Why does R think these numbers ***are*** equal?
In a somewhat bizarre set of circumstances I calculated
x0 <- 0.03580067
x1 <- 0.03474075
y0 <- 0.4918823
y1 <- 0.4474461
dx <- x1 - x0
dy <- y1 - y0
xx <- (x0 + x1)/2
yy <- (y0 + y1)/2
chk <- yy*dx - xx*dy + x
On Tue, 2 Aug 2011, David Winsemius wrote:
On Aug 1, 2011, at 11:14 PM, zhenjiang xu wrote:
Hi,
I have a for loop to make 2 types of plots and I'd like to save one
type of plots to a pdf file and the other to another pdf file. How can
I control which plot will be saved to which pdf? Thanks
On Mon, 1 Aug 2011, Paul Rodriguez wrote:
Hello R experts,
I'm trying to test R in a shared memory environment in which addressable memory
is aggregrated to about 600G.
However, I get an error of 'too many elements' specified when I try creating a
45K x 100K matrix.
I tried running R with a
On Aug 1, 2011, at 11:14 PM, zhenjiang xu wrote:
Hi,
I have a for loop to make 2 types of plots and I'd like to save one
type of plots to a pdf file and the other to another pdf file. How can
I control which plot will be saved to which pdf? Thanks
Why not give them file names that identify t
David, I'm so appreciate!
Sincerely,
Zoe
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Hello R experts,
I'm trying to test R in a shared memory environment in which addressable memory
is aggregrated to about 600G.
However, I get an error of 'too many elements' specified when I try creating a
45K x 100K matrix.
I tried running R with a --max-nsize=500 option, but got the
Simplest to make two consecutive loops.
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with
/Software/Embedded Co
Hi,
I have a for loop to make 2 types of plots and I'd like to save one
type of plots to a pdf file and the other to another pdf file. How can
I control which plot will be saved to which pdf? Thanks
--
Best,
Zhenjiang
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Why can't you simply draw a 3 - D plot. The third variable being categorical
can expressed as a bar.
Regards,
Indrajit
From: Andrew McCulloch
To: r-help@r-project.org
Sent: Monday, August 1, 2011 3:14 PM
Subject: [R] Plotting question
Hi,
I use R to draw my
Am Montag, den 01.08.2011, 12:43 -0400 schrieb R. Michael Weylandt :
> I've only got a 20 minute layover, but three quick remarks:
>
> 1) Do a sanity check on your data size: if you want a million walks of
> a thousand steps, that already gets you to a billion integers to
> store--even at a very l
In addition to what David said:
On Mon, Aug 1, 2011 at 6:57 PM, zoe_zhang <1987.zhan...@gmail.com> wrote:
> Dear All,
> Sorry to bother
> I want to write a function in R using if
> Say I have a dataset x,
> if x[i]<0, then x[i]=x[i],
> if x[i]>0, then x[i]=0
>
> for example, x=-3:3,
> then using t
On Mon, 1 Aug 2011, Iasonas Lamprianou wrote:
Thank you, it works! And the estimates (as well as the standard errors)
seem to be more reasonable now, compared to the "normal" Poisson model.
Thank you. However, I tried to find a manual? (although I did manage to
find the paper published in the
On Aug 1, 2011, at 6:57 PM, zoe_zhang wrote:
Dear All,
Sorry to bother
I want to write a function in R using if
Say I have a dataset x,
if x[i]<0, then x[i]=x[i],
if x[i]>0, then x[i]=0
for example, x=-3:3,
then using the function, x becomes [-3,-2,-1,0,0,0,0]
Just use logical indexing
x[ x>
On Aug 1, 2011, at 20:12 , monk wrote:
> dear all,
>
> i have a quite simple question, i want to fill up a Matrix like done in the
> following function,
> but the performance is very bad for large dimensions
> is there a way to do this like with apply or something similar?
>
>
> makeMatrix <-
Dear All,
Sorry to bother
I want to write a function in R using if
Say I have a dataset x,
if x[i]<0, then x[i]=x[i],
if x[i]>0, then x[i]=0
for example, x=-3:3,
then using the function, x becomes [-3,-2,-1,0,0,0,0]
I write the codes as follows,
gjr=function(x)
{lena=length(x)
for(i in 1:lenx)
Hi All,
I am looking for some help figuring out what is causing an error in my
attempt to fit a regularized logistic regression (specifically finding the
optimal lambda value using cv.glmnet).
Running the following command:
RegLR_CV<-cv.glmnet(x=train.sub.clean[,-c(431)],y=as.factor(train.sub$fi
Well stated, Duncan, and I plead guilty, though I did try to weasel
out with caveats.
Perhaps I may plead down to a lesser sentence or probation by saying
that I was offering what I still believe to be appropriate advice for
a general strategy for handling this sort of plotting issue; but that
as
Dear all,
Thanks again for your help. I looked at a polynomial regression of 12th order
and the sinus regression and to me the polynomial model seems to fit better.
What do you think?
Thanks a lot,
Mariannne
time <-
c(0.15,0.30,0.45,1.00,1.15,1.30,1.45,2.00,2.15,2.30,2.45,3.00,3.15,3.30,3.45
Hi Matt,
On Mon, Aug 1, 2011 at 4:47 PM, Matt Curcio wrote:
> Greetings all,
> I am getting this error that is driving me nuts... (not a long trip, haha)
>
> I have a set of files and in these files I want to calculate ttests on
> rows 'compareA' and 'compareB' (these will change over time there
To be specific, I only need to get rid of 2 NYSE holidays:
Washington's Birthday and Good Friday.
Is there a way to reduce the vector of NYSE holidays in timeDate by
throwing out those two?
Thank you!
Dimitri
On Mon, Aug 1, 2011 at 4:24 PM, R. Michael Weylandt
wrote:
> Don't know if this is suff
On 11-08-01 11:48 AM, Bert Gunter wrote:
IMHO:
On Mon, Aug 1, 2011 at 7:51 AM, Duncan Murdoch wrote:
On 11-08-01 5:44 AM, Andrew McCulloch wrote:
Hi,
I use R to draw my graphs. I have 100 points on a simple xy-plot. The
points are
distinguished by a third variable which is categorical with
Greetings all,
I am getting this error that is driving me nuts... (not a long trip, haha)
I have a set of files and in these files I want to calculate ttests on
rows 'compareA' and 'compareB' (these will change over time there I
want a variable here). Also these files are in many different
directo
Don't know if this is sufficiently slick for this list (which never fails to
impress me with quick and elegant solutions) but I would point out to you that
GF is the only NYSE holiday falling in March or April so it shouldn't be hard
to discard it if desired.
Michael Weylandt
On Aug 1, 2011,
Just to clarify - I realize that "major" is subjective here. Maybe I
should say "most common".
But maybe there is a way for me to select from a list of all NYSE
holidays and flag only some of them?
Just not sure how to do it...
Thanks!
Dimitri
On Mon, Aug 1, 2011 at 3:45 PM, Dimitri Liakhovitski
But if you do mean to divide by max(x), I'll also vote for the prior
ROI <- function(x) {
if (length(x)==1) return(NA)
r=c(x[1], diff(x))/max(x)
return(r)}
As being about as quick and elegant as this can be done in R.
M
On Aug 1, 2011, at 4:07 PM, "R. Michael Weylandt "
wrote:
> Just jumpin
> But why not just
> c(x[1], diff(x))/max(x)
So simple! Thank you ever so much Berend.
Best wishes,
Ben
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Sent from the R help mailing list a
Just jumping into this, but does the ROC(x, type="discrete") function of either
the TTR or caTools (can't remember which) work if you need a prebuilt function?
Also, why are you dividing by the max value? That seems a funny way to
calculate ROC...
On Aug 1, 2011, at 3:14 PM, bjmjarrett wrote:
It looks like I had been missing an update needed for Ubuntu systems. All I
needed was the following. Thank you.
update.packages(lib.loc = "/usr/local/lib/R/site-library")
On Mon, Aug 1, 2011 at 2:39 PM, Jean V Adams wrote:
>
> That's interesting. My function matrix() looks like this:
>
> fun
I have a function to calculate the rate of increase (the difference between
the value and the previous value divided by the total number of eggs in a
year) of egg production over the course of a year:
rate <- function(x){
storage <- matrix(nrow=length(x),ncol=1)
storage[1,] <- x[1] /
thanks a lot , that will do the trick
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Hi Everyone,
When i try to install a package using
> install.packages("agricolae")
--- Please select a CRAN mirror for use in this session ---
|
The cursor keeps blinking i dont get a popup menu to choose a CRAN mirror? Is
it due to my proxy server settings? I tried to echo $http_proxy ,
On 7/31/2011 6:24 PM, Alexandre Aguiar wrote:
Em Domingo 31 Julho 2011, você escreveu:
My memory is that this question gets asked every few months and one of
the stock answers is to use the function 'package.skeleton' in the
utils package as a starting point.
Got that from docs. And actually I
bjmjarrett wrote:
>
> ...
> rate <- function(x){
> storage <- matrix(nrow=length(x),ncol=1)
> ifelse(length(x)==1,storage[1,] <- NA,{
> storage[1,] <- x[1]/max(x)
> for(i in 2:length(x)){
> p <- i-1
> storage[i,] <- ((x[i] - x[p]) / max(x))
>
Hello!
I am trying to identify which ones of a vector of dates are US
holidays. And, ideally, which is which. And I do not know (a-priori)
which dates those should be.
I have, for example:
x<-seq(as.Date("2011-01-01"),as.Date("2011-12-31"),by="day")
(x)
I think chron should help me here - but ma
Actually Sara's method fails if the insertion is after the first or before
the last column:
>x <- data.frame(A=1:3, B=1:3, C=1:3, D=1:3, E=1:3)
>newcol <- 4:6
>cbind(x[,1], newcol, x[,2:ncol(x)])
x[, 1] newcol B C D E
1 1 4 1 1 1 1
2 2 5 2 2 2 2
3 3 6 3 3 3 3
> cbi
Most certainly you can speed it up:
X <- exp(-abs(row(X) - col(X)) / (3*b))
diag(X) <- a
should do what you want. This is called
'vectorization' and is discussed lots of
places -- for instance, in the two documents
mentioned below in my signature.
On 01/08/2011 19:12, monk wrote:
dear all,
That's interesting. My function matrix() looks like this:
function (data = NA, nrow = 1, ncol = 1, byrow = FALSE, dimnames = NULL)
{
if (is.object(data) || !is.atomic(data))
data <- as.vector(data)
.Internal(matrix(data, nrow, ncol, byrow, dimnames, missing(nrow),
missi
Yes, even if I only run the command matrix(0,30,10) I get the error. I am
running R with Ubuntu 10.10 (maverick) with R version:
R version 2.13.1 (2011-07-08)
When I check the function matrix, I can see that it is only passing five
arguments to the function .Internal() (shown below).
function (
Making use of the row() and col() functions speeds things up a bit.
makeMatrix2 <- function(a, b, dim) {
X <- matrix(NA, ncol=dim, nrow=dim)
X <- exp( (-1*abs(row(X) - col(X)))/(3*b) )
diag(X) <- a
X
}
system.time(makeMatrix(1, 2, 1000))
system.time(makeMat
See the footer of this and every R-help message.
In particular, that DLL is not used by R itself, so this is probably
something called from a third-party package.
A number of packages used to use that DLL (which is rather out of
date), but no longer, so is your R actually current (the posting
dear all,
i have a quite simple question, i want to fill up a Matrix like done in the
following function,
but the performance is very bad for large dimensions
is there a way to do this like with apply or something similar?
makeMatrix <- function(a, b,dim) {
X=matrix(0,ncol=dim,nrow=dim)
Thanks Sarah and David.
Yes, but note this:
> z <- data.frame(a=1:2,b=3:4)
> z
a b
1 1 3
2 2 4
> newdat <- 5:6
> cbind(z[,1],newdat,z[,2])
newdat
[1,] 1 5 3
[2,] 2 6 4
> cbind.data.frame(z[,1],newdat,z[,2])
z[, 1] newdat z[, 2]
1 1 53
2 2 6
Merik,
You did get an answer to the question, and it's even included in the material
below.
What doesn't work for you in Ista's suggestion?
id<- c(1,1,1,1,1,2,2,2,3,3,3)
month <- c(1, 1, 2, 3, 6, 2, 3, 6, 1, 3, 5)
value <- c(10, 12, 11, 14, 16, 12, 10, 8, 14, 11, 15)
dat.tmp <- data.frame(id
On Mon, Aug 1, 2011 at 1:43 PM, Sarah Goslee wrote:
> Bert,
>
> On Mon, Aug 1, 2011 at 1:27 PM, Bert Gunter wrote:
>> Folks:
>>
>> I consider my reply below rather clumsy: One has to keep track of
>> index numbers other than that which is inserted and must separately
>> change column names. Is th
Bert,
On Mon, Aug 1, 2011 at 1:27 PM, Bert Gunter wrote:
> Folks:
>
> I consider my reply below rather clumsy: One has to keep track of
> index numbers other than that which is inserted and must separately
> change column names. Is there as "essentially better" way to do this,
> either via base R
Bert,
On Mon, Aug 1, 2011 at 1:17 PM, Bert Gunter wrote:
> Doesn't work -- you lose column names.
But I don't lose column names:
> x <- data.frame(A=1:3, B=1:3, C=1:3, D=1:3, E=1:3)
> x
A B C D E
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
> newcol <- 4:6
> cbind(x[,1:2], newcol, x[,3:ncol(x)])
A B
Not when I do it.
> a <- data.frame(A=1:10, B=11:20, D=31:40, E=41:50)
> a
A B D E
1 1 11 31 41
2 2 12 32 42
3 3 13 33 43
4 4 14 34 44
5 5 15 35 45
6 6 16 36 46
7 7 17 37 47
8 8 18 38 48
9 9 19 39 49
10 10 20 40 50
> b <- cbind(a[,1:2], C=21:30, a[,3:4])
> b
A B C D
Since I didn't get an answer to this question, I'm rephrasing my question in
simpler terms:
I have a dataframe and I want to split it based on the levels of one of its
columns, and apply a function to each section of the data. Output of the
function may be drawing a plot, returning a value, what
Folks:
I consider my reply below rather clumsy: One has to keep track of
index numbers other than that which is inserted and must separately
change column names. Is there as "essentially better" way to do this,
either via base R or via an R package. I leave it to you to define
"essentially better.
Searching R Graphical Manual (http://www.oga-lab.net/RGM2/, mirror
http://www.oga-lab.net/RGM2/) shows possible candidates in packages circular
(windrose), IDPmisc (plot.rose), climatol (rosavent), openair (windRose),
and oce (as.windrose).
--
David L Ca
Doesn't work -- you lose column names.
Try this instead:
yourframe[,30:51] <- cbind( newcolumn,yourframe[,30:50])
Adjust column names after via:
names(yourframe) [30:51] <- c(newcolname,names(yourframe[30:50])
Cheers,
Bert
On Mon, Aug 1, 2011 at 10:10 AM, Sarah Goslee wrote:
> x <- cbind(x[,
Providing the data will help, but the first thing I noted is that you have more
columns (variables) than rows (cases). PCA will return a maximum of (the number
of columns) or (the number of rows-1) whichever is less. With 84 columns and 66
rows means you can get no more than 65 components. If th
x <- cbind(x[,1:29], newcolumn, x[,30:ncol(x)])
On Mon, Aug 1, 2011 at 12:59 PM, Bansal, Vikas wrote:
> Dear all,
>
> I have a very simple question.I have data frame of 50 columns and i want to
> insert a column in 30th position.But i do not want to delete that column.Is
> it possible to includ
Y'know, you aren't likely to get many responses with this kind of request. Why
don't you go read the posting guidelines and come back with:
R version info
Sample data
Actual commands used, so we can reproduce the problem
---
Robert,
What code did you run to get that error?
Do you get the error if the only code that you run is ...
matrix(0, 30, 10)
You gave three arguments to matrix, which requires none, but can take up
to five.
In the function matrix there is a call to .Internal(matrix) which requires
7 argum
Dear all,
I have a very simple question.I have data frame of 50 columns and i want to
insert a column in 30th position.But i do not want to delete that column.Is it
possible to include a column in between, so that new values are in 30th column
and 30 th column is now 31st and 31st is 32nd..
Hello,
I am having a problem with the function matrix. Specifically, when I pass
three arguments (two more being instantiated in the function), I get the
following error message:
Error in matrix(0, 30, 10) :
5 arguments passed to .Internal(matrix) which requires 7
I looked into it, and someon
Hi Tina,
That is quite a bit of missingness, especially considering the sample
size is not large to begin with. This would make me treat *any*
result cautiously. That said, if you have a reasonable idea what the
mechanism causing the missingness is or if from additional variables
in your study,
the answer was indeed in subtle differences, and 'str' did help.
Problem is solved.
Thanks everybody for comments which was all very useful.
Best,
On Mon, Aug 1, 2011 at 12:25 PM, jim holtman wrote:
> What you "see" and what the data really is may be two different
> things. You should have at
I've only got a 20 minute layover, but three quick remarks:
1) Do a sanity check on your data size: if you want a million walks of a
thousand steps, that already gets you to a billion integers to store--even at a
very low bound of one byte each, thats already 1GB for the data and you still
have
On Aug 1, 2011, at 12:17 PM, world peace wrote:
Hi Guys,
working on a "merge" for 2 data frames.
Using the command:
x <- merge(annotatedData, UCSCgenes, by.x="names",
by.y="Ensembl.Gene.ID", all.x=TRUE)
names and Ensembl.Gene.ID are columns with similar elements from the x
and y data frames
Dan,
If the variables you are merging by are character variables, there may be
subtle differences that you haven't noticed, e.g., capitalization or
spacing. You can look for differences by listing off the unique values:
table(c(annotatedData$names, UCSCgenes$Ensembl.Gene.ID))
Jean
`·.,, ><
What you "see" and what the data really is may be two different
things. You should have at least enclosed an 'str' of the two data
frames; even better would be a subset of the data using 'dput'. Most
likely your problem is that your data is not what you 'expect' it to
be.
On Mon, Aug 1, 2011 at
Try this: had to add extra names to your data since it was not clear
how it was organized. Next time use 'dput' to enclose data.
> x <- read.table(textConnection(" index time key date values
+ 13732 27965 DATA.Q211.SUM.Index04/08/11 1.42
+ 13733 27974 DATA.Q211.SUM.Index
Dear R users,
I am a new R user and something stops me when I try to write a academic
article. I want to make a nomogram to predict the risk of prostate cancer
(PCa) using several factors which have been selected from the Logistic
regression run under the SPSS. Always, a calibration plot is needed
Hi Guys,
working on a "merge" for 2 data frames.
Using the command:
x <- merge(annotatedData, UCSCgenes, by.x="names",
by.y="Ensembl.Gene.ID", all.x=TRUE)
names and Ensembl.Gene.ID are columns with similar elements from the x
and y data frames.
annotatedData has 8909 entries, so has x(as expec
Kindly do not attach questions in a separate document.
Install and read the documentation for the R rms package, and see handouts
at http://biostat.mc.vanderbilt.edu/rms
Frank
sytangping wrote:
>
> Dear R users,
>
> I am a new R user and something stops me when I try to write a academic
> arti
Hi Margaux,
Check the row.names and col.names arguments of write.table.
See ?write.table
write.table (dat, file = "/path/to/my/data.txt", sep = " ",
col.names=FALSE, row.names=FALSE)
HTH,
Ivan
Le 8/1/2011 17:18, Margaux Keller a écrit :
Hi,
I'm trying to create an abbreviated data file fr
What do you think about this?
apply(data, 3, '[', indices)
On Mon, Aug 1, 2011 at 4:38 AM, Jannis wrote:
> Dear R community,
>
>
> I have a general question regarding indexing in multidiemensional arrays.
>
> Imagine I have a three dimensional array and I only want to extract on
> vector along
IMHO:
On Mon, Aug 1, 2011 at 7:51 AM, Duncan Murdoch wrote:
> On 11-08-01 5:44 AM, Andrew McCulloch wrote:
>>
>> Hi,
>>
>> I use R to draw my graphs. I have 100 points on a simple xy-plot. The
>> points are
>> distinguished by a third variable which is categorical with 10 levels. I
>> have
>> bee
On Aug 1, 2011, at 5:01 AM, Przemek Jura wrote:
Dear group,
I experience some problems with gam() function after R update to
version 2.13.1
The function in both gam and mgcv packages stopped to work. Before,
with the same code I used, everything was fine.
Reports like this often turn out
Hello,
I am having a problem with the function matrix. Specifically, when I pass
three arguments (two more being instantiated in the function), I get the
following error message:
Error in matrix(0, 30, 10) :
5 arguments passed to .Internal(matrix) which requires 7
I looked into it, and someon
Hi,
I'm trying to create an abbreviated data file from a larger version. I can
use the subset command to create a value for this data:
dat <-subset(raw.data, select=c(SNP, Pvalue))
> head (dat)
SNP Pvalue
1 rs11 0.6516
2 rs12 0.3311
3 rs13 0.5615
but when I try to write.table using:
w
plot(1:10, pch=letters[1:10])
On Mon, Aug 1, 2011 at 4:44 AM, Andrew McCulloch wrote:
> Hi,
>
> I use R to draw my graphs. I have 100 points on a simple xy-plot. The
> points are
> distinguished by a third variable which is categorical with 10 levels. I
> have
> been plotting x against y and usin
Dear Contributors
thanks for any help you can provide. I searched the threads
but I could not find any query that satisfied my needs.
This is my database:
index time values
13732 27965 DATA.Q211.SUM.Index04/08/11 1.42
13733 27974 DATA.Q211.SUM.Index05/10/11 1.45
1
On Aug 1, 2011, at 10:50 AM, Duncan Murdoch wrote:
On 11-08-01 5:38 AM, Jannis wrote:
Dear R community,
I have a general question regarding indexing in multidiemensional
arrays.
Imagine I have a three dimensional array and I only want to extract
on
vector along a single dimension from
Hi again,
I have tried playing around with the code given to me by Alan and Jim, thank
you for the code but unfortunatelyI can't seem to get either of them to
work... Alans does not work with the sample data and Jims is giving the
error :
Error in radial.grid(labels = labels, label.pos = labe
On Aug 1, 2011, at 3:04 AM, Dimitris.Kapetanakis wrote:
Thanks a lot for the help.
Actually, I am using a mac which (R for Mac OS X GUI 1.40-devel
Leopard
build 32-bit (5751)) but I think I can find access on windows 7 64-
bit.
I don't think that was what Holtman was advising. You just ne
Dear David and Hans- Werner,
Thank you very much for your help. I would like to compare now if a
polynomial or the sinus model fits better. How can I see R-squared or
the F- Statistic for the sinus regression, so as to be able to compare
it with the polynomial model?
Thanks a lot and have a nice ev
On 11-08-01 5:44 AM, Andrew McCulloch wrote:
Hi,
I use R to draw my graphs. I have 100 points on a simple xy-plot. The points are
distinguished by a third variable which is categorical with 10 levels. I have
been plotting x against y and using gray scales to distinguish the level of the
categori
Dear R-help
We are getting an error message `jpeg62.dll missing'.
We are running Windows 7 64-bit, from a Mac using Boot Camp.
Do you know of this error message, and can you give us help trying to
resolve the problem?
many thanks
Rocky
Rocky Hyacinth
Technician
Department of Archaeology
Univer
On 11-08-01 5:38 AM, Jannis wrote:
Dear R community,
I have a general question regarding indexing in multidiemensional arrays.
Imagine I have a three dimensional array and I only want to extract on
vector along a single dimension from it:
data<- array(rnorm(64),dim=c(4,4,4))
result<- data[1
Dear all,
this must have been a temporary problem. In this case I assume that the
build cycle did not finish in time, i.e., binaries were synced to the
staging area although not all were built.
best,
stefan
On 07/31/2011 05:52 PM, David Winsemius wrote:
On Jul 31, 2011, at 11:26 AM, Michae
Dear all,
I have been attempting to use multiple imputation (MI) to handle missing data
in my study. I use the mice package in R for this. The deeper I get into this
process, the more I realize I first need to understand some basic concepts
which I hope you can help me with.
For example, let u
Dear Mailing-list
I used hclust to make a dendrogram of 2613 leafs. I also have a list
with the names of certain labels which are of interest and I would like
to visualize their appearance within the dendrogram. I found an example
how to use dendrapply to colour the labels but the problem is that
Hi,
I use R to draw my graphs. I have 100 points on a simple xy-plot. The points
are
distinguished by a third variable which is categorical with 10 levels. I have
been plotting x against y and using gray scales to distinguish the level of the
categorical variable for each point. It looks ok to
Dear group,
I experience some problems with gam() function after R update to version 2.13.1
The function in both gam and mgcv packages stopped to work. Before, with the
same code I used, everything was fine.
The function from gam package yields following warning:
Residual degrees of freedom are
Dear group,
I experience some problems with gam() function after R update to version 2.13.1
The function in both gam and mgcv packages stopped to work. Before, with the
same code I used, everything was fine.
The function from gam package yields following warning:
Residual degrees of freedom are
Thanks a lot for the help.
Actually, I am using a mac which (R for Mac OS X GUI 1.40-devel Leopard
build 32-bit (5751)) but I think I can find access on windows 7 64-bit. What
I am trying to do is a maximization through grid search (because I am not
sure that any of the optim() methods works suff
Dear R community,
I have a general question regarding indexing in multidiemensional arrays.
Imagine I have a three dimensional array and I only want to extract on
vector along a single dimension from it:
data<- array(rnorm(64),dim=c(4,4,4))
result <- data[1,1,]
If I want to extract m
Yes, that's what I was looking for.
Many thanks,
Samuel
-Original Message-
From: S Ellison [mailto:s.elli...@lgcgroup.com]
Sent: 01 August 2011 15:16
To: Samuel Le; r-h...@stat.math.ethz.ch
Subject: RE: formula used by R to compute the t-values in a linear regression
> -Original Me
Exactly.
My formula holds only for k=1, this is how I generated it.
Do you have any references concerning the " rather more careful algorithms"?
Thanks,
Samuel
-Original Message-
From: peter dalgaard [mailto:pda...@gmail.com]
Sent: 01 August 2011 14:45
To: Samuel Le
Cc: r-h...@stat.math.
> -Original Message-
> [mailto:r-help-boun...@r-project.org] On Behalf Of Samuel Le
> Subject: [R] formula used by R to compute the t-values in a
> linear regression
> I was wondering if someone knows the formula used by the
> function lm to compute the t-values.
Typing
summary.lm
I
Yes! Would you mind filing an issue so I dont forget?
Hadley
On Friday, July 29, 2011, Stavros Macrakis wrote:
> Perfect! Thanks!
> By the way, I see that, unlike base rbind, it does not work for vectors
and lists:
> rbind(c(a=1),c(b=2)) => matrix(1:2,2,1,dimnames=list(NULL,"a"))
== as.matr
On Aug 1, 2011, at 15:27 , Samuel Le wrote:
> Hello,
>
>
>
> I was wondering if someone knows the formula used by the function lm to
> compute the t-values.
>
>
>
> I am trying to implement a linear regression myself. Assuming that I have K
> variables, and N observations, the formula I a
On Aug 1, 2011, at 9:27 AM, Samuel Le wrote:
Hello,
I was wondering if someone knows the formula used by the function lm
to compute the t-values.
I am trying to implement a linear regression myself. Assuming that I
have K variables, and N observations, the formula I am using is:
For t
Hello,
I was wondering if someone knows the formula used by the function lm to compute
the t-values.
I am trying to implement a linear regression myself. Assuming that I have K
variables, and N observations, the formula I am using is:
For the k-th variable, t-value= b_k/sigma_k
With b_k
Greetings all,
Thanks for all your help so far.
Let me give a better idea of what I am doing. I have hundreds of
files that I need to plow thru with a t-test and correlation test.
BTW, 'tempA' and tempB' are simply columns of numbers from a gene-chip
experiment that spits out dna 'amounts'. So I h
Hi Peter,
Many thanks. It worked.
Regards
Ogbos
On 1 August 2011 14:05, Peter Ehlers wrote:
> On 2011-08-01 03:32, ogbos okike wrote:
>
>> Dear All,
>> I am trying to put 10^-8 st km^-2day^-1 on x-axis of my plot. I tried
>> using
>> : ylab = expression(paste("st / ", plain(km)^2, " / day")) to
Am Sonntag, den 31.07.2011, 23:32 -0500 schrieb R. Michael Weylandt :
> Glad to help -- I haven't taken a look at Dennis' solution (which may be far
> better than mine), but if you do want to keep going down the path outlined
> below you might consider the following:
I will try Dennis’ solution ri
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