Displaying 20 results from an estimated 20000 matches similar to: "contr.treatments query"
2008 Oct 11
2
R vs SPSS contrasts
Hi Folks,
I'm comparing some output from R with output from SPSS.
The coefficients of the independent variables (which are
all factors, each at 2 levels) are identical.
However, R's Intercept (using default contr.treatment)
differs from SPSS's 'constant'. It seems that the contrasts
were set in SPSS using
/CONTRAST (varname)=Simple(1)
I can get R's Intercept to match
2006 Aug 22
1
summary(lm ... conrasts=...)
Hi Folks,
I've encountered something I hadn't been consciously
aware of previously, and I'm wondering what the
explanation might be.
In (on another list) using R to demonstrate the difference
between different contrasts in 'lm' I set up an example
where Y is sampled from three different normal distributions
according to the levels ("A","B","C")
2010 Aug 29
2
glm prb (Error in `contrasts<-`(`*tmp*`, value = "contr.treatment") : )
glm(A~B+C+D+E+F,family = binomial(link = "logit"),data=tre,na.action=na.omit)
Error in `contrasts<-`(`*tmp*`, value = "contr.treatment") :
contrasts can be applied only to factors with 2 or more levels
however,
glm(A~B+C+D+E,family = binomial(link = "logit"),data=tre,na.action=na.omit)
runs fine
glm(A~B+C+D+F,family = binomial(link =
2012 Oct 27
1
contr.sum() and contrast names
Hi!
I would like to suggest to make it possible, in one way or another, to
get meaningful contrast names when using contr.sum(). Currently, when
using contr.treatment(), one gets factor levels as contrast names; but
when using contr.sum(), contrasts are merely numbered, which is not
practical and can lead to mistakes (see code at the end of this
message).
This issue was discussed quickly in 2005
2008 Aug 26
2
options("contrasts")
Code:
> options("contrasts")
$contrasts
factor ordered
"contr.treatment" "contr.poly"
I want to change the first entry ONLY, without retyping "contr.poly". How do
I do it? I have tried various possibilities and cannot get anything to work.
I found out that the response to options("contrasts") has class
2005 Jul 13
1
Name for factor's levels with contr.sum
Good morning,
I used in R contr.sum for the contrast in a lme model:
> options(contrasts=c("contr.sum","contr.poly"))
> Septo5.lme<-lme(Septo~Variete+DateSemi,Data4.Iso,random=~1|LieuDit)
> intervals(Septo5.lme)$fixed
lower est. upper
(Intercept) 17.0644033 23.106110 29.147816
Variete1 9.5819873 17.335324 25.088661
Variete2 -3.3794907 6.816101 17.011692
Variete3
2012 Oct 05
1
Setting the desired reference category with contr.sum
Hi,
I have 6 career types, represented as a factor in R, coded from 1 to 6. I
need to use the effect coding (also known as deviation coding) which is
normally done by contr.sum, e.g.
contrasts(career) <- contr.sum(6)
However, this results in the 6th category being the reference, that is being
coded as -1:
$contrasts
[,1] [,2] [,3] [,4] [,5]
1 1 0 0 0 0
2 0 1 0
2013 Apr 27
1
Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) :
i am getting the following error
Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) :
contrasts can be applied only to factors with 2 or more levels
can any on e suggest how to rectify
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2010 Sep 15
1
contr.sum, model summaries and `missing' information
Hi,
I have a dataset with a response variable and multiple factors with more
than two levels, which I have been fitting using lm() or glm(). In
these fits, I am generally more interested in deviations from the global
mean than I am in comparing to a "control" group, so I use contr.sum()
as the factor contrasts. I think I'm happy to interpret the
coefficients in the model summary
2009 Jan 23
1
Interpreting model matrix columns when using contr.sum
With the following example using contr.sum for both factors,
> dd <- data.frame(a = gl(3,4), b = gl(4,1,12)) # balanced 2-way
> model.matrix(~ a * b, dd, contrasts = list(a="contr.sum", b="contr.sum"))
(Intercept) a1 a2 b1 b2 b3 a1:b1 a2:b1 a1:b2 a2:b2 a1:b3 a2:b3
1 1 1 0 1 0 0 1 0 0 0 0 0
2 1 1 0 0 1 0
2002 Dec 01
1
generating contrast names
Dear R-devel list members,
I'd like to suggest a more flexible procedure for generating contrast
names. I apologise for a relatively long message -- I want my proposal to
be clear.
I've never liked the current approach. For example, the names generated by
contr.treatment paste factor to level names with no separation between the
two; contr.sum simply numbers contrasts (I recall an
2010 Apr 21
5
Bugs? when dealing with contrasts
R version 2.10.1 (2009-12-14)
Copyright (C) 2009 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
R is free software and comes with ABSOLUTELY NO WARRANTY.
You are welcome to redistribute it under certain conditions.
Type 'license()' or 'licence()' for distribution details.
Natural language support but running in an English locale
R is a collaborative project with
2004 Aug 20
1
drop1 with contr.treatment
Dear R Core Team
I've a proposal to improve drop1(). The function should change the
contrast from the default ("treatment") to "sum". If you fit a
model with an interaction (which ist not signifikant) and you
display the main effect with
drop1( , scope = .~., test = "F")
If you remove the interaction, then everything's okay. There is
no way to fit a
2011 May 11
1
Help with contrasts
Hi,
I need to build a function to generate one column for each level of a factor
in the model matrix created on an arbitrary formula (instead of using the
available contrasts options such as contr.treatment, contr.SAS, etc).
My approach to this was first to use the built-in function for
contr.treatment but changing the default value of the contrasts argument to
FALSE (I named this function
2005 Apr 13
2
multinom and contrasts
Hi,
I found that using different contrasts (e.g.
contr.helmert vs. contr.treatment) will generate
different fitted probabilities from multinomial
logistic regression using multinom(); while the fitted
probabilities from binary logistic regression seem to
be the same. Why is that? and for multinomial logisitc
regression, what contrast should be used? I guess it's
helmert?
here is an example
2007 May 17
1
model.matrix bug? Nested factor yields singular design matrix.
Hi all,
I believe this is a bug in the model.matrix function.
I'd like a second opinion before filing a bug report.
If I have a nested covariate B with multiple values for
just one level of A, I can not get a non-singular design
matrix out of model.matrix
> df <- data.frame(A = factor(c("a", "a", "x", "x"), levels = c("x",
2007 Oct 09
2
fit.contrast and interaction terms
Dear R-users,
I want to fit a linear model with Y as response variable and X a categorical variable (with 4 categories), with the aim of comparing the basal category of X (category=1) with category 4. Unfortunately, there is another categorical variable with 2 categories which interact with x and I have to include it, so my model is s "reg3: Y=x*x3". Using fit.contrast to make the
2011 May 18
1
Need expert help with model.matrix
Dear experts:
Is it possible to create a new function based
on stats:::model.matrix.default so that an alternative factor coding is used
when the function is called instead of the default factor coding?
Basically, I'd like to reproduce the results in 'mat' below, without having
to explicitly specify my desired factor coding (identity matrices) in the
'contrasts.arg'.
dd
1999 May 05
1
Ordered factors , was: surrogate poisson models
For ordered factor the natural contrast coding would be to parametrize by
the succsessive differences between levels, which does not assume equal
spacing
of factor levels as does the polynomial contrasts (implicitly at least).
This requires the contr.cum, which could be:
contr.cum <- function (n, contrasts = TRUE)
{
if (is.numeric(n) && length(n) == 1)
levs <- 1:n
2010 Jul 07
6
forcing a zero level in contr.sum
I need to use contr.sum and observe that some levels are not statistically different from the overall mean of zero.
What is the proper way of forcing the zero estimate? It seems the column corresponding to that level should become a column of zeros.
Is there a way to achieve that without me constructing the design matrix?
Thank you.
Stephen Bond
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