Folks,
The betareg package appears to be polished and works well. But I would
like to look at the exact formulas for the underlying model being
estimated, the likelihood function, etc. E.g. if one has to compute
\frac{\partial E(y)}{\partial x_i}, this requires careful calculations
through these f
A few days ago, I had asked this question. Consider this situation:
> x1 <- runif(100); x2 <- runif(100); y <- 2 + 3*x1 - 4*x2 + rnorm(100)
> m1 <- summary(lm(y ~ x1))
> m2 <- summary(lm(y ~ x2))
> m3 <- summary(lm(y ~ x1 + x2))
You have estimated 3 different "competing" models, and suppose you
wa
The R code I just mailed out had a small error in it. This one
works. Now what one needs is a way to get decimal alignment in LaTeX
tabular objects.
x1 <- runif(100); x2 <- runif(100); y <- 2 + 3*x1 - 4*x2 + rnorm(100)
m1 <- summary(lm(y ~ x1))
m2 <- summary(lm(y ~ x2))
m3 <- summary(lm(y ~ x1 + x
Consider this situation:
> x1 <- runif(100); x2 <- runif(100); y <- 2 + 3*x1 - 4*x2 + rnorm(100)
> m1 <- summary(lm(y ~ x1))
> m2 <- summary(lm(y ~ x2))
> m3 <- summary(lm(y ~ x1 + x2))
Now you have estimated 3 different "competing" models, and suppose you
want to present the set of models in one
Folks,
A few days ago, I had asked a question on this mailing list about
making a contour plot where a function z(x,y) is evaluated on a grid
of (x,y) points, and the data structure at hand is a simple table of
(x,y,z) points. As usual, R has wonderful resources (and subtle
complexity) in doing th
Folks,
The contour() function wants x and y to be in increasing order. I have
a situation where I have a grid in x and y, and associated z values,
which looks like this:
x y z
[1,] 0.00 20 1.000
[2,] 0.00 30 1.000
[3,] 0.00 40 1.000
[4,] 0.00 50 1.0
A few minutes ago I had asked why this didn't seem to work:
# Simulate from probit model --
x1 <- 2*runif(5000)
x2 <- 5*runif(5000)
ystar <- 7 + 3*x1 - 4*x2 + rnorm(5000)
y <- cut(ystar, breaks=c(-100, -5, 0, 5, 100))
table(y)
library(MASS)
summary(polr(y ~ x1 + x2, method="probit"))
A little thi
This part, a vanilla probit, works perfectly --
# Simulate from probit model --
x1 <- 2*runif(5000)
x2 <- 5*runif(5000)
ystar <- 7 + 3*x1 - 4*x2 + rnorm(5000)
y <- as.numeric(ystar>0)
table(y)
# Estimation using micEcon::probit()
library(micEcon)
summary(probit(y ~ x1 + x2))
#
Suppose one has
x <- c(1, 2, 7, 9, 14)
y <- c(71, 72, 77)
How would one write an R function which alternates between elements of
one vector and the next? In other words, one wants
z <- c(x[1], y[1], x[2], y[2], x[3], y[3], x[4], y[4], x[5], y[5])
I couldn't think of
Folks,
I'm doing fine with using orthogonal polynomials in a regression context:
# We will deal with noisy data from the d.g.p. y = sin(x) + e
x <- seq(0, 3.141592654, length.out=20)
y <- sin(x) + 0.1*rnorm(10)
d <- lm(y ~ poly(x, 4))
plot(x, y, type="l"); lines(x, d$fitted.values, col=
I solved the problem in one more (and more elegant) way. So here's the
program again.
Where does R stand on the Anderson-Goodman test of 1957? I hunted
around and nobody seems to be doing this in R. Is it that there has
been much progress after 1957 and nobody uses it anymore?
# Problem statement
On Sun, Jan 22, 2006 at 01:47:00PM +1100, [EMAIL PROTECTED] wrote:
> If this is a real problem, here is a slightly tidier version of the
> function I gave on R-help:
>
> transitionM <- function(name, year, state) {
> raw <- data.frame(name = name, state = state)[order(name, year), ]
> raw01 <-
Folks,
I am holding a dataset where firms are observed for a fixed (and
small) set of years. The data is in "long" format - one record for one
firm for one point in time. A state variable is observed (a factor).
I wish to make a markov transition matrix about the time-series
evolution of that sta
Folks,
Based on
http://www.biostat.wustl.edu/archives/html/s-news/1999-06/msg00125.html
I thought I should experiment with using survreg() to estimate tobit
models.
I start by simulating a data frame with 100 observations from a tobit model
> x1 <- runif(100)
> x2 <- runif(100)*3
> ystar <- 2
I am doing
> library(rpart)
> m <- rpart("y ~ x", D[insample,])
> D[outsample,]
y x
8 0.78391922 0.579025591
9 0.06629211 NA
10 NA 0.001593063
> p <- predict(m, newdata=D[9,])
Error in model.frame(formula, rownames, variables, varnames, extras,
extraname
Folks,
I have placed an example of a self-contained R program later in this
mail. It generates a file inflation.pdf. When I stare at the picture,
I see the "X label string" and "Y label string" sitting lonely and far
away from the axes. How can these distances be adjusted? I read ?par
and didn't f
On Wed, Sep 28, 2005 at 08:23:59AM +0100, Prof Brian Ripley wrote:
> I've not seen a reply to this, nor ever seen it.
> Please make a reproducible example available (do see the posting guide).
It was a mistake on my part. Just in case others are able to
recognise the situation, what was going on w
I have a situation with a large dataset (3000+ observations), where
I'm doing lags as regressors, where I get:
Call:
lm(formula = rj ~ rM + rM.1 + rM.2 + rM.3 + rM.4)
Residuals:
1990-06-04 1994-11-14 1998-08-21 2002-03-13 2005-09-15
-5.64672 -0.59596 -0.041430.554128.18229
Coeffi
Folks,
I am using R 2.1.1 on Apple OS X 10.3.
Earlier, I used to say
$ sudo R
> update.packages()
and all the packages used to get installed.
For several weeks, I noticed that nothing has been coming through. I
used the R-for-Mac graphics console and I find that there are many
packages where
When using get.hist.quote(), I find the dates are broken. This is with
R 2.1.1 on Mac OS X `panther'.
> library(tseries)
Loading required package: quadprog
'tseries' version: 0.9-27
'tseries' is a package for time series analysis and computational
finance.
See 'library(help="tse
I have a data frame with one column "x":
> str(data)
`data.frame': 20 obs. of 1 variable:
$ x: num 0.0495 0.0986 0.9662 0.7501 0.8621 ...
Normally, I know that the notation dataframe[indexes,] gives you a new
data frame which is the specified set of rows. But I find:
> str(data[1:10,])
num
I have written two functions which do useful things with panel data
a.k.a. longitudinal data, where one unit of observation (a firm or a
person or an animal) is observed on a uniform time grid:
- The first function makes lagged values of variables of your choice.
- The second function makes g
I'm in a situation where I say:
> predict(m.rpart, newdata=D[N1+t,])
0 1
173 0.8 0.2
which I interpret as meaning: an 80% chance of "0" and a 20% chance of
"1". Okay. This is consistent with:
> predict(m.rpart, newdata=D[N1+t,], type="class")
[1] 0
Levels: 0 1
But I'm puzzled at the fol
On Mon, Jul 11, 2005 at 08:27:40AM -0700, Rob J Goedman wrote:
> Ajay,
>
> After installing both setRNG (2004.4-1, source or binary) and dse
> (2005.6-1, source only), it works fine.
Thanks! :-) Now dse1 works, but I get:
> library(dse2)
Warning message:
replacing previous import: acf in: name
I have a situation where this is fine:
> if (length(x)>15) {
clever <- rr.ATM(x, maxtrim=7)
} else {
clever <- rr.ATM(x)
}
> clever
$ATM
[1] 1848.929
$sigma
[1] 1.613415
$trim
[1] 0
$lo
[1] 1845.714
$hi
[1] 1852.143
But this variant, using ifelse(),
> > I have a program which is doing a few thousand runs of lm(). Suppose
> > it is a simple model
> > y = a + bx1 + cx2 + e
> >
> > I have the R object "d" where
> > d <- summary(lm(y ~ x1 + x2))
> >
> > I would like to obtain Var(x2) out of "d". How might I do it?
> >
> > I can, of course,
I have a program which is doing a few thousand runs of lm(). Suppose
it is a simple model
y = a + bx1 + cx2 + e
I have the R object "d" where
d <- summary(lm(y ~ x1 + x2))
I would like to obtain Var(x2) out of "d". How might I do it?
I can, of course, always do sd(x2). But it would be much
It is a very nice touch that optim() offers SANN (simulated annealing)
as a random search algorithm.
The R community already has genoud - an implementation of a genetic
algorithm for search.
Wouldn't it be neat if optim() would additionally offer method="GA"
where it internally uses code from gen
I learned R & MLE in the last few days. It is great! I wrote up my
explorations as
http://www.mayin.org/ajayshah/KB/R/mle/mle.html
I will be most happy if R gurus will look at this and comment on how
it can be improved.
I have a few specific questions:
* Should one use optim() or should one
> CAN YOU TELL ME HOW TO FIT FIXED-EFFECTS MODEL WITH R? THANK YOU!
Ordinary lm() might suffice.
In the code below, I try to simulate a dataset from a standard
earnings regression, where log earnings is quadratic in experience,
but the intercept floats by education category - you have 4 intercep
I wrote a simple log likelihood (for the ordinary least squares (OLS)
model), in two ways. The first works out the likelihood. The second
merely calls the first, but after transforming the variance parameter,
so as to allow an unconstrained maximisation. So the second suffers a
slight cost for one
Thanks to Gabor for setting me right. My code is as follows. I found
it useful for learning optim(), and you might find it similarly
useful. I will be most grateful if you can guide me on how to do this
better. Should one be using optim() or stats4::mle?
set.seed(101) # F
> well ESS has such a facility.
>
> However, I think Mathematica has a super scheme: unbalanced brackets
> show up
> in red, making them obvious.
>
> This is particularly good for spotting wrongly interleaved brackets, as
> in
>
> ([ blah di blah )]
>
>
>
> in which case both opening brac
I am using R 2.1 on Apple OS X.
When I get the ">" prompt, I find it works well with emacs commandline
editing. Keys like M-f C-k etc. work fine.
The one thing that I really yearn for, which is missing, is bracket
matching When I am doing something which ends in it is really
useful to have e
Folks,
I'm in a situation where I do a few thousand regressions, and some of
them are bad data. How do I get back an error value (return code such
as NULL) from lm(), instead of an error _message_?
Here's an example:
> x <- c(NA, 3, 4)
> y <- c(2, NA, NA)
> d <- lm(y ~ x)
Error in lm.fit(x, y, o
Yesterday, I had asked for help on the list. Brian Ripley and Bruno
Falissard had most kindly responded to me. Here is the solution.
> factorlabels <- c("School", "College", "Beyond")
> # 1 2 3
> education.man <- c(1,2,1,2,1,2,1,2) # PROBLEM: Level "
I'm in this situation:
factorlabels <- c("School", "College", "Beyond")
with data for 8 families:
education.man <- c(1,2,1,2,1,2,1,2) # Note : no "3" values
education.wife <- c(1,2,3,1,2,3,1,2) # 1,2,3 are all present.
My goal is to create this table:
Should I be worried? The installation seems to go through fine and
apparently nothing is broken. The errors I repeatedly get are like this:
g++ -no-cpp-precomp -I/Library/Frameworks/R.framework/Resources/include -I/usr/
local/include -DUNIX -DOPTIM -DNONR -fno-common -g -O2 -c unif.cpp -o unif.
R is so smart! I found that when you switch a column from integer to
factor, the memory consumption goes down rather impressively.
Now I'd like to learn more. How does R do this? What does R do? How do
I learn more?
I got to thinking: If I was really smart, I'd see that a factor with 2
levels req
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