I think you are looking for this:
x=rnorm(6,10,1)
e=rnorm(6,0,1)
y=x+e
plot(y~x,xlim=c(min(x)-2,max(x)+2),ylim=c(min(y)-2,max(y)+2))
text(x,y,pos=1,labels=c("Prof","CEO","Janitor","Admin","Farmer","Fire
Chief"))
HTH,
Daniel
-
cuncta stricte discussurus
Dear R-users,
I am using the simple plot() function and i cannot find a way to add the point
label (i would like to plot the points and up to each point the point name). I
found a way to do it with n.plot but then the point is not plotted, only the
label. Do you know if there is a parameter for
Thanks Jim,
Further from this; the data I have doen't appear to be that complete for
some species e.g
result$MOO
1992 2000 2002
3060 1300 1300
How can I change this list into a time series so that I can create a
meaningful plot e.g.
1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2
Hi Roslina,
I believe that you can ignore the warning.
Alternatively, you may add a very small random noise to pairs with ties, i.e.
something like
xobs[which(duplicated(xobs))] <- xobs[which(duplicated(xobs))] +
1.0e-6*sd(xobs)*rnorm(length(which(duplicated(xobs
Regards,
Moshe.
--- On Tu
On Oct 12, 2009, at 10:10 PM, Yan Li wrote:
Hi, there,
I have 5 datasets. I would like to choose a basis spline with same
knots in
GAM function in order to obtain same basis function for 5 datasets.
Moreover, the basis spline is used to for an interaction of two
covarites.
It is possibl
I am sure that there is a much easier way of doing this, but this is
what I have figured out so far. I would like to run a script that
downloads USGS data and parses it, and run this with as a cron process
every night at x o'clock and append the output to a file.
Can I run it like the following:
On 13/10/2009, at 2:55 PM, joshua wells wrote:
I'm using the function gee from the library(gee)
gee(Y~X,id=clust.id,corstr="exchangeable",b=tmc$coef,family=binomial
(link=logit),silent=T)
Every time it runs, it dutifully prints out
Beginning Cgee S-function, @(#) geeformula.q 4.13 9
joshua wells wrote:
>
>
> Unfortunately, i run thousands of these and the output gets extremely
> lengthy. Is there any way to suppress this printout in R?
>
> Thank you,
>
> Josh
>
>
One function that comes to mind is sink() which allows you to red
On Oct 12, 2009, at 8:20 PM, Jason Gasper wrote:
Greetings,
I have what I hope is a simple question. I would like to change my
contour interval on the vis.gam( plot.type="contour") in the mgcv
package. Is this a situation where I need to modify the function or
is there a default value
Hi, there,
I have 5 datasets. I would like to choose a basis spline with same knots in
GAM function in order to obtain same basis function for 5 datasets.
Moreover, the basis spline is used to for an interaction of two covarites.
I used "cr" in one covariate, but it can only smooth w.r.t 1 covari
wk yeo wrote:
>
>
> Hi, all,
>
> My objective is to split a dataframe named "cmbine" according to the value
> of "classes". After the split, I will take the first instance from each
> class and bin them into a new dataframe, "df1". In the 2nd iteration, I
> will take the 2nd available instanc
On Oct 12, 2009, at 7:06 PM, Stropharia wrote:
Hi everyone,
i'm having a problem extracting objects out of functions i've
created, so i
can use them for further analysis. Here's a small example:
# ---
test <- function(i, j){
x <- i:j
y <- i*j
Thanks a lot Tony and Daniel for making that clear.
best,
Steve
Stropharia wrote:
>
> Hi everyone,
>
> i'm having a problem extracting objects out of functions i've created, so
> i can use them for further analysis. Here's a small example:
>
> # ---
> test <- functi
Hi there,
Can anyone suggest some packages in R doing variable selections in
predictive modeling besides randomForest? Faster, better.
Any also in clustering analysis?
Thanks,
Weiwei
--
Weiwei Shi, Ph.D
Research Scientist
GeneGO, Inc.
"Did you always know?"
"No, I did not. But I believed..."
I'm using the function gee from the library(gee)
gee(Y~X,id=clust.id,corstr="exchangeable",b=tmc$coef,family=binomial(link=logit),silent=T)
Every time it runs, it dutifully prints out
Beginning Cgee S-function, @(#) geeformula.q 4.13 98/01/27
user's initial regression estimate
wk yeo wrote:
>
>
> Hi, all,
>
> My objective is to split a dataframe named "cmbine" according to the value
> of "classes". After the split, I will take the first instance from each
> class and bin them into a new dataframe, "df1". In the 2nd iteration, I
> will take the 2nd available instanc
Hi, all,
My objective is to split a dataframe named "cmbine" according to the value of
"classes". After the split, I will take the first instance from each class and
bin them into a new dataframe, "df1". In the 2nd iteration, I will take the 2nd
available instance and bin them into another new
Greetings,
I have what I hope is a simple question. I would like to change my
contour interval on the vis.gam( plot.type="contour") in the mgcv
package. Is this a situation where I need to modify the function or is
there a default value I can change?
Thanks
_
Hi,Can I use "reorder" function with barchart as in dotchart? Here are some
codes which do not work for me. Thanks
Chetty
___
a1c.cast$bmi.cat.reordered[a1c.cast$eth!="Other"]
<-with(a1c.cast[a1c.cast$eth!="Other",],reorder(bmi.cat.ordered[a1c.cast$eth!="Other"],
BP.FN.RATE
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Stropharia
> Sent: Monday, October 12, 2009 4:07 PM
> To: r-help@r-project.org
> Subject: [R] Re use objects from within a custom made function
>
>
> Hi everyone,
>
> i'm havin
test$x doesn't evaluate the function, you want something like test(1,2)$x, e.g.:
test <- function(i, j){
x <- i:j
y <- i*j
z <- i/j
return(list(x=x,y=y,z=z))
}
test(1,2)$x
[1] 1 2
test(1,2)$y
[1] 2
test(1,2)$z
[1] 0.5
Or if you want to avoid evaluating y
Hi everyone,
i'm having a problem extracting objects out of functions i've created, so i
can use them for further analysis. Here's a small example:
# ---
test <- function(i, j){
x <- i:j
y <- i*j
z <- i/j
return(x,y,z)
}
# -
Hi r-users,
I would like to use Kolmogorov smirnov test but in my observed data(xobs) there
are ties. I got the warning message. My question is can I do something about
it?
ks.test(xobs, xsyn)
Two-sample Kolmogorov-Smirnov test
data: xobs and xsyn
D = 0.0502, p-value = 0.924
al
Helmer,
You can just place another boxplot on top of the first like this:
boxplot(x)
axis(2,tck=1,col.ticks='grey',lty=5 )
boxplot(x, col="white", add=TRUE)
-Peter Ehlers
Helmer Belbo wrote:
Dear R people,
I wonder how to hide tick lines behind other figures in a plot, e.g.
in a boxpl
I'd like to generate on a single device multiple plots, each of which
contains two plots. Essentially, I've got sub-plots which consist of
two tracks, the upper one displaying gene expression data, and the
lower one mapping position. I'd like to display four of these
two-track sub-plots on one devi
On Mon, Oct 12, 2009 at 4:03 PM, Robert Kern wrote:
> On 2009-10-12 15:45 PM, Peng Yu wrote:
>>
>> Hi,
>>
>> I'm wondering what is the general way to define __hash__. I could add
>> up all the members. But I am wondering if this would cause a
>> performance issue for certain classes.
>
> Unless if
Hi George,
Your problem is not with xyplot, but with the NA occurrences in
your data. Try adding
subset = {!is.na(MSE)},
to your xyplot call, or (better), subset the data before
calling xyplot.
-Peter Ehlers
George Kalema wrote:
Hi Peter (and anyone else willing to help me out),
Many than
One other way that is sometimes useful is to use the
tick.number argument as in:
xyplot(yield ~ nitro, data=Oats,
scales=list(x=list(tick.number=4)), subset=Variety=="Victory"
)
This is especially handy if you want to just tick/label
every other value.
-Peter Ehlers
Jacob Wegelin wrote:
Try: mapply(..., SIMPLIFY = FALSE)
On Mon, Oct 12, 2009 at 5:25 PM, Magnus Torfason wrote:
> I have a list of isometric structures, and I want to change the same part of
> each structure in the list, assigning one element of a vector to each of
> these parts.
>
> In other words, I want to achiev
Hmmm, after thinking about it, better see
?mapply
SIMPLIFY logical; attempt to reduce the result to a vector or
matrix?
Try mapply( ..., SIMPLIFY=FALSE )
HTH,
Chuck
On Mon, 12 Oct 2009, Magnus Torfason wrote:
I have a list of isometric structures, and I wan
On Mon, 12 Oct 2009, Magnus Torfason wrote:
I have a list of isometric structures, and I want to change the same part of
each structure in the list, assigning one element of a vector to each of
these parts.
In other words, I want to achieve the following:
l <- list( list(a=1,b=2), list(a=3
I have a list of isometric structures, and I want to change the same
part of each structure in the list, assigning one element of a vector to
each of these parts.
In other words, I want to achieve the following:
> l <- list( list(a=1,b=2), list(a=3,b=4))
> unlist(lapply(l, "[[", "a"))
[1] 1 3
Hi Catherine,
Assuming your variables are in a dataframe called myData, some
variation of the following may be what you want:
library(ggplot2)
myData.m <- melt(myData, measure.vars=c("Y1", "Y2"))
qplot(X, value, colour=variable, shape=groupf3,
facets=groupf1~groupf2, geom="point", data=myData.m)
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of eugen pircalabelu
> Sent: Monday, October 12, 2009 1:06 PM
> To: David Winsemius
> Cc: R-help
> Subject: Re: [R] crosstabulation and unlist function
>
> Hello,
> First of all, th
Instead of creating a lot of variables, work with a list. Since you
did not supply any test data, I will make a guess at the code would be
like:
result <- lapply(split(newdataset, newdataset$target_species), function(.spec){
tappy(.spec$total_hook_num, .spec$year, sum)
})
You can summarize f
Dear everyone,
sorry to bother you. I have a big data, suppose it has 200 groups and each
group has 100 data. So the data have 2 observations in total. The
variables are
groupf1 groupf2 groupf3 X Y1 Y2
1 11 1 0.5 0.5
groupf1, groupf2 and groupf3 are
Dear R users,
The code below splits a dataset by "target_species" and then further splits
each target species group by year. Finally the sum of a variable called
"total_hook_num" is calculated and then assigned a name based on
(paste(a$target_species[1],a$year[1],sep="").
newsplit<-split(newd
What you're really saying is that you don't care about the distinction between
"aa", "bb" and "cc". In that case, a different arrangement of the data will be
more useful:
library (reshape )
df.melt <- melt ( df, id.var = "dd")
with ( df.melt, table ( dd, value ) )
Eric
- Original messag
try this: (?get)
j<- get(paste("list",substring(input,3,3),sep=""))
On Mon, Oct 12, 2009 at 4:14 PM, Maxwell Reback wrote:
> Hi all. I'd like to define an object within a function based on an
> argument to that function.
> Specifically, I've got:
>
> do.something<-function(input){
> id<-substri
On 13/10/2009, at 9:14 AM, Maxwell Reback wrote:
Hi all. I'd like to define an object within a function based on an
argument to that function.
Specifically, I've got:
do.something<-function(input){
id<-substring(input,3,3)
j<-list1
if(id==2)j<-list2
if(id==3)j<-list3
if(id==4)j<-list4
...}
In
>> What principle is at work?
>
> A strange one called "standard non-standard evaluation"; see
>
> http://developer.r-project.org/nonstandard-eval.pdf
>
> for a nice overview by Thomas Lumley.
>
>
> ?xyplot says:
>
> data: For the 'formula' method, a data frame containing values (or
>
Hi all. I'd like to define an object within a function based on an
argument to that function.
Specifically, I've got:
do.something<-function(input){
id<-substring(input,3,3)
j<-list1
if(id==2)j<-list2
if(id==3)j<-list3
if(id==4)j<-list4
...}
Instead of all these if() arguments, I was hoping to us
Hello,
First of all, thank you David for your reply, but sadly this is not what i
wanted (i am sorry for not being more specific about my problem!)
aa<-c(1:5)
bb<-c(NA,2,NA,4,5)
cc<-c(1,2,NA,4,NA)
dd<-c("A","B","B","A","C")
table(unlist(df[,1:3]))
> df
aa bb cc dd
1 1 NA 1 A
2 2 2
On Oct 12, 2009, at 3:25 PM, David Winsemius wrote:
On Oct 12, 2009, at 2:36 PM, eugen pircalabelu wrote:
Hello R-users,
My toy example:
aa<-c(1:5)
bb<-c(NA,2,NA,4,5)
cc<-c(1,2,NA,4,NA)
dd<-c("A","B","B","A","C")
df<-data.frame(aa,bb,cc,dd=as.factor(dd))
table(unlist(df[,1:3]))
Can anyone
On Oct 12, 2009, at 2:36 PM, eugen pircalabelu wrote:
Hello R-users,
My toy example:
aa<-c(1:5)
bb<-c(NA,2,NA,4,5)
cc<-c(1,2,NA,4,NA)
dd<-c("A","B","B","A","C")
df<-data.frame(aa,bb,cc,dd=as.factor(dd))
table(unlist(df[,1:3]))
Can anyone point me to what function let's me do a crosstabulation
Others have show how to use the segments function, but this can also be done
using the original abline function along with the clip function. Here is an
example:
plot( iris$Petal.Width, iris$Petal.Length,
col=c('red','green','blue')[iris$Species])
tmp <- levels(iris$Species)
tmp2 <- p
Dear R people,
I wonder how to hide tick lines behind other figures in a plot, e.g.
in a boxplot.
# Sample code:
x<- c(rep(4,50),rep(5,20),rep(6,50),rnorm(20,5,1))
boxplot(x)
axis(2,tck=1,col.ticks='grey',lty=5 )
# end of sample code
The tick lines is put on top of the box-plot, but I would li
The problem is that ifelse does not work the way you might think (see
value section of ?ifelse) and basically should not be used with three
zoo objects unless the three arguments to ifelse have the same time
index.
We can get that effect by using na.pad = TRUE in your diff call:
TradedRate <-
On 10/12/2009 08:30 PM, Deepayan Sarkar wrote:
On Mon, Oct 12, 2009 at 9:32 AM, Jacob Wegelin wrote:
When we call a lattice function such as xyplot, to what extent does
the "data" designation cause the function to look inside the "data"
for variables?
In the examples below, the "subset" argume
?write.table is probably a good starting point.
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of ws
> Sent: Monday, October 12, 2009 1:39 PM
> To: r-h...@stat.math.ethz.ch
> Subject: [R] Using tabs in output instead of spaces
>
Is there a way to have xtabs and friends use tabs in their output to separate
columns rather than spaces? It would be great for importing into other
software.
TIA!
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEA
Thanks Tony (and others). Setting oma corrects the problem. Mark
Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry
Indiana University School of Medicine
15032 Hunter Court, Westfield, IN 46074
(317) 490-5129 Work, & Mobile & VoiceMail
(317) 399-1219 Skype No Voicemail please
On Mon
Hello R-users,
My toy example:
aa<-c(1:5)
bb<-c(NA,2,NA,4,5)
cc<-c(1,2,NA,4,NA)
dd<-c("A","B","B","A","C")
df<-data.frame(aa,bb,cc,dd=as.factor(dd))
table(unlist(df[,1:3]))
Can anyone point me to what function let's me do a crosstabulation between
table(unlist(df[,1:3])) and df$dd?
I wan
On Mon, Oct 12, 2009 at 9:32 AM, Jacob Wegelin wrote:
> When we call a lattice function such as xyplot, to what extent does
> the "data" designation cause the function to look inside the "data"
> for variables?
>
> In the examples below, the "subset" argument understands that
> "Variety" is a vari
Try dir.create
On Mon, Oct 12, 2009 at 3:00 PM, Peng Yu wrote:
> Hi,
>
> Besides calling shell command mkdir by system(), I'm wondering if
> there is a buildin command in R to make a new directory.
>
> Regards,
> Peng
>
> __
> R-help@r-project.org maili
?dir.create
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Peng Yu
> Sent: Monday, October 12, 2009 1:01 PM
> To: r-h...@stat.math.ethz.ch
> Subject: [R] mkdir in R?
>
> Hi,
>
> Besides calling shell command mkdir by system(
Deepayan will correct me if I'm wrong, but I'm pretty sure that the answer
is that it looks in the frame in the data argument only for variables in the
formula argument. Note that the fact that it also works for the subset
argument is explicitly mentioned therein:
subset: logical or integer index
Interesting, I hadn't tried this but it probably explains why
navigation to different regions of a layout is neither documented nor
advisable.
Yet another alternative is to use Grid graphics. In particular,
1- lattice or ggplot2 provide ways to arrange several plots in a
rectangular layout, with
Hi,
Besides calling shell command mkdir by system(), I'm wondering if
there is a buildin command in R to make a new directory.
Regards,
Peng
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PLEASE do read the posting gu
gauti wrote:
>
>
> I have been doing series of linear regression models lm(). In this case
> the execution time and memory usage becomes a huge issue. I have therefore
> been trying to speed the process and limit the memory usage.
>
>
Have a look at package biglm.
Dieter
--
View this m
This only works if all the plots are the same size and the defaults are used
for the margins. Try it with different sized figure regions in layout, the
added lines don't match at the end.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.4
drlucyasher wrote:
>
>
> The questionnaire has a section which contains a particular issue and then
> questions which are related to this issue (and potentially to each other):
> 1) importance of the issue (7 ordinal categories from -3 to +3)
> 2) impact of the impact (7 ordinal categroies fro
Since you are using LaTeX, you might consider creating the plots all in 1 step,
but use the tikzdevice, this provides LaTeX (pgf) commands to create the plot
that you can then insert LaTeX commands to built up the plot bit by bit.
You could also look at using the subplot function in the Teachin
Hey Mark,
The text is actually there -- I can just see the bottom of the 'y' and
the 'p' in my plotting window. You can move the text down (into the
plot) with the argument line. E.g.:
mtext(text = "my test plots", side = 3, outer = TRUE, line=-2)
Hope that helps...
-
Simon Bonner
Post-Doctor
Mark Kimpel wrote:
>
> I'm trying to use mtext to create a main title over multiple plots. Below
> is
> a simple self-contained example and my sessionInfo (I should note I've
> also
> tried this with R-2.8.1 with the same results). When I execute the code
>
> ...
>
>
>
Thanks for your nice
On Oct 12, 2009, at 1:41 PM, Tony Plate wrote:
Try playing around with the "oma" setting in par() -- it sets the
outer margins, which by default are zero.
The following shows the mtext label for me, using the windows device:
par(mfrow=c(2,2))
par("oma")
[1] 0 0 0 0
par("oma"=c(0,0,2,0))
Try playing around with the "oma" setting in par() -- it sets the outer
margins, which by default are zero.
The following shows the mtext label for me, using the windows device:
par(mfrow=c(2,2))
par("oma")
[1] 0 0 0 0
par("oma"=c(0,0,2,0))
for (i in 1:4) plot(0:1,0:1)
mtext(text = "my test
I'm trying to use mtext to create a main title over multiple plots. Below is
a simple self-contained example and my sessionInfo (I should note I've also
tried this with R-2.8.1 with the same results). When I execute the code
chunk below, I get the plots, but no title. I've tried this using the scre
On Oct 12, 2009, at 5:24 AM, sdlywjl666 wrote:
Dear all£¬
Is there some functions to estimate the spectrum by the fft of
autocorrelation?
The blindingly obvious search terms: fft spectrum
autocorrelation ... in two hops to what appears to be a function to
this:
Auto And/Or Cross
Dear all,
I would like to visualise when days are rainy or dry in bar/boxplots.
Therefore I've tried to combine raingauge data (boxplots) and percentage
of raingauges with 0mm measurements (barplot). See attachment
mei2004.pdf (if it came through).
I've come this far:
barplot(dcp0[monthindex], c
When we call a lattice function such as xyplot, to what extent does
the "data" designation cause the function to look inside the "data"
for variables?
In the examples below, the "subset" argument understands that
"Variety" is a variable in the data.
But the "scales" argument does not understand t
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of tdm
> Sent: Monday, October 12, 2009 12:22 AM
> To: r-help@r-project.org
> Subject: Re: [R] field index given name.
>
>
> Thanks - would never have guessed that. I eventually go
You seem to have gotten an extraneous list item called "Chromosome" to
which an empty string has been assigned. What happens if you issue
this command:
ann2 <- data.frame( Name=ann$Name, Position=ann$Position,
Chr.num=factor(ann$Chr.num) )
(I took the liberty of making Chr.num into a fact
fft, specturm, and there is an auto correlation function (I don't
remember which one).
On Mon, Oct 12, 2009 at 4:24 AM, sdlywjl666 wrote:
> Dear all，
> Is there some functions to estimate the spectrum by the fft of
> autocorrelation?
> Is there Parzen's lag window in R ?Thank you!
> Best wish
I have been asked to analyse some questionnaire data- which is not data I'm
that used to dealing with. I'm hoping that I can make use of the nabble
expertise (again).
The questionnaire has a section which contains a particular issue and then
questions which are related to this issue (and potentia
Hi there,
I need to build up a tktext-widget that contains a longer text than
the tktext-widget actually is. So what I mean is, that the tktext
window is of width=100 and the text in it has a length greater 100.
But I don't want the window to just wrap the line, but to belong to
scrollba
When i tried running Zip<-sqldf("select a.*,b.s_lv from Zip a inner join lin
b on a.S=B.S")
Error: cannot allocate vector of size 15.6 Mb
and with following warning
Warning messages:
1: In as.list.data.frame(X) :
Reached total allocation of 1024Mb: see help(memory.size)
2: In as.list.data.frame
As the error says, you have different row numbers in your variables.
The variable $Chromosome has no values.
try :
ann <- data.frame( ann [-3] )
Cheers
Joris
On Mon, Oct 12, 2009 at 8:29 AM, Ilyas . wrote:
> i have two RData files,,i want to print them to check the format of the
> tables in the
Hi all
I have been doing series of linear regression models lm(). In this case the
execution time and memory usage becomes a huge issue. I have therefore been
trying to speed the process and limit the memory usage.
Here follows part of the code do give better understanding of what I am d
Hi,
I'm having an issue when using diff and ifelse on a zoo object.
x.Date <- as.Date("2003-02-01") + c(1, 3, 7, 9, 14) - 1
x <- zoo(rnorm(5), x.Date)
x.POS <- c(0,0,0,1,1)
x<- merge(x,x.POS)
x
x x.POS
2003-02-01 -0.1858136 0
2003-02-03 -1.3188533
Dear all£¬
Is there some functions to estimate the spectrum by the fft of
autocorrelation?
Is there Parzen's lag window in R ?Thank you!
Best wishes!
Wang
[[alternative HTML version deleted]]
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Hi Peter (and anyone else willing to help me out),
Many thanks for your help. Having used your code plus a few other
modifications, I only get the points plotted but without the two lines. I
just cannot figure out what the problem is.
My code is as follows:
library(lattice)
datos2 <- subset(datos
i have two RData files,,i want to print them to check the format of the
tables in these files,,,i can load both the files and can read it as well
> load('ann.RData')
> str(ann)
List of 4
$ Name : chr [1:561466] "rs3094315" "rs12562034" "rs3934834"
"rs9442372" ...
$ Position : int [1:561466
Hi, I would like to be able to develop a function to plot an igraph object
with lattice (trellis type displays will be usefull for grouping etc).
Anyway, I mostly feeble
Igraph requires that you convert two columns of data two an igraph object
and to be able to plot the graph...I have tried a ve
Sample Size
λ=5     α =
4      β = 3
Min.
1St Qu.
Median
Mean
3rd Qu.
Max.
100
0.00
1.740638
4.040032
4.433828
5.607589
22.450405
5
To save for posterity, I include an email from Joris who's advice solved my
problem. Lession learned is to check your data before passing it along to a
function.
Cheers,
Roman
-- Forwarded message --
From: joris meys
Date: Mon, Oct 12, 2009 at 4:01 PM
Subject: Re: [R] Error: cann
Dear List,
I am trying to connect from R 2.9.2 on Win XP SP3 to a remotely
installed PostgreSQL DB (8.3.7 on Ubuntu Server 9.04). Everything
seems to be properly installed, as I can connect to the DB from within
Excel and RKWard (running on another machine).
But regarding R on the Win XP, I
Hi, Michael,
Seems like all you need is aggregate and rbind:
x <- aggregate(saw.aggr.data["value"],
saw.aggr.data[c("conversion.type", "filteredID", "metric")],
sum)
x$bucketID <- "combined"
y <- rbind(saw.aggr.data, x)
Is this what you need?
--sundar
On Mon, Oct
That's the file I needed, and the problem I expected. The factor you
specify is not a factor, but a numerical variable. Even more, if you
tabulate it, you have 2 times the values "1", "2" and "5", and all
other values only once. That's what gives you the error.
p is an internal variable of the fun
On 12-Oct-09 13:33:17, Ted Harding wrote:
> On 12-Oct-09 13:24:01, Terry Therneau wrote:
>> -- begin included
>> Greetings!
>> I want to follow the evolution of a Nelder-Mead function
>> minimisation (a function of 2 variables). Hence each simplex
>> will have 3 vertices.
>>
>> Th
Vibe Henriette Skov wrote:
> Dear R-mailing list
>
> Hope you can help me. I am using R for windows to analyze my 107
> HGU133Plus2.0 chips, however, R chrash when I try to use ReadAffy(). I
> want to buy a computer that can handle all these arrays, do you know how
> big a computer I need to buy?
On 12-Oct-09 13:24:01, Terry Therneau wrote:
> -- begin included
> Greetings!
> I want to follow the evolution of a Nelder-Mead function
> minimisation (a function of 2 variables). Hence each simplex
> will have 3 vertices.
>
> Therefore I would like to have a function which can o
I think you are missing the point. You have 4 zero death counts
associated with much higher person years of exposure followed by 4
death counts in the thousands associated with lower degrees of
exposures. It seems unlikely that these are real data as there are not
cohorts that would exhibit
-- begin included
Greetings!
I want to follow the evolution of a Nelder-Mead function
minimisation (a function of 2 variables). Hence each simplex
will have 3 vertices.
Therefore I would like to have a function which can output
the coordinates of the 3 vertices after each new simp
Dear Peter,
Thanks for the input. The zero rates in some strata occurs because
sampling depended on case status: In Finland only 50% of the non-cases
were sampled, while all others were sampled with 100% probability.
Best
Laust
On Sat, Oct 10, 2009 at 11:02 AM, Peter Dalgaard
wrote:
> Sorry, fo
Dear R-mailing list
Hope you can help me. I am using R for windows to analyze my 107
HGU133Plus2.0 chips, however, R chrash when I try to use ReadAffy(). I
want to buy a computer that can handle all these arrays, do you know how
big a computer I need to buy?
Best,
Skov, Denmark
[[altern
Hi All,
I have a questions about associative list mappings in R, and if they are
possible?
I have data in the form show below, and want to make a new 'bucket' variable
called combined. Which is the sum of the control and the exposed metric
values
This combined variable is a many to many matching
I'm basically put off by the question itself. Plotting a 4-dimensional
graph is rather "complicated" if the world has only 3 dimensions. A
4-dimensional representation is typically a movie (with time as the
4th dimension). You could try to project a heatmap on a 3D surface
graph, but I doubt this w
On Oct 12, 2009, at 4:07 AM, ogbos okike wrote:
Good morning to you. I have about 4 different lines in one plot. I
have used
legend to indicate the colour of each plot. But the box contain the
legend
covers part of the lines thereby blurring the legend. There are some
spaces
in the plot th
On Oct 12, 2009, at 3:22 AM, tdm wrote:
Thanks - would never have guessed that. I eventually got the
following to do
what I want...
colprob <- array(dim=NCOL(iris))
for(i in 1:NCOL(iris)){
+ colprob[i]=
+ ifelse(names(iris)[i] == 'Species',1,0.5)
+ }
colprob
[1] 0.5 0.5 0.5 0.5 1.0
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