On Mon, Apr 23, 2012 at 02:33:22PM -0700, cyclondude wrote:
> Yes. That is what I was looking for. Is there a simple way to (in this
> scenario)
>
>
> > out[[1]]
>
> v1 v2
> 1 a 1
> 4 a 2
> 7 a 3
>
> > a <- out[[1]]
>
> for each one?
Hi.
If you want to generate variable nam
Hi,
I am a new user and this is a basic question. I have read the
posting guidelines too. So if this more appropriate for a statistics
forum I will ask them.
I have a socket server and client and the bytes are piped to
graphite which is a graphing library. The intention is to under
On 23.04.2012 20:55, Brian Diggs wrote:
> On 4/23/2012 9:24 AM, Matthias Rieber wrote:
>> Hello,
>>
>> I've some problem with the ggplot2. Here's a small example:
>>
>> [...]
>> Is it wrong to use geom_bar with that kind of data? I could avoid this
>> issue when I cast the data.frame, but I like to
The scatter plot is easy:
plot(pH1 ~ pH2, data = OBJ)
When you say a loess for each -- how do you break them up? Are there
repeat values for pH1? If so, this might be hard to do in base
graphics, but ggplot2 would make it easy:
library(ggplot2)
ggplot(OBJ, aes(x = pH1, y = pH2)) + geom_point() +
Wrong list. Post on r-sig-mixed-models.
-- Bert
On Mon, Apr 23, 2012 at 8:46 PM, arun wrote:
> Hi,
>
> I have some difficulty in figuring out whether I am doing correct or not.
>
>
> A brief introduction about the work: It is a light/dark choice test
> conducted in insect larvae. The response i
Hi,
I have some difficulty in figuring out whether I am doing correct or not.
A brief introduction about the work: It is a light/dark choice test
conducted in insect larvae. The response is binary (0- present in dark
area, 1-present in light area) and the experiment is run for 15 min, so
th
Dear all,
I need to do some calculation where the code used are below. I get
error message when I choose k to be large, say greater than 25.
The error message is
"Error in integrate(temp, lower = 0, upper = 1, k, x, rho, m) :
the integral is probably divergent".
Can anyone give some help on res
Hi folks.
If I have the following in my "data"
eventpH1pH2
14.0 6.0
24.3 5.9
34.1 6.1
44.0 5.9
and on and on. for about 400 events
Is there a way I can get R to plot event vs. pH1 and event vs. pH2 and
then do a loess o
> From: dwinsem...@comcast.net
> To: leanneheis...@hotmail.com
> Subject: Re: [R] change color scheme in mvpart
> Date: Mon, 23 Apr 2012 19:20:06 -0400
>
> The fact that you are posting in HTML indicates that a) you have
> either not read the Posting Guide or b) do not know enough about your
> mai
It depends how you want to ensure that condition -- if you just want
to censor at an upper bound of 6k, this is easy:
pmin(zmsample, 6000)
If you want to sample "as before" but it just happens to all be less
than 6000 -- that's not really a rigorous statement, but just go with
it -- intuitively,
It is indeed the fact you're plotting factors, but unless you say what
"as intended" is, it's hard to provide exactly what you're seeking.
Perhaps this will help though:
X <- factor(sample(letters[1:5], 15, TRUE))
Y <- rnorm(15)
dats <- data.frame(X, Y)
plot(Y ~ X, data = dats) # No good
plot(X
My bad, I was thinking of rds (?saveRDS). RData and rda are mentioned under
?data as alternate file extensions for the same data format. Sorry for posting
without checking first.
---
Jeff NewmillerThe
Lists of numbers of length ~1000 are no problem for the wilcox.test()
function (Mann-Whitney is a special case) if you leave the default
exact = NULL.
The choice of test is all yours.
Michael
On Mon, Apr 23, 2012 at 12:39 PM, aoife doherty wrote:
> Hello
> I have two lists of numbers, each list
Hi,
I have a question on generating random variables based on zipf-mandelbrot
distribution.
So when I execute the following lines:
ZM = lnre ("zm", alpha = 2/3, B=0.1)
zmsample = rlnre (ZM, n =100)
zmsample
It generates 100 random values based on a zipf-mandelbrot distribution as
below. But ho
I'd like to fit a multinomial log-linear model for 4 categories of the form:
log[(P(D=i | x)/P(D=0 | x)] = alpha + beta_i x for i = 1,2,3.
Is there a way to impose such a constraint in the multinom function of nnet
or another function of some library?
regards,
Hosotsubo
--
Moritaka Hosotsubo
Hi Jeff.
Can you point me toward the documentation for "rda saves content of
one variable, with no associated name"? I don't seem to find it in
?save ?load etc.
Thanks,
Ista
On Mon, Apr 23, 2012 at 5:58 PM, Jeff Newmiller
wrote:
> No, RData saves both the variable name and corresponding content
Hello everyone, I am currently using the mvpart package and would like to
change the color scheme it uses, and was hoping someone could help me out. All
of the papers I have found have used a grayscale but I can't seem to figure out
how they did that! Currently, mvpart plots barplots in a repea
All,
I am trying to estimate the parameters of a bivariate Von Mises distributions.
I am looking for somebody to point me in the direction of an R package or
function that does this. I have noted the existing packages that allow for
obtaining the density values once the parameters have been es
Hi, r-help members.
I have a question about summing two density distributions. I have two
samples from which I've estimated hazard parameters for a Gompertz mortality
model. With those parameters, I can calculate the PDF (survival function
times hazard function) of ages-at-death in a birth cohor
No, RData saves both the variable name and corresponding content of multiple
variables. rda saves content of one variable, with no associated name. The
latter allows for greater flexibility in importing the data later into
different working environments, the former is convenient for recreating
I recommend not putting POSIXlt vectors in data frames because of memory use
and added complexity of the resulting data frame. That is, use
colClasses = c('character', 'POSIXct', 'POSIXct')
instead. The POSIXlt values will still be created as temporary variables for
reading in, but the data
There are, but it's generally considered better style to keep them all
in a single list and use lapply() if you want to do things to each
element.
Michael
On Mon, Apr 23, 2012 at 5:33 PM, cyclondude wrote:
> Yes. That is what I was looking for. Is there a simple way to (in this
> scenario)
>
>
Petr,
Thank you very much this works. A little more tweaking and I'll have what
I need.
Thanks
Steve Friedman Ph. D.
Ecologist / Spatial Statistical Analyst
Everglades and Dry Tortugas National Park
950 N Krome Ave (3rd Floor)
Homestead, Florida 33034
steve_fried...@nps.gov
Office (305) 224
On Apr 23, 2012, at 5:21 PM, R. Michael Weylandt wrote:
That's not the ifelse() that's the for loop returning NULL
(everything's a function!). If you put the assignment inside you'll
get expected behavior.
x <- (for(i in 1:5) i) # Strange
for(i in 1:5) x<- i # Normal (but notice you only get t
Yes. That is what I was looking for. Is there a simple way to (in this
scenario)
> out[[1]]
v1 v2
1 a 1
4 a 2
7 a 3
> a <- out[[1]]
for each one?
Thanks!
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That's not the ifelse() that's the for loop returning NULL
(everything's a function!). If you put the assignment inside you'll
get expected behavior.
x <- (for(i in 1:5) i) # Strange
for(i in 1:5) x<- i # Normal (but notice you only get the last value
because previous ones are overwritten)
Michae
Hi
> EMU1993<-(for (i in 1:nrow(data)){
+ ifelse(year==1992,sum(avgflowEMU),0)
+ })
>EMU1993
NULL
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One option is to subtract the continuous variable from y before doing
the regression (this works with any regression package/function). The
probably better way in R is to use the 'offset' function:
formula = I(log(data$AB.obs + 1, 10)-log(data$SIZE,10)) ~
log(data$SIZE, 10) + data$Y
formula = log
Here is a method that uses negative look behind:
> tmp <- c('mutation','nonmutated','unmutated','verymutated','other')
> grep("(? wrote:
> Hello All,
>
> Started out awhile ago trying to select columns in a dataframe whose names
> contain some variation of the word "mutant" using code like:
>
> n
Dear All,
I have just submitted a new version of the QCAGUI package on CRAN, it
should be propagated in a couple of days.
This version is nothing but a quick update to the latest Rcmdr base
package, and works (as usual) with the QCA package up to version 0.6-5
For the later versions of the QCA pa
On Mon, Apr 23, 2012 at 04:02:45PM -0400, steve_fried...@nps.gov wrote:
>
> Hi,
>
>
> The following script (which I did not develop) is used to calculate and
> plot a skewed normal curve. The script currently requires the user to
> input six parameters, rather than reading these directly from a
On Mon, Apr 23, 2012 at 10:58:26AM -0700, cyclondude wrote:
> Hello, very new to R, playing with tables, and I am trying to do
>
> x <- subset(data, columnlabel == x)
>
> for every element in my column that I could find by using
>
> table (data [,"columnlabel"])
Hi.
The following may be clos
On 23-04-2012, at 21:37, phillip03 wrote:
> Thank you!
>
> Do you know why ifelse() sometimes returns "NULL" ?
Please provide a reproducible example for this phenomenon.
Berend
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listi
Thank you!
Do you know why ifelse() sometimes returns "NULL" ?
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R-help@r-project.org m
This is really a job for a database, and Excel is not a database (even
though many think it is). I have some clients that I have convinced
to create an Access database rather than use Excel (still MS product
so it can't be that scary, right?). They were often a little
reluctant at first because t
Hi,
The following script (which I did not develop) is used to calculate and
plot a skewed normal curve. The script currently requires the user to
input six parameters, rather than reading these directly from a file.
I've been spinning wheels here, trying to figure out how to modify the
script
Hi,
On 24 April 2012 05:00, Duncan Murdoch wrote:
> On 23/04/2012 10:49 AM, Adam Wilson wrote:
>>
>> I routinely write graphics into multi-page PDFs, but some graphics (i.e.
>> plots of large spatial datasets using levelplot()) can result in enormous
>> files. I'm curious if there is a better wa
gt; > foo
>a b
> 1 1 a
> 2 2 a
> 3 3 a
> 4 4 a
> 5 5 a
> 6 6 a
> 7 7 a
> 8 8 a
> 9 9 a
> 10 10 a
> > foo$date <- '20120423'
> > foo
>a b date
> 1 1 a 20120423
> 2 2 a 20120423
> 3 3 a 20120423
>
> If I want to make a new data.frame where it is the NONEURO avgflows. how
> do I do that ?
>
Exactly like above, but without the negation (the exclamation mark).
You must also start to use the help system, for instance:
?"!"
Rui Barradas
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Are they the same with .RData being the newer format? Thanks,
...Tao
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented,
On 4/23/2012 9:24 AM, Matthias Rieber wrote:
Hello,
I've some problem with the ggplot2. Here's a small example:
--8<--
library(ggplot2)
molten<- data.frame(date=c('01','01','01','01',
'02','02','02','02'),
channel=c('red','red','blue','blue',
This little example might help.
> foo <- data.frame(a=1:10, b=letters[1:0])
> foo
a b
1 1 a
2 2 a
3 3 a
4 4 a
5 5 a
6 6 a
7 7 a
8 8 a
9 9 a
10 10 a
> foo$date <- '20120423'
> foo
a b date
1 1 a 20120423
2 2 a 20120423
3 3 a 20120
see file.rename()
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 4/22/12 1:25 AM, "sagarnikam123" wrote:
>i want to cut file from e.g. "abc" folder & put it into another location
>with folder name e.g. "xyz"
>how should i
Hi RUI
Thank you so much ! I know I have a lot to learn : / sorry for that.
If I want to make a new data.frame where it is the NONEURO avgflows. how do
I do that ?
Ph
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Hello, very new to R, playing with tables, and I am trying to do
x <- subset(data, columnlabel == x)
for every element in my column that I could find by using
table (data [,"columnlabel"])
I'd appreciate any useful help and I'm sorry if I didn't get the terminology
perfect. Thanks.
--
Vie
Dear R-helpers,
I would like to test if the slope corresponding to a continuous variable in
my model (summary below) is different than one.
I would appreciate any ideas for how I could do this in R, after having
specified and run this model?
Many thanks,
Mark Na
Call:
lm(formula = log(data$A
Hi Bert,
Yes, code like:
x <- names(yourdataframe)
grepl("muta",x) & !grepl("nonmuta|unmuta",x)
works perfectly.
Thanks very much for your help.
Paul
--- On Mon, 4/23/12, Bert Gunter wrote:
> From: Bert Gunter
> Subject: Re: [R] Selecting columns whose names contain "mutated" except whe
This might do it for you:
for (i in fileNames){
input <- read.table(i, .)
# you might want to use regular expressions to extract just the date.
input$fileName <- i
write.table(i, )
}
On Mon, Apr 23, 2012 at 12:29 PM, Shivam wrote:
> Hi,
>
> I am relatively new to R. Have
Building in Berend's suggestions I think this function should work for most
people (I'm going to wrap it into a package but figured people may want to
grab this directly):
# Please see http://www.netlib.org/lapack/double/dggev.f for a description
of inputs and outputs.
Rdggev <- function(A,B,JOBVL
On Apr 23, 2012, at 12:53 PM, phillip03 wrote:
So how would I use the lm() to estimate b_0 and b_1 for example
My Y_i and Y_j are data observations how does the lm() use my
data.frame ?
Have your read "An Introduction to R". Section 11 would appear to
cover all the needed topics.
So how would I use the lm() to estimate b_0 and b_1 for example
My Y_i and Y_j are data observations how does the lm() use my data.frame ?
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Hello,
phillip03 wrote
>
> Thank you Rui
>
> Can you help me with my ifelse problem - I would like to add a list to my
> data.frame where avgflow in those rows where ONLY my country pair both are
> in euro
>
First of all, try to use better (much, much better) code writing.
What are you trying
Dear useRs,
I have used using the excellent mgcv package (version 1.7-12) to
create a generalized additive model (gam) including random effects -
represented with s(...,bs="re") - on the basis of dialect data.
My model contains two random-effect factors (Word and Key - the latter
representing a s
We are aware that glmnet_1.7.3 does not pass for windows
and are looking into the problem. It has something to do
with the gcc compiler being slightly different on
windows versus linux/mac platforms. As soon as we have
resolved the issue, we will post a new version to CRAN
Trevor Hastie
--
On Apr 23, 2012, at 11:48 AM, Thomas Levine wrote:
I'm loading a nicely formatted csv file.
#!/usr/bin/env Rscript
kpi <- read.csv(
# This is a dump of the username, date_joined and last_login
columns
# from the auth_user Django table.
'data/2012-04-23.csv',
But maybe ... (see below)
-- Bert
On Mon, Apr 23, 2012 at 9:25 AM, Paul Miller wrote:
> Hello Dr. Winsemius,
>
> Unfortunately, I also have terms like "krasmutated". So simply selecting
> words that start with "muta" won't work in this case.
>
> Thanks,
>
> Paul
>
>
> --- On Mon, 4/23/12, David
On Apr 23, 2012, at 12:25 PM, Paul Miller wrote:
Hello Dr. Winsemius,
Unfortunately, I also have terms like "krasmutated". So simply
selecting words that start with "muta" won't work in this case.
You are aware that negative indexing can be used with grep aren't you?
--
David.
Thanks,
Below.
-- Bert
On Mon, Apr 23, 2012 at 9:10 AM, Paul Miller wrote:
> Hello All,
>
> Started out awhile ago trying to select columns in a dataframe whose names
> contain some variation of the word "mutant" using code like:
>
> names(KRASyn)[grep("muta", names(KRASyn))]
>
> The idea then would be
On 23/04/2012 10:49 AM, Adam Wilson wrote:
I routinely write graphics into multi-page PDFs, but some graphics (i.e.
plots of large spatial datasets using levelplot()) can result in enormous
files. I'm curious if there is a better way. For example:
#First, make some data:
library(lattice)
d=exp
On 23/04/2012 17:33, Duncan Murdoch wrote:
On 23/04/2012 12:05 PM, Jocelyn Ireson-Paine wrote:
I want to invoke R on a Linux Web server from Java, in order to analyse
data in ways that would take too long to code, and run too slowly, in
Java. In particular, to do correspondence analyses. To this
Hello
I have two lists of numbers, each list is ~800 numbers long. I want to know
if the two lists are significantly different from each other.
Could anyone suggest what library in R to use?
I think maybe the mann-whitney test, as it is not parametric, but i am
unsure if it is suitable as my list
Hi,
I am relatively new to R. Have scourged the help files and the www but
havent been able to get a solution.
I have around 250 csv files, one file for each date. They have columns of
all types, numeric, string etc. The name of each file is the date in the
form of 'mmdd'. There is no column
I'm running into an unexpected error using the glmnet and Matrix packages.
I have a matrix that is 8 million rows by 100 columns with 75% of the
entries being zero. When I run a vanilla glmnet logistic model on my server
with 300 GB of RAM, the task completes in 20 minutes:
> x # 8 million x 100
Thanks Steve, your explanation of the "==" is very helpful, and I will look up
?by. Ben
-Original Message-
From: S Ellison [mailto:s.elli...@lgcgroup.com]
Sent: Mon 4/23/2012 3:19 AM
To: Ben Neal; r-help@r-project.org
Subject: RE: Splitting a dataframe by character vector
> -Or
I'm loading a nicely formatted csv file.
#!/usr/bin/env Rscript
kpi <- read.csv(
# This is a dump of the username, date_joined and last_login columns
# from the auth_user Django table.
'data/2012-04-23.csv',
colClasses = c('character')
)
print(kpi[sample(nro
Hello,
jeff6868 wrote
>
> Hi Sarah,
>
> Thank you for your answer.
> Yes I know that my proposition is not necessary the better way to do it.
> But my problem concerns only big gaps of course (more than half a day of
> missing data, till several months of missing data).
> I've already filled s
I routinely write graphics into multi-page PDFs, but some graphics (i.e.
plots of large spatial datasets using levelplot()) can result in enormous
files. I'm curious if there is a better way. For example:
#First, make some data:
library(lattice)
d=expand.grid(x=1:1000,y=1:1000)
d$z=rnorm(nrow(d)
Hello,
I am having a problem where code that plots lines using a different data
frame plots bars with the current data frame (I am intended to plot lines).
The code specifies lines (see below), so I can't figure out why the results
are bars. I suspect that it may have something to do with the fact
On 23/04/2012 12:05 PM, Jocelyn Ireson-Paine wrote:
I want to invoke R on a Linux Web server from Java, in order to analyse
data in ways that would take too long to code, and run too slowly, in
Java. In particular, to do correspondence analyses. To this end, I've
installed R version 2.15.0 on my
Hello Dr. Winsemius,
Unfortunately, I also have terms like "krasmutated". So simply selecting words
that start with "muta" won't work in this case.
Thanks,
Paul
--- On Mon, 4/23/12, David Winsemius wrote:
> From: David Winsemius
> Subject: Re: [R] Selecting columns whose names contain "mu
On Apr 23, 2012, at 12:10 PM, Paul Miller wrote:
Hello All,
Started out awhile ago trying to select columns in a dataframe whose
names contain some variation of the word "mutant" using code like:
names(KRASyn)[grep("muta", names(KRASyn))]
The idea then would be to add together the various
Hello All,
Started out awhile ago trying to select columns in a dataframe whose names
contain some variation of the word "mutant" using code like:
names(KRASyn)[grep("muta", names(KRASyn))]
The idea then would be to add together the various columns using code like:
KRASyn$Mutant_comb <- rowSum
I want to invoke R on a Linux Web server from Java, in order to analyse
data in ways that would take too long to code, and run too slowly, in
Java. In particular, to do correspondence analyses. To this end, I've
installed R version 2.15.0 on my Web host's x86_64 GNU/Linux machine, and
tried usi
Hi Rui,
Yes you're right. It's me again ^^
This post is the last part (I hope) of my job. You helped me a lot last time
for the correlation matrices.
I have to leave my work now, so I'll check and test your proposition
tomorrow. But it makes no doubt that it'll help me a lot again.
I'll tell you
On Mon, Apr 23, 2012 at 6:33 AM, Ekta Jain wrote:
> Hello,
> Can anyone please suggest any packages in R that can be used to overlay
> gene expression data on SNP (affymetrix) copy number ?
>
Hi Ekta,
If you mean visually, as Steve suggested, you could try packages like
ggbio, Gviz, Rcytoscape..
This thread reveals that R has some holes in the solution of some of the linear algebra
problems that may arise. It looks like Jim Ramsay used a quick and dirty approach to the
generalized eigenproblem by using B^(-1) %*% A, which is usually not too successful due to
issues with condition of B a
On Apr 23, 2012, at 8:29 AM, wwreith wrote:
> Consider the following generic code for a survival model
>
> survobj<-Surv(data$Time,data$Satisfactory)
> survmodel<-survreg(survobj~x1+x2+x3+x4+x5+x6, data=data, dist="weibull")
> survsum<-summary(survmodel)
> survsum
>
> My question: Does anyone kn
Another possibility for visual display of several kinds of data is RCytoscape,
for which an example can be seen here:
http://rcytoscape.systemsbiology.net/versions/current/gallery/TCGA/subnet.TCGA.02.0014.png
This portrays
1) gene expression (green: under-expression; red: over-expression)
2
Yes, the (start, stop] formalism is the easiest way to deal with time
dependent data.
Each individual only needs to have sufficient data to describe them, so
for if id number 4 is in house 1, their housemate #1 was eaten at time
2, and the were eaten at time 10, the following is sufficient dat
Hi there,
To see the results of my clustering graphically I was using clusplot. But it
only provides a look at the two most important components of the dataset.
I recently found the Mclust() function which produces very nice colored pair
plots for the clustered dataset.
see Graph: http://www.stat
On 23-04-2012, at 15:20, Eiko Fried wrote:
> Hello,
>
> I'm working with RStudio, which does not display enough lines in the
> console that I can read the summary of my (due to the covariance-matrix
> rather long) model. There are no ways around this, so I guess I need to
> export the summary in
Consider the following generic code for a survival model
survobj<-Surv(data$Time,data$Satisfactory)
survmodel<-survreg(survobj~x1+x2+x3+x4+x5+x6, data=data, dist="weibull")
survsum<-summary(survmodel)
survsum
My question: Does anyone know what exactly survreg() does differently if
x1<-factor(dat
Hello,
I'm working with RStudio, which does not display enough lines in the
console that I can read the summary of my (due to the covariance-matrix
rather long) model. There are no ways around this, so I guess I need to
export the summary into a file in order to see it ...
I'm new to R, and "R sa
I could not reply directly to the initial thread with the same title.
There are two sorts of Robust PCA, those that were devised before the
recent string of Low Rank approaches and then the new set of algorithms
that provide robust PCA in light of sparse but potentially large
errors/outliers (typi
Thanks for saying. :)
This morning I tried out some things with this code, and different variants
with it.
Works nice.
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On 04/22/2012 05:00 AM, r-help-requ...@r-project.org wrote:
I am trying to run Weibull PH model in R.
Assume in the data set I have x1 a continuous variable and x2 a
categorical variable with two classes (0= sick and 1= healthy). I fit the
model in the following way.
Test=survreg(Surv(t
Rpart has a built in cross-validation proceedure. the xerror and so on
are based on a ten fold cross validation, I believe. Read the manual
for more results. If you plot the fit with post and then look at the
postscript file it should have a misclassification rate under the
terminal nodes.
On 23.04.2012 12:42, phillip03 wrote:
Hi
I am looking for a efficient way to estimate all parameters in your
data.frame set using a specific function:
for example
ln(T)=b_0 + b_1*ln(Y_i*Y_j) + b_2*ln()+ ... + etc.
Sounds like you are looking for lm().
Uwe Ligges
Thanks,
Ph
--
View thi
Hi
I am looking for a efficient way to estimate all parameters in your
data.frame set using a specific function:
for example
ln(T)=b_0 + b_1*ln(Y_i*Y_j) + b_2*ln()+ ... + etc.
Thanks,
Ph
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http://r.789695.n4.nabble.com/OLS-Estimating-tp4580055p4580055.html
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Hi Sarah,
Thank you for your answer.
Yes I know that my proposition is not necessary the better way to do it. But
my problem concerns only big gaps of course (more than half a day of missing
data, till several months of missing data).
I've already filled small gaps with the interpolation that you
Dear all,
I have tow (several) bivariate distributions with a known mean and
variance-covariance structure (hence a known density function) that I would
like to compare in order to get an intersect that tells me something about "how
different" these distributions are (as t-statistics for univa
Hi, Jorge!
Help me please! I made a classification tree (rpart package) according to my
train data and set of variables. How can I validate my test data? I want to
check if the test data will classified properly by the same tree. Thanks
cheers
Maria
Tue, 10 Apr 2012 08:44:39 -0700 (PDT) Ð¾Ñ "J
Dear all,
I've been getting different results from the sparse and dense version
of model.Matrix when used with sparse contrasts and interactions
between factors. The same happens when using model.matrix and
sparse.model.matrix. When calculating list.contrasts I get the same
results for sparse and
Hi,
So I've been figuring out how to use igraph in R and like it for it's speed
and simplicity. Now I have a graph built from an edgelist where vectors have
a $name attribute. I have another dataframe with attributes tied to a
vector ID, which is the same as the $name attribute of vectors represe
Small mistake in my subset example. I mean to remove the `$ID` part at
the end so that you can play with the whole subset-ted data.frame, and
not just get back the ID column. Instead of this:
R> interesting <- subset(DataFile, log2 >= 7)$ID
do this:
R> interesting <- subset(DataFile, log2 >= 7)
Hi,
On Mon, Apr 23, 2012 at 7:33 AM, Ekta Jain wrote:
> Hello,
> Can anyone please suggest any packages in R that can be used to overlay gene
> expression data on SNP (affymetrix) copy number ?
I guess you mean visually? If so, I'd suggest skimming through the
vignettes of the following package
Hello,
Can anyone please suggest any packages in R that can be used to overlay gene
expression data on SNP (affymetrix) copy number ?
Thanks,
Ekta
Senior Research Associate
Bioinformatics Department
Jubilant Biosys Pvt Ltd,
#96, Industrial Suburb, 2nd Stage
Yeshwantpur, Bangalore 560 022
Ph No :
Hi,
Even your example should show why this is a bad way to fill in missing weather
data: you end up with a sequence for station 1 of 1, 2, 10, 4 even though
that's certainly wrong because Station 2 is reliably 7 units above Station 1.
"Correlated" doesn't mean "identical."
There are other bett
Hi everyone.
I have a question about a work on R I have to do for my job.
I have temperature data coming from 70 weather stations. One data file
corresponds to one station for one year (so 70 files for one year). Each
file looks like this (important: each file contains NAs):
time
> -Original Message-
> I am just trying to split a dataframe of 750 observations of
> 29 variables by "Site", which is a vector in the dataframe
> with five text names (ex. PtaCaracol).
A couple of methods.
i) First, look up ?split, which chops your data frame into a list of five data
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