On Mon, Apr 11, 2011 at 10:49 PM, Erin Hodgess erinm.hodg...@gmail.com wrote:
Dear R People:
I have a data frame with the following column names:
names(funky)
[1] UHD.1 UHD.2 UHD.3 UHD.4 L..W..1 L..W..2 L..W..3
[8] L..W..4 B..W..1 B..W..2 B..W..3 B..W..4 W..B..1 W..B..2
[15]
Hi Erin,
Please read ?grep. It is clearly not the function you want (neither
is strsplit() either really). This does what you want and you can
modify for upper/lower case if you need it. Also note that regular
expressions exist separate from R, so while : may have seemed
natural to select a
On Mon, 11 Apr 2011, ty ty wrote:
Hello, dear experts. I don't have much experience in building
regression models, so sorry if this is too simple and not very
interesting question.
Currently I'm working on the model that have to predict proportion of
the debt returned by the debtor in some
On Mon, Apr 11, 2011 at 12:49 AM, Jeff Stevens stev0...@gmail.com wrote:
Many thanks, Peter. This works brilliantly, and I prefer to have the
labels assigned outside of panel function as well.
You could also consider using which.packet(). You haven't explicitly
told us how the labels are
I am trying to optimize a nested function using nlminb. This throws out an
error that y is missing. Can someone help me with the correct syntax?? Thank
you.
test1 - function(x,y)
{
sum - x + y
return(sum)
}
test2 - function(x,y)
{
sum - test1(x,y)
sumSq - sum*sum
return(sumSq)
}
does some know iterative methods in R ? thank you!
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R-help@r-project.org mailing
P.S.: It is bad taste to call a variable 'sum' because it is a R function.
Ciao!
mario
On 12-Apr-11 08:33, vioravis wrote:
I am trying to optimize a nested function using nlminb. This throws out an
error that y is missing. Can someone help me with the correct syntax?? Thank
On Mon, Apr 11, 2011 at 9:27 PM, Szumiloski, John
john_szumilo...@merck.com wrote:
Dear useRs,
I have a longitudinal experiment with several treatment groups, ~20 subjects
per group, ~6 timepoints and a continuous dependent variable. I have been
successfully been using lattice::xyplot with
On Apr 12, 2011, at 08:45 , Achim Zeileis wrote:
On Mon, 11 Apr 2011, ty ty wrote:
Hello, dear experts. I don't have much experience in building
regression models, so sorry if this is too simple and not very
interesting question.
Currently I'm working on the model that have to predict
On Tue, 12 Apr 2011, peter dalgaard wrote:
On Apr 12, 2011, at 08:45 , Achim Zeileis wrote:
On Mon, 11 Apr 2011, ty ty wrote:
Hello, dear experts. I don't have much experience in building
regression models, so sorry if this is too simple and not very
interesting question.
Currently I'm
On Mon, Apr 11, 2011 at 3:53 AM, baptiste auguie
baptiste.aug...@googlemail.com wrote:
Yes, very sorry about this -- I had subconsciously ignored the
hypothetical possibility that anyone wouldn't have ggplot2 loaded in
their .Rprofile ;)
Replacing mpg with beaver1 (datasets) should be more
It is not clear (at least to me) what exactly you want. You want two forest
plots in one graph but apparently not side-by-side or one on top and the
other on the bottom. So, you want to superimpose them? How do you want to do
that without creating an illegible mess? Or do you want one graph,
Hi hnlidexi
On 10 April 2011 02:47, 李德洗 hnlid...@foxmail.com wrote:
Dear all,
I want to finished my paper by latent class Stochastic Frontier
Analysis , but i can not find the package, is there anyone that
may help me
As far as I know, there is no R function/package that estimates latent
rcom has its own mailing list.
Please subscribe at rcom.univie.ac.at
and post all questions there.
On 4/12/2011 1:04 AM, array chip wrote:
Yes, I did, and no error message. And comRegisterRegistry() returns NULL, not
sure if that matters
John
On 2011-04-11 13:29, Felix Nensa wrote:
Hi,
I am using nls to fit a non linear function to some data but R keeps giving
me singular gradient matrix at initial parameter estimates errors.
For testing purposes I am doing this:
### R code ###
x- 0:140
y- 200 / (1 + exp(17 - x)/2) * exp(-0.02*x)
Dear all,
I'm running a coxph model of the form:
coxph(Surv(Start, End, Death.ID) ~ x1 + x2 + a1 + a2 + a3)
Within this model, I would like to compare the influence of x1 and x2 on the
hazard rate.
Specifically I am interested in testing whether the estimated coefficient
for x1 is equal (or not)
On 11-04-12 2:33 AM, vioravis wrote:
I am trying to optimize a nested function using nlminb. This throws out an
error that y is missing. Can someone help me with the correct syntax?? Thank
you.
See ?nlminb. The parameters to optimize should be given as a vector in
a single parameter. Other
On 2011-04-11 09:44, PhDGuy wrote:
Hello,
I am using the function simple.violinplot from the package UsingR.
I have some outliers in my dataset so that the distribution has very long
tails.
As a result, the y-axis of the output of simple.violinplot extends to very
large values. I would like to
On 04/12/2011 12:29 AM, dirknbr wrote:
I am looking for good examples of visualising a tabulation using
plot(table()) maybe with colour coding or indexing.
Hi Dirk,
A basic function for this is addtable2plot in the plotrix package. No
color coding, but you can have row and column names.
Jim
Thanks Duncan and Mario. It works fine now.
On Tue, Apr 12, 2011 at 3:11 PM, Duncan Murdoch murdoch.dun...@gmail.comwrote:
On 11-04-12 2:33 AM, vioravis wrote:
I am trying to optimize a nested function using nlminb. This throws out an
error that y is missing. Can someone help me with the
Thank you for your reply. I would like to have two forest plots one on top and
the other on the bottom. I am using R version 2.12.2 (32-bit) Windows. The code
you sent me still plotting two windows one after the other?
Best wishes and many thanks,
Samor
--- On Tue, 12/4/11, Viechtbauer
Hi there,
I really tried hard to understand and find my own solution, but now I
think I have to ask for your help.
I already developed some script code for my problem but I doubt that it
is correct.
I have the following problem:
Image you develop a logistic regression model with a binary
Hi Duncan,
I've just read the help for the plot3d function you've written and tried out
the example code given at the bottom of the page. I then modified the function
by adding an aspect ratio of 0.1 and ran it again, which crashed my R. I am
using 2.12.2 on Windows, if that tells you
Dear R-users,
I would like to use optim( ) to minimize a function which depends on 4
parameters: 2 vectors, a scalar, and a matrix.
And I have a hard to define the parameters at the beginning of the function,
and then to call optim. Indeed, all the examples I have seen dont treat cases
where
Hi Michael,
One way to work out you problem is with bootstrap methods. The following is
a toy example that may be help you for you:
# test data...
library(survival)
set.seed(1007)
x - runif(50)
mu - c(rep(1, 25), rep(2, 25))
test1 - data.frame(Time = qsurvreg(x, mean = mu, scale= 0.5,
You said that you do NOT want to use par(mfrow=c(2,1)). Why not?
Isn't this (below) what you want?
library(metafor)
data(dat.bcg)
windows(height=8, width=6, pointsize=10)
par(mfrow=c(2,1))
dat - escalc(measure=RR, ai=tpos, bi=tneg, ci=cpos, di=cneg, data=dat.bcg)
forest(dat$yi, dat$vi,
On 2011-04-12 02:50, Peter Ehlers wrote:
On 2011-04-11 09:44, PhDGuy wrote:
Hello,
I am using the function simple.violinplot from the package UsingR.
I have some outliers in my dataset so that the distribution has very long
tails.
As a result, the y-axis of the output of simple.violinplot
Hi all,
I find myself sometimes in the situation where I lapply over a list and
in the particular function I'd like to use the name and or position of
the respective list item. What I usually do is to use mapply on the list
and the names of the list / a position list:
o - list(A=1:3, B=1:2, C=1)
Hi:
Another possible approach (untested) would be to compare the two models
m1 - coxph(Surv(Start, End, Death.ID) ~ x1 + x2 + a1 + a2 + a3)
m0 - coxph(Surv(Start, End, Death.ID) ~ I(x1 + x2) + a1 + a2 + a3)
anova(m0, m1)
This should be able to test H_0: beta_1 = beta_2. If you want to test that
rvohen bingbingzhang87 at 126.com writes:
does some know iterative methods in R ? thank you!
Please read the posting guide and ask a more specific
question; then someone may be able to help.
Ben Bolker
__
R-help@r-project.org mailing list
as.table(as.matrix(X))
Uwe Ligges
On 11.04.2011 22:54, dirknbr wrote:
Ok it looks like this
x yz
a 12 12 34
b 34 34 35
c 56 78 0
where the numbers are counts of cases
I can read it in, but how do I tell R it's a table?
Dirk
--
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Ignacio Quintero ignacioquinterom at gmail.com writes:
I am making some DNA distances and I would like to use dist.dna as matrix in
R, but this command does not include models like GTR...
Is there some command or alternative using dist.dna in ape package for
models not included like GTR?
I have a categorical variable in a dataframe similar to the following...
cat
1
1
3
2
4
I need to convert it to 4 dichotemous variables for each observations like...
cat1cat2cat3cat4
1 0 0 0
1 0 0 0
0 0 1 0
0 1 0
I don't have access to that article, but just reading the abstract, it
should be quite easy to do by writing a wrapper function that calls
randomForest(). I've done so with random projections before. One
limitation to methods like these is that they only apply to all numeric
data.
Andy
On 11.04.2011 21:03, Douglas Bates wrote:
The first thing to do is try another mirror. The official (or as
official as we ever get about anything) U.S. mirror is
http://cran.us.R-project.org
They tend to be very good about updating. Presently the source
package for plyr is at version 1.5
Dear list,
I have a fitted nlme object from which I want to produce estimates of
the (marginal) likelihood for new data, given the fitted model. I am trying to
cross validate a number of nonlinear mixed effects models, and I would
like to find a way to calculate the marginal likelihood for
Hello
Im struggling on something... I have one continuous variable (A), and I need
to explain it with 4 factors, and maybe one continuous covariate.
And of course, my variable A is not normal at all (it's a duration in
seconds, whole numbers).
What can I do? I would know how do deal with it if
Since R is built for statistics almost all is based on iterative methods
converging towards something.
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Sent from the R help mailing list archive at Nabble.com.
hi:
here is one solution:
cat-as.factor(c(1,1,3,2,4))
model.matrix(~cat-1,cat)
cbind(cat,model.matrix(~cat-1,cat))
Andrija
On Tue, Apr 12, 2011 at 2:17 PM, Shane Phillips sphill...@lexington1.netwrote:
I have a categorical variable in a dataframe similar to the following...
cat
1
1
3
2
Hi Sadaf,
Out of curiosity, what sorts of things have you tried to fix this?
For example, after playing around with this a bit, if I remove your
eps parameter from your `ranges` list, it works.
Perhaps you should try tweaking the values you pick for your
parameters. You don't even have to put
Hi Shane,
An alternative is:
cat-as.factor(c(1,1,3,2,4))
outer(cat, levels(cat), ==)+0
Cheers,
Pablo
- Original Message -
From: andrija djurovic djandr...@gmail.com
To: Shane Phillips sphill...@lexington1.net
Cc: r-help@r-project.org
Sent: Tuesday, April 12, 2011 2:32 PM
Subject:
Im struggling on something... I have one continuous variable (A), and I
need
to explain it with 4 factors, and maybe one continuous covariate.
And of course, my variable A is not normal at all (it's a duration in
seconds, whole numbers).
What can I do? I would know how do deal with it if I
Another approach is to use the rms package:
require(rms)
f - cph(Surv(...) ~ x1 + x2 + ...)
contrast(f, list(x1=1, x2=0), list(x1=0, x2=1))
Frank
djmuseR wrote:
Hi:
Another possible approach (untested) would be to compare the two models
m1 - coxph(Surv(Start, End, Death.ID) ~ x1 + x2
Hi Deepayan
Many thanks for bringing the which.packet() function to my
attention--I missed it in the book. I do prefer that the panel
function applies the labels to the data rather than adding them
directly to the data in advance.
Jeff
On Tue, Apr 12, 2011 at 8:50 AM, Deepayan Sarkar
The problem was on the ArcGIS 10 end. The program corrupted my layers. I
re-created the layer and now I get the same answer. Thank you all for your
insight.
Alexis
PS, the scale is in terms of owl territories and therefore 1.3 km is really
huge.
--
View this message in context:
Hi,
I've a list of list.
I want to extract an element by the rownames.
I can extract it by:
data[[1]][[1]][[4]][1]
But I want to exctract it by a command like this:
data[[1]][[B0]][[smac]][[cont]][1]
It's possible?
Thanks,
Alfredo
str(data)
List of 1
$ :List of 4
..$ :List of 4
Thank you very much for your suggestion...that works perfectly.
Thanks,
Andrew
On Mon, Apr 4, 2011 at 5:46 PM, Prof Brian Ripley rip...@stats.ox.ac.uk wrote:
On Mon, 4 Apr 2011, Andrew Yee wrote:
This has to do with using pipe() and grep and read.csv()
I have a .csv file that I grep using
On 12.04.2011 14:52, Ian Davis wrote:
Uwe,
I really do appreciate the help from you and Douglas on resolving this
problem. However, I respectfully disagree on a few points:
- install.packages() was broken and failing to install successfully.
install.packages() works well, it is *not*
Dear all,
I have performed a simple logistic regression using the lrm function from
the Design library. Now I want to plot the summary, or make a nomogram. I
keep getting a datadist error: options(datadist= m.full ) not created with
datadist.
I have tried to specify datadist beforhand (although I
I trained a linear svm and did classification. looking at the model I
have, with a binary response 0/1, the decision values look like this:
head(svm.model$decision.values)
2.5
3.1
-1.0
looking at the fitted values
head(svm.model$fitted)
1
1
0
So it looks like anything less than or equal 0 is
Dear list,
i'm checking the residuals plots of a gam model after a processus of model
selection. I found the best model, all my terms are significant, the
r-square and the deviance explained are good, but I have strange residuals
plots:
http://dl.dropbox.com/u/1169100/gam.check.png
Here is my original script.
subject=1:1000
treat=rbinom(1*1000,1,.13)
gender=rbinom(1*1000,1,.5)
eth=runif(1*1000, min=1, max=4)
cogat=rnorm(1*1000, 100, 16)
map=rnorm(1*1000, 200, 9)
growth=0
simtest=data.frame (subject=subject, treat=treat, gender=gender,
eth=round(eth,digits=0),
Hello,
It appears you are using the igraph package. You can get rid of the arrows
using a few different methods.
You can make the graph undirected:
new.graph - as.undirected(old.graph)
plot(new.graph)
Or you can simply specify the size of the arrows as zero using either
Hello!
Below is my exmample. myref is my reference data frame with columns a and b.
temp is my data with column c analogous to column a in myref.
I am trying to create a new variable b - in temp - that matches
values from b in myref to values in c. If you look at the resulting
data frame (temp -
On 12.04.2011 17:08, Ian Davis wrote:
2011/4/12 Uwe Ligges lig...@statistik.tu-dortmund.de
mailto:lig...@statistik.tu-dortmund.de
On 12.04.2011 14:52, Ian Davis wrote:
I really do appreciate the help from you and Douglas on
resolving this
problem. However, I
2011/4/12 Uwe Ligges lig...@statistik.tu-dortmund.de
On 12.04.2011 14:52, Ian Davis wrote:
I really do appreciate the help from you and Douglas on resolving this
problem. However, I respectfully disagree on a few points:
- install.packages() was broken and failing to install successfully.
Uwe,
I really do appreciate the help from you and Douglas on resolving this
problem. However, I respectfully disagree on a few points:
- install.packages() was broken and failing to install successfully. This
is why I was tracking down the URL via the CRAN website.
- I think it's crazy to
Hello,..
Apologies for the newbie question but...
I have a matrix R, and I know that *B %*% t(b) = R*
*I'm trying to solve for B *(aka. 'factoring the correlation matrix' I
think)
Please help!
I've read that 'to solve for B we define the eigenvalues of R and then
apply the techniques of
I'm sure this must be trivial, but I'm a novice with R and can't work
out how to handle the axes when I am constructing multiple plots on a
page and try to return to a previous one to put multiple data sets it.
A simple example:
---
x- 1:10
y- (1:100)*3
par(mfcol=c(2,1))
plot(x)
plot(y)
First of all I should say this email is more of a general statistics questions
rather than being specific to using R but I'm hoping that this may be of
general interest.
I have a dataset that I would really like to use PCA on and have been using the
package pcaMethods to examine my data. The
Thanks
Im not sure about the gamma, but a survival analysis seems appropriate, but
does it work for factors and continuous covariates? Do you have to verify
some conditions beforehand?
--
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Thanks, That is a nice one. Is there any option that I can plot the pooled
estimate?
Regards,
Samor
--- On Tue, 12/4/11, Viechtbauer Wolfgang (STAT)
wolfgang.viechtba...@maastrichtuniversity.nl wrote:
From: Viechtbauer Wolfgang (STAT) wolfgang.viechtba...@maastrichtuniversity.nl
Subject: RE:
This probably is not ideal, but this works on a list of mine..
## so you can see the structure of my list
str(srMT)
List of 4
$ mode : chr discrete
$ ks.stat : chr mean
$ observed :List of 4
..$ filter: num [1:13, 1:4] 0.213 0.207 0.144 0.311 0.24 ...
.. ..- attr(*,
I have two matrices A and B
dim (A)
[1] 30380 104
dim(Bt)
[1] 30380 63
I want to combine both A and B to matrix C where
dim(C)
[1] 30380 167
How do I do that ?
Regards,
Pankaj Barah
Department of Biology,
Norwegian University of Science Technology (NTNU)
Realfagbygget,
Hello.
I found a solution that may interest others.
Recall that my problem was how to use R to decompose a matrix into the
product of a matrix and its transpose. or, symbolically:
For known matrix M (3x3 matrix) and unknown matrix F and its
transpose t(F)
where F * t(F) = M
Then, can we have the ERROR message, please?
Otherwise the only explanation I can guess is that a mirror grabs the
contents of a repository exactly in the second the repository is updated and
that is unlikely, particularly if more than one mirror is involved.
Isn't one possible explanation
?cbind should solve your problem
2011-04-12 07:54 keltezéssel, pankaj borah írta:
I have two matrices A and B
dim (A)
[1] 30380 104
dim(Bt)
[1] 3038063
I want to combine both A and B to matrix C where
dim(C)
[1] 30380 167
How do I do that ?
Regards,
Pankaj Barah
On 04/12/2011 10:54 AM, pankaj borah wrote:
I have two matrices A and B
dim (A)
[1] 30380 104
dim(Bt)
[1] 3038063
I want to combine both A and B to matrix C where
dim(C)
[1] 30380 167
How do I do that ?
Assuming that Bt is the transpose of B and you want C = [A|Bt] you
Dear Lucy,
not an R-related response at all, but if it's questionnaire data, I'd
probably try to do dimension reduction in a non-automated way by defining
a number of 10 or so meaningful scores that summarise your questions.
Dimension reduction is essentially about how to aggregate
the given
Dimitri,
It isn't clear to me exactly what you are trying to do, but this might
be closer.
Note the stringsAsFactors argument I added to data.frame: I don't think you
are likely to want factors for this application. Also, it's a bad idea
to create a
variable named c since that is the name of a
There are easier solutions. Suppose you have a matrix A, such as:
### Use the info from lm() help
ctl - c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt - c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group - gl(2,10,20, labels=c(Ctl,Trt))
weight - c(ctl, trt)
lm.D9 - lm(weight ~
Is anyone aware of a fast way of doing fisher's exact test for a series of 2
x 2 tables in R? The fisher.test is really slow if n1=1000 and n2 = 1000.
If you don't require exact two-sided p-values (determined according to a
likelihood criterion as in fisher.test), you can use the vectorised
Try this.
===
x- 1:10
y- (1:100)*3
par(mfcol=c(2,1))
plot(x, type=o)
plot(y)
===
--- On Tue, 4/12/11, James Annan jdan...@jamstec.go.jp wrote:
From: James Annan jdan...@jamstec.go.jp
Subject: [R] multiple lines on multiple plots
To:
Shane,
Does this work?
# Your simulated data
subject=1:1000
treat=rbinom(1*1000,1,.13)
gender=rbinom(1*1000,1,.5)
eth=runif(1*1000, min=1, max=4)
cogat=rnorm(1*1000, 100, 16)
map=rnorm(1*1000, 200, 9)
growth=0
simtest=data.frame (subject=subject, treat=treat, gender=gender,
Hi,
On Tue, Apr 12, 2011 at 10:54 AM, Saeed Abu Nimeh sabun...@gmail.com wrote:
I trained a linear svm and did classification. looking at the model I
have, with a binary response 0/1, the decision values look like this:
head(svm.model$decision.values)
2.5
3.1
-1.0
looking at the fitted
BTW,
The same solution can be found using SVD (Singular Value Decomposition)
example,
## Define the matrix that we want to decompose into the product of a
matrix and its transform
M-matrix(c(0.6098601, 0.2557882, 0.1857773,
0.2557882, 0.5127065, -0.1384238,
I'm thinking this isn't what you want.. but also:
data.frame((srMT[[3]][1]))[b,][2]
filter.vel2
MK_SP10.2503257
SB1_SP1 0.2075117
SB4_SP1 0.2358855
B77S wrote:
This probably is not ideal, but this works on a list of mine..
## so you can see the structure of my list
For non-square matrices, see qr and svd.
Spencer
On 4/12/2011 8:43 AM, Doran, Harold wrote:
There are easier solutions. Suppose you have a matrix A, such as:
### Use the info from lm() help
ctl- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt-
Thank you, Sarah. This seems to be working:
a=c(ba ba,ca ca,da da, lake lake, a, lake lake, b,lake
of,lama ca, a,lama ca, b,ma ma)
b=c(ba ba,ca ca,OTHER, lake lake, a, lake lake, b,lake
of,lama ca, a,lama ca, b,OTHER)
myref-data.frame(a=a, b=b)
myref$a-as.character(myref$a)
Hi Peter,
thank you for your reply. Now I see, that P3 is indeed redundand.
But with the simplified model...
fit = nls(yeps ~ p1 / (1 + exp(p2 - x)) * exp(p4 * x))
...nls still produces the same error.
Any ideas?
Felix
2011/4/12 Peter Ehlers ehl...@ucalgary.ca
On 2011-04-11 13:29, Felix
I gave you a solution for the triangular matrix. Can you explain why that is
not what you need?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Shawn Koppenhoefer
Sent: Tuesday, April 12, 2011 11:37 AM
To: r-help@r-project.org
The curve is caused by the zeroes in your data.
raw.residual = response - fitted
so if response=0 then
raw.residual = -fitted.
You get a curve, rather than a straight line, on the fitted vs residual
plot because the residuals are standardised in a way that also depends
on the fitted value.
Instead of trying to go back to a previous plot, gather up all the
data for the plots and generate each one with the appropriate data.
This is much easier than trying to keep track of what the dimensions
are. Also if the data you want to add is outside the plot, then you
have issues with
When a function I have stop()s, I'd like it to return its evaluation
frame, but not halt execution of the script. In experimenting with this,
I became confused with dump.frames. From ?dump.frames:
If ‘dump.frames’ is installed as the error handler, execution will
continue even in
Use a more realistic starting point instead of the default one:
fit - nls(yeps ~ p1 / (1 + exp(p2 - x)) * exp(p4 * x),
start=list(p1=410,p2=18,p4=-.03))
This works for me:
fit
Nonlinear regression model
model: yeps ~ p1/(1 + exp(p2 - x)) * exp(p4 * x)
data: parent.frame()
p1
Hi Mario,
yes works great. Thanks!
2011/4/12 Mario Valle mva...@cscs.ch
Use a more realistic starting point instead of the default one:
fit - nls(yeps ~ p1 / (1 + exp(p2 - x)) * exp(p4 * x),
start=list(p1=410,p2=18,p4=-.03))
This works for me:
fit
Nonlinear regression model
model:
chirine wolley wolley.chirine at hotmail.com writes:
Dear R-users, I would like to use optim( ) to minimize a function
which depends on 4 parameters: 2 vectors, a scalar, and a matrix.
And I have a hard to define the parameters at the beginning of the
function, and then to call optim.
Please follow posting guide and provide a simple completely self-contained
example showing the error. Then I'll take a look.
Frank
Sander wrote:
Dear all,
I have performed a simple logistic regression using the lrm function from
the Design library. Now I want to plot the summary, or make
Or perhaps this as an example of using lines() rather than just getting a line
and dot output in the upper graph.
x - 1:10
y - (1:100)*3
z - rnorm(100, 150, sd= 75)
a - rnorm(10,mean=5, sd= 2.5)
par(mfcol=c(2,1))
plot(x)
lines(a, col= red)
plot(y)
lines(z , col=blue)
--- On Tue,
Dear colleagues,
I am trying to get someone to use R on MS Windows with the browser
disabled. My question is how does he/she get to use R-help which goes
off the browser (and correspondingly complains about the inability to
start firefox, etc). In linux, which is what I use, this is not a
I was given a list of parameter estimates from my boss. She wants to predict
the
dependent variable fsshen beyond jan 2011, the last observation, through
December 2011, giving the prediction intervals (90%). I don't know if I have
the
complete information to do this. So my question(s) is can
try:
options(help_type = 'text')
?options
If this works, you can create a site profile (A default is created
automatically in windows, if I remember correctly) where you can set this
to run in each session.
--
Jonathan P. Daily
Technician - USGS Leetown
el.romaro at gmail.com writes:
Hi Duncan,
I've just read the help for the plot3d function you've written
and tried out the example code given at the
bottom of the page. I then modified the function by adding
an aspect ratio of 0.1 and ran it again, which
crashed my R. I am using
Dear R Helpers,
I am trying to write a character value to the row of a data frame and am
running into a problem that I don't have when I do this for numeric
arguments. For example, the following works just fine:
test-data.frame(number=numeric(1))
test[1,]-.5
test
number
10.5
But the
Hi John,
The error arises because the hold data frame contains factors. This
happens by default when creating data frames with character data as
you did. Factors have a set number of levels and new values cannot be
assigned to them that fall outside their specified levels, hence the
error. The
That column of your data frame contains a factor, rather than character
values. You don't tell us how you created the data frame, but you might
be interested in the stringsAsFactors option to data.frame() and read.table().
Or, if you do actually want a factor for that column, you can use factor()
Hi Zeda,
The short answer to your question is no. All you have are the
parameter estimates with no information on variability which you would
need to create a prediction interval. Given the great ease of fitting
models in R, I would offer to refit the model for my boss, and then
(with the model
I will do that. but I may be back. Thanks.
From: Joshua Wiley jwiley.ps...@gmail.com
Cc: r-help@r-project.org
Sent: Tue, April 12, 2011 1:59:20 PM
Subject: Re: [R] Predictkion interval using regression coefficients
Hi Zeda,
The short answer to your question
Anybody use fgui? I find it very handy.
I have a question about an argument to the call to gui in the first
example. The argument is argFilter, my question has to do with specifying
multiple filters. The argument in that example is
argFilter=list(flname={{Text files} {.txt}}) ) ##
I am running a regression equation and i want to enter in 12 IV then stepwise
enter 8 variables and not have an origin.
DV is shfl. I
want to enter in the following 12 independent dummy variables
ajan
bfeb
cmar
dapr
emay
fjun
And then I want to enter in a stepwise fashion
slag6
slag6
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