Murray,
Here is a general paradigm I tend to use for such problems. It extends
to fairly general model sequences, including different responses, c
First a couple of tiny, tricky but useful functions:
subst - function(Command, ...) do.call(substitute, list(Command,
list(...)))
abut -
Markus Gesmann writes:
Murray,
How about creating an empty list and filling it during your loop:
mod - list()
for (i in 1:6) {
mod[[i]] - lm(y ~ poly(x,i))
print(summary(mod[[i]]))
}
All your models are than stored in one object and you can use lapply
to
This (below) also runs into trouble if you try to predict with new data
since you have no rule for re-constructing the formula. Also, you can't
plot the term as a single contributor to the linear predictor with
termplot().
I'm sure given enough ingenuity you can get round these two, but why
Gorjanc Gregor asks:
Hello!
I would like to use constrain to sum coeficients of a factor to 0
instead
of classical corner contraint i.e. I would like to fit a model like
lm(y ~ 1 + effectA + effectB)
and say get parameters
intercept
effectA_1
effectA_2
effectB_1
effectB_2
Earl F. Glynn asks:
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Earl F. Glynn
Sent: Tuesday, 15 August 2006 8:44 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Help with workaround for: Function '`[`' is not in
thederivatives table
# This works
Mike Wolfgang asks:
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Mike Wolfgang
Sent: Wednesday, 23 August 2006 1:31 PM
To: R-help list
Subject: [R] glm inside one self-defined function
Hi list,
I've searched in R-help and found some related discussions but still
could
As a minor footnote to both of these, I would add that both assume
that all the columns of the dataset are numeric. It doesn't cost much
to generalize it to cover any matrix structure, of any mode:
constantColmuns - function(Xmat)
which(apply(Xmat, 2, function(z) length(unique(z)) == 1))
Nongluck Klibbua reports:
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Nongluck
Klibbua
Sent: Tuesday, 29 August 2006 11:12 AM
To: R-help@stat.math.ethz.ch
Subject: [R] singular matrix
Dear R-users,
I try to use solve to get the solution
I cannot resist a very brief entry into this old and seemingly
immortal issue, but I will be very brief, I promise!
Amasco Miralisus suggests:
As I understood form R FAQ, there is disagreement among Statisticians
which SS to use
Benjamin,
A couple of points:
fit - glm(y ~ x, gaussian(link = log))
does NOT fit a model with a lognormal response distribution. It fits a
non-linear regression model with an ordinary gaussian response
distribution. This model has constant variance, whereas the lognormal
model (which you
When Haydn was asked about his 100+ symphonies he is reputed to have
replied sunt mala bona mixta which is kind of dog latin for There are
good ones and bad ones all mixed together. It's certainly the same
with R packages so to continue the latin motif: caveat emptor
The R engine, on the other
This is always tricky. Here is a work-around.
Try asking for the Hessian with the original fit:
fm - polr(factor(y) ~ lx, data = ord.dat, Hess=T)
summary(fm)
Call:
polr(formula = factor(y) ~ lx, data = ord.dat, Hess = T)
Coefficients:
Value Std. Error t value
lx 2.420614 0.8146359
lm works for multivariate responses
rlm does not - check what the help file says about the response.
That's about it, really.
Bill Venables.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Markku
Mielityinen
Sent: Thursday, 24 March 2005 5:20 PM
To:
rev(expand.grid(k = 1:5, j = 1:4, i = 1:3))
i j k
1 1 1 1
2 1 1 2
3 1 1 3
4 1 1 4
5 1 1 5
6 1 2 1
7 1 2 2
8 1 2 3
...
55 3 3 5
56 3 4 1
57 3 4 2
58 3 4 3
59 3 4 4
60 3 4 5
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Ronnen Levinson
Your message doesn't help us very much. You haven't said what kind of
calculation it is you want to do, and that certainly matters. For
example, for some kinds of computations the solution you started below
would work fine:
M - matrix(1:16, 4, 4)
is.na(diag(M)) - TRUE
M
[,1] [,2] [,3]
is.na(diag(M)) - TRUE
cmeans - colMeans(M, na.rm = TRUE)
csd - apply(M, 2, sd, na.rm = TRUE)
cvar - csd ^2
(or
cvar - apply(M, 2, var, na.rm = TRUE)
)
Using 'kmeans' on a matrix but 'ignoring the diagonal entries' just
doesn't make sense as it stands, so I can't help you there.
V.
You didn't try very hard.
Try this, look at it and think about it:
jj - scale(sample(1:100, 10))
qqnorm(jj)
abline(0, 1)
Rather than abline, however, most people, however, would use
qqline(jj)
in which case you don't need the scaling.
V.
-Original Message-
From: [EMAIL
Remember this is just a diagnostic procedure and all you are really
looking for is whether the normal scores plot is pretty straight. The
reason for anchoring the guiding line at the quartiles is really the
same reason that boxplots use quartiles for the central chunk. You
don't want the guiding
This is possible if x and z are orthogonal, but in general it doesn't
work as you have noted. (If it did work it would almost amount to a way
of inverting geenral square matrices by working one row at a time, no
going back...)
It is possible to fit a bivariate regression using simple linear
Hi Anne,
Here's one suggestion, use a simple panel function:
cols - c(red, green3, blue)
with(iris,
pairs(iris[, -5], main = Andersons Iris Data - 3 species,
panel = function(x, y, ...)
points(x, y, pch = (2:4)[Species], col = cols[Species], ...)
))
Bill Venables
For three of the beta distribution functions in R the parameters
defining the distribution are alpha, beta and 'ncp'. Pretty trivially,
there is no bijection between these three and the mean and variance, but
for the special case of ncp = 0, I think there is.
Rather than just write it down, it's
At the cost of breaking the thread I'm going to change your subject and
replace 'Principle' by 'Principal'. I just can't stand it any longer...
OK, here is how I would solve your other problems. First put
wh - c(USA, New Zealand, Dominican Republic,
Western Samoa, Cook Islands)
Couple of points:
* If you provide relative frequencies for the binomial response, you
need also to give weights so that the actual counts can be
reconstructed. This is what the warning message is telling you: if you
reconstruct the counts using the default (unity) weights, the counts are
not
Hi Spencer,
I think if you have a problem that needs a lot of symbolic manipulation
you are probably better off driving it directly from something like
Maple or Mathematica (I prefer maple, actually) than trying to drive it
from R. It just gets too clumsy. On the other hand it is very handy
While symbolic computation is handy, I actually think a more pressing
addition to R is some kind of automatic differentiation facility,
particularly 'reverse mode' AD, which can be spectacular. There are
free tools available for it as well, though I don't know how well
developed they are. See:
This particular call to exportData is probably equivalent in effect to
the R call:
write.table(MU.Cost,
file = paste(C:/RAUDSL/S,
as.character(MU.Cost$Run.Id[1]),
.,as.character(MU.Cost$MU.Id[1]),
Getting a picture like this is pretty easy. e.g.
x - y - seq(-5, 5, len = 200)
X - expand.grid(x = x, y = y)
X - transform(X, z = dnorm(x, -2.5)*dnorm(y) - dnorm(x, 2.5)*dnorm(y))
z - matrix(X$z, nrow = 200)
persp(x, y, z, col = lightgoldenrod, border = NA,
theta = 30, phi = 15, ticktype =
Alexandre,
I have a couple of remarks to make, not all of which you might find
immediately helpful, I regret to say.
* The choice between using predictors linearly or in factor versions is
a modelling choice that is in no way specific to multinom. It is a
general aspect of modelling that has to
You should look at
?[
and look very carefully at the drop argument. For your example
sw[, 1]
is the first component of the data frame, but
sw[, 1, drop = FALSE]
is a data frame consisting of just the first component, as
mathematically fastidious people would expect.
This is a
str - as.character(d$name)
should do the trick. (damn... my shift key just broke as well...)
bill venables.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Cuichang Zhao
Sent: Saturday, 16 April 2005 2:00 PM
To: r-help@stat.math.ethz.ch
Subject: [R]
First put
day.names - c(sun, mon, tue, wed, thu, fri, sat)
then
days - factor(as.character(days), levels = day.names)
will ensure the ordering you want.
Bill Venables.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Eric C. Jennings
Sent: Saturday,
This is how I'd write the formula for use with nls/nlme:
y ~ b41*(x - 1) + b42*(x^2 - 1) +
ifelse((a41 - x) = 0, b43*(a41 - x)^2, 0) +
ifelse((a42 - x) = 0, b44*(a42 - x)^2, 0)
This is a direct translation from your funny foreign-looking code below
that probably makes it clear what's going
You need to think about it just a bit harder.
[Hint: what happens if you leave out the first 'which' and just make
ids - (d[, 1] 0)
does it work then...?]
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Werner Wernersen
Sent: Monday, 18 April 2005
Andrew Criswell asks:
Hello All:
Should I conclude from this discussion that there is no
practical
means by which nested generalized mixed models can be compared
from
output produced through glmmPQL or GLMM?
[WNV] The picture is, in my view, not as bleak as
Here's a bit of a refinement on Ted's first suggestion.
N - 1
graphics.off()
par(mfrow = c(1,2), pty = s)
for(k in 1:20) {
m - (rowMeans(matrix(runif(M*k), N, k)) - 0.5)*sqrt(12*k)
hist(m, breaks = FD, xlim = c(-4,4), main = k,
prob = TRUE, ylim = c(0,0.5), col =
Ronggui asks:
: -Original Message-
: From: [EMAIL PROTECTED]
: [mailto:[EMAIL PROTECTED] On Behalf Of ronggui
: Sent: Saturday, 23 April 2005 5:41 PM
: To: r help
: Subject: [R] question about about the drop1
:
:
: the data is :
: table.8.3-data.frame(expand.grid(
:
: -Original Message-
: From: [EMAIL PROTECTED]
: [mailto:[EMAIL PROTECTED] On Behalf Of
: michael watson (IAH-C)
: Sent: Friday, 22 April 2005 7:47 PM
: To: r-help@stat.math.ethz.ch
: Subject: [R] Anova - interpretation of the interaction term
:
:
: Hi
:
: So carrying on my use of
Luis Fernando Chaves asks:
: -Original Message-
: From: [EMAIL PROTECTED]
: [mailto:[EMAIL PROTECTED] On Behalf Of Luis
: Fernando Chaves
: Sent: Sunday, 24 April 2005 11:29 AM
: To: [EMAIL PROTECTED]
: Subject: [R] A question on the library lme4
:
:
: Hi,
:
: I ran the following
To me this is not at all surprising.
If you read the help info for eigen it says clearly that calculating the
eigenvectors is the slow part. So it is entirely likely that completely
different algorithms will be used if you are asking for only the
eigenvalues, or if you are asking for both
instead of as.character(substitute(arg)) use deparse(substitute(arg))
: -Original Message-
: From: [EMAIL PROTECTED]
: [mailto:[EMAIL PROTECTED] On Behalf Of Shigeru Mase
: Sent: Saturday, 30 April 2005 1:24 PM
: To: r-help@stat.math.ethz.ch
: Subject: [R] How to extract function
Are you looking for a unique solution or families of solutions?
Can't you turn a root-finding problem for a system of equations
with a unique solution into an optimisation problem, anyway?
E.g. You want to solve
f1(x) = g1
f2(x) = g2
...
Why not optimise L(x) = (f1(x) - g1)^2 + (f2(x) -
No, for regression trees collinearity is a non-issue, because it is not a linear
procedure. Having variables that are linearly dependent (even exactly so) merely
widens the scope of choice that the algorithm has to make cuts.
I'm not sure what you mean by Multicollinear variables should appear
Adrian Dragulescu asks:
Hello all,
This is the error message that I get.
hyp.res - nls(log(y)~log(pdf.hyperb(theta,X)), data=dataModel,
+ start=list(theta=thetaE0),
+ trace=TRUE)
45.54325 : 0.100 1.3862944 -4.5577142 0.0005503
3.728302 :
Goodness Patrick, this must surely qualify for the obfuscated R competition finals. I
love it!
There are two solutions I can think of with do.call and here they are:
x - 1
x2 - runif(10)
x12 - c(x, x2)
do.call(cbind, lapply(x12, as.name))
x x2
[1,] 1 0.99327265
[2,] 1
Alternatively, if you can arrive at a sensible ordering of the levels
you can declare them ordered factors and make the computation feasible
once again.
Bill Venables.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Torsten Hothorn
Sent: Monday, 5 April
sub$mm - factor(sub$mm)
is the simplest way to change a single factor in this way. To do a
whole data frame you just need a loop:
drop_unused_levels - function(data)
as.data.frame(lapply(data, function(x) if(is.factor(x))
factor(x) else x ))
Here's your example again, but witn a
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Sent: Sunday, 15 January 2006 11
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-Original
That solution for the case 'with gaps' merely omits transitions where
the transition information is not for a single time step. (Mine can be
modified for this as well - see below.)
But if you know that a firm went from state i in year y to state j in
year y+3, say, without knowing the
Yes Bert, this time you are missing something (unusually) ...
As Brian Ripley pointed out 'dnorm' is in the derivative table, *but*
only as a function of one variable. So if you want to find the
derivative of
dnorm(x, mean, sigma)
you have to write it as 1/sigma * dnorm((x - mu)/sigma). Here
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-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Jacob van Wyk
Sent: Friday, 10 November 2006 3:33 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Command Line Prompt Symbol
Hi
I run R in Windows.
Is there a simple way of changing
Use factors with specified levels.
lev - letters[1:4]
table(factor(letters[1:4], levels = lev),
factor(letters[c(1:3,3)], levels = lev))
a b c d
a 1 0 0 0
b 0 1 0 0
c 0 0 1 0
d 0 0 1 0
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On
Here is a possibility:
test - expand.grid(id = 1:2, sex = c('male', 'female'))
sapply(test, class)
id sex
integer factor
test - transform(test, sex = as.character(sex))
sapply(test, class)
id sex
integer character
But I am surprised at the reason you
Here is an elementary way of doing it:
dat
url time somethingirrelevant visits
1 www.foo.com 1:00 xxx100
2 www.foo.com 1:00 yyy 50
3 www.foo.com 2:00 xyz 25
4 www.bar.com 1:00 xxx200
5 www.bar.com 1:00
Lars from space [sic] asks:
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of downunder
Sent: Monday, 25 December 2006 12:31 PM To: r-help@stat.math.ethz.ch
Subject: [R] Higher Dimensional Matrices
Hi all.
I want to calculate partial
Hi Erin,
You would be safe on a machine that represented floating point numbers
in base 10, and I haven't seen one of those for such a long time... All
modern machines use base 2 for floating point numbers.
The moral of the story is not to believe what you see printed. The
number you see
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Sent: Monday, 1 January
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-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Osorio
But this is using paste() the wrong way round. A better way would be
join - function(x) do.call(paste, c(as.data.frame(x), sep = \r))
which(join(mat1) %in% join(mat2))
[1] 8 13 16 19 24
This is essentially the technique used by duplicated.data.frame
Bill Venables
-Original Message-
Jonathon Kopecky asks:
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Jonathon Kopecky
Sent: Tuesday, 30 January 2007 5:52 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Need to fit a regression line using orthogonal residuals
I'm trying to fit a simple
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From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Bob Green
Sent: Monday, 26 March 2007 6:30 AM
To: r-help@stat.math.ethz.ch
Subject: [R] resolving
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-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of John Sorkin
Sent: Monday, 26 March 2007 12:19 PM
To: r-help
I take it all subjects are measured at the same time points, or
Hotelling's T^2 becomes rather messy.
The essential difference lies in the way the variance matrix is
modelled. The usual repeated measures model would model the variance
matrix as equal variances and equal covariances, i.e. with
There is a statistical problem and a code problem.
The statistical problem is that if your 'x' has a mean that depends
non-trivially on predictors, then you would not expect its distribution
ignoring predictors to be normal. You would expect the residuals after
modelling to be normal. Basically
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, Cleveland, 4163
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Subject: [R] scan a directory
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...but with 500 variables and only 20 'entities' (observations) you will
have 481 PCs with dead zero eigenvalues. How small is 'smaller' and how
many is a few?
Everyone who has responded to this seems to accept the idea that PCA is
the way to go here, but that is not clear to me at all. There
Here is a step by step explanation.
The way you present the data is as species (rows) by sites (columns)
data frame
dim(species_x_sites)
[1] 17 20
There are in fact only 19 sites as one of the columns of the data frame
is the species name:
names(species_x_sites)
[1] Species Cuba
AB - with(B, subset(A, coords.x1 %in% X1))
AB
coords.x1 coords.x2
0 542250.9 3392404
7 541512.5 3394722
8 541479.3 3394878
9 538903.4 3395943
18 543274.0 3389919
19 543840.8 3392012
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[EMAIL PROTECTED]
Sent: Tuesday, 10 July 2007 9:25 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Within matrix
Hi all,
I am working on cluster, I
The message from this cute little data set is very clear. Consider
fm - aov(resp ~ A*B + A/C, mydata)
drop1(fm, test = F)
Single term deletions
Model:
resp ~ A * B + A/C
Df Sum of Sq RSS AIC F value Pr(F)
none 65.540 47.261
A:B 216.132
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for the upper tail:
1-pt(1.11, 9)
[1] 0.1478873
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Nair, Murlidharan
T
Sent: Thursday, 2 August 2007 4:43 AM
To: r-help@stat.math.ethz.ch
Subject: [R] t-distribution
If I have a calculated t can I get the
: +61 7 3286 7700
mailto:[EMAIL PROTECTED]
http://www.cmis.csiro.au/bill.venables/
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Ben Bolker
Sent: Thursday, 2 August 2007 4:57 AM
To: r-help@stat.math.ethz.ch
Subject: Re: [R] t-distribution
Bill.Venables
: +61 7 3286 7700
mailto:[EMAIL PROTECTED]
http://www.cmis.csiro.au/bill.venables/
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of João Fadista
Sent: Wednesday, 5 September 2007 11:51 PM
To: r-help@stat.math.ethz.ch
Subject: [R] length of a string
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