Hi
how do we get intercepts standard error. I'm using the package pls.
i got the coefficient but not able to get the stabdard error
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Dear All,
I am facing a problem in calling an user defined R function from Java
through JRI. The user defined R function does a loess normalization on micro
array data ( find in the limma package of BioConductor) and the last 2 lines
of the R Code is :
MA <- normalizeWithinArrays(RG, method="loes
n <- 10
P1 <- runif(n)
P2 <- runif(n)
P3 <- P1 + P2 + runif(n)/100
P4 <- P1 + P2 + P3 + runif(n)/100
mydata <- data.frame(cbind(P1,P2,P3,P4))
mydata[1,1] <- 8
mydata[3,1] <- -5
mydata[2,3] <- -6
mydata[7,3] <- 7
f=function(z){quantile(z, c(0.01, 0.99)) }
temp1 <- lapply(mydata, f)
temp1
$P1
Dear list members,
what is the reason that one obviously can't do arithmetic operations on
zoo members with different index positions?
> require(zoo)
> z <- zoo(c(1,1,1),order.by=c(1,2,3))
> z
1 2 3
1 1 1
> z[1] + z[1]
1
2
> z[1:2] + z[1:2]
1 2
2 2
> z[1] +
Hi All:
If I download yahoo data by getSymbols() in R, the date column gets
accompanied along with the downloaded data. There is no column header for
the date column to access separately.
What is the way to eliminate the date column?
If I want to draw a xy scatter plot with the downloaded price (
Please update your R (and probably your RGtk2: you did not tell us its
version), as the posting guide asked you to do before posting.
On Sun, 23 Oct 2011, Aref Nammari wrote:
Hello,
I hope this is the right place to ask for help with a problem I am
having with RGtk2 installation with R on Win
Dear R People:
I've been trying to get R Portable from sourceforge.net all day today
and there is a problem accessing the site.
Has anyone else run into that, please?
Thank you!
Sincerely,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of
And the plyr version of this would be (using DF as the data frame name)
## transform method, mapping length(runtime) to all observations
## similar to David's results:
library('plyr')
ddply(DF, .(time, partitioning_mode), transform, n = length(runtime))
# or equivalently, the newer and somewhat fa
On Sun, Oct 23, 2011 at 11:48 AM, Robert McDonald wrote:
> I think your question is answered by
>>
>> http://cran.r-project.org/web/packages/fOptions/ChangeLog
>>
>> 2010-04-23 chalabi
>>
>>* ChangeLog, DESCRIPTION: updated DESCR and ChangeLog
>>* src/085A-LowDiscrepancy.f: fixed
On Oct 23, 2011, at 7:37 PM, apeer wrote:
Does your point about proportionality also hold for ordinality? In
other
words, if I have several X variables that do not behave in an ordinal
fashion with Y, should I still include them in the full model? My
understanding or perhaps misunderstandin
Hello,
I hope this is the right place to ask for help with a problem I am
having with RGtk2 installation with R on Windows XP.
I am running R 2.11.1 and have installed the package RGtk2 from CRAN.
I also have GTK 2.10.11 installed as well as GTK2-runtime 2.22.0. I
have added the environment variab
Does your point about proportionality also hold for ordinality? In other
words, if I have several X variables that do not behave in an ordinal
fashion with Y, should I still include them in the full model? My
understanding or perhaps misunderstanding of the ordinality assumption was
that all X va
Another package to consider, especially if your dataframe is large, is
'data.table':
> tp <- read.table(textConnection(" time partitioning_mode workload runtime
+ 1 1 shardingquery 607
+ 2 1 shardingquery 85
+ 3 1 shardingquery 52
+
On Oct 23, 2011, at 4:18 PM, > wrote:
Hello,
I have a list of dates in which I am going use for a time series
analysis. I want to break these dates up into 2 week intervals and
count the number of times a date appears in this interval.
For example from Nov. 19, 2000 to Dec 2 ,2000 with t
Hello,
I am trying to do a manova test in r, and have used the "manova" function to
test differences between two dependent variables. The results were
significant for the whole model, but the sources I've read say that in order
to do a post-hoc multiple comparison, I have to do separate aovs for e
I am a very new user of R and I need some suggestions on how perform a
diptest.
I dowloaded the package diptest. following instructions given by the file I
attach R I performed a diptest on the dataset statfaculty.
However I do not manage to do it with my dataset that consists of a single
column
You also did unaccounted for stepwise selection. Regarding the proportional
odds assumption, if you assessed it correctly, something that is not
operating proportionally would have to be associated with the outcome for at
least one cutoff of Y, so you could say that you are doing reverse screening
Hello all,
I really appreciate how helpful the people in this list are. Would it be too
much to ask to send a small script to have it peer-reviewed? to make sure I am
not making blatant mistakes? The script takes an experiment.dat as input and
generates system Throughput using ggplot2. It works
Hi,
When I plot text and use cex to change the text size, I notice that the cex
multiplier is not exact. It looks as if the real size of text can take only
certain discrete values. Is there a workaround to get text to follow the cex
value more closely, or at least to be able to figure out what the
Hello,
I have a list of dates in which I am going use for a time series analysis. I
want to break these dates up into 2 week intervals and count the number of
times a date appears in this interval.
For example from Nov. 19, 2000 to Dec 2 ,2000 with the data listed below I want
to return
Start_d
Thank you, Michael
This is what i need. It works fine
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__
R-help@r-pr
I had to set it up as a data frame and then it workd beautifully with the
reshape package.
DF<-data.frame(A,B,x)
library(reshape)
cast(DF, A ~ B, fun.aggregate=mean,
margins=c("grand_row", "grand_col"))
Cheers
Tyler
> Date: Sun, 23 Oct 2011 14
This could be done with aggregate but I am unfamiliar with it so I'll give what
I think you want from your message using the library 'reshape' that you'll have
to doneload. If you're problem is large the data.table library would be much
faster.
You haven't really said what you'd like to get
Thank you Dan and Ted for these helpful comments. I will implement this
simple force symmetry code you suggested and make sure I familiarize with
this floating-point calculation problem so I can recognize such issues in
the future.
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Thank you for the suggestions! I'm actually running simulations in R over two
separate networks (and thus will need to smile sweetly (and authentically, of
course!) twice). In one, I am ssh-ing from my Mac laptop into one Mac machine
with 32 cpu. In the other, I am ssh-ing into a cluster and am
On 23-Oct-11 19:03:05, Carl Witthoft wrote:
> You misused nls(). Observe:
>
> x<- 1:101
> y2 <- 5*exp(x/20) + runif(101)/100 # nls will NOT converge for perfect
> data.
>
> nls(y2 ~ exp(A*x), start=list(A=.1))
>
> Nonlinear regression model
>model: y2 ~ exp(A * x)
> data: parent.frame
On Sun, Oct 23, 2011 at 7:46 PM, Molly Davies wrote:
> Hello,
>
> I am running R remotely on my university's network from my laptop (Macbook
> Pro, running leopard, in case this is useful). I have a strict limit on how
> much disk space I can take up on my network account at school, which is
>
You misused nls(). Observe:
x<- 1:101
y2 <- 5*exp(x/20) + runif(101)/100 # nls will NOT converge for perfect data.
nls(y2 ~ exp(A*x), start=list(A=.1))
Nonlinear regression model
model: y2 ~ exp(A * x)
data: parent.frame()
A
0.06709
residual sum-of-squares: 136703
Number of ite
Hello,
I am running R remotely on my university's network from my laptop (Macbook Pro,
running leopard, in case this is useful). I have a strict limit on how much
disk space I can take up on my network account at school, which is insufficient
for the size of some of the objects I need to create
If you just want the same symbol at each point, you could use Weylandt's
approach, tho' personally I think it's tidier to create a new vector
x10 <- x[seq(1,length(x),by=10)] and plot that.
If you would like a different symbol at each point, then take a look at
?text.
From: R. Michael
Suppose I have data like this:
A <- sample(letters[1:3], 1000, replace=TRUE)
B <- sample(LETTERS[1:2], 1000, replace=TRUE)
x <- rnorm(1000)
I can get a table of means via
tapply(x, list(A, B), mean)
and I can add the marginal means to this using cbind/rbind:
main <- tapply(x, list(A,B), mean)
On Oct 23, 2011, at 1:29 PM, Giovanni Azua wrote:
Hello,
This is one problem at the time :)
I have a data frame df that looks like this:
> df <-read.table(textConnection(" time partitioning_mode workload
runtime
+ 1 1 shardingquery 607
+ 2 1 shardingq
Hello,
This is one problem at the time :)
I have a data frame df that looks like this:
time partitioning_mode workload runtime
1 1 shardingquery 607
2 1 shardingquery 85
3 1 shardingquery 52
4 1 shardingquery
Hi Dennis,
Thank you very nice :)
Best regards,
Giovanni
On Oct 23, 2011, at 6:55 PM, Dennis Murphy wrote:
> Hi:
>
> Here's one approach:
>
> # Function to process a list component into a data frame
> ff <- function(x) {
> data.frame(time = x[1], partitioning_mode = x[2], workload = x[3],
On 10/22/2011 08:03 AM, Omphalodes Verna wrote:
Thanks Martin.
Here is my ''updated'' code.
setClass("myClass", representation(ID.r = "numeric", ID.c = "character", DAT =
"matrix"))
to.myClass<- function(ID.r, ID.c, DAT) {
out<- new("myClass", ID.r = ID.r, ID.c = ID.c, DAT = DAT)
re
Hi Ravi,
Thank you for your reply and please excuse my late response.
Plugging w2 = k/w1 from (A) into (B) yields
(C) f(w1') = (w1-w1)^2 + (w2-k/w1)^2
The partial derivative wrt w1' is
(D) df(w1')/ dw1' = -2(w1-w1) + 2(w2-k/w1)*k/(w1')^2
in order for this to be a minimum the f.o.c. df(w
Henri Mone gmail.com> writes:
> I want to fit to my data an exponential function with following
> functional form:
> y=exp(a*x)
>
> I used the function "nls" but this gives me exponential fits with
> following functional form:
> y=l*exp(a*x)
>
> With "l" being an scaling factor. What do I need
Hi:
Here's one approach:
# Function to process a list component into a data frame
ff <- function(x) {
data.frame(time = x[1], partitioning_mode = x[2], workload = x[3],
runtime = as.numeric(x[4:length(x)]) )
}
# Apply it to each element of the list:
do.call(rbind, lapply(
Hi
I have tried to estimate race winning probabilities with mlogit in R. I have
different amount of contestors in the races and mlogit has a bug so that in
those situations the mlogit does not work. So i tried to add dummy contestors
to the race so that every race has an equal amount of conte
I guess I must be misunderstanding the point of checking the ordinality
assumptions prior to fitting a model. Are you saying that a response
variable that does not behave in an ordinal fashion can still be included in
the initial and final model?
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Dear R Users, Beginners and Experts,
I want to fit to my data an exponential function with following functional form:
y=exp(a*x)
I used the function "nls" but this gives me exponential fits with
following functional form:
y=l*exp(a*x)
With "l" being an scaling factor. What do I need to change in
>
> I think your question is answered by
>
> http://cran.r-project.org/web/packages/fOptions/ChangeLog
>
> 2010-04-23 chalabi
>
>* ChangeLog, DESCRIPTION: updated DESCR and ChangeLog
>* src/085A-LowDiscrepancy.f: fixed sobol RVS on 64 bit platform
>* ChangeLog, DESCRIPTION
R. Michael Weylandt gmail.com> writes:
> It seems like the relevant plot would depend on what you are trying to
> investigate, but usually a scatterplot would well work for bivariate
> data with no other assumptions needed. I usually find ecdf() plots
> rather hard to interpret without playing ar
That's not reasonable for 2 reasons. First, selecting variables based on
apparent assumption satisfaction is an unexplored technique. Second, you
failed to account for variable selection during resampling validation. You
will need to give the model all CANDIDATE variables and use the bw=TRUE
opt
Hello,
I used R a lot one year ago and now I am a bit rusty :)
I have my raw data which correspond to the list of runtimes per minute (minute
"1" "2" "3" in two database modes "sharding" and "query" and two workload types
"query" and "refresh") and as a list of char arrays that looks like this:
Perhaps to clarify Prof. Ripley's remarks below , the part that you missed
was "symmetric," which your matrix obviously is not.
-- Bert
2011/10/23 Prof Brian Ripley
> On Sun, 23 Oct 2011, Ron Michael wrote:
>
> I think I am missing something with the chol() function. Here is my
>> calculation:
Use regular expressions
?grepl
On Sunday, October 23, 2011, Philipp Fischer wrote:
> Hello,
> I am just starting with R and I am having a (most probably) stupid problem
by creating a new variable in a data.frame based on a part of another
character variable.
>
> I have a data frame like this one
Hello,
I am just starting with R and I am having a (most probably) stupid problem by
creating a new variable in a data.frame based on a part of another character
variable.
I have a data frame like this one:
A B C
AWI-test1 1 i
AWI-test5
Hello
I have a list of row names that needs to be deleted from a data frame. how
do i do that.
one of the columns in the data frame contains the row names as numbers I can
also select by this column(will it be easier?).
Thank you
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Dr. Harrell,
Thanks for your response. The predictor variables I initially included in
the model were based on the x mean plots and whether they exhibited
ordinality and whether they appeared to meet the CR assumptions. Only 7 of
16 potential variables fit that designation and those are the vari
Október 19-től 21-ig irodán kívül vagyok, és az emailjeimet nem érem el.
Sürgős esetben kérem forduljon Kárpáti Edithez (karpati.e...@gyemszi.hu).
Üdvözlettel,
Mihalicza Péter
I will be out of the office from 19 till 21 October with no access to my emails.
In urgent cases please contact Ms. Ed
I think that only works because the rows are ordered and have no
names: try something more like this:
df1 <- data.frame(1:20, 21:40, 31:50)
rownames(df1) <- sample(letters, 20)
toDrop <- sample(rownames(df1), 5)
df1[ !(rownames(df1) %in% toDrop), ]
or alternatively
toKeep <- sample(rownames(df
here is one way
df1 <- data.frame(c(1:20), c(21:40), c(31:50))
list1 <- c(3, 6, 20)
df2 <- df1[-list1,]
hanansela wrote:
>
> Hello
> I have a list of row names that needs to be deleted from a data frame. How
> do i do that?
> one of the columns in the data frame contains the row names as nu
It seems like the relevant plot would depend on what you are trying to
investigate, but usually a scatterplot would well work for bivariate
data with no other assumptions needed. I usually find ecdf() plots
rather hard to interpret without playing around with the data
elsewhere first and I'm not su
On Sun, Oct 23, 2011 at 8:10 AM, Giovanni Azua wrote:
> Hi Josh,
>
> Thank you for your feedback, after lot of trial and error the problem is
> finally solved.
>
> To solve this problem, I tried in this order:
>
> 1) uninstalling the two packages "Matrix" and "lme4" and reinstalling them.
> 2) un
Hi Josh,
Thank you for your feedback, after lot of trial and error the problem is
finally solved.
To solve this problem, I tried in this order:
1) uninstalling the two packages "Matrix" and "lme4" and reinstalling them.
2) uninstalling doBy and reinstalling it with and without 1)
3) upgrading t
Hi,
I have the following data about courses (504) in a university, two
attributes about the proportion of resources used (#resources_used /
#resources_available), namely the average and the standard deviation.
Thus I have:
[1] n=504 rows
[2] 1 id column and 2 attributes
Here's a sample of the data
On Sun, 23 Oct 2011, Ron Michael wrote:
I think I am missing something with the chol() function. Here is my calculation:
mat
[,1] [,2] [,3] [,4] [,5]
[1,] 1 3 0 0 0
[2,] 0 1 0 0 0
[3,] 0 0 1 0 0
[4,] 0 0 0 1 0
[5,] 0 0
That's fantastic, thank you very much, the qnorm option is interesting, I
will have to play around with it.
Many thanks again
Philip
-Original Message-
From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com]
Sent: Sunday, 23 October 2011 10:28 AM
To: Philip Robinson
Cc: r-help@r-pr
I think I am missing something with the chol() function. Here is my calculation:
> mat
[,1] [,2] [,3] [,4] [,5]
[1,] 1 3 0 0 0
[2,] 0 1 0 0 0
[3,] 0 0 1 0 0
[4,] 0 0 0 1 0
[5,] 0 0 0 0 1
> eigen(mat)
$values
[1] 1 1 1
Hi David,
Thanks for the reply,
f=function(x){quantile(x, c(0.25, 0.75),na.rm = TRUE) - matrix(IQR(x,na.rm =
TRUE) * c(1.5), nrow = 1) %*% c(-1, 1)}
Here parameter 1.5 is set for example in the above function as argument, it
can be even more may be 3.0 after analyzing actual data. Here expecta
Just to avoid possible confusion, let me correct a typo
(at step [2] in the example below). Apologies!
-FW: -
Date: Sun, 23 Oct 2011 08:43:27 +0100 (BST)
Sender: r-help-boun...@r-project.org
From: (Ted Harding)
To: r-help@r-project.org
Subject: Re: [R] symmetric matrix multiplication
On
On 23-Oct-11 07:00:07, Daniel Nordlund wrote:
>> -Original Message-
>> From: r-help-boun...@r-project.org
>> [mailto:r-help-boun...@r-project.org]
>> On Behalf Of statfan
>> Sent: Saturday, October 22, 2011 10:45 PM
>> To: r-help@r-project.org
>> Subject: [R] symmetric matrix multiplication
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of statfan
> Sent: Saturday, October 22, 2011 10:45 PM
> To: r-help@r-project.org
> Subject: [R] symmetric matrix multiplication
>
> I have a symmetric matrix B (17x17), and a (17x
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