Ado wrote:
>
>
> My data:
>
> XYSpecies Group
> 0 0A1
> .. A1
> .. A1
> I want to plot X vs Y for each species onto the
I guess "invert" does the trick.
For recording ... example ..
file <- grep("Repurchase Price",file, fixed = TRUE, invert = TRUE)
-Original Message-
From: Santosh Srinivas [mailto:santosh.srini...@gmail.com]
Sent: 14 October 2010 11:28
To: 'r-help'
Subject: Drop matching lines from readL
Dear R-group,
I have some noise in my text file (coding issues!) ... I imported a 200 MB
text file using readlines
Used grep to find the lines with the error?
What is the easiest way to drop those lines? I plan to write back the
"cleaned" data set to my base file.
Thanks.
_
Dear all,
The help page of stem() says that the width argument gives the desired
width of plot. However, I cannot come up with any idea what this width
stays for. This becomes especially difficult if you run these couple
of examples:
The decimal point is 3 digit(s) to the right of the |
0 | 0
R version 2.11.1 on WinXP
How do I get 3 scatterplots with marginal boxplots (from the car
package) onto a single plot?
I have a data frame called bank
> dim(bank)
[1] 46 5
head(bank)
x1x2 x3 x4 pop
1 -0.45 -0.41 1.09 0.45 0
2 -0.56 -0.31 1.51 0.16 0
3 0.06 0.02 1.01 0.40
Hi all,
My apologies if this is a very simple problem. I am very new to R,
having worked a lot previously with Excel. I recently completed a R
course with John Hoenig which introduced me to R.
My data:
XYSpecies Group
0 0A
On Oct 13, 2010, at 11:52 PM, Santosh Srinivas wrote:
I have a question about the output given below after running few
lines of
code. Surely a 101 query!
MF_Data <- read.csv("MF_Data_F.txt", header = F, sep="|")
temp <- head(MF_Data) #Get the sample Data
temp1 <- subset(temp, select= c(V1,V
I have a question about the output given below after running few lines of
code. Surely a 101 query!
MF_Data <- read.csv("MF_Data_F.txt", header = F, sep="|")
temp <- head(MF_Data) #Get the sample Data
temp1 <- subset(temp, select= c(V1,V4,V6)) #where V1, V4, V6 are the col
names .. to Get the rel
On Oct 13, 2010, at 5:20 PM, Julia Lira wrote:
Dear all,
I need to do a loop as following:
#Consider a matrix:
M <- matrix(1, nrow=10, ncol=20)
#Matrices to store the looping results
M1 <- matrix(0, nrow=10, ncol=400)
h <- c(1:20/1000)
#loop
# (I've seen more informative comments)
for (j i
Hi R-helpers , can you please give me more insights on the other data mining
and predictive techniques that the nnet package can be used for?
--
View this message in context:
http://r.789695.n4.nabble.com/nnet-help-tp2993609p2994756.html
Sent from the R help mailing list archive at Nabble.com.
Hi:
This isn't particularly elegant, but I think it works:
# The function to be applied:
f <- function(x, idx) {
n <- length(x)
if(idx[1] < idx[2]) {idx <- seq(idx[1], idx[2])
} else { idx <- c(seq(idx[1], n), seq(1, idx[2])) }
mean(x[idx])
}
# tests
> month.data = matrix(c(3,
Hi Juan,
Yes, you can use EMD to quantify the difference between any pair of
histograms regardless of their shape. The only constraint, at least
the way that I've done it previously, is to have compatible bins. The
original application of EMD was to compare images based on colour
histograms which
On Wed, 13 Oct 2010, David Winsemius wrote:
On Oct 13, 2010, at 4:50 PM, Antonio Paredes wrote:
Hello everyone,
I wanted to ask if there is an R-package to fit the following Poisson
regression model
log(\lambda_{ijk}) = \phi_{i} + \alpha_{j} + \beta_{k}
i=1,\cdots,N (subjects)
j=0,1 (two le
One possible way to treat parameters as "nuisance parameters" is to model them
as random. This gives allows them to have a reduced parametric load.
There are many packages with funcitons to fit glmms. One you may wish to look
at is lme4, which has the lmer fitting function
library(lme4)
fm
thanks Phil, I have your solution and another which I will attempt in the
next day or so and will post results to the list then.
cheers
andy
On Wed, Oct 13, 2010 at 10:30 AM, Phil Spector wrote:
> Andrew -
> I think
>
> answer = replicate(50,{fit1 <- rpart(CHAB~.,data=chabun, method="anova",
On Oct 13, 2010, at 4:50 PM, Antonio Paredes wrote:
Hello everyone,
I wanted to ask if there is an R-package to fit the following Poisson
regression model
log(\lambda_{ijk}) = \phi_{i} + \alpha_{j} + \beta_{k}
i=1,\cdots,N (subjects)
j=0,1 (two levels)
k=0,1 (two levels)
treating the \phi_{i
On Tue, Oct 12, 2010 at 2:39 AM, Chris Howden
wrote:
> I’m working with some very big datasets (each dataset has 11 million rows
> and 2 columns). My first step is to merge all my individual data sets
> together (I have about 20)
>
> I’m using the following command from sqldf
>
> dat
Just to clarify I meant opening it in a new window (and perhaps closing old
frame?).
Thanks again,
Sachin
Sachinthaka
Abeywardana/HO/Al
Hey all,
Suppose a=b^2 for starters. I want to be able to create a graph that
displays a initially and if i was to click on 'a' to show 'b' on the chart
itself. Does anyone know if this is possible in R?
Also as an extension (not necessary as yet) to output the above into a
'html' file.
Thanks,
Hi, is there a package for getting type II or type III tests on mixed models
(lme or lmer), just like what Anova() in car package does for aov, lm, etc.?
Thanks
John
[[alternative HTML version deleted]]
__
R-help@r-project.org mailin
Dear all,
I need to do a loop as following:
#Consider a matrix:
M <- matrix(1, nrow=10, ncol=20)
#Matrices to store the looping results
M1 <- matrix(0, nrow=10, ncol=400)
h <- c(1:20/1000)
#loop
for (j in h){
M1 <- M/(2*j)
}
But this means that the first 20 columns of matrix M
On Oct 13, 2010, at 4:54 PM, Alisa Wade wrote:
Thanks, David for the response.
Unfortunately, that only works for the case where I happen to only
want the last and first month "wrap".
But it doesn't work for any other case, e.g., say I want months
start.month = 4, end.month = 2.
Now
woul
On Oct 13, 2010, at 4:44 PM, Alison Callahan wrote:
Hello all,
I am trying to use cbind to add a named empty column to a Matrix:
outputmatrix <- cbind(outputmatrix,kog_id = seq(0,0,0))
Not sure off the top of my head what the right way might be, but this
should work:
names(outputmatrix)
Try this:
`colnames<-`(cbind(m, kog_id = 0), c(colnames(m), kog_id))
On Wed, Oct 13, 2010 at 5:44 PM, Alison Callahan
wrote:
> Hello all,
>
> I am trying to use cbind to add a named empty column to a Matrix:
>
> outputmatrix <- cbind(outputmatrix,kog_id = seq(0,0,0))
>
> The problem I have is t
Thanks, David for the response.
Unfortunately, that only works for the case where I happen to only want the
last and first month "wrap".
But it doesn't work for any other case, e.g., say I want months start.month
= 4, end.month = 2.
Now
start=4; end=2; year.avg = apply(month.data[, c(start,end)],
Hello everyone,
I wanted to ask if there is an R-package to fit the following Poisson
regression model
log(\lambda_{ijk}) = \phi_{i} + \alpha_{j} + \beta_{k}
i=1,\cdots,N (subjects)
j=0,1 (two levels)
k=0,1 (two levels)
treating the \phi_{i} as nuinsance parameters.
Thank you very much
--
-Ton
Hello all,
I am trying to use cbind to add a named empty column to a Matrix:
outputmatrix <- cbind(outputmatrix,kog_id = seq(0,0,0))
The problem I have is that kog_id is a variable that has a value e.g.
"KOG1234", but I when I try to use this to name the added column, it
is named literally "kog_
On Oct 13, 2010, at 4:17 PM, Erik Iverson wrote:
Hello,
There are probably many ways to do this, but I think
it's easier if you use a data.frame as your object.
The easy solution for the matrix you provide is escaping
me at the moment.
Perhaps using sampling to derive an index?
A[ tapply(1
I know *what* happened (Calc reformatted the data in ways I did not want or
expect). It is not end-of-line conventions; they reformatted the data leaving
the structure intact. As to why/how, that could depend on the sequence of
operations, so I thought to ask here to see if you had collective
On Oct 13, 2010, at 4:10 PM, Alisa Wade wrote:
Perhaps it is just that I don't even know the correct term to search
for,
but I can find nothing that explains how to wrap around from the end
to a
start of a row in a matrix.
For example, you have a matrix of 2 years of data, where rows are
On Wed, Oct 13, 2010 at 10:49 AM, wrote:
> Greetings
> I'm having difficulty witht the strptime function. I can't seem to
> figure a way to strip month (name) and year and create separate columns
> from a column with MM/DD/ formatted dates. Can anyone help?
Try month.day.year in chron:
Try this:
A <- data.frame(V1 = rnorm(100), V2 = runif(100), V3 = rep(c(1,2,3,4,5),20))
do.call(cbind, lapply(aggregate(. ~ V3, A, FUN = sample, size = 5), c))
On Wed, Oct 13, 2010 at 5:01 PM, Mauricio Romero <
mauricio.rom...@quantil.com.co> wrote:
> Hi,
>
>
>
> I want to select a subsample from
Erik- thank you very much. The rbinom worked.
Thanks !
GR
From: Erik Iverson-3 [via R]
Sent: Wed, October 13, 2010 2:24:18 PM
Subject: Re: Coin Toss Simulation
Shiv wrote:
> I am trying a simple toin coss simulation, of say 200 coin tosses. The idea
> From: dwinsem...@comcast.net
> To: bsch...@anest.ufl.edu
> Date: Wed, 13 Oct 2010 14:52:21 -0400
> CC: r-help@r-project.org
> Subject: Re: [R] [OT] (slightly) - OpenOffice Calc and text files
>
>
> On Oct 13, 2010, at 1:13 PM, Schwab,Wilhelm K wrote:
>
Hello,
There are probably many ways to do this, but I think
it's easier if you use a data.frame as your object.
The easy solution for the matrix you provide is escaping
me at the moment.
One solution, using the plyr package:
library(plyr)
A <- data.frame(a = rnorm(100),b = runif(100), c = rep
On Wed, Oct 13, 2010 at 2:16 PM, Bart Joosen wrote:
>
> Hi,
>
> this should be an easy one, but I can't figure it out.
> I have a vector of tests, with their units between brackets (if they have
> units).
> eg tests <- c("pH", "Assay (%)", "Impurity A(%)", "content (mg/ml)")
>
strapply in gsubfn
Perhaps it is just that I don't even know the correct term to search for,
but I can find nothing that explains how to wrap around from the end to a
start of a row in a matrix.
For example, you have a matrix of 2 years of data, where rows are years, and
columns are months.
month.data = matrix(c(3,4
On Oct 13, 2010, at 2:19 PM, Jon Zadra wrote:
Hi,
I've looked all over for a solution to this, but haven't had much
look in specifying what I want to do with appropriate search terms.
Thus I'm turning to R-help.
In the process of trying to write a simple function to rename
individual
Hi,
I want to select a subsample from my data, choosing one record from each
group. I know how to do this with a for.
For example: lets say I have the data:
A=cbind(rnorm(100),runif(100),(rep(c(1,2,3,4,5),20)))
Where the third column is the group variable. Then what I want is to select
5
On Wed, 13 Oct 2010, Schwab,Wilhelm K wrote:
It will get a good look, as will gnumeric - thanks to all!
emacs org-mode can convert your tab delimited file to a 'table' that you
can edit either using org-mode functions OR as plain text by switching to
fundamental mode.
In emacs speak, just
Julia -
Your subscript is out of range because in this loop:
for (t in aeven2){
Mhb0[,t] <- M[,(41+t)]-M[,(41-t)]
Mhb1[,t] <- M[,(42+t)]-M[,(42-t)]
}
t takes the values from 2 to 40 by 2, and you've
declared Mhb0 and Mhb1 as matrices with 20 columns.
So when t reach
R is primarily a function language not a macro language and what you are trying
to do matches more with macro programming.
The best approach is to think functionally and learn the more Rish ways of
doing things (have the function return the changed data and the caller does the
replacement, like
Dear Eriki and all
To run Quantile regression, it is necessary to install the following package in
R:
install.packages("quantreg")
Then, write:
library(quantreg)
And the software will run.
rm(list=ls()) #remove almost everything in the memory
set.seed(180185)
nsim <- 10
m
As an addendum to my question, I'm attempting to apply the solution to the
robust non-linear regression function nlrob from the robustbase package, and
it doesn't work in that situation. I'm getting
allRobustFit <- nlrob(Y ~ (upper)/(1+10^(X-LOGEC50[dset])), data=all
,start=list(upper=max(all$Y),L
Here's a shorter (but more cryptic) one:
> gsub("^([^\\(]+)(\\((.+)\\))?", "\\2", tests)
[1] """(%)" "(%)" "(mg/ml)"
> gsub("^([^\\(]+)(\\((.+)\\))?", "\\3", tests)
[1] "" "%" "%" "mg/ml"
-Matt
On Wed, 2010-10-13 at 14:34 -0400, Henrique Dallazuanna wrote:
> Try th
Note:
My original proposal, not quite right, can be made quite right via:
gsub(".*\\((.*)\\).*||[^()]+", "\\1",tests)
The "||" or clause at the end handles the case where there are no
parentheses in the string.
-- Bert
On Wed, Oct 13, 2010 at 11:16 AM, Bart Joosen wrote:
>
> Hi,
>
> this shou
Julia,
Can you provide a reproducible example? Your code calls the
'rq' function which is not found on my system.
Any paring down of the code to make it more readable would
help us help you better, too.
Julia Lira wrote:
Dear all,
I am trying to run a loop in my codes, but the software r
On Oct 13, 2010, at 1:13 PM, Schwab,Wilhelm K wrote:
Hello all,
I had a very strange looking problem that turned out to be due to
unexpected (by me at least) format changes to one of my data files.
We have a small lab study in which each run is represented by a row
in a tab-delimited fi
Hello,
Jon Zadra wrote:
Hi,
I've looked all over for a solution to this, but haven't had much look
in specifying what I want to do with appropriate search terms. Thus I'm
turning to R-help.
In the process of trying to write a simple function to rename individual
column names in a data fr
One way:
gsub(".*\\(([^()]*)\\).*", "\\1",tests)
Idea: Pick out the units designation between the "()" and replace the
whole expression with it. The "\\1" refers to the "[^()]*
parenthesized expression in the middle that picks out the units.
Cheers,
Bert
On Wed, Oct 13, 2010 at 11:16 AM, Bart
Bart,
I'm hardly one of the lists regex gurus: but this can
get you started...
tests <- c("pH", "Assay (%)", "Impurity A(%)", "content (mg/ml)")
x <- regexpr("\\((.*)\\)", tests)
substr(tests, x + 1, x + attr(x, "match.length") - 2)
Bart Joosen wrote:
Hi,
this should be an easy one, but I c
Dear all,
I have just sent an email with my problem, but I think no one can see the red
part, beacuse it is black. So, i am writing again the codes:
rm(list=ls()) #remove almost everything in the memory
set.seed(180185)
nsim <- 10
mresultx <- matrix(-99, nrow=1000, ncol=nsim)
mresultb <- ma
- which version of Mac OS X?
- Which version of R? (version, architecture)
- Officially provided R or compiled by you? Official R is compiled with
Apple gcc.
- if R was compiled with Apple compiler, who knows what can happen if you
link with code compiled with a non Apple gcc?
- if the code runs o
Dear all,
I have just sent an email with my problem, but I think no one can see the red
part, beacuse it is black. So, i am writing again the codes:
rm(list=ls()) #remove almost everything in the memory
set.seed(180185)
nsim <- 10
mresultx <- matrix(-99, nrow=1000, ncol=nsim)
mresultb <-
Try this:
replace(gsub(".*\\((.*)\\)$", "\\1", tests), !grepl("\\(.*\\)", tests), "")
On Wed, Oct 13, 2010 at 3:16 PM, Bart Joosen wrote:
>
> Hi,
>
> this should be an easy one, but I can't figure it out.
> I have a vector of tests, with their units between brackets (if they have
> units).
> e
It will get a good look, as will gnumeric - thanks to all!
Bill
From: Albyn Jones [jo...@reed.edu]
Sent: Wednesday, October 13, 2010 2:14 PM
To: Schwab,Wilhelm K
Subject: Re: [R] [OT] (slightly) - OpenOffice Calc and text files
emacs shows you exactly w
I don't know much about the iMac. R's .C() passes
R-language integers as pointers to 32-bit ints. If
your iMac is 64-bit and sizeof(long)==8 in your compiler
(pretty common for 64-bit compilers but not so for
Microsoft Windows compilers) then Long[0] will
use the next 64 bits to make an integral
On 13/10/2010 2:19 PM, Jon Zadra wrote:
Hi,
I've looked all over for a solution to this, but haven't had much look
in specifying what I want to do with appropriate search terms. Thus I'm
turning to R-help.
In the process of trying to write a simple function to rename individual
column name
Shiv wrote:
I am trying a simple toin coss simulation, of say 200 coin tosses. The idea
is to have a data frame like so:
Experiment#Number_Of_Heads
1 104
296
3101
So I do:
d <-data.frame(exp_num=c(1,2,3)); /* J
Hi,
I've looked all over for a solution to this, but haven't had much look
in specifying what I want to do with appropriate search terms. Thus I'm
turning to R-help.
In the process of trying to write a simple function to rename individual
column names in a data frame, I ran into the follow
try this:
n <- 3
data.frame(Exp = seq_len(n),
Heads = rbinom(n, 200, 0.5))
I hope it helps.
Best,
Dimitris
On 10/13/2010 7:28 PM, Shiv wrote:
I am trying a simple toin coss simulation, of say 200 coin tosses. The idea
is to have a data frame like so:
Experiment#Number_Of_Heads
Hi,
this should be an easy one, but I can't figure it out.
I have a vector of tests, with their units between brackets (if they have
units).
eg tests <- c("pH", "Assay (%)", "Impurity A(%)", "content (mg/ml)")
Now I would like to hava a function where I use a test as input, and which
returns the
Dear all,
I am trying to run a loop in my codes, but the software returns an error:
"subscript out of bounds"
I dont understand exactly why this is happenning. My codes are the following:
rm(list=ls()) #remove almost everything in the memory
set.seed(180185)
nsim <- 10
mresultx <- ma
Thank you very much for your kind reply.
I used gdb, and it returns a reason "KERN_INVALID_ADDRESS" on a very simple
operation (obtaining the time step for a numerical integration). Please see
bellow.
But, um, I solved it by changing the function's arguments in the C code from
"unsigned long int"
I am trying a simple toin coss simulation, of say 200 coin tosses. The idea
is to have a data frame like so:
Experiment#Number_Of_Heads
1 104
296
3101
So I do:
d <-data.frame(exp_num=c(1,2,3)); /* Just 3 experimen
Hi Rajarshi,
Here is a post I found from Pipeline pilot community help pages:
https://community.accelrys.com/message/3466
Eric
-Original Message-
From: Rajarshi Guha [mailto:rajarshi.g...@gmail.com]
Sent: Wednesday, October 13, 2010 7:52 AM
To: Eric Hu
Cc: r-help@r-project.org
Subject
Peter,
vi is *really* primitive =:0 R is a little late because I tend to do shape
changes prior to invoking R. However, I could load tweak and re-save and then
bring R back into it later. I never would have thought of it. Thanks!
Bill
From: Peter
Albyn,
I'll look into it. In fact, I have a small book on it that I bought in my very
early days of using Linux. I quickly found TeX Maker (for the obvious),
Code::Blocks for C/C++ and I would not have started the move without a working
Smalltalk (http://pharo-project.org/home).
For editing
On Wed, Oct 13, 2010 at 10:13 AM, Schwab,Wilhelm K
wrote:
> Hello all,
>
> I had a very strange looking problem that turned out to be due to unexpected
> (by me at least) format changes to one of my data files. We have a small lab
> study in which each run is represented by a row in a tab-delim
On Oct 13, 2010, at 12:13 PM, Schwab,Wilhelm K wrote:
> Hello all,
>
> I had a very strange looking problem that turned out to be due to unexpected
> (by me at least) format changes to one of my data files. We have a small lab
> study in which each run is represented by a row in a tab-delimite
On Wed, Oct 13, 2010 at 10:13 AM, Schwab,Wilhelm K
wrote:
> Hello all,
>
> I had a very strange looking problem that turned out to be due to unexpected
> (by me at least) format changes to one of my data files. We have a small lab
> study in which each run is represented by a row in a tab-delim
How about emacs?
albyn
On Wed, Oct 13, 2010 at 01:13:03PM -0400, Schwab,Wilhelm K wrote:
> Hello all,
> <.>
> Have any of you found a nice (or at least predictable) way to use OO Calc to
> edit files like this? If it insists on thinking for me, I wish it would
> think in 24 hour time and 4
Try:
ci.auc(all$D,all$pre,m="b")
2010/10/13 zhu yao :
> Dear useRs:
>
> I use pROC package to compute the bootstrap C.I. of AUC.
>
> The command was as follows:
>
> roc1<-roc(all$D,all$pre,ci=TRUE,boot.n=200)
--
Miłego dnia
__
R-help@r-project.org ma
Try this:
library(RDCOMClient)
xl <- COMCreate("Excel.Application")
wk <- xl$Workbooks()$Open("Book1.xlsx")
do.call(cbind, wk$Sheets(1)$Range("B50:C55")$Value())
On Tue, Oct 12, 2010 at 9:17 PM, Jeevan Duggempudi wrote:
> Hello all,
>
> I have a business user who generates monthly reports in M
Hello all,
I had a very strange looking problem that turned out to be due to unexpected
(by me at least) format changes to one of my data files. We have a small lab
study in which each run is represented by a row in a tab-delimited file; each
row identifies a repetition of the experiment and a
> which(list=="C")
[1] 3
See ?which
On 10/13/2010 11:56 AM, lord12 wrote:
If I have a character array:
list = c("A", "B", "C")
how do I access the third element without doing list[3]. Can't I find the
index of "C" using a particular function?
--
Christian Raschke
Department of Econom
If I have a character array:
list = c("A", "B", "C")
how do I access the third element without doing list[3]. Can't I find the
index of "C" using a particular function?
--
View this message in context:
http://r.789695.n4.nabble.com/Extracting-index-in-character-array-tp2994029p2994029.html
Se
This often happens when your C code uses memory that
it did not allocate, particularly when it reads or
writes just a little beyond the end of a memory block.
On some platforms or if you are lucky there is unused
memory between blocks of allocated memory and you don't
see a problem. Other machines
Is this the sort of thing you are looking for?
> f <- function(testName, dataset, ...) {
+ testFunc <- match.fun(paste(testName, "_test", sep=""))
+ testFunc(dataset, ...)
+ }
> m_test <- function(x) "the m test"
> p_test <- function(x) "the p test"
> f("m", 17)
[1] "the m
>From what I read, you want something like this:
myfunction<-function(dataset,arg1,arg2)
{
func = match.fun(arg2)
argument = dataset[, match(paste(arg1,"_test", sep=""), names(dataset))]
result=func(argument)
return(result)
}
On Wed, Oct 13, 2010 at 9:28 AM, Manta wrote:
>
> Thaks for your q
Got it , thanks!
From: David Winsemius [via R]
[mailto:ml-node+2992845-2070125287-199...@n4.nabble.com]
Sent: Tuesday, October 12, 2010 3:28 PM
To: Steve Swope
Subject: Re: Plotting Y axis labels within a loop
On Oct 12, 2010, at 5:54 PM, Steve Swope wrote:
>
> When I plot y axis lab
Thaks for your quick reply Bert, although I doubt it works.
The reason is that the names of the objects of the dataset, all end with the
sufix '_test' and therefore I need to attach/paste/glue this suffix to the
'arg2' of the function. Any other idea?
--
View this message in context:
http://r.7
If I understand you correctly, just arg2(arg1) will work fine...One of
the nice features of R.
Example:
f <- function(dat,fun=mean,...)
{
fun(dat, ...) ## ... allows extra arguments to fun
}
f( rnorm(10), trim=.05) ## trimmed mean
x <- 1:10
f(x, fun = max,na.rm=TRUE)
Cheers,
Bert
On Wed,
On Oct 13, 2010, at 11:44 AM, Jeremy Olson wrote:
Dear All,
I have 4 or 5 contour plots that I need to overlay. Currently they
are maps showing hot and cold areas for specific elements.
Providing paste-able examples is the standard way to present such
problems.
I would like to combine
Also look at textplot in the gplots package and addtable2plot in the plotrix
package.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
>
Dear R community,
I am struggling a bit with a probably fairly simple task. I need to use some
already existing functions as argument for a new function that I am going to
create. 'dataset' is an argument, and it comprises objects named
'mean_test', 'sd_test', 'kurt_test' and so on. 'arg1' tells
I don't know of any tools that would do this generally (RODBC could do it if
there were specific column names in the 1st row and a given column with
information to identify the rows, but this seems unlikely from your
description). A couple of possibilities:
In Excel you can highlight the colum
Dear All,
I have 4 or 5 contour plots that I need to overlay. Currently they are maps
showing hot and cold areas for specific elements.
I would like to combine these plots into one map, where each color will
correspond to a different element and you would be able to see the areas that
each elem
> newmat <- oldmat[ c(rep(FALSE,19),TRUE), ]
Or
> newmat <- oldmat[ seq(20, nrow(oldmat), 20), ]
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-b
hello list,
i'd very much appreciate help with setting up the
contrast for a 2-factorial crossed design.
here is a toy example:
library(multcomp)
dat<-data.frame(fac1=gl(4,8,labels=LETTERS[1:4]),
fac2=rep(c("I","II"),16),y=rnorm(32,1,1))
mod<-lm(y~fac1*fac2,data=dat)
## the
On Oct 13, 2010, at 10:05 AM, Christophe Bouffioux wrote:
Dear R-community,
Using bwplot, how can I put the whiskers at percentile 5 and
percentile 95,
in place of the default position coef=1.5??
Using panel=panel.bwstrip, whiskerpos=0.05, from the package
agsemisc gives
satisfaction, bu
Thanks Giovanni and Achim!
I will try out some of the things you suggested. Let's see how far I get :-)
__
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-
On Tue, Oct 12, 2010 at 8:54 PM, Eric Hu wrote:
> Hi,
>
> I am trying to see if I can use R to perform more rigorous regression
> analysis. I wonder if the fingerprint package is able to handle pipeline
> pilot fingerprints (ECFC6 etc) now.
Currently no - does Pipeline Pilot out put their ECFP's
Greetings
I'm having difficulty witht the strptime function. I can't seem to
figure a way to strip month (name) and year and create separate columns
from a column with MM/DD/ formatted dates. Can anyone help?
Cheers
Kurt
***
Kur
Please take a look at the documentation for lmList regarding
the na.action= argument. It should be a *function*, not TRUE
or FALSE. For example, try
lmList(score ~ childid | spring, data=a, na.action=na.omit)
- Phil Spector
Dear All,
I was trying to use the lmList function to get the lmList graphic but have a
problem creating the graphic. My data has missing values and an error
occurred
as stated below.
###
library(nlme)
> a
schoolid spring score childi
Dear R-community,
Using bwplot, how can I put the whiskers at percentile 5 and percentile 95,
in place of the default position coef=1.5??
Using panel=panel.bwstrip, whiskerpos=0.05, from the package agsemisc gives
satisfaction, but changes the appearance of my boxplot and works with an old
versio
Hi:
On Wed, Oct 13, 2010 at 5:31 AM, dpender wrote:
>
> Dennis,
>
> Thanks for that. The problem now is that I am trying to use it in a for
> loop. Based on the example before, 2 entries are required after H[3] as
> specified by O. The problem is that when inserting values the length of
> the
Hello.
In principle Achim is right, by default vcovHC.plm does things the "Arellano"
way, clustering by group and therefore giving SEs which are robust to general
heteroskedasticity and serial correlation. The problem with your data, though,
is that this estimator is N-consistent, so it is inapp
Hello,
I am aware this may be an obscure problem difficult to advice about, but
just in case... I am calling a C function from R on an iMac (almost shining:
bought by my institution this year) and gives a "memory not mapped" error.
Nevertheless, exactly the same code runs without problem in a pow
1 - 100 of 150 matches
Mail list logo
