Displaying 20 results from an estimated 10568 matches for "treates".
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2012 Jun 07
0
how lm behaves
I was wondering if somebody could explain why I get different results here:
>treats[,2]<-as.factor(treats[,2])
>treats[,5]<-as.factor(treats[,5])
>treats[,4]<-as.factor(treats[,4])
#there are 'c' on more days than I have 'h2o2', where treats[,4] is the day. I only want 'c' that correspond to the same days that I have a 'h2o2' also.
2011 Nov 08
2
why NA coefficients
Hi, I am trying to run ANOVA with an interaction term on 2 factors (treat has 7 levels, group has 2 levels). I found the coefficient for the last interaction term is always 0, see attached dataset and the code below:
> test<-read.table("test.txt",sep='\t',header=T,row.names=NULL)
> lm(y~factor(treat)*factor(group),test)
Call:
lm(formula = y ~ factor(treat) *
2011 Aug 26
3
elegant way to check if 2 values are in 3 columns?
Dear all,
I'm trying to rerun some data linkage exercises in R (they are designed
to be done in SPSS, SAS or STATA)
The exercise in question is to relabel the column "treat" to "1", if
"yearsep" is smaller than 1988 and columns "proc1"-"proc3" contain the
values 56.36 or 59.81.
My pathetic solution to do this in R currently looks like
2011 Apr 29
1
Use nparcomp function from nparcomp library to run post hoc
Dear list,
I tried to use the nparcomp to run some post hoc non-parametric comparison
and got and error.
Error in uniroot(pfct, interval = interval) :
f() values at end points not of opposite sign
Appreciate any comments.
the command line:
>nparcomp(Ulceration~Group,data=test,type='Dunnett',control='Non-treated')
Jun
2003 Sep 30
0
lme vs. aov
Hi,
I have a question about using "lme" and "aov" for the
following dataset. If I understand correctly, using
"aov" with Error term in the formula is equivalent to
using "lme" with default settings, i.e. both assume
compound symmetry correlation structure. And I have
found that equivalency in the past. However, with the
follwing dataset, I got different
2003 Oct 02
0
lme vs. aov with Error term
Hi,
I have a question about using "lme" and "aov" for the
following dataset. If I understand correctly, using
"aov" with an Error term in the formula is equivalent
to using "lme" with default settings, i.e. both assume
compound symmetry correlation structure. And I have
found that equivalency in the past. However, with the
follwing dataset, I got different
2009 Apr 01
4
Recode of text variables
Hi all
I am trying to do a simple recode which I am stumbling on. I figure
there must be any easy way but haven't come across it.
Given data of A","B","C","D","E","A" it would be nice to recode this
into say three categories ie A and B becomes "Treat1", C becomes "Treat
2" and E becomes "Treat 3".
I tried
2003 Oct 01
0
lme vs. aov with Error term again
Hi all,
Sent the following question yesterday, but haven't got
any suggestions yet. So just trying again, can anyone
comment on the problem that I have? Thank you!
-------------
Hi,
I have a question about using "lme" and "aov" for the
following dataset. If I understand correctly, using
"aov" with an Error term in the formula is equivalent
to using
2011 Nov 01
3
factor level issue after subsetting
Dear list,
I cannot figure out why, after sub-setting my data, that particular item
which I don't want to plot is still in the newly created subset (please
see example below). R somehow remembers what was in the original data
set. A work around is exporting and importing the new subset. Then it's
all fine; but I don't like this idea and was wondering what am I missing
here?
Thanks!
2003 Oct 02
0
RE: [S] lme vs. aov with Error term
Hi Bert,
Thanks for the suggestions. I tried lme with different
control parameters, and also tried using "ML", instaed
of "REML", but still got the same answers.
Yes, I hope some gurus on this list could give me some
hints.
Thanks
--- "Gunter, Bert" <bert_gunter at merck.com> wrote:
> But they are close. This is almost certainly a
> numeric issue --
2007 Oct 29
3
Strange results with anova.glm()
Hi,
I have been struggling with this problem for some time now. Internet,
books haven't been able to help me.
## I have factorial design with counts (fruits) as response variable.
> str(stubb)
'data.frame': 334 obs. of 5 variables:
$ id : int 6 23 24 25 26 27 28 29 31 34 ...
$ infl.treat : Factor w/ 2 levels "0","1": 2 2 2 2 1 1 1 2 1 1 ...
$ def.treat :
2010 Aug 11
1
Growth Curves with lmer
Dear all,
I have some growth curve data from an experiment that I try to fit using
lm and lmer. The curves describe the growth of classification accuracy
with the amount of training data t, so basically
y ~ 0 + t (there is no intercept because y=0 at t0)
Since the growth is somewhat nonlinear *and* in order to estimate the
treatment effect on the growth curve, the final model is
y ~ 0 + t +
2004 Mar 29
0
Error term in aov
Hi,
I'm trying to analyse a hierachical design and am running into some
trouble. Clearly I don't fully understand "Error" and I was hoping someone
could set me straight.
We measure percentage algal cover in each of 5 quadrats from each of 16
patches where 4 treatments are randomly allocated to a patch.
First suppose patches are coded 1 to 16. then the following gives the
2008 Jan 30
1
assign column classes when creating a data frame from several vectors
R-helpers,
Thanks in advance for your help. I am an R newbie and I am having trouble
figuring out the easiest/most efficient way to assign classes to columns in
a newly created data frame. R seems to want to convert everything to a
factor when I use the cbind function to compile vectors into a data frame.
ID<-seq(1,10,1)
TREAT<-c(rep("B",5),rep("G",5))
2005 May 26
1
Single factor from interaction.
I have a vague recollection of seeing reference, fairly recently, to
a function that forms a single factor which ``codes'' the interaction
between two (or more?) factors. Do I recollect correctly? It would
be easy enough to roll one's own, but if there is an existing
function it probably does a much cleverer job than I would do.
cheers,
Rolf Turner
rolf at math.unb.ca
2009 Nov 27
0
changing titlessymbols in plot.spm
Hi all,
I'm trying my best in changing the names of my experiments in the spectral
map.
My experiments are from six different time points, two as control and two
treated experiments.
I set have for each time point a different color .
Now I want to change the symbols so that I have for the treated experiments
one and for the controls a different one. But I want only two different
symbols
2008 Apr 13
3
Matched pairs with two data frames
Hi,
I have a frame "treat" and want to find matched pairs in the data frame
"control". In the matched (combined) data frame there should be two
variables (0/1),indicating the "source" of the data (treat or control),
so that it is possibe to set a "filter" (extraxt/select data).
#Here are the dataframes (my real data frames have many variables)
treat <-
2008 Dec 28
1
Random coefficients model with a covariate: coxme function
Dear R users:
I'm new to R and am trying to fit a mixed model
Cox regression model with coxme function.
I have one two-level factor (treat) and one
covariate (covar) and 32 different groups
(centers). I'd like to fit a random coefficients model, with treat and covar
as fixed factors and a random intercept, random
treat effect and random covar slope per center.
I haver a couple of
2003 Nov 16
1
SE of ANOVA (aov) with repeated measures and a bewtween-subject factor
Hallo!
I have data of the following design:
NSubj were measured at Baseline (visit 1) and at 3
following time points (visit 2, visit 3, visit 4).
There is or is not a treatment.
Most interesting is the question if there is a
difference in treatment between the results of visit 4
and baseline. (The other time points are also of
interest.) The level of significance is alpha=0.0179
(because of an
2005 Oct 05
1
Analyses of covariation with lme() or lm()
Hello all!
I have a problem that calls for a better understanding, than mine, of
how lme() uses the random part of the call.
The dataset consists of eleven field trials (Trial) with three
replicates (Block) and four fertiliser treatments (Treat). Analysing for
example yield with lme() is easy:
m1 <- lme(Yield ~ Treat, data=data,
random =~1| Trial/Block)
giving estimates of