On 10/12/2011 03:15 PM, JeffND wrote:
Hi to all,
Suppose we have a group of points on the plane,
at each point, we want to draw a pie chart centered at that point.
...
Hi Jeff,
I think what you are looking for is the floating.pie function (plotrix).
Jim
__
On Oct 12, 2011, at 1:20 AM, Erin Hodgess wrote:
Dear R People:
I have the following set of data
Block[1:5]
[1] "5600-5699" "6100-6199" "9700-9799" "9400-9499" "8300-8399"
and I want to split at the -
strsplit(Block[1:5],"-")
[[1]]
[1] "5600" "5699"
[[2]]
[1] "6100" "6199"
[[3]]
[1] "9
You should read the posting guide again (at all?). Among other
things, it suggests you should show what you have tried doing that did
not work (worst case it shows us you tried to figure it out on your
own before asking other people to freely give their time to solve your
problems, at best case, w
Hi Erin,
this is one way:
Block <- c("5600-5699","6100-6199","9700-9799","9400-9499","8300-8399")
splBlock <- strsplit(Block,"-")
sapply(splBlock, "[", 1)
greetings,
Remko
--
View this message in context:
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Sent from th
Hi to all,
Suppose we have a group of points on the plane,
at each point, we want to draw a pie chart centered at that point.
I have found that pie.labels() function might be useful. This function
allows us
to achieve my goal, but before using the function,
we have to use the function plot(), say
Hi,
We have created a cox regression model (coxph) and exported from R as PMML
using the Rattle package. The resulting PMML is a standard Regression model
in the format y=b1*x1+b2*x2+…….bn*xn:
risk= (0.027968680616809*dropped_Calls)+(0.00777220409115466*helpdesk_Calls)
The PMML is included at t
I am using the normtemp data set in UsingR package. I want to store all
normtemp[,1] from 1 to 65 as x and normtemp[,1] from 66to 130 as y. How do I
write code to do that from here:
> data(normtemp)
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sapply(strsplit(Block[1:5],"-"), function (x) {x[1]})
comes to mind...
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#.
unlist(strsplit(Block[1:5], "-.+$"))
if you are going to want the other pieces later, the most efficient
way depends on the assumptions you can make about your data. If there
are always two elements from the split:
matrix(unlist(strsplit(Block[1:5], "-")), ncol = 2, byrow = TRUE)
## or
do.call("
Hi all,
This is probably a very simple question but I cannot figure out how to do
it. I run the fourthcorner method with my data and would like to adjust the
p values for multiple comparisons using Holm correction. When I run the
fourthcorner I obtain the results in yellow. What do I need to do to
On Tue, 11 Oct 2011, Akram Khaleghei Ghosheh balagh wrote:
Hello ;
I am doing a regression of count data (number of award and there are some
covariates)
I have estiamted the parameters of negative binomial distribuion (lambda is
a function of covaraites, GLM model) by glm.nb function and traini
Dear R People:
I have the following set of data
> Block[1:5]
[1] "5600-5699" "6100-6199" "9700-9799" "9400-9499" "8300-8399"
and I want to split at the -
> strsplit(Block[1:5],"-")
[[1]]
[1] "5600" "5699"
[[2]]
[1] "6100" "6199"
[[3]]
[1] "9700" "9799"
[[4]]
[1] "9400" "9499"
[[5]]
[1] "8300
On Wed, Oct 12, 2011 at 3:29 AM, Yves S. Garret
wrote:
> Hi all,
>
> I recently started learning about Forex and found this O'Reilly book in
> Barnes & Nobles about R. I bought it out of pure curiosity. I like what I
> see. However, I have a question. Has anyone tried to bring these two idea
On 12/10/11 09:11, flokke wrote:
Dear all,
Sorry to bother you with such a stupid question, but I just cannot find the
solution to my problem.
I'd like to use matrix multiplication for meanA and factorial 3.
I use the command meanA%*%factorial 3.
But everything I get is: Error in factorial3 %*%
> m <- matrix(rpois(16, 10), ncol = 4)
> m
[,1] [,2] [,3] [,4]
[1,]9 128 11
[2,] 127 118
[3,] 12788
[4,] 11 1148
> apply(m, 2, sort)
[,1] [,2] [,3] [,4]
[1,]9748
[2,] 11788
[3,] 12 1188
[4,] 12
Hi all,
I have matrix, A. I want to sort the elements in each column in a ascending
order and output the index number their position. For example
A =
[,1] [,2] [,3] [,4]
[1,] 135.87060 97.55820 121.07722 171.625495
[2,] 70.19811 36.64529 50.13501 6.319125
[3,] 26.8
Hi all,
I recently started learning about Forex and found this O'Reilly book in
Barnes & Nobles about R. I bought it out of pure curiosity. I like what I
see. However, I have a question. Has anyone tried to bring these two ideas
together in a financial and trading sense? Are there any libr
Thanks for the help, Carlos and Jean.
What y'all sent worked beautifully. Ultimately I chose to use the rect()
function. Below is my functional code (with a legend thrown in for good
measure):
plot(x = c(1,numYears), y = c(200,300), main = c("Successional Trajectories
for ",startState," Over Time
I have a large amount of NAs.
Does anyone know the which command that will not throw out any rows that
include NAs? I have to use gam/glm for now, but I know of other packages
(RandomForests) that will use the mean of the column.
Anything like this for glms?
Right now this code
glm2 <- glm(PA~
Dear all,
Sorry to bother you with such a stupid question, but I just cannot find the
solution to my problem.
I'd like to use matrix multiplication for meanA and factorial 3.
I use the command meanA%*%factorial 3.
But everything I get is: Error in factorial3 %*% A : non-conformable
arguments
I
I used the 'llsImpute' method from 'pcaMethods' packages in R but couldn't be
satisfied with the accuracy I got. Do anyone know how to use the original
software (like in 'http://www.cc.gatech.edu/~hpark/softwareMVE.html') in
MATLAB?
More specifically, I have a data in .csv format with lots of
Hello ;
I am doing a regression of count data (number of award and there are some
covariates)
I have estiamted the parameters of negative binomial distribuion (lambda is
a function of covaraites, GLM model) by glm.nb function and training
dataset.
Now I want to predict the number of award (for ex
An organization has asked me to comment on the validity of their
recent all-employee survey. Survey responses, by geographic region,
compared
with the total number of employees in each region, were as follows:
ByRegion
All.Employees Survey.Respondents
Region_1735
On Oct 11, 2011, at 4:22 PM, Bert Gunter wrote:
Replicate the row indices?
x[rep(seq_len(nrow(x)), k), ]
Or the column indices.
x[ , rep(seq_len(length(x)), k) ]
--
David,
-- Bert
On Tue, Oct 11, 2011 at 12:55 PM, Martin Batholdy
wrote:
Hi,
is there a way to replicate a data.frame
The source does not document intent. I, too disagree with Dr Ripley on this
point. The library mechanism insists that some attempt at documentation be
included with the source, for good reason.
I would rather the documentation assert intent to support negative rounding
values and later add the
Colleagues,
I am fitting an Emax model using nls. The code is:
START <- list(EMAX=INITEMAX, EFFECT=INITEFFECT,
C50=INITC50)
CONTROL <- list(maxiter=1000, warnOnly=T)
#FORMULA<- as.formula(YVAR ~ EMAX - EFFECT * XVAR^GAMMA
/ (XVAR
which.max is even faster:
dims <- c(1000,1000)
tt <- array(rnorm(prod(dims)),dims)
# which
system.time(
replicate(100, which(tt==max(tt), arr.ind=TRUE))
)
# which.max (& arrayInd)
system.time(
replicate(100, arrayInd(which.max(tt), dims))
)
Best,
Denes
> But it's simpler and probably faster to
On 11-10-11 7:14 PM, Rolf Turner wrote:
On 11/10/11 08:17, Michael Friendly wrote:
On 10/9/2011 6:18 AM, Prof Brian Ripley wrote:
Sometimes it is better not to document things than try to give precise
details which may get changed *and* there will be useRs who misread (and
maybe even file bug
On 11/10/11 08:17, Michael Friendly wrote:
On 10/9/2011 6:18 AM, Prof Brian Ripley wrote:
Sometimes it is better not to document things than try to give precise
details which may get changed *and* there will be useRs who misread (and
maybe even file bug reports on their misreadings). The source
On 12/10/11 08:31, Timothy Bates wrote:
To do matrix multiplication: m x n, the Rows and columns of m must be equal to
the columns and rows of n, respectively.
No. The number of columns of m must equal the number of rows of n,
that's all. The number of *rows* of m and the number of *columns
But it's simpler and probably faster to use R's built-in capabilities.
?which ## note the arr.ind argument!)
As an example:
test <- matrix(rnorm(24), nr = 4)
which(test==max(test), arr.ind=TRUE)
row col
[1,] 2 6
So this gives the row and column indices of the max, from which row and
col
Hi,
With this code you can find row and col names for the largest value applied
to your example:
r.m.tmp<-apply(dat,1,max)
r.max<-names(r.m.tmp)[r.m.tmp==max(r.m.tmp)]
c.m.tmp<-apply(dat,2,max)
c.max<-names(c.m.tmp)[c.m.tmp==max(c.m.tmp)]
It's inmediate how to get the same for the smallest and
In the absence of a reproducible example, a general question induces a
general response. I'd suggest creating a small data frame that
contains the x and y coordinates, a third variable consisting of
expressions representing each fitted model and an indicator of the
group to which the expression is
On Oct 11, 2011, at 3:20 PM, Rob James wrote:
The integration of plot methods for various outputs from rms
packages is a great appreciated aspect of the rms package.
I particularly like to use:
plot(summary(model))
for my own purposes, but... for publication/presentation I need to
modify
check the version of libcurl you have installed. If you have an older
version some of the
options may not be present.
On Sun, Oct 9, 2011 at 10:39 AM, Steven Oliver wrote:
> Hey Guys,
>
> I just started fooling around with the twitteR package in order to get a
> record of all tweets from a s
> Hi,
>
> One way to do that is this (avoiding the use of a for loop):
>
>
> l.txt<- "id category attribute1 attribute2 attribute3 attribute4
> 661 SCHS 43.2 0 56.5 1
> 12202 SCHS 161.7 5.7 155 16
> 1182 SCHS 21.4 0 29 0
> 1356 SSS 8.8182 0.1818 10.6667 0.6667
> 1864 SCHS 443.7273 9.9091 537 4
Replicate the row indices?
x[rep(seq_len(nrow(x)), k), ]
-- Bert
On Tue, Oct 11, 2011 at 12:55 PM, Martin Batholdy
wrote:
> Hi,
>
>
> is there a way to replicate a data.frame like you can replicate the entries
> of a vector (with the repeat-function)?
>
> I want to do this:
>
> x <- data.frame(
Replace "stop()" with "break" to see if that does what you want. (you may also
want to include "cat()" or "warn()" to indicate the early stopping.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> F
Hi,
One way to do that is this (avoiding the use of a for loop):
l.txt<- "id category attribute1 attribute2 attribute3 attribute4
661 SCHS 43.2 0 56.5 1
12202 SCHS 161.7 5.7 155 16
1182 SCHS 21.4 0 29 0
1356 SSS 8.8182 0.1818 10.6667 0.6667
1864 SCHS 443.7273 9.9091 537 46
12360 SOA 6.6364 0 1
Your question as answered by Timothy in your previous thread
http://r.789695.n4.nabble.com/Re-Creating-the-mean-using-algebra-matrix-td3895689.html
flokke wrote:
>
> Dear all,
> Sorry to bother you with such a stupid question, but I just cannot find
> the solution to my problem.
>
> I'd lik
It may be best to either write to the package maintainer (me, as you did) or
post to the group but not both.
Frank
Stan Maydan-2 wrote:
>
> Hello,
>
> does anyone have an example on how to use restricted cubic
> splines function rcs within survfit.cph, if cph (Cox Proportional Hazard
> Regressi
Hi,
is there a way to replicate a data.frame like you can replicate the entries of
a vector (with the repeat-function)?
I want to do this:
x <- data.frame(x, x)
(where x is a data.frame).
but n times.
And it should be as cpu / memory efficient as possible, since n is pretty big
in my cas
Hello,
I'm looking to get the values, row names and column names of the largest and
smallest values in a matrix.
Example (except is does not include the names):
> x <- swiss$Education[1:25]
> dat = matrix(x,5,5)
> colnames(dat) = c('a','b','c','d','c')
> rownames(dat) = c('z','y','x','w','v')
>
Katie:
I would say that this is not an R question, so I would suggest that either
a) You ask it on a statistics help website like stats.stackexchange.com or
b) You consult with someone locally who knows about nonlinear regression
(possibly a statistician, but not necessarily so).
-- Bert
On T
The integration of plot methods for various outputs from rms packages is
a great appreciated aspect of the rms package.
I particularly like to use:
plot(summary(model))
for my own purposes, but... for publication/presentation I need to
modify details like variable names, or the number of sign
To do matrix multiplication: m x n, the Rows and columns of m must be equal to
the columns and rows of n, respectively.
Sent from my iPhone
On 11 Oct 2011, at 06:45 PM, flokke wrote:
> Dear all,
> I wanted to create the mean using a algebra matrix.
> so I tried this one:
>
>> meanAnimal
On Oct 11, 2011, at 1:45 PM, flokke wrote:
Dear all,
I wanted to create the mean using a algebra matrix.
so I tried this one:
meanAnimals <- new3%*%factorial
(Calculates the matrix multiplication of the new3 * factorial).
But I get the following error message:
Error in new3 %*% factorial
On Wed, Oct 12, 2011 at 4:56 AM, ivo welch wrote:
> thanks, josh. in my posting example, I did not need anything except
> coefficients. (when this is the case, I usually do not even use
> lm.fit, but I eliminate all missing obs first and then use solve
> crossprod(y,cbind(1,x)) crossprod(cbind(1
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Dimitris Rizopoulos
> Sent: Tuesday, October 11, 2011 10:43 AM
> To: Doran, Harold
> Cc: r-help@r-project.org
> Subject: Re: [R] stop()
>
> You could use return(), e.g.,
>
> myF
Dear all,
I wanted to create the mean using a algebra matrix.
so I tried this one:
> meanAnimals <- new3%*%factorial
(Calculates the matrix multiplication of the new3 * factorial).
But I get the following error message:
Error in new3 %*% factorial : non-conformable arguments
These are my m
I am a novice in R but using R 2.13.1 in Windows I wish to be able to
calculate the percentage variation in a
zero-inflated negative binomial regression model that is explained by the
two predictors in my model. My response variable was no. of dung-piles per
km and the predictor of excess zero
I am trying to fit a nonlinear regression to infiltration data in order to
determine saturated hydraulic conductivity and matric pressure. The
original equation can be found in Bagarello et al. 2004 SSSAJ (green-ampt
equation for falling head including gravity). I am also VERY new to R and
to non
Carlos Ortega wrote on 10/11/2011 11:30:46 AM:
>
> Hi,
>
> Yes, one way to do that is by using function polygon().
>
> Regards,
> Carlos Ortega
> www.qualityexcellence.es
>
> 2011/10/11 Gabriel Yospin
>
> > Hi R-Help -
> >
> > If I make a plot:
> >
> > numYears = 500
> > plot(x = c(1,numYears
I do not know if stripping down functions is generally recommended,
but it is not too difficult to do if you know that you can make
assumptions. Here is an example (I also found a fast way to convert
the data table to a matrix, again if some assumptions can be made).
Using the stripped down functi
Sarah_R_edu wrote on 10/11/2011 04:57:08 AM:
>
> Hi every one
>
> i have problem in R program to import the data from excel ,
>
> I have done the following:
>
> 1. install.packages("xlsReadWrite")
>
> 2. library(xlsReadWrite)
>
> 3. z<-
read.xls("ReadXls",LTS,colNames=FALSE,sheet,type,f
Thanks, Dimitris. Very helpful on something I *should* know by now.
> -Original Message-
> From: Dimitris Rizopoulos [mailto:d.rizopou...@erasmusmc.nl]
> Sent: Tuesday, October 11, 2011 1:43 PM
> To: Doran, Harold
> Cc: r-help@r-project.org
> Subject: Re: [R] stop()
>
> You could use ret
You could use return(), e.g.,
myFun <- function (x, max.iter = 5) {
for (i in 1:10) {
result <- x + i
iter <- i
if (iter == max.iter) {
return(result)
}
}
result
}
myFun(10, max.iter = 4)
I hope it helps.
Best,
Dimitris
On 10/11/2011 7
Suppose I have a function, such as the toy example below:
myFun <- function(x, max.iter = 5) {
for(i in 1:10){
result <- x + i
iter <- i
if(iter == max.iter) stop('Max reached')
}
result
}
I c
Ok, I see my mistake, just did as you suggest and works. Thanks for the
answer people
Best,
Sergio Rné
El 11 de octubre de 2011 19:03, Sergio René Araujo Enciso <
araujo.enc...@gmail.com> escribió:
> Dear all:
>
> I am having some problems to use the function "sink()". Basically I am
> doing a
so… cleared out, and now it’s working: Must have been an obscure workspace
conflict. Thanks for quick helpful replies
a <- 1:4
assign("a[1]", 2)
> a[1] == 2
[1] FALSE
> get("a[1]") == 2
[1] TRUE
On 11 Oct 2011, at 5:45 PM, Duncan Murdoch wrote:
> On 11/10/2011 12:31 PM, Timothy Bates wrote:
>>
On 11/10/2011 1:03 PM, Sergio René Araujo Enciso wrote:
Dear all:
I am having some problems to use the function "sink()". Basically I am doing
a loop over two files which contain unit-root variables. Then on a loop, I
extract every i element of both files to create an object called z. If z
meets
On Oct 11, 2011, at 1:03 PM, Sergio René Araujo Enciso wrote:
Dear all:
I am having some problems to use the function "sink()". Basically I
am doing
a loop over two files which contain unit-root variables. Then on a
loop, I
extract every i element of both files to create an object called z
Untested, does adding a print() around summary() get it done?
Michael
On Oct 11, 2011, at 1:03 PM, Sergio René Araujo Enciso
wrote:
> Dear all:
>
> I am having some problems to use the function "sink()". Basically I am doing
> a loop over two files which contain unit-root variables. Then on a
Hi,
Inside a loop, you must explicitly wrap your summary() command and anything
else from which you expect output in a print() command.
Sarah
2011/10/11 Sergio René Araujo Enciso :
> Dear all:
>
> I am having some problems to use the function "sink()". Basically I am doing
> a loop over two file
Hi,
Yes, one way to do that is by using function polygon().
Regards,
Carlos Ortega
www.qualityexcellence.es
2011/10/11 Gabriel Yospin
> Hi R-Help -
>
> If I make a plot:
>
> numYears = 500
> plot(x = c(1,numYears), y = c(200,300), xlab = "Time", ylab = "Vegetation
> Class", xlim = c(100,600),
Dear R Fundation,
I am a post-doc researcher at the University of Pisa, Italy.
I apologize for my english and I have to tell you in advance that I
am a very beginner with R.
I used R for fitting dose-response curves (drc package) and for an
ordinary ANOVA (one, two or three factors), including
Hello,
does anyone have an example on how to use restricted cubic
splines function rcs within survfit.cph, if cph (Cox Proportional Hazard
Regression) was done with restricted cubic
splines (which I made to work)?
Thank you.
>
[[alternative HTML version deleted]]
___
Dear all:
I am having some problems to use the function "sink()". Basically I am doing
a loop over two files which contain unit-root variables. Then on a loop, I
extract every i element of both files to create an object called z. If z
meets some requirements, then I perform a unit root test (ADF t
On Oct 11, 2011, at 11:18 AM, Samir Benzerfa wrote:
Hi everyone,
I've got two data sets as below. My question now is: how can I use
Dataset2
as a filter for Dataset1? My goal is just to keep the rows of
Dataset1 where
the first column (Date) matches the Dates in Dataset2.
Perhaps:
m
On Oct 11, 2011, at 6:02 AM, luke1022 wrote:
The following code will get me a curve plot:
cutoff <- seq(1,7,0.25)
Sensitivity <- 1 - pnorm(cutoff, 5, 0.8)
Specificity <- pnorm(cutoff, 3, 1.2)
plot(1-Specificity,Sensitivity,main = "ROC curve",type = "o")
How do I get a gradient of a particular
On 11/10/2011 12:31 PM, Timothy Bates wrote:
In the help for get(), the following example is given:
a<- 1:4
assign("a[1]", 2)
a[1] == 2 #FALSE
get("a[1]") == 2 #TRUE
However, executing that last line for me gives
Error in get("a[1]") : object 'a[1]' not found
What did the second l
Hi,
I want to do a visualisation of a matrix plot made up of several plots of
correlation matrices (using corrplot()). My data is in csv format. Here's an
example:
id,category,attribute1,attribute2,attribute3,attribute4
661,SCHS,43.2,0,56.5,1
12202,SCHS,161.7,5.7,155,16
1182,SCHS,21.4,0,29,0
1356
Hi,
On Tue, Oct 11, 2011 at 12:31 PM, Timothy Bates
wrote:
> In the help for get(), the following example is given:
>
> a <- 1:4
> assign("a[1]", 2)
> a[1] == 2 #FALSE
> get("a[1]") == 2 #TRUE
>
> However, executing that last line for me gives
>
> Error in get("a[1]") : object 'a[1]' n
Dear R-users,
I wanted to inform you that a new package called R2STATS is available,
as a graphical front-end for the glm() and glmer() functions.
The GUI is based on the RGTk2 and gWidgets packages by Michael Lawrence
and John Verzani, and so requires that the GTK+ library be installed
firs
In the help for get(), the following example is given:
a <- 1:4
assign("a[1]", 2)
a[1] == 2 #FALSE
get("a[1]") == 2 #TRUE
However, executing that last line for me gives
Error in get("a[1]") : object 'a[1]' not found
__
R-help@r-project.org
Thanks guys, that's a great help.
Nellie
--
View this message in context:
http://r.789695.n4.nabble.com/question-about-string-to-boor-tp3890983p3894996.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
Hi,
On Tue, Oct 11, 2011 at 12:08 PM, Ben qant wrote:
> Hello,
>
> There has to be a more R'ish way to do this. I have two matrices, one has
> the values I want, but I want to NA some of them. The other matrix has
> binary values that tell me if I want to NA the values in the other matrix. I
> pr
Hello,
There has to be a more R'ish way to do this. I have two matrices, one has
the values I want, but I want to NA some of them. The other matrix has
binary values that tell me if I want to NA the values in the other matrix. I
produce a third matrix based on this. I've also tried apply() passing
thanks, josh. in my posting example, I did not need anything except
coefficients. (when this is the case, I usually do not even use
lm.fit, but I eliminate all missing obs first and then use solve
crossprod(y,cbind(1,x)) crossprod(cbind(1,x)).) this is pretty fast.)
alas, I will need to figure
Hi R-Help -
If I make a plot:
numYears = 500
plot(x = c(1,numYears), y = c(200,300), xlab = "Time", ylab = "Vegetation
Class", xlim = c(100,600), ylim = c(200,300), type="n")
Is there a way to make different parts of the background for the plot
different colors?
For example, I'd like to have th
Thanks for the quick response.
Read the FAQ. If i want to keep the values in R the same as when inputed
should i be converting the data to a different type - i.e. Not numeric?
Sent from my iPhone
On Oct 11, 2011, at 4:46 AM, Jim Holtman wrote:
> FAQ 7.31
>
> Sent from my iPad
>
> On Oct
Yes thank you Gu…
I am just trying to do this as a rough step and will try other
imputation methods which are more appropriate later.
I am just learning R, and was trying to do the for loop and
f-statement by hand but something is going wrong…
This is what I have until now:
*fake array:
age<-
Hi Kenneth
First off, you probably don't need to use xmlParseDoc(), but rather
xmlParse(). (Both are fine, but xmlParseDoc() allows you to control many of
the options in the libxml2 parser, which you don't need here.)
xmlParse() has some capabilities to fetch the content of URLs. Howeve
Hi everyone,
I've got two data sets as below. My question now is: how can I use Dataset2
as a filter for Dataset1? My goal is just to keep the rows of Dataset1 where
the first column (Date) matches the Dates in Dataset2.
I would appreciate any solutions to this issue.
Many thanks!
S.B.
Sounds like what you were hoping for does happen in the function environment
but isn't returned to the global environment. The "proper" fix is to put the
values in a list and return() them as the function output.
Michael
On Oct 11, 2011, at 9:21 AM, Divyam wrote:
> Hello All,
>
> I have a s
Thank you all for your suggestions.
Sharad
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https://st
Hi R-helpers!
Here is my problem:
I have a graph with 3 different facets where there are 3 different
regression line. My goal is to mention separately in each facet each
equation that describes my lines.
So far, I managed to add a line and the same equation to all my facets but
that's not unfor
Hello All,
I have a series of steps that needs to be run many times. Hence I put them
all into a function. There is no problem in function creation, but when I
call the function, the steps are not getting executed or only the first step
gets executed. What possibly could be the reason?
Sample Fun
In your case, it may not be sensible to simply fill missing values by
mean or mode as multiple imputation becomes the norm this day. For
your specific question, na.roughfix in randomForest package would do
the work.
Weidong Gu
On Tue, Oct 11, 2011 at 8:11 AM, francesca casalino
wrote:
> Dear R e
Thanks for mentioning rcorrp.cens which is much more powerful than testing
for differences in C. Likelihood ratio tests would be even more powerful.
Ordinary differences in C index yields a test with power that is too low.
Frank
alanm (Alan Mitchell) wrote:
>
> ?Hmisc::rcorrp.cens
>
> -Alan
>
?Hmisc::rcorrp.cens
-Alan
-Original Message-
From: Eik Vettorazzi [mailto:e.vettora...@uke.de]
Sent: Tue 10/11/2011 2:25 AM
To: Yujie Wang
Cc: r-help@r-project.org
Subject: Re: [R] How to test if two C statistics are significantly different?
Hi Yujie,
there is still a lot of work in p
Dear all,
I am struggling to align "textentry" fields in a Tcl/Tk widget. In the
example below, I'd like to have the boxes aligned.
library(rpanel)
panel <- rp.control(title="title",size=c(100,100))
rp.textentry(panel,var=a,labels="Variable A",
initval="1",pos=list(row=0,column=0
maybe this helps:
http://r.789695.n4.nabble.com/Create-a-map-td3689877.html#a3893581
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R-help@r-
Dear R experts,
I have a large database made up of mixed data types (numeric,
character, factor, ordinal factor) with missing values, and I am
looking for a package that would help me impute the missing values
using either the mean if numerical or the mode if character/factor.
I maybe could use
hi,
I've just figured out how to plot (more) up-to-date poilitical borders in R
in an easy way:
You can get the outline file Borders_MWDB3 from the NASA panoply site
http://www.giss.nasa.gov/tools/panoply/overlays/
Then read it into R, get the indices with a jump over the -180/180 E line,
remove
hi,
see here
http://r.789695.n4.nabble.com/Create-a-map-td3689877.html#a3893581
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Hi every one
i have problem in R program to import the data from excel ,
I have done the following:
1. install.packages("xlsReadWrite")
2. library(xlsReadWrite)
3. z<- read.xls("ReadXls",LTS,colNames=FALSE,sheet,type,form,rowNames=FALSE)
and i got on the result:
Error in read.xls("ReadX
I've installed Rcmdr package and it doesn't run
Here is the error message:
R version 2.9.2 (2009-08-24)
[R.app GUI 1.29 (5464) powerpc-apple-darwin8.11.1]
[Workspace restored from /Users/jfc/Documents/TravauxFR/.RData]
Le chargement a nécessité le package : tcltk
Chargement de Tcl/Tk... termin
I have an R script that consists of a for loop
that repeats a process for many different files.
I want to process this parallely on machine with
multiple cores, is there any package for it ?
Thanks
--
Sandeep R Patil
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The following code will get me a curve plot:
cutoff <- seq(1,7,0.25)
Sensitivity <- 1 - pnorm(cutoff, 5, 0.8)
Specificity <- pnorm(cutoff, 3, 1.2)
plot(1-Specificity,Sensitivity,main = "ROC curve",type = "o")
How do I get a gradient of a particular point on that curve?
Any packages/functions allow
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