On Sat, Sep 19, 2009 at 3:36 PM, di jianing wrote:
> Hello R helpers,
>
> I am producing a figure with dual strips, i.e., x~y | S1 + S2, where S1 and
> S2 are two strips. For example, in figure 2.1 at
> http://lmdvr.r-forge.r-project.org/figures/figures.html.
>
> In this case, I would like to comb
Lina,
check ?density (and do so carefully). R uses a kernel estimate, by default
"Gaussian" if I remember correctly. Values in a certain grid can be found from
the code I sent earlier. I didn't check, but these are most likely just
linearly interpolated by plot.density, and as the grid is suff
On Sat, Sep 19, 2009 at 9:42 AM, Larry White wrote:
> Hi,
>
> I'm trying to create a lattice plot with three xyplots in one vertical
> column. I would like to reduce the vertical space between the charts. My
> code is below. There seems to be a "between" parameter for lattice.options,
> but I can
On Tue, 29 Sep 2009, edche...@gmail.com wrote:
I have chances to work with both linux based and win based R codes.
And as you all know in linux, the file directories use "/" and win
uses "\\". Is there a function like sub or gsub that could
substitute those slashes automatically?
See ?char
On Tue, Sep 29, 2009 at 11:09 PM, Dan Kortschak
wrote:
> Hi Deepayan,
>
> Thanks for that, I had a think (a few hour too late) and came to the
> same conclusion. I had wanted to have vertical bars for each of the data
> points, I have gone to a straight xyplot and this shows other intersting
> inf
On Fri, Sep 18, 2009 at 6:06 AM, baptiste auguie
wrote:
> No box is easy,
>
> bwplot(y~x, data=data.frame(y=rnorm(10),x=sample(letters[1:3],10,repl=T)),
> par.settings=list(axis.line=list(col=NA)))
>
> but that seems to remove all axis lines and ticks as well. You may
> have to define a custom pan
Hi,
You should use this mailing list instead
http://mailman.rz.uni-augsburg.de/mailman/listinfo/stats-rosuda-devel.
... where I just posted something that might help you.
Romain
On 09/28/2009 02:57 PM, ajoec...@gmx.de wrote:
Hello,
I am writing a Java frontend for a selfwritten R program us
On Tue, 29 Sep 2009, Steve Bellan wrote:
Hi,
I'd like to use Arial for the font in the PDF's and TIFF's I produce in R
As ever, it is helpful to explain why you want to do something.
First, because there may be better solutions to the larger problem,
and secondly because the answer may invo
Hi Deepayan,
Thanks for that, I had a think (a few hour too late) and came to the
same conclusion. I had wanted to have vertical bars for each of the data
points, I have gone to a straight xyplot and this shows other intersting
information - maybe I want to be able to do both point and bar plots
(
On Wed, Sep 16, 2009 at 2:02 PM, Dan Kortschak
wrote:
> Hi, I trying to produce a bar chart describing hits to specific bins by
> chromosome for a large data set (I am asking here because
> experimentation with options is precluded due to this - generating the
> figure takes about an hour):
>
> ba
Hi,
I'd like to use Arial for the font in the PDF's and TIFF's I produce
in R on my Mac (running 10.5.5). I've found the following archived
help file on how to do it in Linux (http://tolstoy.newcastle.edu.au/R/e4/help/08/08/19847.html
) but don't understand how to do this on my system. Are
Hello Primoz,
with traditional graphics, you may want to use par (?par), options like mar,
mai, oma etc. may be interesting for you. And for the relation between
y-axis and x-axis, the option asp to function plot (?plot.default) will
help.
Regards, Ulrike
Primoz PETERLIN-2 wrote:
>
> Hello e
Thanks a ton Rolf,
I was using the coefficients as given from summary(fit), which have been
rounded. When I used coef(fit) as you've done, the poly function is the same
now in Octave and returned the R predicted values as expected. Your time is
appreciated,
C
> CC: r-help@r-project.org
> Fro
Hi,
I want to compute the pairwise correlation for about 10,000 genes. In
Matlab there is a function called 'pdsit' that can do this very
efficiently. I am wondering is there a similar function in R?
Thanks,
RT Ye.
__
R-help@r-project.org mailing list
I have chances to work with both linux based and win based R codes. And as you
all know in linux, the file directories use "/" and win uses "\\". Is there a
function like sub or gsub that could substitute those slashes automatically?
Thanks!
Edward Chen
Email: tke...@msn.com
Cell Phone: 510-37
Hey guys,
im sort of a beginner with R, but here's what i need to do.
i need to perform a time series analysis on a set of financial data that i've
been given. im trying to look at the ACF and PACF and fit it to a particular
model (i think its the ARIMA model because i've read that financi
Peng Yu wrote:
> On Tue, Sep 29, 2009 at 3:47 PM, Tobias Verbeke
> wrote:
>> Peng Yu wrote:
>>
>>> I want to compile R with command completion. But I don't find such an
>>> option in configure. Can somebody let me know how to enable command
>>> completion in an R session?
>> AFAIK this is not an o
rgl should let you draw segments (maybe even arrows) in pseudo 3D
between sequential locations. I wonder if you could trick persp into
doing it as well?
scatterplot3d would be another avenue to explore.
On Sep 29, 2009, at 10:39 PM, Jen Maresh wrote:
Hello All -
Any recommendations or su
What about using NULL?
> x<-runif(12)
> y<-runif(12)
> z<-runif(12)
> mydf<-data.frame(cbind(x,y,z))
> names(mydf)
[1] "x" "y" "z"
> mydf$x<-NULL
> names(mydf)
[1] "y" "z"
--- On Tue, 9/29/09, Rolf Turner wrote:
> From: Rolf Turner
> Subject: Re: [R] Deleting a column in a dataframe by name
On Tue, Sep 29, 2009 at 3:47 PM, Tobias Verbeke
wrote:
> Peng Yu wrote:
>
>> I want to compile R with command completion. But I don't find such an
>> option in configure. Can somebody let me know how to enable command
>> completion in an R session?
>
> AFAIK this is not an option you set when comp
Hello All -
Any recommendations or suggestions for neat ways to visualize data
taken from a 3-axis accelerometer? My study species is aquatic, so I
would be interested in movement in the 3 dimensions in addition to
being able to incorporate the time series as well. Is there a package
in R that mig
Dear R Users,
Does anyone know of a package that would enable me to create figures like
the ones below?
Regards,
James
http://www.nabble.com/file/p25673513/Picture1.jpg
http://www.nabble.com/file/p25673513/Picture2.jpg
--
View this message in context:
http://www.nabble.com/3D-bar-graphs-de
Wow! Yes, that is perfect! Thanks!!
On Wed, Sep 30, 2009 at 1:17 PM, jim holtman wrote:
> Is this close to what you want:
>
> x <- read.table(textConnection("locus min max
> K9/10* 70 72
> Pcc40 80 81
> 2F9 84 100
> 4E8/Pca12 96 108
> Pau6 99 103
> Pau20 100 113
> Pau9 103 129
> Pocc1* 104 128
>
Is this close to what you want:
x <- read.table(textConnection("locus min max
K9/10* 70 72
Pcc40 80 81
2F9 84 100
4E8/Pca12 96 108
Pau6 99 103
Pau20 100 113
Pau9 103 129
Pocc1* 104 128
Pau7 105 111"), header=TRUE, as.is=TRUE)
closeAllConnections()
x$row <- seq(nrow(x))
plot(0, type='n', ylim=c(0,m
Hi,
I am trying to connect points on a graph that originate from *different
columns of data*. For each sample I have minimum and maximum data points and
I would like to draw a line connecting these in order to visualize the
spread, as well as where each sample is in relation to the x-axis. So far
Hi chipmaney,
Here is a suggestion:
with(data, tapply(value, group, function(x) sd(x)/sqrt( length(x) ) ) )
* sqrt(fc$fc)
BTW, take a look at
require(fortunes)
fortune('dog')
HTH,
Jorge
On Tue, Sep 29, 2009 at 6:29 PM, chipmaney <> wrote:
>
> #I have these data. Basically, I want to run a ta
Hi Doug,
An interesting site I stumbled on long ago related to your intial
question of data download is at
http://www.mratings.com/cf/compare.htm
which also contains a CSV file version at
http://www.mratings.com/cf/compare.csv
of *64* different ranking systems. Other pages within the site
disc
#I have these data. Basically, I want to run a tapply to calculate the mean
and st. err. by factor.
#The problem is, I want to add a finite correction to the variance prior to
calculating the standard error.
#Now I know I could just do this in 3/4 steps by calculating the var,
applying the correct
On Sep 29, 2009, at 5:53 PM, Antonio Paredes wrote:
Hello everyone,
Can somebody give a hint on how to go about speeding the following
loop:
You could try a loop-less approach:
system.time(
targets <- dat$ycon[i]==0 | dat$ytrt[i]==0
dat$ycon[targets]<-dat$ycon[targets]+0.5
da
Hello everyone,
Can somebody give a hint on how to go about speeding the following loop:
system.time(
for(i in 1:nrow(dat)){
if(dat$ycon[i]==0 || dat$ytrt[i]==0)
dat$ycon[i]<-dat$ycon[i]+0.5
dat$ytrt[i]<-dat$ytrt[i]+0.5
dat$cony[i]<-dat$cony[i]+0.5
dat$trty[i]<-dat$trty[i]+0.5
}
On 30/09/2009, at 10:25 AM, Duncan Murdoch wrote:
On 29/09/2009 4:54 PM, Rolf Turner wrote:
On 30/09/2009, at 9:32 AM, milton ruser wrote:
x=runif(12)
y=runif(12)
w=runif(12)
mydf<-data.frame(cbind(x,y,w))
head(mydf)
mydf<-subset(mydf, select=c(-x,-w))
head(mydf)
But this doesn'
On 29/09/2009 4:54 PM, Rolf Turner wrote:
On 30/09/2009, at 9:32 AM, milton ruser wrote:
x=runif(12)
y=runif(12)
w=runif(12)
mydf<-data.frame(cbind(x,y,w))
head(mydf)
mydf<-subset(mydf, select=c(-x,-w))
head(mydf)
But this doesn't work if NAME1 and NAME2 are ***names***,
as
Stephen Collins wrote:
Say I have a formula Y ~ 1 + X, where X is a categorical variable. A
previous thread showed how to evaluate this model using the mle package
from "stats4" (see below). But, the user had to create the data matrix,
X, including the column of one's for the regression const
Peace on your heart... it is still Tuesday, and Friday is away... :-)
On Tue, Sep 29, 2009 at 4:54 PM, Rolf Turner wrote:
>
> On 30/09/2009, at 9:32 AM, milton ruser wrote:
>
> x=runif(12)
>> y=runif(12)
>> w=runif(12)
>>
>> mydf<-data.frame(cbind(x,y,w))
>> head(mydf)
>>
>> mydf<-subset(mydf,
>
> as.Surv <- function(x) {
> Surv(as.integer(sub("\\+", "", x)),
>as.integer(ifelse(regexpr("\\+", x) > -1, 0, 1)))
> }
>
> > identical(srv, as.Surv(srv.char))
> [1] TRUE
>
But Do NOT use "as.integer" with real data, which may contain fractions of
course. I was too eager to get i
Try this;
cut(x, breaks = c(0, 10, 20, 100), labels = c(0.3, .5, 1))
On Tue, Sep 29, 2009 at 3:11 PM, Jarek Jasiewicz wrote:
> Dear All!
>
> I'm looking for equivalent of Matematica function "Which" which works as
> follows:
>
> z = Which[x<10,0.3, 10<=x<20,0.5, 20<=x<100,1]
>
> where x is a vec
Thanks for the reply! However, what is the code that is used in order to
calculate it.. Is it included in the "randomForest" R function.. Because I
am trying to find the code corresponding to the estimation of the gini
coefficient but I am a bit confused. Could you please help me, by indicating
me
On 30/09/2009, at 9:32 AM, milton ruser wrote:
x=runif(12)
y=runif(12)
w=runif(12)
mydf<-data.frame(cbind(x,y,w))
head(mydf)
mydf<-subset(mydf, select=c(-x,-w))
head(mydf)
But this doesn't work if NAME1 and NAME2 are ***names***,
as the terminology would apply. And that is
Hi Jarek,
Take a look at ?which, ?ifelse and ?recode (car package).
HTH,
Jorge
On Tue, Sep 29, 2009 at 2:11 PM, Jarek Jasiewicz <> wrote:
> Dear All!
>
> I'm looking for equivalent of Matematica function "Which" which works as
> follows:
>
> z = Which[x<10,0.3, 10<=x<20,0.5, 20<=x<100,1]
>
> wh
Peng Yu wrote:
I want to compile R with command completion. But I don't find such an
option in configure. Can somebody let me know how to enable command
completion in an R session?
AFAIK this is not an option you set when compiling.
Did you try to type a letter (say 'l') and press the
TAB key
Hi Paul,
>> I have a data set for which PCA based between group analysis (BGA) gives
>> significant results but CA-BGA does not.
>> I am having difficulty finding a reliable method for deciding which
>> ordination
>> technique is most appropriate.
Reliability really comes down to you thinking
x=runif(12)
y=runif(12)
w=runif(12)
mydf<-data.frame(cbind(x,y,w))
head(mydf)
mydf<-subset(mydf, select=c(-x,-w))
head(mydf)
On Tue, Sep 29, 2009 at 4:27 PM, Rolf Turner wrote:
>
> On 30/09/2009, at 9:15 AM, milton ruser wrote:
>
> May be this:
>>
>> FRAME <- FRAME[-c(NAME1, NAME2)]
>> or
>>
Hello,
> I'm looking for equivalent of Matematica function "Which" which works as
> follows:
>
> z = Which[x<10,0.3, 10<=x<20,0.5, 20<=x<100,1]
>
> where x is a vector
Unless someone happens to be a Mathematica user (very possible), I don't know
how we would answer the question. You give an
On 30/09/2009, at 9:15 AM, milton ruser wrote:
May be this:
FRAME <- FRAME[-c(NAME1, NAME2)]
or
FRAME<- subset(FRAME, select=(-NAME1, -NAME2))
This is ridiculous advice. Try things out before you suggest them.
cheers,
Rolf Turner
May be this:
FRAME <- FRAME[-c(NAME1, NAME2)]
or
FRAME<- subset(FRAME, select=(-NAME1, -NAME2))
bests
milton
On Tue, Sep 29, 2009 at 4:12 PM, Dennis Fisher wrote:
> Colleagues,
>
> Hopefully a simple problem: I want to delete a column with a known name
> from a dataframe. I could write:
>
>
Try this:
FRAME$NAMETODELETE <- NULL
On Tue, Sep 29, 2009 at 5:12 PM, Dennis Fisher wrote:
> Colleagues,
>
> Hopefully a simple problem: I want to delete a column with a known name from
> a dataframe. I could write:
>
> FRAME <- FRAME[, names(FRAME) != NAMETODELETE]
> or
> FRAME
Colleagues,
Hopefully a simple problem: I want to delete a column with a known
name from a dataframe. I could write:
FRAME <- FRAME[, names(FRAME) != NAMETODELETE]
or
FRAME <- FRAME[, !names(FRAME) %in% c(NAME1, NAME2, ETC)]
Is there some simpler means to accomplish this
On 29/09/2009, at 10:54 PM, Steve Jones wrote:
Hi List,
I'm trying to perform some spectral analysis on a time series, but
I've
got a few data points missing, so they have NA values. Unfortunately,
the fft function doesn't seem to like this, and gives a completely
empty
result.
Other th
On Sep 29, 2009, at 3:08 PM, Erik Iverson wrote:
Hello,
I have
srv <- Surv(sample(1:10), sample(0:1, 10, replace = TRUE))
srv
[1] 1 10 2+ 8 6+ 7+ 3 5+ 4+ 9+
srv.char <- as.character(srv)
srv.char
[1] " 1 " "10 " " 2+" " 8 " " 6+" " 7+" " 3 " " 5+" " 4+" " 9+"
Is there an inver
>
> identical(srv, ???(srv.char)) to return TRUE, where ??? is some unknown
> function. I don't think it exists, but maybe I'm wrong. I suppose it
> would be easy enough to roll my own...
>
FYI, This works in my simple case for right-censored data... it takes as input
a vector like c("1", "2+
On 30/09/2009, at 5:34 AM, chris carleton wrote:
Thanks for the response. I'm sorry I didn't provide the code or
data example earlier. I was using the polynomial fitting technique
of this form;
test <- lm(x[,34] ~ I(x[,1]) + I(x[,1]^2) + I(x[,1]^3))
for the original fitting operation. I
-Original Message-
From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
Sent: Tuesday, September 29, 2009 4:52 AM
To: Nair, Murlidharan T
Cc: r-help@r-project.org
Subject: Re: [R] error while plotting
Nair, Murlidharan T wrote:
>
> -Original Message-
> From: Uwe Ligges [
William Dunlap pisze:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Jarek Jasiewicz
Sent: Tuesday, September 29, 2009 11:36 AM
To: Erik Iverson
Cc: R-help@r-project.org
Subject: Re: [R] Equivalent for Matematica function Which
Hello,
I have
srv <- Surv(sample(1:10), sample(0:1, 10, replace = TRUE))
srv
[1] 1 10 2+ 8 6+ 7+ 3 5+ 4+ 9+
srv.char <- as.character(srv)
srv.char
[1] " 1 " "10 " " 2+" " 8 " " 6+" " 7+" " 3 " " 5+" " 4+" " 9+"
Is there an inverse to as.character(srv). That is, I would like
No. The forest object is too large as is. I didn't think it's worth
the extra memory to store them. They were never kept even in the
Fortran/C code.
Andy
From: Chrysanthi A.
> Sent: Monday, September 28, 2009 5:20 PM
> To: r-help@r-project.org
> Subject: [R] how to visualize gini coefficient
On Sep 29, 2009, at 2:36 PM, Jarek Jasiewicz wrote:
well function arguments are in square brackets. z is result (new
vector). I show Matematica syntax, but cannot explain what I expect.
Sorry
The example is wrong because it can be replaced by R cut function.
The arguments are: condition,ac
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Jarek Jasiewicz
> Sent: Tuesday, September 29, 2009 11:36 AM
> To: Erik Iverson
> Cc: R-help@r-project.org
> Subject: Re: [R] Equivalent for Matematica function Which...
>
> wel
On 30/09/2009, at 3:45 AM, Lina Rusyte wrote:
Hello,
Could someone help me please and to tell how to get the probability
from empirical DENSITY (not parametric) for each data value (R
function).
For example, for normal distribution there is such a function like:
“dnorm(q, mean = 0, sd =
Thanks for the response. I'm sorry I didn't provide the code or data example
earlier. I was using the polynomial fitting technique of this form;
test <- lm(x[,34] ~ I(x[,1]) + I(x[,1]^2) + I(x[,1]^3))
for the original fitting operation. I also tried to use;
lm(y ~ poly(x,3,raw=TRUE))
with the
Dear all
I have a data set for which PCA based between group analysis (BGA) gives
significant results but CA-BGA does not.
I am having difficulty finding a reliable method for deciding which ordination
technique is most appropriate.
I have been told to do a 1 table CA and if the 1st axis is>
well function arguments are in square brackets. z is result (new
vector). I show Matematica syntax, but cannot explain what I expect. Sorry
The example is wrong because it can be replaced by R cut function. The
arguments are: condition,action and can be replaced by ste of ifelse
formulas:
Hi,
I want to compile R with command completion. But I don't find such an
option in configure. Can somebody let me know how to enable command
completion in an R session?
Regards,
Peng
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/li
On Tue, Sep 29, 2009 at 8:05 PM, David Winsemius wrote:
>
> On Sep 29, 2009, at 1:43 PM, Carlos Hernandez wrote:
>
> Dear All,
>> I´m using the following code:
>>
>> all1<-gg2[[1]][[1]]; for(i in 1:48){ all1 <- rbind(all1,gg2[[i]][[1]]) }
>>
>
> Looks to me that you would be getting a duplicate c
On Sep 29, 2009, at 1:49 PM, Martin Batholdy wrote:
hi,
on the mac you have the opportunity to start a script directly
instead of opening it in the script editor.
So when I doubleclick on an .R - file the code is executed
immediately.
I haven't found such an option in the preferences of
One option, see ?assign and write a wrapper around it, using the "pos"
argument.
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of David Young
> Sent: Tuesday, September 29, 2009 10:58 AM
> To: r-help@r-project.org
> Subject:
well thanks, when I post the mail I thought I got too simple example
which may be really replaced by "cut", but I thought about little more:
let say:
z = Which[x<10,x/3, 10<=x<20,0.5, 20<=x<100,x^2/(x-1)]
where there are both values and formula
sorry for mismatch and thanks for quick answer
Ja
Hi,
I guess you want ?assign
See also this page for a working example,
http://wiki.r-project.org/rwiki/doku.php?id=guides:assigning-variable-names
HTH,
baptiste
2009/9/29 David Young :
> Hello All,
>
> I'm a new R user and have a question about what in SAS would be called
> macro variable sub
Hello All,
I'm a new R user and have a question about what in SAS would be called
macro variable substitution. Below is some R code that doesn't work,
but I think it will illustrate what I'd like to do.
readfunc<-function(x) {
x <<- read.table(paste(x,".csv",sep=""), header=TRUE,sep=",")
}
rea
Dear All!
I'm looking for equivalent of Matematica function "Which" which works as
follows:
z = Which[x<10,0.3, 10<=x<20,0.5, 20<=x<100,1]
where x is a vector
I can replace it with custom function with set of ifelse but I'm looking
for simpler and faster (much faster) solution
best wishes
On Sep 29, 2009, at 1:43 PM, Carlos Hernandez wrote:
Dear All,
I´m using the following code:
all1<-gg2[[1]][[1]]; for(i in 1:48){ all1 <- rbind(all1,gg2[[i]]
[[1]]) }
Looks to me that you would be getting a duplicate copy of the first
matrix, but aside from that what problems are you expe
Felix,
Thanks, that did the trick! Lattice is a lot less intuitive than basic
plotting!
Also, another person suggested using gap.plot from the plotrix package to put a
break in the graph. I am surprised Lattice doesn't have something similar
since it seems like a common problem when you have
plot(y~x, type='p', col=3, xlim=c(50,1000), ylim=c(0,80), xlab='', ylab='',
data=example)
abline(a=11, b=0.04, col = 2)
--
Thanks,
A.B.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/lis
Thank you for your quick reply! It works perfectly!
On Tue, Sep 29, 2009 at 7:51 PM, Henrique Dallazuanna wrote:
> Try this;
>
> do.call(rbind, sapply(gg2, '[', 1))
>
> On Tue, Sep 29, 2009 at 2:43 PM, Carlos Hernandez
> wrote:
> > Dear All,
> > I´m using the following code:
> >
> > all1<-gg2[[1
On Tue, Sep 29, 2009 at 7:56 AM, Douglas Bates wrote:
> An interesting, and topical, example of multivariate data for
> classroom illustrations are the American college football rankings.
> Starting at the end of October (or "week 8", the 8th week of the
> football season) a set of rankings called
On Tue, Sep 29, 2009 at 6:47 PM, chipmaney wrote:
>
> I have a dataset. Initially, it has 25 levels for a certain factor,
> Description.
>
> However, I then subset it, because I am only interested in 2 of the 25
> factors. When I subset it, I get the following. The vector lists only the
> two fa
Dear All,
I´m using the following code:
all1<-gg2[[1]][[1]]; for(i in 1:48){ all1 <- rbind(all1,gg2[[i]][[1]]) }
to create a new matrix that contains all the matrices in a list called gg2.
gg2 is a list that looks like
>> gg2
[[1]]
[[1]][[1]]
[[2]]
[[2]][[1]]
.
.
.
[[48]]
[[48]][[1]]
Is the
Try this;
do.call(rbind, sapply(gg2, '[', 1))
On Tue, Sep 29, 2009 at 2:43 PM, Carlos Hernandez wrote:
> Dear All,
> I´m using the following code:
>
> all1<-gg2[[1]][[1]]; for(i in 1:48){ all1 <- rbind(all1,gg2[[i]][[1]]) }
>
> to create a new matrix that contains all the matrices in a list call
hi,
on the mac you have the opportunity to start a script directly instead
of opening it in the script editor.
So when I doubleclick on an .R - file the code is executed immediately.
I haven't found such an option in the preferences of the R program for
windows.
Is there also a way to do s
I have a dataset. Initially, it has 25 levels for a certain factor,
Description.
However, I then subset it, because I am only interested in 2 of the 25
factors. When I subset it, I get the following. The vector lists only the
two factors, yet there remain 25 levels:
> Quadrats.df$Description
If you want to see how Venables and Ripley get their lda distances,
then this is a quick path to the (uncommented) source:
1) > methods(lda)
[1] lda.data.frame* lda.default*lda.formula*lda.matrix*
2) since you call involved a formula I looked first at:
> getAnywhere(lda.formula)
A singl
A alternative in cases where is there NA's shoul be:
sapply(sapply(df, summary), '[', 1:7)
On Tue, Sep 29, 2009 at 2:28 PM, William Dunlap wrote:
>> -Original Message-
>> From: r-help-boun...@r-project.org
>> [mailto:r-help-boun...@r-project.org] On Behalf Of Henrique
>> Dallazuanna
>> S
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Henrique
> Dallazuanna
> Sent: Tuesday, September 29, 2009 9:57 AM
> To: Ashta
> Cc: R help
> Subject: Re: [R] Summary
>
> Try this:
>
> sapply(xc, summary)
This fails if ther
Try this:
sapply(xc, summary)
On Tue, Sep 29, 2009 at 12:42 PM, Ashta wrote:
> My data is called xc and has more than 15 variables.
>
>
> When I used summary(xc) it gave me the detail description of each
> variable.
>
>
>
> Summary(xc)
>
>
>
> Y1 x1
On 9/29/2009 11:57 AM, Gábor Csárdi wrote:
On Tue, Sep 29, 2009 at 5:36 PM, Duncan Murdoch wrote:
Gábor Csárdi wrote:
Dear All,
I have the following in a .Rd file:
...
human readable (not binary) format. The format itself is like
the following:
\preformatted{
\# vertex1
Thanks David,
Yes, I am talking about the MASS package.Thank you for pointing out that
these scale the same. My question is, how do I get from the V1 data:
V1
1 164.4283
2 166.2492
3 170.5232
4 156.5622
5 127.7540
6 136.7704
7 136.3436
to the other set of data:
+ 1 -2.3769280
+ 2 -2.704
Thanks, Uwe.
You code does what I would like to have!
Best,
Mike
-Original Message-
From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
Sent: Tuesday, September 29, 2009 1:36 AM
To: Ping-Hsun Hsieh
Cc: r-help@r-project.org
Subject: Re: [R] plot error -- figure margins too large
That code is way more than a _minimal_ example, and its not
_reproducible_ either, so just a comment:
Have you considered creating a vector rather than separate "event"
variables?
R_ev[1:16] <- R_event > (rx[1] + 300*(0:15)& R_event <= (rx[1] +
300*(1:16)
?cut # would also appear to b
Dear Thierry,
Thank you very much for the fast reply.
Is there a way not to plot outliers in the ggplot2 boxplots? In a way
that the scale of the y-axis is rendered as if there was no outlier?
(not just the the ouliers removed with the same scale of the
y-axsis...)
[ example for a boxplot: qpl
How about this:
> x <- c("sadfdsaf 24353245", 'wqerwqer 6577', 'xzcv 6587')
> sub("^([^[:space:]]+)[[:space:]].*", "\\1", x)
[1] "sadfdsaf" "wqerwqer" "xzcv"
>
On Tue, Sep 29, 2009 at 12:03 PM, wrote:
> Through converting a miRNAs file from FASTA to character format I get a
> vector which lo
On 29-Sep-09 16:03:31, mau...@alice.it wrote:
> Through converting a miRNAs file from FASTA to character format I get
> a vector which looks like the following:
>
>> nml
> [1] "hsa-let-7a MIMAT062 Homo sapiens let-7a"
> [2] "hsa-let-7b MIMAT063 Homo sapiens let-7b"
Not sure exactly what you want, but does using 'ylim' do what you want:
plot(1:5, sample(100:200,5), type='b', ylim=c(0,450))
On Tue, Sep 29, 2009 at 12:08 PM, chipmaney wrote:
>
> I have read all the help on axis scales, which seems to be much harder to
> deal with than it should be, given how
On Tue, 29 Sep 2009, mau...@alice.it wrote:
Through converting a miRNAs file from FASTA to character format I get a vector
which looks like the following:
nml
[1] "hsa-let-7a MIMAT062 Homo sapiens let-7a"
[2] "hsa-let-7b MIMAT063 Homo sapiens let-7b"
[3] "hsa-let-7c MIMAT064
Hi,
On Sep 29, 2009, at 12:03 PM, wrote:
Through converting a miRNAs file from FASTA to character format I
get a vector which looks like the following:
nml
[1] "hsa-let-7a MIMAT062 Homo sapiens let-7a"
[2] "hsa-let-7b MIMAT063 Homo sapiens let-7b"
[3] "hsa-let-7c MIMAT064
I have read all the help on axis scales, which seems to be much harder to
deal with than it should be, given how common the need to alter axes.
Anywayyeah, suppress the axis using yaxt then call and define the new
scale using axis().
The problem is, this doesn't change the actual axis, only
jim holtman wrote:
Add a dummy argument to the function:
randomSamples<-lapply(1:2000,function(dummy){
+ meta_comp[sample(nrow(meta_comp),nTimes),]})
Or
df = data.frame(x=1:10, y=10:1)
replicate(10, df[sample(nrow(df), 5),], simplify=FALSE)
Martin
On Mon, Sep 28, 2009 at 9:39 PM, ewaters
Through converting a miRNAs file from FASTA to character format I get a vector
which looks like the following:
> nml
[1] "hsa-let-7a MIMAT062 Homo sapiens let-7a"
[2] "hsa-let-7b MIMAT063 Homo sapiens let-7b"
[3] "hsa-let-7c MIMAT064 Homo sapiens let-7c"
On Tue, Sep 29, 2009 at 5:36 PM, Duncan Murdoch wrote:
> Gábor Csárdi wrote:
>>
>> Dear All,
>>
>> I have the following in a .Rd file:
>> ...
>> human readable (not binary) format. The format itself is like
>> the following:
>> \preformatted{
>> \# vertex1name
>> verte
something like this should work (not exactly sure what you mean by 'compare')
result <- lapply(List1, function(.lst1){
lapply(List2, function(.lst2, .lst1){
all(.lst2 %in% .lst1)
}, .lst1 = .lst1)
})
On Tue, Sep 29, 2009 at 10:34 AM, Christina Rodemeyer
wrote:
> Hi guys,
>
> I
I have the following code which I wanted to convert using for loop
previous code:
R_ev1 <- R_event[R_event > (rx[1] + 300*0)& R_event <= (rx[1] + 300*1)]
R_ev2 <- R_event[R_event > (rx[1] + 300*1)& R_event <= (rx[1] + 300*2)]
R_ev3 <- R_event[R_event > (rx[1] + 300*2)& R_event <= (rx[1] + 300*3)]
My data is called xc and has more than 15 variables.
When I used summary(xc) it gave me the detail description of each
variable.
Summary(xc)
Y1x1 x2
x3 ..
Min. :0. Min. : 1.000 Min. : 1.000 Min. : 1.000
1st Qu
1 - 100 of 150 matches
Mail list logo
