On Mon, 12-Oct-2009 at 10:54AM +1100, Steven Kang wrote:
| Hi all,
|
| I require 2 RMySQL libraries in order to query from a database.
You mean you require 2 RMySQL packages. We must be pedantic to answer
your question.
|
| 'RMySQL_0.7-4' (newest version) results in an error when more than 1
And try also search
statsrus
The first hit shall be Paul Johnsons's howto's which helped me several
years ago especially with basic issues.
Regards
Petr
r-help-boun...@r-project.org napsal dne 10.10.2009 18:00:19:
I'm just learning R (I don't know any other programming languages),
Hi,
How do I access the index number of a field given I only know the field
name?
eg - I want to set the probability of the field 'species' higher than the
other fields to use in sampling.
colprob - array(dim=NCOL(iris))
for(i in 1:NCOL(iris)){colprob[i]=0.5}
colprob[iris$species] = 1 #this
On Mon, 12 Oct 2009, Patrick Connolly wrote:
On Mon, 12-Oct-2009 at 10:54AM +1100, Steven Kang wrote:
| Hi all,
|
| I require 2 RMySQL libraries in order to query from a database.
You mean you require 2 RMySQL packages. We must be pedantic to answer
your question.
|
| 'RMySQL_0.7-4' (newest
Hi Phil
Try the following
which(names(iris)=='Species')
[1] 5
HTH
Schalk Heunis
On Mon, Oct 12, 2009 at 8:53 AM, tdm ph...@philbrierley.com wrote:
Hi,
How do I access the index number of a field given I only know the field
name?
eg - I want to set the probability of the field 'species'
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
Schalk Heunis-2 wrote:
Hi Phil
Try the
Hi R-users,
I would like to ask question related to error output.
If an error comments come out, then the program will automatically stop.
I want to ask , how I can still continue the program even though there is an
error comment?
var=VAR(Canada,p=3,type=const)
for (j in 1:nrow(com))
{
This may be of some help:?try
On Mon, Oct 12, 2009 at 9:42 AM, Arif Chandra arif.chan...@hotmail.comwrote:
Hi R-users,
I would like to ask question related to error output.
If an error comments come out, then the program will automatically stop.
I want to ask , how I can still continue the
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 that are empty and large enough to accommodate the legend
Hi R Users, When I use package nlme for linear model with random
effects, there exists errors and I don't know the data structure of lme.
Here is my data:
Rice-data.frame(Yield=c(8,7,4,9,7,6,9,8,8,8,7,5,9,9,5,7,7,8,8,8,4,8,6,4,8,8,9),
You did not specify the random effects. Try something like
Rice.lme-lme(Yield ~ Variety + Stand,data=Rice, random = ~1|Block)
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research
I write to thank all those who have helped me to fix this problem.
Best regards
Ogbos
2009/10/7 ogbos okike ogbos.ok...@gmail.com
Good morning. I wish to plot two data on the same axis. I tried plot(x,y,
type = l) for the first and tried to use lines or points(x,y, lty = 2, col
= 4) to add or
Dear List,
today I turn to you with a next problem. I'm trying to compare species
richness between various datasets (locations) using species accumulation
curves (Chapter 4, page 54 in Tree diversity
analysishttp://www.worldagroforestry.org/treesandmarkets/tree_diversity_analysis.aspby
Kindt
On 10/12/2009 01:53 AM, zhijie zhang wrote:
Thanks. I think there may be no easy method to achive it.
library(lattice)
barchart(Titanic, scales = list(x = free),auto.key = list(title
=Survived),layout=c(4,1),horizontal = FALSE)
The above method generates four graphs, two graphs in the left
one example is to plot two vectors against each other and then, to plot the CI
(confidence interval) of each point on the same graphics. The first could be
done by plot and the second by plotCI in gplots package but how to plot on the
same R graphic device?
Look forward to your reply,
--- On
On 10/12/2009 08:47 PM, carol white wrote:
one example is to plot two vectors against each other and then, to plot the CI
(confidence interval) of each point on the same graphics. The first could be
done by plot and the second by plotCI in gplots package but how to plot on the
same R graphic
Dear Roman,
could you give us the trace given by traceback() ? I suspect the error
is resulting from the permutations and/or jackknife procedure in the
underlying functions specaccum and specpool.
You can take a look at the package R.huge, but that one is deprecated
already. There are other
Hi,
I need to compare non-linear fittings of 2 different experimental
distributions. I use nls() to the fit the 2 distribution and it works
pretty well.
The statistical comparison of 2 non linear fits is well described in
the book Fitting Models to Biological Data using Linear and
Hello joris,
this is the traceback() output. Hopefully you can make some sense out of it.
Thank you for the tips as well (R.huge looks promising)!
traceback()
7: vector(integer, length)
6: integer(nbins)
5: tabulate(bin, pd)
4: as.vector(data)
3: array(tabulate(bin, pd), dims, dimnames = dn)
2:
Hi Roman,
that throws a different light on the problem. It goes wrong from the
start, so it has little to do with the bootstrap or jackknife
procedures. R.huge won't help you either.
Likely your error comes from the fact that factor1 is not an
argument of the function accumcomp. the argument is
Hi Joris,
thanks for spotting that one. This little mistake has gotten in when I was
trying desperate things with the analysis (factor1 is used in diversitycomp).
Nevertheless, here is the result:
poacc2 - accumcomp(PoCom, y=PoEnv, factor=HM_sprem, method=exact)
Error in if (p == 1) { :
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
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
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
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
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
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
-- 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
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
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 output
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?
Hi
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 function
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,
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
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
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] 742429
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 -
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 0
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
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
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
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
-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 got the
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 that nitro
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],
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
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
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 plots,
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))
for
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 example
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
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
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 from -3
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
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 message
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 mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting
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
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
?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(), I'm
Try dir.create
On Mon, Oct 12, 2009 at 3:00 PM, Peng Yu pengyu...@gmail.com 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
__
On Mon, Oct 12, 2009 at 9:32 AM, Jacob Wegelin jacob.wege...@gmail.com 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
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 want to find out
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,
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
?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
On 10/12/2009 08:30 PM, Deepayan Sarkar wrote:
On Mon, Oct 12, 2009 at 9:32 AM, Jacob Wegelinjacob.wege...@gmail.com 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 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 -
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
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 -
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
between
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 point me to
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 2 B
3 3 NA
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 use
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
more
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
...}
Instead
try this: (?get)
j- get(paste(list,substring(input,3,3),sep=))
On Mon, Oct 12, 2009 at 4:14 PM, Maxwell Reback mreb...@gmail.com 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){
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 message -
-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, thank you
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)
-Ista
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
#
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),
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
Try: mapply(..., SIMPLIFY = FALSE)
On Mon, Oct 12, 2009 at 5:25 PM, Magnus Torfason zulutime@gmail.com 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
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:
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
On Mon, Oct 12, 2009 at 4:03 PM, Robert Kern robert.k...@gmail.com 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.
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
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
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
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)
}
#
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)$z
[1] 0.5
Or if you want to avoid evaluating your
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Stropharia
Sent: Monday, October 12, 2009 4:07 PM
To: r-help@r-project.org
Subject: [R] Re use objects from within a custom made function
Hi everyone,
i'm having a
Hi,Can I use reorder function with barchart as in dotchart? Here are some
codes which do not work for me. Thanks
Chetty
___
a1c.cast$bmi.cat.reordered[a1c.cast$eth!=Other]
-with(a1c.cast[a1c.cast$eth!=Other,],reorder(bmi.cat.ordered[a1c.cast$eth!=Other],
BP.FN.RATE,median
Greetings,
I have what I hope is a simple question. I would like to change my
contour interval on the vis.gam( plot.type=contour) in the mgcv
package. Is this a situation where I need to modify the function or is
there a default value I can change?
Thanks
Hi, all,
My objective is to split a dataframe named cmbine according to the value of
classes. After the split, I will take the first instance from each class and
bin them into a new dataframe, df1. In the 2nd iteration, I will take the 2nd
available instance and bin them into another new
wk yeo wrote:
Hi, all,
My objective is to split a dataframe named cmbine according to the value
of classes. After the split, I will take the first instance from each
class and bin them into a new dataframe, df1. In the 2nd iteration, I
will take the 2nd available instance and bin
I'm using the function gee from the library(gee)
gee(Y~X,id=clust.id,corstr=exchangeable,b=tmc$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
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...
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 - function(i,
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
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