if you set parameter simplify=TRUE, it returns a vector of the ragged
mean. In your case,
mean_rating - tapply(ratok$value, ratok$project_id , mean,simplify = TRUE)
df-data.frame(ID=dimnames(mean_rating)[[1]], mean=mean_rating)
Weidong Gu
On Sun, Oct 16, 2011 at 9:53 AM, Helene Schreyer
Hi:
Try this:
ratok - data.frame(Id = rep(1:3, 3:1), value = c(2, 3, 4, 2, 1, 5))
aggregate(value ~ Id, data = ratok, FUN = mean)
Id value
1 1 3.0
2 2 1.5
3 3 5.0
aggregate() returns a data frame with the Id variable and mean(value).
HTH,
Dennis
On Sun, Oct 16, 2011 at 6:53 AM,
pre-predict(ozonea,groupA,type=terms,terms=NULL,newdata.guaranteed=FALSE,na.action=na.pass)
yeah!
but I don't know how to only show the value of s(ratio,bs=cr)
--
View this message in context:
http://r.789695.n4.nabble.com/Question-about-GAMs-tp3900848p3903753.html
Sent from the R help mailing
predict.gam is returning a matrix with a named column for each term.
Select the appropriate column. Example below
library(mgcv)
n-200;sig - 2
dat - gamSim(1,n=n,scale=sig)
b - gam(y~s(x0)+s(I(x1^2))+s(x2)+offset(x3),data=dat)
newd -
hi! I hope all of you can help me this question
for example GAMs:
ozonea-gam(newozone~
pressure+maxtemp+s(avetemp,bs=cr)+s(ratio,bs=cr),family=gaussian
(link=log),groupA,methods=REML)
formula(ozonea)
newozone ~ pressure + maxtemp + s(avetemp, bs = cr) + s(ratio,bs = cr)
#formula of gams
I am not an expert on this, but there is a way to check this. You can predict
from a gam using predict(ozonea, newdata=...). In the newdata argument you
can specify the X-values of interest to you. Thus, you can compare if your
predictions are the same when predicted directly from the gam or when
I'd be inclined to use predict(ozonea,type=terms) to extract the
estimates of s(ratio,bs = cr)
that you need. But do you really want newozone - s(ratio,bs=cr) when
you've used a log link?
best,
Simon
On 10/13/2011 09:05 AM, pigpigmeow wrote:
hi! I hope all of you can help me this question
I'm confused...
I type ..
predict.gam(ozonea,type=s(ratio,bs=cr))
pressure maxtemp s(avetemp) s(ratio)
1 -0.0459102290 -0.185178463 0.263358446 -0.164558673
2 -0.0286464652 -0.194731320 0.199315027 0.727823293
30.0478073459 -0.013227033 0.002228896 0.342373202
4
On Thu, 2011-10-13 at 08:20 -0700, pigpigmeow wrote:
I'm confused...
I type ..
predict.gam(ozonea,type=s(ratio,bs=cr))
That is not a valid 'type'; normally you'd use `type = terms` or `type
= iterms`, depending on whether you want (any) standard errors to
include the uncertainty about the
On Oct 13, 2011, at 11:20 AM, pigpigmeow wrote:
I'm confused...
I type ..
predict.gam(ozonea,type=s(ratio,bs=cr))
pressure maxtemp s(avetemp) s(ratio)
1 -0.0459102290 -0.185178463 0.263358446 -0.164558673
2 -0.0286464652 -0.194731320 0.199315027 0.727823293
3
Hi All,
I am a relative newbie to R and have the following problem I was trying to
solve. I had taken a look at the 'sample selection' package but was having
trouble applying it to my use case and was wondering if anyone out there had
done something similar and could share code or
Thanks guys, that's a great help.
Nellie
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Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
for the 'boxplots' using 5%, 25%,75%, 95%
limits as we have with the 'boxplots'.
Tom
- Original Message -
From: Hadley Wickham had...@rice.edu
Date: Tuesday, October 4, 2011 10:23 am
Subject: Re: [R] Question about ggplot2 and stat_smooth
To: Thomas Adams thomas.ad...@noaa.gov
Hello!
So I am handling this problem with some arrays grp1-grp7, I want to write a
loop to avoid tedious work, but I don't know how to transform string to
boor?
For example I used
i=1
paste(grp,i, sep=)
I only got grp1 instead of grp1, which can't be manipulate using mean() or
other function.
On Oct 10, 2011, at 12:52 PM, song_gpqg wrote:
Hello!
So I am handling this problem with some arrays grp1-grp7, I want to
write a
loop to avoid tedious work, but I don't know how to transform string
to
boor?
For example I used
i=1
paste(grp,i, sep=)
?get
e.g.
get( paste(grp,i, sep=) )
Hi Nellie,
hope I got you right. I guess you want something like that.
for (i in 1:7) {
oneOfNelliesArray - eval(parse(text=paste(grp,i, sep=)))
anyFunction(oneOfNelliesArray)
}
Paste() just returns you a string. But you want R to evaluate the
expression. So you have to parse it and tell
On 11/10/11 08:21, Christoph Molnar wrote:
Hi Nellie,
hope I got you right. I guess you want something like that.
for (i in 1:7) {
oneOfNelliesArray- eval(parse(text=paste(grp,i, sep=)))
anyFunction(oneOfNelliesArray)
}
Paste() just returns you a string. But you want R to evaluate the
Dennis Hadley,
This does exactly what I need — thank you so much!
Regards,
Tom
On 10/4/11 5:34 PM, Dennis Murphy wrote:
Hi Hadley:
When I tried your function on the example data, I got the following:
dd- data.frame(year = rep(2000:2008, each = 500), y = rnorm(4500))
g- function(df, qs =
On Mon, Oct 3, 2011 at 12:24 PM, Thomas Adams thomas.ad...@noaa.gov wrote:
I'm interested in creating a graphic -like- this:
c - ggplot(mtcars, aes(qsec, wt))
c + geom_point() + stat_smooth(fill=blue, colour=darkblue, size=2, alpha
= 0.2)
but I need to show 2 sets of bands (with different
Hi,
I applied a linear mixed effect model in my data using the nlme package.
lme2-lme(distance~temperature*condition, random=~+1|trial, data) and then
anova.
I want to ask if it is posible to get the least squares means for the
interaction effect and the corresponding 95%ci. And then plot this
Below.
On Tue, Oct 4, 2011 at 7:34 AM, Panagiotis p...@hi.is wrote:
Hi,
I applied a linear mixed effect model in my data using the nlme package.
lme2-lme(distance~temperature*condition, random=~+1|trial, data) and then
anova.
I want to ask if it is posible to get the least squares means
Bert Gunter gunter.berton at gene.com writes:
Below.
On Tue, Oct 4, 2011 at 7:34 AM, Panagiotis pat2 at hi.is wrote:
Hi,
I applied a linear mixed effect model in my data using the nlme package.
lme2-lme(distance~temperature*condition, random=~+1|trial, data) and then
anova.
I
where I can
specify the limits calculated for the 'boxplots' using 5%, 25%,75%, 95% limits
as we have with the 'boxplots'.
Tom
- Original Message -
From: Hadley Wickham had...@rice.edu
Date: Tuesday, October 4, 2011 10:23 am
Subject: Re: [R] Question about ggplot2 and stat_smooth
To: Thomas
...@rice.edu
Date: Tuesday, October 4, 2011 10:23 am
Subject: Re: [R] Question about ggplot2 and stat_smooth
To: Thomas Adams thomas.ad...@noaa.gov
Cc: R-help forum r-help@r-project.org
On Mon, Oct 3, 2011 at 12:24 PM, Thomas Adams thomas.ad...@noaa.gov
wrote:
I'm interested in creating
# Function to compute quantiles and return a data frame
g - function(d) {
qq - as.data.frame(as.list(quantile(d$y, c(.05, .25, .50, .75, .95
names(qq) - paste('Q', c(5, 25, 50, 75, 95), sep = '')
qq }
You could cut out the melt step by making this return a data frame:
g -
Hi Hadley:
When I tried your function on the example data, I got the following:
dd - data.frame(year = rep(2000:2008, each = 500), y = rnorm(4500))
g - function(df, qs = c(.05, .25, .50, .75, .95)) {
data.frame(q = qs, quantile(d$y, qs))
}
ddply(dd, .(year), g)
ddply(dd, .(year), g)
year
the limits calculated for the 'boxplots' using 5%, 25%,75%, 95% limits
as we have with the 'boxplots'.
Tom
- Original Message -
From: Hadley Wickhamhad...@rice.edu
Date: Tuesday, October 4, 2011 10:23 am
Subject: Re: [R] Question about ggplot2 and stat_smooth
To: Thomas Adamsthomas.ad
I'm interested in creating a graphic -like- this:
c - ggplot(mtcars, aes(qsec, wt))
c + geom_point() + stat_smooth(fill=blue, colour=darkblue, size=2,
alpha = 0.2)
but I need to show 2 sets of bands (with different shading) using 5%,
25%, 75%, 95% limits that I specify and where the heavy
Hi:
I would think that, at least in principle, this should work:
a - ggplot(mtcars, aes(qsec, wt))
a + geom_point() + stat_smooth(fill=blue, colour=darkblue, size=2,
level = 0.9, alpha = 0.2) +
stat_smooth(fill = 'blue', colour = 'darkblue', size = 2,
Andrés,
Thank you for your help, but that does not capture what I'm looking for. I need
to be able to control the
shaded bound limits and they need to be coincident.
Tom
On 10/3/11 3:37 PM, Andrés Aragón wrote:
Hi,
Try some like this:
c- ggplot(mtcars, aes(qsec, mpg, colour=factor(cyl)))
Hi,
Try some like this:
c - ggplot(mtcars, aes(qsec, mpg, colour=factor(cyl)))
c + stat_smooth(aes(group=cyl))+stat_smooth(aes(fill=factor(cyl)))+geom_point()
Andrés AM
2011/10/3, Thomas Adams thomas.ad...@noaa.gov:
I'm interested in creating a graphic -like- this:
c - ggplot(mtcars,
*An:* Samir Benzerfa; r-help
*Betreff:* Re: [R] Question concerning Box.test
** **
Plaintext data looks like this:
P - structure(list(X77.BANK = c(0, 0, 0, 0.003181659, -0.006386799,
0.028028724, -0.015347692, -0.015910002, 0.00322897, -0.013062473,
0, -0.03809005, 0.021189299
Hi everyone,
I've got a question concerning the function Box.test for testing
autocorrelation in my data.
My data consist of (daily) returns of several stocks over time (first
row=time, all other rows=stock returns). I intend to perform a Box-Ljung
test for my returns (for each stock).
Did you try regular apply? If you have univariate input, there's no reason
to use the multivariate mapply. Or more generally:
apply(P[-1,],1,function(p) Box.test(p)$p.value)
Michael
On Tue, Sep 27, 2011 at 4:45 AM, Samir Benzerfa benze...@gmx.ch wrote:
Hi everyone,
I've got a question
Weylandt [mailto:michael.weyla...@gmail.com]
*Gesendet:* Dienstag, 27. September 2011 13:12
*An:* Samir Benzerfa
*Cc:* r-help@r-project.org
*Betreff:* Re: [R] Question concerning Box.test
** **
Did you try regular apply? If you have univariate input, there's no reason
to use
and columns.
Actually, I want to perform the test for each column and not row.
** **
*Von:* R. Michael Weylandt [mailto:michael.weyla...@gmail.com]
*Gesendet:* Dienstag, 27. September 2011 17:51
*An:* Samir Benzerfa; r-help
*Betreff:* Re: [R] Question concerning Box.test
I encounters some codes in ggplot2 manual and confused with one of its lm
syntax. The code is here:
library(ggplot2) d - subset(diamonds, carat 2.5 rbinom(nrow(diamonds), 1,
0.2) == 1) d$lcarat - log10(d$carat) d$lprice - log10(d$price) detrend -
lm(lprice ~ lcarat, data = d) d$lprice2 -
阮铮 rz1991 at foxmail.com writes:
I encounters some codes in ggplot2 manual and confused with one of
its lm syntax.
[snip]
mod - lm(lprice2 ~ lcarat * color, data = d)
# *** what puzzled me is the
last statement marked with ***. How does R deal with lcarat * color,
since color is not of
.
Do you understand my intention?
THX, SB
-Ursprüngliche Nachricht-
Von: Jean-Christophe BOUËTTÉ [mailto:jcboue...@gmail.com]
Gesendet: Freitag, 16. September 2011 17:31
An: Samir Benzerfa
Cc: r-help@r-project.org
Betreff: Re: [R] question concerning the acf function
Hi Samir
I have a question about what is the best function or package to use to
do the following in R:
minimize Q = g W g' where respect to \gamma
where g is n x 1 vector, W is a n x n matrix and is given,
and g = 1 / T * sum_from_i=1_to_1=60 [ (1-R[i+1] * ( E[i+1]/E[i ]
)^(\gamma) ) ]
On 17-Sep-11 01:20:53, Rolf Turner wrote:
On 17/09/11 01:19, Michael Friendly wrote:
SNIP
Trying to interpret associations in complex loglinear models
from tables of parameter estimates is like trying to extract
sunlight from a cucumber. You have to squeeze very hard, and
then are
Hi everyone,
I've got a question concerning the function acf(.) in R for calculating the
autocorrelation in my data.
I have a table with daily returns of several stocks over time and I would
like to calculate the autocorrelation for all the series (not only for one
time series). How can I
Hi Yana
Trying to interpret associations in complex loglinear models from tables
of parameter estimates is like trying to extract sunlight from a
cucumber. You have to squeeze very hard, and then are usually unhappy
with the quality of the sunlight.
Instead, you can visualize the
Hi,
you did not supply a reproducible example. We do not know what your
data nor your code looks like.
Please follow the recommandations found at the bottom of this email!
You're more likely to get a quick and meaningful reply.
JC
2011/9/16 Samir Benzerfa benze...@gmx.ch:
Hi everyone,
I've
Would not something like
V = apply(P,2,acf, plot=FALSE)
V = sapply(V,`[[`,acf)
work? I'm pretty sure acf() doesn't return any sort of residuals so you'll
have to calculate those on your own.
Hope this helps,
Michael Weylandt
On Fri, Sep 16, 2011 at 9:20 AM, Jean-Christophe BOUËTTÉ
Benzerfa
Cc: r-help@r-project.org
Betreff: Re: [R] question concerning the acf function
Hi,
you did not supply a reproducible example. We do not know what your
data nor your code looks like.
Please follow the recommandations found at the bottom of this email!
You're more likely to get a quick
Hi!
I am doing a permutation test on 7 data randomly selected into two groups,
one has 4 and the other 3. The purpose is to list all the 35 possible
outcomes and calculate p-value. I googled the function perm.test() and not
sure if my usage is correct. So I set a=c(7 data), b=(1,1,1,1,2,2,2)
Hope you don't find this a rethorical question, but why did you prefer
to google the function instead of reading the fine manual?
Em 16/9/2011 16:07, song_gpqg escreveu:
Hi!
I am doing a permutation test on 7 data randomly selected into two groups,
one has 4 and the other 3. The purpose is to
On 17/09/11 01:19, Michael Friendly wrote:
SNIP
Trying to interpret associations in complex loglinear models from
tables of parameter estimates is like trying to extract sunlight from
a cucumber. You have to squeeze very hard, and then are usually
unhappy with the quality of the sunlight.
Dear R gurus,
I am looking for a way to fit a predictive model for a contingency table which
has counts. I found that glm( family=poisson) is very good for figuring out
which of several alternative models I should select. But once I select a model
it is hard to present and interpret it,
On Sep 15, 2011, at 4:33 PM, Yana Kane-Esrig wrote:
Dear R gurus,
I am looking for a way to fit a predictive model for a contingency
table which has counts. I found that glm( family=poisson) is very
good for figuring out which of several alternative models I should
select. But once I
Hi,
I have one basic doubt. Suppose X ~ N(50,10).
I need to calculate Probability X = 50.
dnorm(50, 50, 10) gives me
[1] 0.03989423
My understanding is (which is bit statistical or may be mathematical) on a
continuous scale, Probability of the type P(X = .) are nothing but
1/Infinity
-project.org] On Behalf Of Vincy Pyne
Sent: 14 September 2011 12:24
To: r-help@r-project.org
Subject: [R] Question regarding dnorm()
Hi,
I have one basic doubt. Suppose X ~ N(50,10).
I need to calculate Probability X = 50.
dnorm(50, 50, 10) gives me
[1] 0.03989423
My understanding is (which
Ellison and Mr Mark for your
guidance.
Regards
Vincy
--- On Wed, 9/14/11, S Ellison s.elli...@lgcgroup.com wrote:
From: S Ellison s.elli...@lgcgroup.com
Subject: RE: [R] Question regarding dnorm()
To: Vincy Pyne vincy_p...@yahoo.ca, r-help@r-project.org
r-help@r-project.org
Received: Wednesday
Hi All,
I have a quick question on random forests. Simply, I am not sure how to read
the values of independent variables related to the highest value of a response
variable from all trees generated from random forests. It is easy to do this in
a single regression tree. But I am not clear how
Hello all, I what to print the banner plot that is output from the mona
method in the cluster package but the problem is I dont want to print all
that red ink. Here is an example:
data(animals)
ma - mona(animals)
ma
## Plot similar to Figure 10 in Struyf et al (1996)
plot(ma)
I can change the
Hi:
Here are a couple of ways to do this. The barplot() attempt is pretty
close to the original banner plot except that the color is transparent
to the (white) background. The ggplot2 graph is similar, but the code
may be less familiar if you don't use the package. In the latter, the
variable
If you read the help file on the step() function
?step
you will see a reference to BIC under the description of the k= argument.
This suggests that you could try:
BIC.fitted = step(glm.fit, k=log(dim(dat)[1]))
Jean
Andra Isan wrote on 09/07/2011 06:12:19 PM:
Hi All,
After
Hi All,
After fitting a model with glm function, I would like to do the model selection
and select some of the features and I am using the step function as follows:
glm.fit - glm (Y ~ . , data = dat, family = binomial(link=logit)) AIC_fitted
= step(glm.fit, direction = both)
I was wondering is
Hi - How can I 'manually' reproduce the results in 'pred1' below? My attempt
is pred_manual, but is not correct. Any help is much appreciated.
library(splines)
set.seed(12345)
y - rgamma(1000, shape =0.5)
age - rnorm(1000, 45, 10)
glm1 - glm(y ~ ns(age, 4), family=Gamma(link=log))
dd -
On Sep 6, 2011, at 5:08 PM, Axel Urbiz wrote:
Hi - How can I 'manually' reproduce the results in 'pred1' below?
Well, not by constructing the prediction function from a different
data basis that the glm function gets.
My attempt
is pred_manual, but is not correct. Any help is much
As others will tell you: you need to provide a reproducible example.
What are p1, u1, u2 ?
quote
Dear all,
I have the following problem with the uniroot function. I want to
find roots for the fucntion Fp2 which is defined as below.
Fz -
Dear all,
I have the following problem with the uniroot function. I want to find
roots for the fucntion Fp2 which is defined as below.
Fz - function(z){0.8*pnorm(z)+p1*pnorm(z-u1)+(0.2-p1)*pnorm(z-u2)}
Fp - function(t){(1-Fz(abs(qnorm(1-(t/2)+(Fz(-abs(qnorm(1-(t/2)}
Fp2 -
Dear all,
I forgot to mention the values for the parameters in the first email.
u1 - -3
u2 - 4
alpha - 0.05
p1 - 0.15
Thank you very much.
2011/9/3 li li hannah@gmail.com
Dear all,
I have the following problem with the uniroot function. I want to find
roots for the fucntion
On 03.09.2011 18:31, li li wrote:
Dear all,
I forgot to mention the values for the parameters in the first email.
u1- -3
u2- 4
alpha- 0.05
p1- 0.15
Thank you very much.
2011/9/3 li lihannah@gmail.com
Dear all,
I have the following problem with the uniroot function. I want
a stats and not
an R question; (b) it might be a good idea to review a stats text,
or even http://en.wikipedia.org/wiki/Bayesian_information_criterion ,
since this is a pretty basic question.
__
R-help@r-project.org mailing list
https://stat.ethz.ch
. stats.stackexchange.com), since it's a stats and not
an R question; (b) it might be a good idea to review a stats text,
or even http://en.wikipedia.org/wiki/Bayesian_information_criterion ,
since this is a pretty basic question.
__
R-help@r-project.org
On 09/02/2011 08:48 AM, John Sorkin wrote:
I believe when using BIC one needs to compare nested models
This is wrong. Hypothesis tests rely on nested models; information
criteria do not.
--
Patrick Breheny
Assistant Professor
Department of Biostatistics
Department of Statistics
University
On Fri, 2 Sep 2011, Patrick Breheny wrote:
On 09/02/2011 08:48 AM, John Sorkin wrote:
I believe when using BIC one needs to compare nested models
This is wrong. Hypothesis tests rely on nested models; information criteria
do not.
Actually, this is off-topic on this list. But blanket
Inline:
On Fri, Sep 2, 2011 at 8:09 AM, Patrick Breheny patrick.breh...@uky.edu wrote:
On 09/02/2011 08:48 AM, John Sorkin wrote:
I believe when using BIC one needs to compare nested models
This is wrong. Hypothesis tests rely on nested models; information criteria
do not.
Yes, indeed.
On 09/02/2011 11:26 AM, Prof Brian Ripley wrote:
This is wrong. Hypothesis tests rely on nested models; information criteria
do not.
Actually, this is off-topic on this list. But blanket statements are
often themselves untrue: there are hypothesis tests of non-nested
models (most famously due
Hi All,
In order to compare two different logistic regressions, I think I need to
compare them based on their BIC values, but I am not sure if the smaller BIC
would mean a better model or the reverse is true?
Thanks a lot,Andra
[[alternative HTML version deleted]]
Hello All,
I have a data frame called train.data which has 100 columns. I am using an
statistical package to learn a model and the format of using that package is as
follows:
ex5 - stepFlexmix(cbind(y,1-y)~x|id2, data=NPreg, k=2,
model=FLXMRglm(family=binomial), nrep=5)
the
uf_mike michael.parent at ufl.edu writes:
Hi, all! I'm new to R but need to use it to solve a little problem I'm having
with a paper I'm writing. The question has a few components and I'd
appreciate guidance on any of them.
1. The most essential thing is that I need to generate some
Gordon Robertson grobert...@bcgsc.ca
on Wed, 24 Aug 2011 22:21:22 -0700 writes:
I'm fairly new to the silhouette functionality in the
cluster package, so apologize if I'm asking something
naive. If I run the 'agnes(ruspini)' example from the
silhouette section of the
Thanks!
This problem isn't uniquely defined. Are you willing to generate more samples
than you need and then throw away extreme values? Or do you want to 'censor'
extreme values (i.e. set values = 1 to 1 and values =7 to 7)?
I'd like the retain a normal distribution so I wouldn't want to
Michael Parent michael.parent at ufl.edu writes:
Thanks!
This problem isn't uniquely defined. Are you
willing to generate more samples than you need and then throw
away extreme values? Or do you want to 'censor'
extreme values (i.e. set values = 1 to 1 and values =7 to 7)?
I'd like
and at a specific prevalence (say, 5%). Is there a
simple way to take the initial data set and randomly replace 5% of values
with NA missing values?
Thanks, I appreciate any guidance folks can offer. :-)
--
View this message in context:
http://r.789695.n4.nabble.com/R-question-generating-data-using-MASS
I'm fairly new to the silhouette functionality in the cluster package, so
apologize if I'm asking something naive.
If I run the 'agnes(ruspini)' example from the silhouette section of the
cluster package vignette, and assign colours to clusters, two clusters have
what appear to be incorrect
Hello,
I am trying to write some code that dumps R objects to the harddisk in a binary
format so they can be quickly re-used later. Goal is to save time. The objects
may be quite large (e.g. classes for a GUI). I was thinking that save() and
load() would be suitable for this (until now I only
The problem I think is in your unmarshal. 'load' will load the object
into the local environment, not the global. You have to explicitly
return it, that means you have to know the name that its was 'save'd
by;
unmarshal - function(xdr) {
object - strsplit(strsplit(xdr, \\.)[[1]][[1]], /)
Cc: R Mailing List r-help@r-project.org
Sent: Thursday, August 25, 2011 4:07 PM
Subject: Re: [R] Question about object permanence/marshalling
The problem I think is in your unmarshal. 'load' will load the object
into the local environment, not the global. You have to explicitly
return
jholt...@gmail.com
To: Albert-Jan Roskam fo...@yahoo.com
Cc: R Mailing List r-help@r-project.org
Sent: Thursday, August 25, 2011 4:07 PM
Subject: Re: [R] Question about object permanence/marshalling
The problem I think is in your unmarshal. 'load' will load the object
into the local
Hello All,
I am trying to use FlexMix package for my two linear regression models:
http://finzi.psych.upenn.edu/R/library/flexmix/doc/regression-examples.pdf
y1 = w1 * x1 + w2 *x2 + w3 * x3 +beta1
y2 = w4 * x4 + w4 *x4 + w4 * x4 +beta2
and I would like to combine these two regression models
Hello all,
May be silly question, but what exactly is beta parameter in functions like
regmixEM from mixtools package?
I mean, how to determine this beta, if i have a set of metrics for each
case? Is there a function for that? I have try to put NULL at this
parameter, but function just do not
Hi,
I have a silly question regarding the usage of two commands: read.table and
gregexprï¼
For read.table, if I read a matrix and set header = T, I found that all the
dash (-) becomes dots (.)
A = read.table(Matrix.txt, sep = \t, header = F)
A[1,1]
# A-B-C-D.
A = read.table(Matrix.txt, sep =
Hi Jack,
yes there is. see ?read.table for option check.names
and to the 2nd task . is a special character in regular expressions,
so mask it or don't use regular expressions:
gregexpr([.],A.B.C.D) #or
gregexpr(.,A.B.C.D,fixed=T)
cheers.
Am 17.08.2011 15:03, schrieb Jack Luo:
Hi,
I have a
Hi!
You can try import the file with header = F, and after inform that the first
row is a header.
On this post is some idea:
http://stackoverflow.com/questions/2293131/reading-first-row-as-header-is-easy-what-gives-with-two-rows-being-the-header
On Wed, Aug 17, 2011 at 10:03 AM, Jack Luo
Thank, Eik, it works!
-Jack
On Wed, Aug 17, 2011 at 9:19 AM, Eik Vettorazzi
e.vettora...@uke.uni-hamburg.de wrote:
Hi Jack,
yes there is. see ?read.table for option check.names
and to the 2nd task . is a special character in regular expressions,
so mask it or don't use regular
Hi,
After I read an xlsx file into the work space:
A - read.xlsx(B.xls, header = T, check.names = F)
There are several headers with the names like:
colnames(A) [1:4]
# [1] A 1B
[3] C 2 D
I can get the content of column 2 and column 4 easily by
A$B
Hi Jack,
You need to quote non-syntactic names.
A$`A 1`
A$'A 1'
A$A 1
should all work, with the first form being the recommended one.
Best,
Ista
On Wed, Aug 17, 2011 at 1:45 PM, Jack Luo jluo.rh...@gmail.com wrote:
Hi,
After I read an xlsx file into the work space:
A - read.xlsx(B.xls,
Hi All,
I would like to create a matrix in R but I dont know the size of my matrix. I
only know the size of the columns but not the size of the rows. So, is there
any way to create a dynamic matrix of size NULL by n_cols? and then add to that
matrix?
I know for a vector, I can do this: x= NULL
On Aug 10, 2011, at 8:23 PM, Andra Isan wrote:
Hi All,
I would like to create a matrix in R but I dont know the size of my
matrix. I only know the size of the columns but not the size of the
rows. So, is there any way to create a dynamic matrix of size NULL
by n_cols? and then add to
On 11/08/11 13:11, David Winsemius wrote:
On Aug 10, 2011, at 8:23 PM, Andra Isan wrote:
Hi All,
I would like to create a matrix in R but I dont know the size of my
matrix. I only know the size of the columns but not the size of the
rows. So, is there any way to create a dynamic matrix of
On Aug 10, 2011, at 9:22 PM, Rolf Turner wrote:
On 11/08/11 13:11, David Winsemius wrote:
On Aug 10, 2011, at 8:23 PM, Andra Isan wrote:
Hi All,
I would like to create a matrix in R but I dont know the size of
my matrix. I only know the size of the columns but not the size of
the rows.
On 11/08/11 13:27, David Winsemius wrote:
On Aug 10, 2011, at 9:22 PM, Rolf Turner wrote:
On 11/08/11 13:11, David Winsemius wrote:
On Aug 10, 2011, at 8:23 PM, Andra Isan wrote:
Hi All,
I would like to create a matrix in R but I dont know the size of my
matrix. I only know the size of
On Aug 10, 2011, at 10:03 PM, Rolf Turner wrote:
On 11/08/11 13:27, David Winsemius wrote:
On Aug 10, 2011, at 9:22 PM, Rolf Turner wrote:
On 11/08/11 13:11, David Winsemius wrote:
On Aug 10, 2011, at 8:23 PM, Andra Isan wrote:
Hi All,
I would like to create a matrix in R but I dont
I just signed up for R Help, so please let me know if I'm doing this
incorrectly.
I've been working with R for a couple of weeks, attempting to process
microarray data. This means 20,000+ rows of data to work with x 24
columns. I am trying to produce heatmaps and found that the best
computer
Hi Tyler --
On 08/08/2011 12:54 PM, Tyler Gruhn wrote:
I just signed up for R Help, so please let me know if I'm doing this
incorrectly.
Better to provide an informative subject line -- microarray heatmap
memory use ?
I've been working with R for a couple of weeks, attempting to process
On Sat, Aug 06, 2011 at 09:37:22PM +0200, Petr Savicky wrote:
On Fri, Aug 05, 2011 at 11:47:49PM +0530, Ron Michael wrote:
Hi all, I have happened to work on MS .NET for sometime now, and I found
that this language offers RNG what is called as Donald E. Knuth's
subtractive random number
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