep(c(USA,GB,France),each=40),stringsAsFactors=FALSE)
library(plyr)
ldply(dlply(dat1,.(country),lm,formula=income~0+age),function(x) coef(x))
# country
There is a documented bug in R 2.15.2 and R 2.15.3. The developers
decided not to fix this with an update, but it has been fixed with R
3.0.0 (perhaps also with R 3.0.1, which I have not tried myself). So,
upgrading R should take care of this problem.
HTH,
Ranjan
On Sun, 19 May 2013 16:17:30
On May 19, 2013, at 5:31 AM, CHEN, Cheng wrote:
Hi gurus,
I have a big data frame df, with columns named as :
age, income, country
what I want to do is very simpe actually, do
fitFunc-function(thisCountry){
subframe-df[which(country==thisCountry),];
fit-lm(income~0+age,
Hi,
xl- vector(list,7)
for(i in 5:7){##loop over numeric vector
xl[[i]] - rnorm(i)
}
xl
#[[1]]
#NULL
#
#[[2]]
#NULL
#
#[[3]]
#NULL
#
#[[4]]
#NULL
#
#[[5]]
#[1] 0.3266762 0.4316069 1.2290551 -0.6725783 1.6159861
#
#[[6]]
#[1] -2.8560618 -0.5694743 -0.7325862 1.6786160 0.3883842
On May 19, 2013, at 18:08 , Liviu Andronic wrote:
Dear all,
I encountered this strange behaviour with loops and lists. Consider this:
xl - list()
for(i in 5:7){##loop over numeric vector
xl[[i]] - rnorm(i)
}
xl
[[1]]
NULL
[[2]]
[1] -0.4448192 -1.3395014
[[3]]
[1] 1.3214195
Also ?tapply
But if you are a beginner who has done no homework -- i.e. you have
made no effort to learn basics with The Intro to R tutorial or other
online tutorial -- then you probably won't be able figure it out. We
expect some minimal effort by posters. If you have done such homework,
then it
On Sun, May 19, 2013 at 7:02 PM, peter dalgaard pda...@gmail.com wrote:
(The first example really had 2:3, not 5:7, right?)
Indeed. I simplified the example mid-email.
The essential bit is that to assign to the 2nd element of a list, it needs to
have at least two elements:
Thanks for the
On May 19, 2013, at 16:22 , Jinsong Zhao wrote:
Hi there,
I have the following code:
z - matrix(c(A, A, B, B, C, C, A, B, C), ncol = 3)
apply(z, 2, table, c(A, B, C))
which give correct results.
However, the following code:
apply(z[,1,drop=FALSE], 2, table, c(A, B, C))
which
Hi,
May be this helps:
lev1- unique(as.vector(z))
lapply(lapply(as.data.frame(z),factor,levels=lev1),table,lev1)
#$V1
#
# A B C
#A 1 1 0
#B 0 0 1
#C 0 0 0
#
#$V2
#
# A B C
#A 0 0 0
#B 1 0 0
#C 0 1 1
#
#$V3
#
# A B C
#A 1 0 0
#B 0 1 0
#C 0 0 1
#or
On Sun, May 19, 2013 at 10:33 AM, mary mary.d...@libero.it wrote:
Sorry in the previuos message I've inverted: I would like generate ...
same eigenvectors and different eigenvalues
Try this:
m0 - 5 + matrix(c(.05, -.05, -.05, .05), 2)
m0.values - eigen(m0)$values
with(eigen(matrix(rnorm(4),
On Sun, May 19, 2013 at 4:57 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Sun, May 19, 2013 at 10:33 AM, mary mary.d...@libero.it wrote:
Sorry in the previuos message I've inverted: I would like generate ...
same eigenvectors and different eigenvalues
Try this:
m0 - 5 +
Dear useRs,i have two distance matrices A and B. The distances in A are linked
with distances in B. What i want to do is the following...There are some points
in A where distances are zero(those points are apart from usual diagonal of
zero values).Whereas, in B the distances at those points are
Hello,
Try the following.
set.seed(27846)
A - matrix(sample(0:2, 16, TRUE), ncol = 4)
B - matrix(1:16, ncol = 4)
zeros - which(A == 0, arr.ind = TRUE)
B[zeros] - 0
Hope this helps,
Rui Barradas
Em 19-05-2013 22:05, eliza botto escreveu:
Dear useRs,i have two distance matrices A and B. The
Thanks Rui,It worked..
Eliza
Date: Sun, 19 May 2013 22:27:41 +0100
From: ruipbarra...@sapo.pt
To: eliza_bo...@hotmail.com
CC: r-help@r-project.org
Subject: Re: [R] coping zeros from matrix to another
Hello,
Try the following.
set.seed(27846)
A - matrix(sample(0:2, 16, TRUE), ncol =
On Sun, May 19, 2013 at 5:04 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Sun, May 19, 2013 at 4:57 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Sun, May 19, 2013 at 10:33 AM, mary mary.d...@libero.it wrote:
Sorry in the previuos message I've inverted: I would like
May be this helps:
set.seed(28)
mat1- matrix(sample(0:50,20,replace=TRUE),nrow=5)
A- dist(mat1)
A[2:4]- 0
set.seed(35)
mat2- matrix(sample(1:50,20,replace=TRUE),nrow=5)
B- dist(mat2)
B[which(A==0)]-0
B
# 1 2 3 4
#2 49.07138
#3 0.0
Thankyou very much Arun,it has even been simplified..
Eliza
Date: Sun, 19 May 2013 14:49:43 -0700
From: smartpink...@yahoo.com
Subject: Re: [R] coping zeros from matrix to another
To: eliza_bo...@hotmail.com
#or
just
B[A==0]-0
- Original Message -
From: arun
Hello, i can't find a test for elliptical symmetry In R, is there any package
that can save me the trouble ?Thank you in advanceSincerely yoursSleiman
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__
Dear All,
This is a data relating leg shaking on differenntre treatments. I reused
several individuals so I want to know 1) if there are significative
differences on shaking per treatment and 2) if the reused individuals
presented some effect or significative variation.
Nevertehless when I make
On 05/20/2013 10:52 AM, Luis Fernando García Hernández wrote:
Dear All,
This is a data relating leg shaking on differenntre treatments. I reused
several individuals so I want to know 1) if there are significative
differences on shaking per treatment and 2) if the reused individuals
presented
I would say that the OP should seek local statistical help, as he
appears to be out of his statistical depth. This would appear to be a
mixed effects models-type setup, but a local statistical expert would
be much better able to judge what the goals of the study were and what
sort of approach,
I'm very new to R, and I'm having trouble figuring out a 3d surface
plot of the data. I typically have something like this:
0.1
0.2
0.3
0.4
0.5
0.001
40960.16
40960.16
40960.16
40960.16
40960.16
0.0025
40960.16
40960.16
40960.16
40960.16
40960.16
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