Even when I try to predict y values from x, let's say I want to predict y at
x=0. Looking at the graph from the provided syntax, I would expect y to be
about 0.85. Is this right:
predict(mylogit,newdata=as.data.frame(0),type="response")
# I get:
Warning message:
'newdata' had 1 rows but variable
Finn,
>> But when I use 'principal' I do not seem to be able to get the same
>> results
>> from prcomp and princomp and a 'raw' use of eigen:
< ...snip... >
>> So what is wrong with the rotations and what is wrong with 'principal'?
I would say that nothing is wrong. Right at the top of the hel
Here's a solution., maybe not the most elegant but works.
df.r = df1[, c(3:5)]; # restricted data
nNonZero = apply(df.r!=0, 1, sum);
one = nNonZero==1;
oneZero = nNonZero==2;
whichOne = apply(df.r[one, ]!=0, 1, which);
whichZero = apply(df.r[oneZero, ]==0, 1, which);
colNames = colnames(df.r);
Hello,
I am not sure where to begin with this problem or what to search for
in r-help. I just don't know what to call this.
If I have 5 columns, the first 2 are the x,y, locations and the last
three are variables about those locations.
x<-seq(1860,1950,by=10)
y<-seq(-290,-200,by=10)
ANN<-c(3,0,0
On Jan 26, 2011, at 10:52 PM, Ahnate Lim wrote:
Dear R-help,
I have fitted a glm logistic function to dichotomous forced choices
responses varying according to time interval between two stimulus. x
values
are time separation in miliseconds, and the y values are proportion
responses for one
Hello to all again :)
Again, I have an issue with my homework. I was troubling myself for two
days, and I still can't get it :(
So, if someone can help me, I would be so grateful...
HOMEWORK 1.
I need to draw function,
random walk take the length of walk, and its in 2D. As a result, I have
Just when I think I'm starting to learn
Statement z1 works, statement z doesn't. Why doesn't z work and what do I do
to fix it ? Clearly the problem is with the first NA, but I would think it's
handled through the loop vectorization.
y1 <- rnorm(20, 0, .013)
y1
[1] -0.0068630836 -0.01011
Dear R-help,
I have fitted a glm logistic function to dichotomous forced choices
responses varying according to time interval between two stimulus. x values
are time separation in miliseconds, and the y values are proportion
responses for one of the stimulus. Now I am trying to extrapolate x value
Hello,
In the documentation for agnes in the package 'cluster', it says that NAs are
allowed, and sure enough it works for a small example like :
> m <- matrix(c(
1, 1, 1, 2,
1, NA, 1, 1,
1, 2, 2, 2), nrow = 3, byrow = TRUE)
> agnes(m)
Call:agnes(x = m)
Agglomerative coefficient: 0.1614168
Eric
Your problem lies in the way cumprod deals with NAs
If you look at ?cumprod you will see
"An NA value in x causes the corresponding and following elements of the
return value to be NA"
Not sure what behaviour you want to see on encountering an NA (ignore it,
restart the cumprod process, .
Hi there,
yet on the topic of greek letters and pdf plotting: when I run the
following code
pdf(file="temp.pdf")
mu=seq(from=-pi, to=pi, length=100)
plot(mu, sin(mu^2),
type="l",
xlab=expression(mu%in%(list(-pi,pi))),
ylab=expression(sin(mu^2)),
main=expression((list(mu,sin(mu
Dear All
I try to use Hilbert Huang Transformation in R.
How I can do.
Many Thanks.
--
Jumlong Vongprasert Assist, Prof.
Institute of Research and Development
Ubon Ratchathani Rajabhat University
Ubon Ratchathani
THAILAND
34000
[[alternative HTML version deleted]]
__
On Jan 26, 2011, at 8:07 PM, Folkes, Michael wrote:
I'm trying to fathom how to answer two example problems (3.3.2 &
3.3.3) in:
Krishnamoorthy. 2006. "handbook of statistical distributions with
applications"
The first requires calculating single trial probability of success
for a binomia
I'm trying to fathom how to answer two example problems (3.3.2 & 3.3.3) in:
Krishnamoorthy. 2006. "handbook of statistical distributions with applications"
The first requires calculating single trial probability of success for a
binomial distribution when we know:
trial size=20, successes k=4, P(
Tena koe David
?polygon
should help you.
HTH
Peter Alspach
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of David Hervas Marin
> Sent: Thursday, 27 January 2011 12:02 p.m.
> To: R-Help
> Subject: [R] Colour area under
Tena koe Wendy
which(test=='')
Should do it.
HTH
Peter Alspach
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Wendy
> Sent: Thursday, 27 January 2011 10:59 a.m.
> To: r-help@r-project.org
> Subject: [R] Find the empt
First off, thank you for the help with the global environment.
I have however attempted to run the code and am now presented with a new
error which is
"Error in formula.default(eval(parse(text = x)[[1L]])) : invalid formula"
and am not sure what to make of it. I have tried a few different work a
A bit of a newbee to R and factor rotation I am trying to understand
factor rotations and their implementation in R, particularly the
GPArotation library.
I have tried to reproduce some of the examples that I have found, e.g., I
have taken the values from Jacksons example in "Oblimin Rotati
I am new to R and am interested in using the program to fit quantile
regression models to data collected from a multi-stage probability
sample of the US population. The quantile regression package, rq, can
accommodate person weights. However, it is not clear to me that
boot.rq is appropriate for
Hi all,
I have a factor array, and some of the elements are empty. How would I
return the index number of the empty elements. For example,
test<-factor(c('A','','B','C','E'))
> test
[1] A B C E
Levels: A B C E
I would like the result equal to 2.
Thank you,
Wendy
--
View this message in con
Hello,
I have this code to plot a certain normal distribution and represent the pnorm
value for a certain "x":
x<-300
xx <- seq(2.5,7.5, by=0.1)
yy <- dnorm(xx,5.01,0.77)
d<-signif(pnorm(log(x), 5.01,0.77),4)
xpts <- round(exp(0:8))
par(bg = "antiquewhite")
plot(xx,yy, type="l", col="blue", lw
Not sure exactly what you mean by, writing a table with several rows per
file.
If what you want to do is write output to an external file, adding to it as
your loop progresses, then look at the functions
sink()
cat()
And their file and append arguments.
If what you want to do is appe
For the last point (cluttered text), look at spread.labels in the plotrix
package and spread.labs in the TeachingDemos package (I favor the later, but
could be slightly biased as well). Doing more than what those 2 functions do
becomes really complicated really fast.
--
Gregory (Greg) L. Snow
Hello all,
I wrote a small function to add labels for outliers in a boxplot.
This function will only work on a simple boxplot/formula command (e.g:
something like boxplot(y~x)).
Code + example follows in this e-mail.
I'd be happy for any suggestions on how to improve this code, for example:
-
It's looking for an object named "b" in the frame of the function.
There is none. You need to specify
get(response, pos = parent.frame())
## or maybe even
get(response, pos= globalenv())
HOWEVER, this is **exactly** why you shouldn't do this! Instead of
passing in the name of the object, pass i
I am having trouble with variable scoping inside/outside functions. I
want to use get() to grab a named and quoted variable as an input to a
function, however the function can't find the variable when it is entered
into the function call, only when it is named in the main environment.
I obviously
Your faith in our ability to read your mind is apparently much higher than our
actual ability to do so. What is the nature of your data? what question are
you trying to answer? What type of equation do you want? What do you mean by
better? Better than what?
Maybe the esp package (pre-alpha) c
On 11-01-26 3:43 PM, zerfetzen wrote:
Thanks Duncan, that helps. It successfully displays what I'm looking for,
but it is not executing it. In a previous code chunk, it notes the time it
took to run something, and in the successive code chunk, it runs something
else where the previous time is
Dear useRs,
This is to announce the RGtk2Extras R package is available on CRAN in version
0.5.1.
This package provides useful extras for R programmers who wish to create
graphic user interfaces. It is based on GTK, using Michael Lawrence's RGtk2
package and John Verzani's gWidgets, and some id
Dear colleagues, I have the following dataset. It is modelled on the data
included in Box-Seteffenheiser and Jones "Event History Modelling"
Using the following code, I try to find the baseline hazard function
haz_1<-muhaz(bpa$time, bpa$censored, subset=(bpa$year=="2010" | bpa$ban=="1"),
min.t
Thanks Duncan, that helps. It successfully displays what I'm looking for,
but it is not executing it. In a previous code chunk, it notes the time it
took to run something, and in the successive code chunk, it runs something
else where the previous time is now a parameter, but I'd like it to
nume
... or perhaps just break things up with assignments and do it in stages.
-- Bert
On Wed, Jan 26, 2011 at 12:52 PM, David Winsemius
wrote:
> I remember something about the degree of nesting of ifelse calls being
> limited to 7 deep (???) that makes me worry about this approach. You may
> want t
Dear R-helpers,
I can not invoke maxent() in Mac OSX. Could you give me any directions
on that? Thank you in advance.
Here is my info:
# (1) the error
> me <- maxent(predictors, occtrain, factors='biome')
me <- maxent(predictors, occtrain, factors='biome')
Error in .jcall(mxe, "S", "fit", c("au
Thanks Dennis this is what I was looking for.
On Wed, Jan 26, 2011 at 4:14 AM, Dennis Murphy wrote:
> Hi:
>
> Here's one approach:
>
> f <- function(df) {
> rs <- with(na.exclude(df), tapply(y, strata, sum)/tapply(x, strata,
> sum))
> u <- transform(subset(df, is.na(y)), y = x * rs[s
I remember something about the degree of nesting of ifelse calls being
limited to 7 deep (???) that makes me worry about this approach. You
may want to look at the arules package or the data.table package or
the sqldf package for approaches that are specifically constructed
with this sort
Yes. That is exactly what I would like to have running. Here is the first
attempt I made at using a nested ?ifelse statement for one of the retirement
plans. The variables are all there but with different names. ageYOSstart is
ageFedStart, SCDCivLeave is srvCompDT. I haven't gotten this working.
On 26/01/2011 3:27 PM, Giles Crane wrote:
Surprising behavior:
Most R users are aware that print()
must be used inside functions to gain
output on the console.
Apparently, print() is sometimes required
when interactively using the console.
For example, the followingmay be
entered into the R cons
If I understand you correctly, you want ?ifelse, which works on the
full logical vectors of rules applied to the variables, not
ifelse, which works on only a single logical.
-- Bert Gunter
On Wed, Jan 26, 2011 at 12:18 PM, KATSCHKE, ADRIAN CIV DFAS
wrote:
> All,
>
> I would like to apply a s
Surprising behavior:
Most R users are aware that print()
must be used inside functions to gain
output on the console.
Apparently, print() is sometimes required
when interactively using the console.
For example, the followingmay be
entered into the R console with different results.
sample(1:8,8)
All,
I would like to apply a set of rules to each row of the sample data set
below. The rule sets are the guidelines for determining an individual's
date for retirement eligibility. The rules are found in this document,
http://www.opm.gov/feddata/RetirementPaperFinal_v4.pdf. I am only
interested i
On Jan 26, 2011, at 2:48 PM, David Winsemius wrote:
On Jan 26, 2011, at 2:42 PM, Akram Khaleghei Ghosheh balagh wrote:
Hello ;
How could I choose subvector of a vctor. for example if
v=c(1,2,5,0,1), how
could I chose the (1,2) or (1,2,5).
?"["
v[ c(1,2,5) ]
I didn't notice at first t
On Jan 26, 2011, at 2:06 PM, Melanie Zoelck wrote:
Hi,
I am relatively new to R and have a question regarding code. I have
a data set which has data organised by location (site names, which
are factors). I now want to add a new variable Region (this will be
non numerical, as it will be n
On Jan 26, 2011, at 2:42 PM, Akram Khaleghei Ghosheh balagh wrote:
Hello ;
How could I choose subvector of a vctor. for example if
v=c(1,2,5,0,1), how
could I chose the (1,2) or (1,2,5).
?"["
v[ c(1,2,5) ]
thanks;
[[alternative HTML version deleted]]
__
Hi,
I am relatively new to R and have a question regarding code. I have a data set
which has data organised by location (site names, which are factors). I now
want to add a new variable Region (this will be non numerical, as it will be
names). Each region will contain locations. So for example:
Hello ;
How could I choose subvector of a vctor. for example if v=c(1,2,5,0,1), how
could I chose the (1,2) or (1,2,5).
thanks;
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-hel
I would leave off the as.character(). With it you get
things like:
> f <- rpart(Kyphosis ~ ., kyphosis[-3])
> as.list(f$call)$data
kyphosis[-3]
> as.character(as.list(f$call)$data)
[1] "[""kyphosis" "-3"
Expressions like quote(kyphosis[-3]) are much
easier to analyze as expressio
Exactly what I needed Henrique,
Thank you.
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
---
Hi all,
I am using a discrete Hidden Markov Model with discrete observations in
order to detect a sequence of integers. I am using the "hmm.discnp" package.
I am using the following code:
signature <- c(-89, -98, -90, -84, -77, -75, -64, -60, -58, -55, -56, -57,
-57, -63, -77, -81, -82, -91, -85
Hello All.
I am using the image2 plot function in the popbio package to create 6
elasticity analyses. I am trying to reduce the number of significant digits
that displays -- from 3 digits to 2. I tried rounding my original matrices,
but one is comprised primarily of zeros, and the cut / breaks op
Try this:
as.character(as.list(fit1$call)$data)
On Wed, Jan 26, 2011 at 4:12 PM, Tal Galili wrote:
> Another (similar) question,
>
> If I now want to know the name of the "data" argument used, is there an
> easy way for me to access it?
>
> I'm trying to use something like:
> eval(parse(text =
Harrell's wiki/website has material on so-called "dynamite plots"
http://biostat.mc.vanderbilt.edu/twiki/bin/view/Main/DynamitePlots
Ben Bolker has a page on them as well:
http://emdbolker.wikidot.com/blog:dynamite
--
David.
On Jan 26, 2011, at 1:26 PM, Joshua Wiley wrote:
Hi,
You wil
;^)
you're right: we'll add it soon. Keep an eye on R-forge.
All the best,
Giovanni
-- original message
Message: 89
Date: Tue, 25 Jan 2011 17:14:00 -0800 (PST)
From: zhaowei
To: r-help@r-project.org
Subject: Re: [R] how to get loglik parameter from splm package?
Me
Hi,
You will find package "sos" has some handy functions for searching for
functions/packages:
###
install.packages("sos")
require(sos)
findFn("error.bars")
###
Perhaps the "psych" package has the error.bars() function you are thinking of?
As a si
Note that all.vars(terms(fit)) only looks at the
formula in the terms object and throws away all
the analysis done by rpart's call to terms(formula,data).
Here is a contrived example of that approach failing:
> ageThreshold <- 50
> fit3 <- rpart(Kyphosis=="present" ~ (Age>ageThreshold) + log(N
Dear all,
I am trying to add error bars on a boxplot but have encountered an error as
indicated below. Is there a package I need to install or a library I have to
load before this goes please.
Thanks for any idea.
Ogbos
x<-replicate(20,rnorm(50))
boxplot(x,notch=TRUE,main="Notched boxplot with er
Another (similar) question,
If I now want to know the name of the "data" argument used, is there an easy
way for me to access it?
I'm trying to use something like:
eval(parse(text = all.vars(terms(fit1))[1]))
Which (of course) wouldn't work, since the response variable is only
available in the d
No. Any valid seed should work. In this case, train() should on;y be
using it to determine which training set samples are in the CV or
bootstrap data sets.
Max
On Wed, Jan 26, 2011 at 9:56 AM, Neeti wrote:
>
> Thank you so much for your reply. In my case it is giving error in some seed
> value f
Thanks Henrique, exactly what I was looking for.
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
-
Hi,
x and y are being picked up from your global environment, not from the
x and y in dataset. Here is a version that seems to work:
rm.outliers = function(dataset,var1, var2) {
dataset$varpredicted = predict(lm(as.formula(paste(var1, var2,
sep=" ~ ")), data=dataset))
dataset$varstdres =
Try this:
all.vars(terms(fit1))
all.vars(terms(fit2))
On Wed, Jan 26, 2011 at 3:33 PM, Tal Galili wrote:
> Hello all,
>
> I wish to extract the terms from an rpart object.
> Specifically, I would like to be able to know what is the response variable
> (so I could do some manipulation on it).
>
Take a look at the output of
terms(fit2)
In particular
tm <- terms(fit2)
attr(tm, "response")
is 1 if there is a response and
variables <- as.list(attr(tm, "variables"))[-1]
variables[[1]]
gives the response expression if there is one.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap ti
The answers to these questions can be found either by looking at the
code or by reviewing what has been said about this in prior postings
to R-help.
--
David.
On Jan 26, 2011, at 10:40 AM, saray wrote:
Although I saw this issue being discussed many times before, I still
did not find the a
Hello all,
I'm was going through the help for
?rpart.object
And noticed some inconsistencies, Some might be a mistake in the help file
and some might be my misunderstanding.
The help in the section:
value -> frame (first paragraph), states that:
> yval, the fitted value of the response at each
Hello all,
I wish to extract the terms from an rpart object.
Specifically, I would like to be able to know what is the response variable
(so I could do some manipulation on it).
But in general, such a method for rpart will also need to handle a "." case
(see fit2)
Here are two simple examples:
f
Hi All,
I am trying to combine two forest plots on the same page using the "forestplot"
function in the rmeta package. Once I use the par() function to combine my
plots on the same page, I find that my two plots are overlaying each other.
Does anyone have any suggestions on how to fix this?
T
Although I saw this issue being discussed many times before, I still
did not find the answer to:
why does R can not calculate p-values for data with ties (i.e. -
sample with two or more values the same)?
Can anyone elaborate some details about how does R calculate the p-
values for the Kolmogorov
Hi,
I have a few lines of code that will remove outliers for a regression test
based on the studentized residuals being above or below 3, -3. I have to do
this multiple times and have attempted to create a function to lessen the
amount of copying, pasting and replacing.
I run into trouble with
Hello,
I have some question on chron
I currently doing this
t1 <- chron(,"11:30:00")
t2 <- chron(,"11:45:00")
tt <- seq(t1,t2,by=times("00:00:01"))
tt has 901 values (15 minutes * 60 secs) and then
x1 <- rnorm(1:901)
x2 <- rnorm(1:901)
x3 <- rnorm(1:901)
df <- data.frame(tt, x1, x2, x3)
I
I hope this is what you are looking for. you will have to add your own colors
and such.
year = c(1966:2008)
tempur =
c(2.9,4.5,1.9,1,2.9,4.3,3.9,4.3,4.9,4.4,4.5,2,2.8,-.4,2.3,3,.3,1.7,3.3,.8,-1.1,.8,4.9,5.2,4.9,1.5,3.7,3.6,3.2,4.8,2.3,2.5,5.2,5.3,4.9,3.2,3.6,3.9,4.8,4.3,3.7,5.8,4.9)
indx =
c(1,1.
Dear R-folks:
I have a group of data ('x' and 'y' axis), but I'd like to know how to draw a
graph which would fits my data in a better way, I also need its equation. Could
you give me any R-rutine ideas?.
I thank you in advance for your kind support.
Hadley and Dennis:
THANK YOU THANK YOU! This is exactly what I was looking for.
Ryan
On Wed, Jan 26, 2011 at 5:27 AM, Dennis Murphy wrote:
> > Hi:
> >
> > Here are two more candidates, using the plyr and data.table packages:
> >
> > library(plyr)
> > ddply(X, .(x, y), function(d) length(unique
Great. Thank you, Peter!
-Original Message-
From: Peter Ehlers [mailto:ehl...@ucalgary.ca]
Sent: Tuesday, January 25, 2011 7:26 PM
To: Jim Moon
Cc: r-help@r-project.org
Subject: Re: [R] write.table -- maintain decimal places
On 2011-01-25 17:22, Jim Moon wrote:
> Thank you for the respo
I am using:
"R version 2.11.1 (2010-05-31)"
It is good to know that it works in 2.12.1
Jim
-Original Message-
From: Peter Ehlers [mailto:ehl...@ucalgary.ca]
Sent: Tuesday, January 25, 2011 5:57 PM
To: Jim Moon
Cc: r-help@r-project.org
Subject: Re: [R] write.table -- maintain decimal plac
Hello there,
Straight to the point: it seems that CairoPDF from package "Cairo"
cannot handle greek letters from expression(). For example,
> eta = seq(from=-pi, to=pi, length=100)
> f = sin(eta)^2
> pdf(file = "temp_pdf.pdf")
> plot(eta, f, type="l", main=expression(f(eta)==sin(eta)^2),
> xlab=
Why do you need the line to overlay the bars? Which bars are touched by the
line is just a quirk of scaling and could easily change with the scales. All
the overlay does is to make it harder to read, why not jut have 2 panels
aligned on the x-axis but with the line plot above the bar plot?
--
Hi Larry,
If I understand correctly, your barplot() call dispatches to the
method function barplot.default() to do the work. Looking at the
definition of that function and your specific call, it seems that
around line 51 of barplot.default(), the value of the width argument
is truncated:
width <
Thank you so much for your reply. In my case it is giving error in some seed
value for example if I set seed value to 357 this gives an error. Does train
have some specific seed range?
--
View this message in context:
http://r.789695.n4.nabble.com/Train-error-subscript-out-of-bonds-tp3234510p323
I checked it using:
Sys.getenv("PATH")
And the output includes the PATH to the GTK2 installation (it's the last
item in the following list):
"C:\\Windows\\system32;C:\\Windows;C:\\Windows\\System32\\Wbem;C:\\Windows\\System32\\WindowsPowerShell\\v1.0\\;C:\\Program
Files (x86)\\Common Files\\Ulea
Good morning,
This problem was already addressed in a previous post:
https://stat.ethz.ch/pipermail/r-help/2009-February/187244.html
In the call to draw.key() use
'vp=viewport(x=unit(0.1,"npc"),y=unit(0.1,"npc"))'.
Prior to calling viewport() make sure grid package is loaded.
Apologies for clut
--- begin inclusion ---
Response variable: survival (death)
Factor 1: treatment (4 levels)
Factor 2: sex (male / female)
Random effects 1: person nested within day (2 people did the
experiment over 2 days)
Random effects 2:
you were caught by the '=' versus '==' error ;-)
and Henrique's elegant one-liner avoids the problem altogether.
-- David
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Henrique Dallazuanna
Sent: Wednesday, January 26, 2011 6:00 A
Your GTK+ installation is not being found: check your PATH.
On Wed, 26 Jan 2011, Tal Galili wrote:
Hello Prof Brian Ripley, Yihui and Tom,
Thank you for your suggestions. It seemed to have made some differences in
the error massages - but rggobi still fails to load.
Steps taken:
1) I removed
Sort of. It lets you define a grid of candidate values to test and to
define the rule to choose the best. For some models, it is each to
come up with default values that work well (e.g. RBF SVM's, PLS, KNN)
while others are more data dependent. In the latter case, the defaults
may not work well.
M
acocac gmail.com> writes:
>
>
> Hi!! Im doing my graduated work in Onion Curves Growth with Nonlinear Models,
> I'm amateur in R so i have doubt how i put or program next models,
>
> http://r.789695.n4.nabble.com/file/n3236748/96629508.png
>
> Also, i cant derivate for Gauss Model, and Rich
On 01/26/2011 05:08 AM, Pascal A. Niklaus wrote:
> Dear all,
>
> My apologies for re-posting this question, but I have not found any
> solution to my problem so far. I also think that my post may have been
> overseen due to the posting time and high traffic on this list.
>
> I experience a proble
Begin forwarded message:
From: David Winsemius
Date: January 26, 2011 8:32:30 AM EST
To: Alaios
Subject: Re: [R] MAtrix addressing
On Jan 26, 2011, at 7:58 AM, Alaios wrote:
Unfortunately right now is convoluted... by I was trying to find
some solution.
Bring again this picture in fron
Hello Prof Brian Ripley, Yihui and Tom,
Thank you for your suggestions. It seemed to have made some differences in
the error massages - but rggobi still fails to load.
Steps taken:
1) I removed the old GTK (through the uninstall interface)
2) I ran library(RGtk2) which downloaded the new GTK-ru
On Wed, Jan 26, 2011 at 5:27 AM, Dennis Murphy wrote:
> Hi:
>
> Here are two more candidates, using the plyr and data.table packages:
>
> library(plyr)
> ddply(X, .(x, y), function(d) length(unique(d$z)))
> x y V1
> 1 1 1 2
> 2 1 2 2
> 3 2 3 2
> 4 2 4 2
> 5 3 5 2
> 6 3 6 2
>
> The function
Dear all,
My apologies for re-posting this question, but I have not found any
solution to my problem so far. I also think that my post may have been
overseen due to the posting time and high traffic on this list.
I experience a problem in implementing a S4 class in a package and would
apprec
Hello all and thanks for the help.
I am analyzing a dataset using MetaMDS and I would like to project some
extra samples onto the plot such that the extra samples do not play a role
in defining the axes. I have been thinking of different ways of doing this
and I was wondering if anyone had a sug
I would look into ggplot2. I use this quite frequently to do what you
are talking about, and also for most of my plotting. Hadley has done
a wonderful job with this package.
kindest regards,
Stephen
On Jan 26, 2011, at 3:48 AM, Claudia Paladini wrote:
Dear R-users,
I used x.out=sign
On Jan 26, 2011, at 5:38 AM, pdb wrote:
There seems to be 2 functions call ecdf...
http://lib.stat.cmu.edu/S/Harrell/help/Hmisc/html/ecdf.html
Apparently that used to be its name and that some time in the
intervening 10 years the name was changed to Ecdf. The Statlib
repository has rath
Hello everybody,
i need some help to display text as label in my barchart
the label is the combination of x value + text
text= calculated percentage => per
it display properly the x value
but, wrongly repeats the text of the fisrt level LangueTXT factor on the
second
any solution?
Thanx very much
On Jan 26, 2011, at 2:47 AM, Alaios wrote:
The reason is the following image
http://img545.imageshack.us/i/maptoregion.jpg/
In the picture above you will find the indexes for each cell.
Also you will see that I place that matrix inside a x,y region that
spans from -1 to 1. I am trying to wri
On 11-01-25 8:22 PM, zerfetzen wrote:
>
> Hi,
> Is it possible in Sweave to put \Sexpr{} inside<<>>? This is a bad
> example, but here goes:
>
> <>
> Age<- 5
> @
>
> <<>>
> x<- \Sexpr{Age}
> @
>
> I'm trying to get it to display x<- 5, rather than x<- Age. It's
probably
> so obvious I'm going
Hi,
This isn't an issue with is.na, you get the same if you use
aa = c(1,1,0,1,1,1,1,1,1)
bb = abs(aa - 1)
xtabs(aa~x1+x2)
xtabs(bb~x1+x2)
it is because you do not have any data in (1,1), i.e. there is no case where x1
= 1 and x2 = 1 so xtabs is putting a zero in that cell
Hope this helps
Mar
Try this:
unique(transform(df, quantity = ave(quantity, client, date, name, FUN =
sum), branch = NULL))
On Wed, Jan 26, 2011 at 8:39 AM, analys...@hotmail.com <
analys...@hotmail.com> wrote:
> I have
>
> > df
> quantity branch client date name
> 110 1 1 2010-01-01 o
Dear R-users,
I used x.out=sign1(data,makeplot=TRUE) from the package mvoutlier to detect
multivariate outliers.
I would like to label the points in the resulting plot with the row names of
my data set. But none of my attempts does lead to a result. Can anybody help
me, please?
There isn't combination of c(1, 1), so is NA:
tapply(y, list(X1, X2), sum)
On Wed, Jan 26, 2011 at 9:04 AM, René Holst wrote:
> Hello,
>
> I have observed the following odd behavior of "is.na( )" and hope someone
> can give me an explanation
> Example:
> X1=rep(1:2,5)[-1]
> X2=rep(1:5,rep(2,5))
If you have a large data frame, one option is package data.table. Try the
following:
library(data.table)
dt <- data.table(df)
dt[, list(qsum = sum(quantity)), by ='client, date, name']
client date name qsum
[1,] 1 2010-01-01 one 45
[2,] 1 2011-01-01 one 4500
[3,] 2
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