Hi
Here's an example of relevel used to relevel and combine groups
InsectSprays2-InsectSprays
levels(InsectSprays2$spray)
levels(InsectSprays2$spray)-list(new1=c(A,C),YEPS=c(B,D,E),LASTLY=F)
levels(InsectSprays2$spray)
InsectSprays2
So for you try...
levels (Data1$Site) -
Dear all
How does one resolve the variance between the actual availability of data
sets in the default implementation with those mentioned in the
documentation?
I am unable to attach some of the datasets, even though help() is available
for the same datasets. For example, ToothGrowth is a dataset
In the following data.frame there are 6 columns, but 7 are written to
the CSV file.
install.packages(pmlr)
library(pmlr)
data(enzymes)
write.table(enzymes, sep=,, eol=\n,file=albert.csv)
__
R-help@r-project.org mailing list
Hi
In the following data.frame there are 6 columns, but 7 are written to
the CSV file.
install.packages(pmlr)
library(pmlr)
data(enzymes)
write.table(enzymes, sep=,, eol=\n,file=albert.csv)
see row.names option in ?write.table
Regards
Petr
On 15.08.2011 06:33, Sierra Bravo wrote:
Dear all
How does one resolve the variance between the actual availability of data
sets in the default implementation with those mentioned in the
documentation?
I am unable to attach some of the datasets, even though help() is available
for the same
data(ToothGrowth)
makes it available in your workspace.
data(ToothGrowth)
ls()
[1] ToothGrowth
str(ToothGrowth)
'data.frame': 60 obs. of 3 variables:
$ len : num 4.2 11.5 7.3 5.8 6.4 10 11.2 11.2 5.2 7 ...
$ supp: Factor w/ 2 levels OJ,VC: 2 2 2 2 2 2 2 2 2 2 ...
$ dose: num 0.5 0.5
Yup.
Read ?write.csv and note the row.names argument.
---
Jeff Newmiller The . . Go Live...
DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer
I like using saveObject() and loadObject() from package R.utils because
you can assign the data to a new object.
Like this:
library(R.utils)
saveObject(df, file=file.Rbin)
df2 - loadObject(file.Rbin)
HTH,
Ivan
Le 8/12/2011 21:38, peter dalgaard a écrit :
On Aug 12, 2011, at 21:12 , Ed Heaton
Dear Michael,
I adjusted the code a bit;
d1 = DataSplitsCore(Data1[(1:nr1)], Data1[-(1:nr1)], alpha = alpha, level =
level + 1)
d2 = DataSplitsCore(Data2[(1:nr2)], Data2[-(1:nr2)], alpha = alpha, level =
level +1)
The code stops when palpha, which is what I wanted. Only problem is that it
By the way, I indeed want to get the data returned.
From: marinadewo...@hotmail.com
To: michael.weyla...@gmail.com
CC: r-help@r-project.org
Subject: RE: [R] Splitting data
Date: Mon, 15 Aug 2011 10:32:04 +0200
Dear Michael,
I adjusted the code a bit;
d1 =
Hello,
just to follow up a question from last week. Here what I've done so far (here
an example):
library(MCMCpack)
Y=c(15,14,23,18,19,9,19,13)
X1=c(0.2,0.6,0.45,0.27,0.6,0.14,0.1,0.52)
X2a=c(17,22,21,18,19,25,8,19)
X2b=c(22,22,29,34,19,26,17,22)
X2 - function()runif(length(X2a), X2a, X2b)
Thanks Uwe, Dennis.
However, I'm unable to make progress...this is what I get:
data(ToothGrow)
Warning message:
In data(ToothGrow) : data set 'ToothGrow' not found
ToothGrow
Error: object 'ToothGrow' not found
TIA
s.b.
--
View this message in context:
Thanks!
On Mon, Aug 15, 2011 at 12:55 AM, Jeff Newmiller
jdnew...@dcn.davis.ca.us wrote:
Yup.
Read ?write.csv and note the row.names argument.
---
Jeff Newmiller The . . Go Live...
DCN:jdnew...@dcn.davis.ca.us
That would be because ToothGrow is not ToothGrowth.
---
Jeff Newmiller The . . Go Live...
DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries
On 15.08.2011 10:26, Sierra Bravo wrote:
Thanks Uwe, Dennis.
However, I'm unable to make progress...this is what I get:
data(ToothGrow)
Warning message:
In data(ToothGrow) : data set 'ToothGrow' not found
ToothGrow
Error: object 'ToothGrow' not found
What happens if you type
-Original Message-
christopher stratton
Sent: 14 August 2011 22:22
Subject: [R] PCA Using prcomp()
From the results
generated by prcomp(), is there a way to print a matrix
showing the contributions of the original variables to each
PC?
Sounds like you're looking for the
Rebekka Schibli rebekka_schibli at gmx.ch writes:
I am using the function optim and I get the error message
ABNORMAL_TERMINATION_IN_LNSRCH. Reason for this
could be a scaling problem. Thus, I used parscale in order to scale the
parameters. But I still have the
error message. For example,
Dear R-Users
My problem is quite simple: I need to use a fitted model to predict the next
point (that is, just one single point in a curve).
The data was divided in two parts: identification (x and y - class matrix)
and validation (xt and yt - class matrix). I don't use all values in x
Thanks Jeff, Uwe. The last case was indeed because of a typo, where I typed
ToothGrow instead of ToothGrowth. Many apologies...!
s.b.
--
View this message in context:
http://r.789695.n4.nabble.com/Missing-datasets-2-13-1-tp3743896p3744249.html
Sent from the R help mailing list archive at
Hello...
just some additional thoughts:
Maybe I can try it in a simple way with a repeated lm-regression, like:
Y=c(15,14,23,18,19,9,19,13)
X1=c(0.2,0.6,0.45,0.27,0.6,0.14,0.1,0.52)
X2a=c(17,22,21,18,19,25,8,19)
X2b=c(22,22,29,34,19,26,17,22)
X2 - function()runif(length(X2a), X2a, X2b)
for
HI there,
Consider a data set like this:
x - data.frame(a=1:10, b=11:20, t=c(1,1,1,2,2,2,3,3,3,3))
x
a b t
1 1 11 1
2 2 12 1
3 3 13 1
4 4 14 2
5 5 15 2
6 6 16 2
7 7 17 3
8 8 18 3
9 9 19 3
10 10 20 3
Here x$t is a vector of integers that represent a moment
in time. I
On 15/08/11 20:26, Sierra Bravo wrote:
Thanks Uwe, Dennis.
However, I'm unable to make progress...this is what I get:
data(ToothGrow)
Warning message:
In data(ToothGrow) : data set 'ToothGrow' not found
ToothGrow
Error: object 'ToothGrow' not found.
You need to spell it correctly! It's
On 15/08/11 10:44, Uwe Ligges wrote:
On 15.08.2011 10:26, Sierra Bravo wrote:
Thanks Uwe, Dennis.
However, I'm unable to make progress...this is what I get:
data(ToothGrow)
Warning message:
In data(ToothGrow) : data set 'ToothGrow' not found
ToothGrow
Error: object 'ToothGrow' not
Thanks for that Gabor, it works fine from the development version
you've pointed to.
There is in addition a performance issue: the following benchmark ran
in under 0.2s in the previous version, now consistently shows elapsed
time over 14s on a Xeon with Windows. It's unaffected if I use the
Just to inform:
I posted that before in R-sig-ecology but as it might be interesting also for
other useRs, I post it also to the general r-user list:
Hello Alexandre,
thank you very much. I also found another way to extract summarizing
information from lm results over e.g. 1000 repeated model
Wir sind bis am 20. August in den Ferien und werden keine e-mails beantworten.
Bei dringenden Fällen melden Sie sich bei Stefanie von Felten
steffi.vonfel...@oikostat.ch
We are on vacation until 20. August. In urgent cases, please contact Stefanie
von Felten steffi.vonfel...@oikostat.ch
*I have the following function:*
/plot_mi_time = function(mdata, miname) {
mdata2 = mdata[row.names(hakat) == miname, ]
print(mdata2)
xcoords - c(1,1,2,2,3,3,4,4,5,5,6,6)
plot(c(xcoords), mdata2, xaxt=n, ylab=Expression, xlab=Time(h), ,
main=miname)
axis(1, at=xcoords,
I believe it's a problem in your variable name hsa-miR-98. R wants to
interpret that as hsa less miR less 98. Since you don't have a variable
called hsa, R (rightly) complains.
Call ls() and see what R thinks your variables are named -- that will
hopefully make the problem evident.
Hope this
Two things,
1) You need to use square brackets in this case because model is an object,
not a function.
2) You probably want to use a list object to store a whole bunch of model
objects so you'll want double brackets.
model = list(NULL)
for (i in 1:1000) {
model[[i]] - lm(Y~X1+X2())
}
This,
On 11-08-15 7:55 AM, rmje wrote:
*I have the following function:*
/plot_mi_time = function(mdata, miname) {
mdata2 = mdata[row.names(hakat) == miname, ]
You passed the expression hsa-miR-98 as miname. The simplest fix is to pass
hsa-miR-98
instead. A more complicated alternative is to
I can't help with the questions on pooling regressions, but for the question
about data-frame usage, I'll note two things.
One,
x = data.frame(y = 1:3, z = 4:6)
is.data.frame(x)
TRUE
is.data.frame(t(x))
FALSE
is.data.frame(as.data.frame(t(x)))
TRUE
Two, you'll need to check that your row
Hello all,
I have some points for x and f(x). I want to fit the best curev through
these points and find the value of x and f(x) for which the gradient is -1.
Anyone knows if this can be done in R?
--
Thanks,
Jim.
[[alternative HTML version deleted]]
Hello,
I am using poLCA to run standard latent class analyses. Is there any way that I
can get parameters for each predictors, in terms of their association with the
latent variable ?
Thank you,
David Joubert
[[alternative HTML version
There is an error in your call to `optim'. The`lower' bound is incorrect:
lower=c(rep(-1,n),1e-5,1e-5,1e5)
This should be:
lower=c(rep(-1,n),1e-5,1e-5,1e-5)
I am not sure if this fixes the problem, but it is worth a try.
I do not understand your scaling. From the lower and upper bounds it
Tyler,
Tried your solution: levels (Data1$Site - list(Fw = c(AB), Est = c(DE)))
but still got a NULL response to str(Data1) and an alternating list of Fw, Est,
Fw, Est under Site when looked at in the Data editor in R console. The use of
the levels function would seem to be appropriate
Try recode from Deducer package.
Sent from my BlackBerry® smartphone from !DEA
-Original Message-
From: B Jessop deel...@hotmail.com
Sender: r-help-bounces@r-project.orgDate: Mon, 15 Aug 2011 11:36:07
To: r-help@r-project.org
Subject: [R] FW: Renaming levels of a factor in a dataframe
Well, it looks like the standard advice here applies to both poster
and responder: try reading the Help (carefully): ?factor
Example:
z - factor(letters[1:3])
z
[1] a b c
Levels: a b c
factor(z,labels=letters[4:6])
[1] d e f
Levels: d e f
factor(z,labels=letters[3:1])
[1] c b a
Levels: c b a
You might want to compare the brackets in your code and in Tyler's
Ivan
Le 8/15/2011 16:36, B Jessop a écrit :
Tyler,
Tried your solution: levels (Data1$Site- list(Fw = c(AB), Est = c(DE)))
but still got a NULL response to str(Data1) and an alternating list of Fw, Est, Fw, Est
under
Hello,
How can I read a xlsx file using xlsx package?
Thanks
Albert
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Hi,
I have a question concerning the selection of data. Let's say that given
list h created below, I would like to select a section of the 1999 matrix.
For a case (rownames and colnames) I would like to select the cells that
have a value 0. So for case 8025
8025 8026 8027
802511
Two ways to do it, both of which come down to forcing R to recognize ( as
a string element than an actual parenthesis:
x = data.frame(1:3, 4:6)
colnames(x) = c((Intercept),Slope)
x$(Intercept)
or x[,(Intercept)] # Works with or without the comma in slightly different
ways -- figure out which
ls(hakat)
[1] X15h X18h X1h X21h X4h X9h
--
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Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
?read.xlsx
Le 8/15/2011 17:19, albert coster a écrit :
Hello,
How can I read a xlsx file using xlsx package?
Thanks
Albert
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
It workes using
But what is this:
p - plot_mi_time(hakat, hsa-miR-100)
X1h X4h X9h X15h X18h X21h
hsa-miR-100 384 1038 1503 1511 2603 2899
*Error in xy.coords(x, y, xlabel, ylabel, log) :
'x' and 'y' lengths differ*
--
View this message in context:
Problem solved!
--
View this message in context:
http://r.789695.n4.nabble.com/Plot-from-function-tp3744428p3744695.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
Morning All:
Having problems installing this one into my R setup and wondered if
anyone on the list had the same problem and found a workable solution.
First to the system particulars:
OS is the Hardy Heron edition of Ubuntu Linux
R version is 2.13.1
Java installation is sun-java6
Next up is
Brian,
I believe I had a similar problem. Make sure you have JDK (*sun-java6-jdk)
* installed as well.
Here is a link to a JDK install guide
http://www.cyberciti.biz/faq/howto-ubuntu-linux-install-configure-jdk-jre/
On Mon, Aug 15, 2011 at 10:18 AM, Brian Lunergan ff...@ncf.ca wrote:
Morning
Reading data with variable column widths.
Here are several lines of a txt data set I would like to read.
The number of variables is fixed at 13 . The problem is how to read the
first variable when it can contain blank space-- for example Alabama
(Seasonally Adjusted) , St. Clair, etc.
Alabama
On 15/08/2011 11:47 AM, R Saba wrote:
Reading data with variable column widths.
Here are several lines of a txt data set I would like to read.
The number of variables is fixed at 13 . The problem is how to read the
first variable when it can contain blank space-- for example Alabama
(Seasonally
Eduardo;
I think you would be more successful if you put your data in a
dataframe, offered it to lm with column names only in the formula,
and then used the newdata argument with predict with the column names
matching the column names in the original data.
--
David.
On Aug 15, 2011, at
Dear R folks,
I am doing some calculations over an array using sweep and apply.
# Sample Data (from help 'addmargins')
Aye - sample(c(Yes, Si, Oui), 177, replace = TRUE)
Bee - sample(c(Hum, Buzz), 177, replace = TRUE)
Sea - sample(c(White, Black, Red, Dead), 177, replace = TRUE)
(A - table(Aye,
On 11-08-15 12:21 PM, David Winsemius wrote:
Eduardo;
I think you would be more successful if you put your data in a
dataframe, offered it to lm with column names only in the formula,
and then used the newdata argument with predict with the column names
matching the column names in the original
Hello!
I'd like to have a function to draw correct grid while using log axis with
xyplot from lattice package. Right now I have the following code inside of my
panel function:
lim - current.panel.limits()
v - latticeExtra:::logTicks(2^lim$xlim, loc=1)
h - latticeExtra:::logTicks(2^lim$ylim,
Whats going on here?
df-data.frame(x=1:10,y=1:10)
ggplot()+geom_point(data=df,aes(x=x,y=y)) ## this is the normal usage
right?
ggplot()+geom_point(data=df,aes(x=df[,1],y=df[,2])) ## but I can also feed
it column indices
ggplot()+geom_point(aes(x=df[,'x'],y=df[,'y'])) ## or column names.
##
I prefer RODBC and odbcConnectExcel this way I can query subsets with SQL.
Mikhail
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
Behalf Of Ivan Calandra
Sent: Monday, August 15, 2011 10:28 AM
To: r-help@r-project.org
Subject: Re:
Tyler,
My apology. It does help to type correctly and eliminate unnecessary spaces in
commands. The following does work correctly:
levels(Data1$Site) - list(Fw = c(AB), Est = c(DE))
Thanks very much,
BJ
From: deel...@hotmail.com
To:
Hello all,
I have what I think is a simple question but I've been unable to solve it. I
have the following string:
A[states=1]:[rate=2]425, B[states=3]:[rate=5]500
I would like to combine the two expressions in the [], so that only one set
of [] is present after each letter, so that I have the
Dear R-users,
I have an anual info of gross product and I would like to disaggregate
to trimestral data.
Can I import a matlab library of Quilis? (
http://www.mathworks.com/matlabcentral/fileexchange/24438-temporal-disaggregation-library
)
Thanks,
Sebastián.
On Mon, Aug 15, 2011 at 10:34 AM, Rebecca Gray atlas...@gmail.com wrote:
Hello all,
I have what I think is a simple question but I've been unable to solve it. I
have the following string:
A[states=1]:[rate=2]425, B[states=3]:[rate=5]500
I would like to combine the two expressions in the
Hi Peter,
Well the problem is that I need to also move the : to the front of the
bracketed expression, although I like the simplicity of your statement. I
cannot replace all of the [ with :[ because I am actually just using a
small example from a large file, and there are many more [ that are
Hi there
Many thanks. I will try to follow what you two said.
Meanwhile I use the dirty solution that I have just come across.
Cheers
Ed
-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
Sent: Monday, August 15, 2011 1:36 PM
To: David Winsemius
Cc:
Hi Albert,
Another option would be the following:
install.packages('gdata')
require(gdata)
?read.xlsx
HTH,
Jorge
On Mon, Aug 15, 2011 at 11:19 AM, albert coster wrote:
Hello,
How can I read a xlsx file using xlsx package?
Thanks
Albert
[[alternative HTML version deleted]]
Hi, the calibrate.cph() function in rms package generate calibration curve for
Cox model on the same dataset where the model was derived using bootstrapping
or cross-validation. If I have the model built on dataset 1, and now I want to
produce a calibration curve for this model on an
On Aug 15, 2011, at 6:09 AM, mdvaan wrote:
Hi,
I have a question concerning the selection of data. Let's say that
given
list h created below, I would like to select a section of the 1999
matrix.
For a case (rownames and colnames) I would like to select the cells
that
have a value 0.
Is it possible to smooth an ecdf plot and get a probability density plot? I
have about 8000 points and I was hoping to get a density curve instead of a
histogram.
Jeff
__
R-help@r-project.org mailing list
Hi:
Thank you for the reproducible example and expected result. Here's one approach:
library('plyr')
x - data.frame(a=1:10, b=11:20, t=c(1,1,1,2,2,2,3,3,3,3))
x$sdif - cumsum(with(x, a - b))
subfun - function(d) tail(d[c('t', 'sdif')], 1)
ddply(x, 't', subfun)
t sdif
1 1 -30
2 2 -60
3 3
Hi Jeff,
Take a look at ?density
HTH,
Jorge
On Mon, Aug 15, 2011 at 2:38 PM, Jeffrey Joh wrote:
Is it possible to smooth an ecdf plot and get a probability density plot?
I have about 8000 points and I was hoping to get a density curve instead of
a histogram.
Jeff
On Mon, Aug 15, 2011 at 6:51 AM, Giles giles.heyw...@amberalpha.com wrote:
Thanks for that Gabor, it works fine from the development version
you've pointed to.
There is in addition a performance issue: the following benchmark ran
in under 0.2s in the previous version, now consistently shows
Dear Dr. Gilbert, Dr. Varadhan and all R-help list members,
I'm using the function BBsolve( ) and I have some questions on the stopping
criterion maxit and noimp specified in the option control. Here
are two such examples I'm having problem with.In these two examples, the
function
... and ?densityplot in the lattice package.
Please also note that you could have avoided posting just by typing
help.search(density) at the RGui prompt. Please use R's Help
facilities before posting.
-- Bert
On Mon, Aug 15, 2011 at 11:44 AM, Jorge Ivan Velez
jorgeivanve...@gmail.com wrote:
Hi
R-help -
This code iterates over a function with 2 free parameters to find a list of
values (which are the number of incorrect predictions for a computational
model). I want to find the values of i,e when there is the minimum number
of incorrect predictions. In other words, the value of i and e
Hi:
Try this:
df-data.frame(x=1:10,y=1:10)
plot.fun.one - function(dff, x.var, y.var)
print(ggplot(dff, aes_string(x = x.var, y = y.var)) + geom_point() )
plot.fun.two - function(dff, x.var, y.var) {
x - names(dff)[x.var]
y - names(dff)[y.var]
print(ggplot(dff, aes_string(x = x, y =
Hi all,
I'm extracting the name of the term in a regression model that
dropterm specifies as the least significant one, and I'm assigning
this name to an object. However, when I use update(), it ignores this
object. Is there a way I can make it not ignore it? A reproducible
example is below:
Does anyone know how could I get the significant codes from mixed model
output fitted with a GEE package?
The output I got is the following:
GEE: GENERALIZED LINEAR MODELS FOR DEPENDENT DATA
gee S-function, version 4.13 modified 98/01/27 (1998)
Model:
Link: Logit
Hello,
I've a question concerning the display of interval data.
A sample dataset where X is an interval between Xa and Xb
which should be displayed:
Y=c(15,14,23,18,19,9,19,13)
Xa=c(17,22,21,18,19,25,8,19)
Xb=c(22,22,29,34,19,26,17,22)
X = (Xa+Xb)/2
It's easily possible to plot the mean of
Hello All,
I am currently using the vennDiagram function in the limma pkg to
construct venn diagrams. I would like to change the size of the circles
based upon the n of each set (if that makes sense). Does anyone have any
ideas of how to do this?
Thanks,
Phil
[[alternative HTML
HI there, I have been trying to use a code posted on R help to be able to
calculate area under the curve for complicated data points and there seems to
be an issue with the code: no b object found. I am not a good R user and
can't find were the problem is. Any help? Thanks!!
This is the code (
Hi all,
I'm having problems installing plotrix. I tried installing it through
install.packages, and from the unix command line, but each time it seems to
stall when it is installing the help indices.
has anyone had this same problem, is this package still maintained ?
any help?
thanks
I have a dataset and a list of labels. I simply want to apply the labels to the
variables, all at once. The only way I was able to do it was using a loop:
for (i in 1:length(data)) label(data[,i]) - data.labels[i]
I'd like to find the non-loop way to do it, using apply or the like... Any
On Mon, Aug 15, 2011 at 3:53 PM, Monsieur Do nonaupourr...@yahoo.ca wrote:
I have a dataset and a list of labels. I simply want to apply the labels to
the variables, all at once. The only way I was able to do it was using a loop:
for (i in 1:length(data)) label(data[,i]) - data.labels[i]
Build a prediction function using 'Function' that gets applied to set2.
Calibrate and validate.
--
David
Sent from my iPhone
On Aug 15, 2011, at 11:31 AM, array chip arrayprof...@yahoo.com wrote:
Hi, the calibrate.cph() function in rms package generate calibration curve
for Cox model on
Sverre:
As your reproducible example shows, the problem is that the string
y1:y2 is not an acceptable term of the formula. One way round this
is simply convert the non-string part of the formula to a string, pase
it with your term, and then reconvert it to a formula:
Alison Waller alison.waller at embl.de writes:
I'm having problems installing plotrix. I tried installing it
through install.packages, and from the
unix command line, but each time it seems to stall when
it is installing the help indices.
has anyone had this same problem, is this
Why re-invent? Use lrm in rms.
--
David
Sent from my iPhone
On Aug 15, 2011, at 10:07 AM, Mariana Varela mariana.var...@glasgow.ac.uk
wrote:
HI there, I have been trying to use a code posted on R help to be able to
calculate area under the curve for complicated data points and there
Hi:
If you're asking about the p-values, here's a reproducible example
from the gee package:
library('gee')
m - gee(breaks ~ tension, id=wool, data=warpbreaks, corstr=exchangeable)
# Nosing around the summary method return object:
names(summary(m))
[1] callversion
That works! Thanks.
On Mon, Aug 15, 2011 at 5:09 PM, Bert Gunter gunter.ber...@gene.com wrote:
Sverre:
As your reproducible example shows, the problem is that the string
y1:y2 is not an acceptable term of the formula. One way round this
is simply convert the non-string part of the formula to
is there a R function that produces calibration curve on an independetn data
automatically, just like what calibrate() does on the training data itself?
Thanks
John
From: Comcast dwinsem...@comcast.net
Cc: r-help@r-project.org r-help@r-project.org
Sent:
I have been using the mlogit package but can't seem to figure out how to make
constraints on the beta coefficients.
For example, I would like to force that two of my beta's are equal to each
other.
Thanks in advance.
Jonah
__
R-help@r-project.org
Hi:
Here's one approach using the ggplot2 package:
Y=c(15,14,23,18,19,9,19,13)
Xa=c(17,22,21,18,19,25,8,19)
Xb=c(22,22,29,34,19,26,17,22)
dd - data.frame(Y, Xa, Xb)
ggplot(dd) +
geom_segment(aes(x = Xa, xend = Xb, y = Y, yend = Y,
colour = Xb - Xa), size = 2) +
Dear R-users
I need to fit a nonlinear model to a piece of data. The model to be fitted
uses past values of the input and the ouput - something like
y(k) ~ f(y(k-1),y(k-2),u(k),u(k-1) ) (k is time index). As far as I
know I could use earth(MARS), nnet and etc but I am not sure how to
Hello,
I'm pretty new to R. Basically, how do I speed up the for loop below. Or
better yet, get rid of the for loop all together.
objective: plot two data sets column against column by index. These data
sets have alot NA's. Some columns are all NA's. I need the plots to overlay.
I don't like the
This may not be helpful, but this sounds difficult enough that you
should work with a local statistician rather than trying to get remote
consulting here.
-- Bert
On Mon, Aug 15, 2011 at 3:00 PM, Eduardo M. A. M.Mendes
emammen...@gmail.com wrote:
Dear R-users
I need to fit a nonlinear model
I don't know the answer in general, but for the specific constraint of
two coefficients being the same, I would assume that you should create
a new covariate which is the sum of the two individual ones and fit
this single combined covariate instead of the two separate ones.
Cheers,
Bert
On Mon,
Hi:
Here's one way, using the following reproducible example.
# Method 1: the variable names are the same in each data frame
# Create two separate data frames
ds1 - data.frame(x1 = rnorm(10), x2 = rnorm(10), x3 = rnorm(10))
ds2 - data.frame(x1 = rnorm(10), x2 = rnorm(10), x3 = rnorm(10))
# Melt
Well, of course that doesn't work for categorical covariates (duhhh!)
-- so I'll just stop at my first clause, I don't know. Sorry.
I would suggest that a better specification of the model and the
constraints may elicit better and faster responses.
-- Bert
On Mon, Aug 15, 2011 at 3:53 PM, Bert
Hi Rolf:
Maybe. But I'm not sure whether the OP wants two levels of a single
variable to have the same coefficient, or two different categorical
variables in some way, or two different numeric variables, or...
Maybe it's obvious, but I thought it fairer to the OP to make clear
that I was not a
I followed a couple threads from the archives and from
stackoverflow.com, and would like to know: just what is ... ? What I
mean by this is,for example, from the point of view of a user running a
function in debug mode, is ... an object, or does it exist in the
current environment as some
Hello list,
## I have been doing the following process to convert data from one
form to another for a while but it occurs to me that there is probably
an easier way to do this. I am often given data that have column names
which are actually data and I much prefer dealing with data that are
sorted
?reshape
You have your data in a wide format, but you want it in a long format.
reshape can convert it both ways.
Mikhail
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
Behalf Of Sam Albers
Sent: Monday, August 15, 2011 6:58 PM
To:
Dear folks --
There are a number of functions -- I am thinking of str() as an example --
that produce text as a side-effect, rather then returning it. Is there any
way to send the text produced by such functions into a character variable?
Any suggestions would be greatly appreciated.
andrewH
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