On 02/09/2011 06:49 PM, Jim Lemon wrote:
On 02/08/2011 09:14 PM, anoopmj wrote:
Hi,
I am trying to plot several columns in different graphs in columns and
rows.
The first column of my data file is the time, and the third is the
'elevation angle' with 19 different numbers. I would like to plot
On 02/08/2011 09:14 PM, anoopmj wrote:
Hi,
I am trying to plot several columns in different graphs in columns and rows.
The first column of my data file is the time, and the third is the
'elevation angle' with 19 different numbers. I would like to plot the rest
of the columns with the x axis as
Hello, Maria,
take a look at
?errbar
in the package Hmisc.
Hth -- Gerrit
Dear all
I have a dataset of how metal concentrations change through time. I have
made a plot of date versus metal concentration. However I want to add
error bars in the plot.
Could you help me?
Thanks
Maria
On Tue, Feb 08, 2011 at 07:50:26PM -0800, Frenando wrote:
>
> Hello World!
> I'm working on my thesis right now (something about financial immunization)
> I'm currently working in the basics, doing a matrix that lists the present
> value (or weight) of every combination of coupon rate and term to
Dear all
I have a dataset of how metal concentrations change through time. I have made a
plot of date versus metal concentration. However I want to add error bars in
the plot.
Could you help me?
Thanks
Maria
[[alternative HTML version deleted]]
___
Hi all,
I'm trying to add updated data to an existing time series where an
overlap exists. I need to give priority to the update data.
My script runs every morning to collect data the updated data. There
are quite often varied lengths, so once off solutions identifying rows
to solve this example wo
Thanks for the reply.
The original test was without the quote. The quote come into play when I try to
solve this problem changing from " to '. However, after restarting my PC the
as.Date(36525, origin="1900-01-01")
gaves me the correct output
[1] "2000-01-02"
After loading several libraries I
On Tue, Feb 8, 2011 at 11:43 PM, Gabor Grothendieck
wrote:
> On Tue, Feb 8, 2011 at 4:12 PM, Anthony Lawrence Nguy-Robertson
> wrote:
>> I am interested in testing two similar nls models to determine if the lines
>> are statistically different when fitted with two different data sets; one
>> corn
On Wed, 02-Feb-2011 at 04:10AM -0800, kv wrote:
|>
|> Is there any reason to expect a problem ?
|> i'm running this script on the cluster down the hall:
I don't think multicore is what you need. It's for a single multicore
machine. It works fine on my dual-core machine. For a cluster, you
nee
Hi Mark,
If you have a function that, for a given file name, returns the data
in the way you want, then you can use 'dir' or 'paste' to generate a
list of names, which you can read in in one go with apply:
my_read <- function(filename)
{
...
}
filenames <- dir(directory,pattern)
filenames <- pa
Hello World!
I'm working on my thesis right now (something about financial immunization)
I'm currently working in the basics, doing a matrix that lists the present
value (or weight) of every combination of coupon rate and term to maturity,
this is the code I have right now which is giving me a "
Thank you for your kindness, but ive done what you've said and the problem
remains. What im doing is pretty straightforward,
>data
response pred1 pred2
1 1 01
2 0 00
3 1 00
4 1 11
5 1 01
6 0 11
7 1 1
Hi R Users,
Thanks in advance.
I am using R-2.12.1 on Windows XP.
May I request you to assist me for the following please.
Is there any R-package or R Script to extract content(s) from web.
Once again, thank you very much for the time you have given.
Regards,
Deb
[[
On Tue, Feb 8, 2011 at 4:12 PM, Anthony Lawrence Nguy-Robertson
wrote:
> I am interested in testing two similar nls models to determine if the lines
> are statistically different when fitted with two different data sets; one
> corn, another soybean. I know I can do this in linear models by testing
Exactly right. I use the phrase to catch the unwary's attention. I
think the effect is properly placed on the y-axis.
IIRC, Ben Bolker (or was it Bert Gunter?) has also commented in the R-
help or r-devel pages this curious inversion of functional meaning.
--
David
On Feb 8, 2011, at 10:36
David,
Please allow me to digress a lot here. You are one of the few (inlcuding yours
truly!) that uses the phrase "shallow learning curve" to indicate difficulty of
learning (I assume this is what you meant). I always felt that "steep learning
curve" was incorrect. If you plotted the amount
On Feb 8, 2011, at 9:11 PM, kirtau wrote:
I am working on a function that will remove outliers for regression
analysis.
I am stating that a data point is an outlier if its studentized
residual is
above or below 3 and -3, respectively. The code below is what i have
thus
far for the funct
Hi!
Here is a newbie question, please: what is the difference between a generic
and a method?
Thanks,
Laura
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read
Hi,
On Tue, Feb 8, 2011 at 6:18 PM, Robinson, David G wrote:
> I am experimenting with parallel processing using foreach and seem to be
> missing something fundamental. Cool stuff. I've gone through the list and
> seen a couple of closely related issues, but nothing I've tried seems to
> work.
>
Dr. Fox,
Thank you for the reply. The discrepancy between lm and the sem package as
well as AMOS makes sense given how the se's are calculated for linear
models. This is a small dataset with non-normal data (part of the reason for
my last comment) which might contribute to the relatively large disc
I'm working on getting this to work - need to figure out how to extract
pieces properly.
In the mean time, I may have figured out an alternate method to group
the factors by the following:
> stems139$SpeciesF <- factor(stems139$Species)
> stems139GLM <- glm(Stems ~ Time*SizeClassF*Species, f
I am experimenting with parallel processing using foreach and seem to be
missing something fundamental. Cool stuff. I've gone through the list and
seen a couple of closely related issues, but nothing I've tried seems to
work.
I know that the results from foreach are combined, but what if there is
I am working on a function that will remove outliers for regression analysis.
I am stating that a data point is an outlier if its studentized residual is
above or below 3 and -3, respectively. The code below is what i have thus
far for the function
x = c(1:20)
y = c(1,3,4,2,5,6,18,8,10,8,11,13,14
Bert,
Thanks for the input. I was hoping for an easy answer, but as life is,
there usually isn't one. I will find a statistician here on campus that
might be able to help. Just so you know, the data is remote sensing data
that is an average of 9 measurements on one day. However, the data set
Quick and dirty answer:
Try ?paste in conjunction with a "for" loop which iterates over
i:length(Files), and you can paste "i" onto the end of your filename, then
add the data to cd=list(), cd[i], then "unlist" the data or
as.data.frame(cd).
I'm sure there are more elegant alternatives.
Ross
--
Dear Ned,
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Ned Dochtermann
> Sent: February-08-11 6:07 PM
> To: r-help@r-project.org
> Subject: [R] SEM: question regarding how standard errors are calculated
>
> Sorry if this qu
Hi,
I've got the following code which seems to work fine for a single
file if I specify the file name explicitly in the code. What I need to
do is run it on all the files in the directory tested and augment the
data frame I'm building to have more results columns.How can I do
that?
Here's th
Hello,
I don't really understand you question but if you want to run the same
code four or five times on the same dataset you could write it into a
for loop where yourread in your incomplete dataset back in at the
beginning. A better practice is to change the name of a dataset when
you make chang
Nonlinear models are an entirely different kettle of fish then linear
models. You need to specify exactly how the different crops affect the
parameters in your growth model. I suggest you consult a local
statistician for help (this is not an R question).
Incidentally, depending on the nature of yo
Sorry if this question has been asked previously, I searched but found
little. There also doesn't seem to be a dedicated SEM list-serv so hopefully
this will find its way to the appropriate audience.
In discussing SEM with a colleague I mentioned that a model they were
fitting in AMOS was equiv
For the sake of simplicity I've made your data example
idd d1 d2 d3 d4
1 a d b c
2 a d v h
3 c b v NA
4 q v NA f
df <- read.table('clipboard', header = TRUE,na.strings="NA")
df
Dear Andrew Halford
Here just merely a suggestion of R newbie without sufficient statistical
background
MCMCpoisson from package MCMCpack
Could you please lately post the answer to your very interesting
question when you will find it.
Wit best regards
Denis
У Аўт, 08/02/2011 у 14:38
Hey thanks for all the help,
I did end up using layout, par, mar to reduce the labelling area and combine
it with other plots.
I also now understand NULL was the wrong argument to place in the field.
Thanks again
--
View this message in context:
http://r.789695.n4.nabble.com/Removing-X-and-Y
Federico Bonofiglio gmail.com> writes:
>
You're confused by the formatting differences betwen
columns. The first column uses 'standard' notation, the
second and third use 'scientific' notation -- see comments
inline.
> > GLU<-lme(gluc~rt*cd4+sex+age+rf+nadir+pharmac+factor(hcv)+factor(hbs)+
xyplot is a lattice plotting command, text is a base plotting command. The 2
types don't play well together without extra work. The base command plot with
text is probably the easiest, or you can just use plot and pch:
> with(iris, plot(Sepal.Width, Sepal.Length, pch=c('s','e','i')[Species] )
Have a look at
?par
especially the "mar" argument, to get rid off the boundaries reserverd
for labelling.
Am 08.02.2011 21:36, schrieb poolmunch:
>
> Hello, had a quick search on the site but no luck.
>
> Heres my problem, when attempting to xlab = NULL and ylab = NULL it displays
>
> x[[1L]]
If you want to expand the area that the graph takes up (using the space that
the labels would have been in) look at the "mar" part of ?par.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-h
Hi:
One approach would be to use dlply() from the plyr package to generate the
models and assign the results to a list, something like the following:
library(plyr)
# function to run the GLM in each data subset - the argument is a generic
data subset d
gfun <- function(d) glm(Stems ~ Time, data =
Mike,
On Feb 8, 2011, at 3:24 PM, Mike Williamson wrote:
> Hello,
>
>This is mostly to developers, but in case I missed something in my
> literature search, I am sending this to the broader audience.
>
>
> - Are there any plans in the works to make "time" classes a bit more
> friendly
On Feb 8, 2011, at 4:54 PM, Nicolas Gutierrez wrote:
Hi All,
I'm trying to delete a row from my dataframe "pop" without changing
the indexing (column 0) as follows:
>pop
id birth size xloc yloc weight energy gonad consumed
1 136 13 34 43 0 18 00
2 2
What happens if you use the "newdata" argument name instead of "data"?
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On
There's a much shorter way. You don't need that ugly h() with all those $
and potential for bugs !
Using the original f :
dt[,lapply(.SD,f),by=key(dt)]
grp1 grp2 grp3 a b d
xxx 1.00 81.00 161.00
xxx 10.00 90.00 17
On Feb 8, 2011, at 3:36 PM, poolmunch wrote:
Hello, had a quick search on the site but no luck.
Heres my problem, when attempting to xlab = NULL and ylab = NULL it
displays
Those are the default settings:
?plot.default
... , xlab = NULL, ylab = NULL, ...
xlab a label for the x
Contrast the behaviour of these two statements:
plot(x,y,xlab=NULL,ylab=NULL)
plot(x,y,xlab='',ylab='')
In other words, use xlab='' to supress the label, not NULL.
- Phil Spector
Statistical Computing Facility
Nicolas -
I don't think it can be done automatically, but
you can use
row.names(pop) = 1:nrow(pop)
after deleting the column(s) to restore consecutive
numbers for the row names.
- Phil Spector
Statistical Comp
Hi,
there is nothing to complain?!
0.0148 < 6.24e-02=0.0624 < 1.1e-01 = 0.11
same for the 5th line.
so I didn't get the point.
Am 08.02.2011 22:14, schrieb Federico Bonofiglio:
> Hi folks...
>
> check this out..
>
>> GLU<-lme(gluc~rt*cd4+sex+age+rf+nadir+pharmac+factor(hcv)+factor(hbs)+
> + h
maxsilva uc.cl> writes:
>
>
> Thank you for your answer. But I still have the problem; for example, if i
> have data for 10 months, estimate the parameters of my logit model using 10
> months of data, and then use
>
> predictions<-predict(model,data=1monthonly,family = binomial(link =
> logit)
On Feb 8, 2011, at 3:42 PM, Schatzi wrote:
I would like to create a plot of y vs x with different treatments
where the
points are actually the letter of the treatment. Here is the code:
A<-as.matrix(rnorm(10,10))
B<-as.matrix(rnorm(10,9.5))
C<-as.matrix(rnorm(10,10.5))
Y<-as.matrix(rnorm(3
Hi All,
I'm trying to delete a row from my dataframe "pop" without changing the
indexing (column 0) as follows:
>pop
id birth size xloc yloc weight energy gonad consumed
1 136 13 34 43 0 18 00
2 236 10 39 38 0 18 00
3 3
Mike -
apply turns its argument into an array (in this case a
matrix), which can't accomodate dates. To operate on the
rows of a data frame, your best bet is probably to use a
for loop. There's no reason to dread loops in R -- just try
to avoid growing objects inside a loop by preallocatin
> Date: Tue, 8 Feb 2011 17:34:15 +
> From: marie.guil...@bordeaux.inra.fr
> To: r-help@r-project.org
> Subject: [R] Recuperate Spectrum() amplitude
>
> Dear list,
> I put the code I use here to understand if the difference comes from the
>
anna freni sterrantino yahoo.it> writes:
> I've got this error while running
>
> example(Glm)
>
> library("rms")
> > example(Glm)
[snip]
> Glm> f <- Glm(counts ~ outcome + treatment, family=poisson())
> Error in Design(eval(mf, parent.frame())) :
> dataset dd not found for options(datadist
Hi:
Try some modification of the following toy example:
df <- data.frame(names = as.character(replicate(100, paste(sample(letters,
3), collapse = ''))),
x1 = rnorm(100), x2 = rnorm(100))
tgts <- sample(df$names, 20)# target list of names
df[df$names %in% tgts, ] # yiel
On Feb 8, 2011, at 4:14 PM, Federico Bonofiglio wrote:
Hi folks...
check this out..
GLU<-lme(gluc~rt*cd4+sex+age+rf+nadir+pharmac+factor(hcv)
+factor(hbs)+
+ haartd+hivdur+factor(arv),
+ random= ~rt|id, na.action=na.omit)
intervals(GLU)$fixed
lower est.
in ?plot it says that type ="n" displays everything but the actual points, so
that you can display the characters on top using the text function
--
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Sent from the R help mailin
Here's an example:
x = c(1,2,3,4,5);
y = rnorm(5);
labels = LETTERS[1:5];
plot(x,y, type = "n") # This sets up the plot but doesn't actually
plot the points
text(x,y, labels) # This adds labels to positions (x,y)
HTH,
Peter
On Tue, Feb 8, 2011 at 12:42 PM, Schatzi wrote:
>
> I would like to
xyplot(Y~X,groups = TRT,type=n) to plot the space but not the points,
then use the text function to display the characters.
Try ?text in R
--
View this message in context:
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Sent from the R help mailing l
I am interested in testing two similar nls models to determine if the
lines are statistically different when fitted with two different data
sets; one corn, another soybean. I know I can do this in linear models
by testing for interactions. See Introductory Statistics with R by
Dallgaard p212-21
Hello, had a quick search on the site but no luck.
Heres my problem, when attempting to xlab = NULL and ylab = NULL it displays
x[[1L]] and X[[2L]] as labels on the axis.
plot(USSenate[,1:2],col=c("blue","grey","red")[unclass(USSenate$Senate)],pch=unclass(USSenate$Senate),
type="n",xaxt="n", y
It is a lot prettier than mine too.
Thanks Jari.
> rrarefy
function (x, sample)
{
if (length(sample) > 1 && length(sample) != nrow(x))
stop("length of 'sample' and number of rows of 'x' do not match")
sample <- rep(sample, length = nrow(x))
colnames(x) <- colnames(x, do.NULL
Thank you for your answer. But I still have the problem; for example, if i
have data for 10 months, estimate the parameters of my logit model using 10
months of data, and then use
predictions<-predict(model,data=1monthonly,family = binomial(link =
logit),type="response")
I still get the rpedict
Prof Brian Ripley wrote:
>
> Well, no, it doesn't (it plots on the screen device). So exactly how
> are you producing the plot?
>
I am trying to produce a visualization of the character network, like this:
> plot(g, layout=layout.fruchterman.reingold, vertex.color="black",
> vertex.size=2.0,
Hi folks...
check this out..
> GLU<-lme(gluc~rt*cd4+sex+age+rf+nadir+pharmac+factor(hcv)+factor(hbs)+
+ haartd+hivdur+factor(arv),
+ random= ~rt|id, na.action=na.omit)
> intervals(GLU)$fixed
lower est. upper
(Intercept) 67.3467070345 7.362307e+01
Dear Marius,
Try this:
plot.list = lapply(1:10,
function(i) xyplot(i~i,type="p",xlim=c(0,11),panel=function(...) {
panel.xyplot(...); panel.abline(v=i)})
)
plot.list[[3]]
I imagine it will work for Mr Luftjammer, too.
Rex
-Original Message-
From: r-help-boun...@r-project.org [ma
try xts package and functions:
reclass
period.max
period.min
best,
daniel
2011/2/8 Mike Williamson
> Hello,
>
>This is mostly to developers, but in case I missed something in my
> literature search, I am sending this to the broader audience.
>
>
> - Are there any plans in the works to ma
I'm having a hard time figuring out how to group results by certain
factors in R. I have data with the following headings:
[1] "Time" "Plot" "LatCat""Elevation" "ElevCat"
"Aspect""AspCat""Slope"
[9] "SlopeCat" "Species" "SizeClass" "Stems"
and I'm trying to use a G
try the code below,
if you use different upper and lower limits you might get different parameter
estimates
HTH, Victor
standard.curve<-data.frame(conc=c(50, 25, 12.5, 6.25, 3.125, 1.563, 0.781,50,
25, 12.5, 6.25, 3.125, 1.563, 0.781),absorb=c(1.918, 1.251, 1.104, 0.719,
0.403, 0.177, 0.083,1.9
Hi dears,
I do
> CHOL<-lme(chol~rt*cd4+sex+age+rf+nadir+pharmac+factor(hcv)+factor(hbs)+
haartd+hivdur+factor(arv),
random= ~rt|id, na.action=na.omit)
...runs sweet,..then
try a multicomparisons approach for the categorical rf
> summary(glht(CHOL, linfct=mcp(rf="Tukey")))
*
Error in model
I would like to create a plot of y vs x with different treatments where the
points are actually the letter of the treatment. Here is the code:
A<-as.matrix(rnorm(10,10))
B<-as.matrix(rnorm(10,9.5))
C<-as.matrix(rnorm(10,10.5))
Y<-as.matrix(rnorm(30,13))
X<-rbind(A,B,C)
nA<-matrix("A",10,1)
On 2/8/2011 3:13 PM, Duncan Murdoch wrote:
expression(paste(italic( "Q. marilandica "), " volume"))
Thank you.
--
"If people sat outside and looked at the stars each night, I'll bet they'd live a
lot differently."-
Bill Watterson
__
R-help@r-project
Great. Thanks. Here is the new code:
xyplot(Y~X,groups = TRT,type="n")
text(x,y,labels=TRT)
--
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Sent from the R help mailing list archive at Nabble.com.
___
On 08/02/2011 4:07 PM, john polo wrote:
Hello useRs,
I'm trying to put italics and regular font into a title and am having
some trouble. The title I would like is " /Q. marilandica/ Volume ".
Use
expression(paste(italic( "Q. marilandica "), " volume"))
Duncan Murdoch
If I use
main=bquote(i
Hello useRs,
I'm trying to put italics and regular font into a title and am having
some trouble. The title I would like is " /Q. marilandica/ Volume ".
If I use
main=bquote(italic( "Q. marilandica ", " volume"))
I get just " Q. marilandica " in italics in the title.
If i use
main=bquote(italic
Phil,
Thanks! I am not sure why min & max sometimes work & other times don't
work. In fact, running the script as you had it did work for me. For
arrays that I had, it wasn't working and still isn't. For the moment, I
won't worry about it. It seems maybe it is a hidden bug somewhere (or I
You are comparing B[1] to A[1]^2, then B[2] to A[2]^2 etc. so for the AB pairs
you are doing the comparison. If that is what you want, then it is fine.
But if you want to compare the 1st B to all the A's, then the 2nd B to all the
A's, etc. Then it does not do this (but the expand.grid functi
How do you anticipate the number of doctors affecting the proportions?
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On
Dennis Murphy gmail.com> writes:
>
> Hi:
>
> Did you try putting the offset in the model formula, as in
>
> bigglm( y ~ offset(z) + x, ...) ?
>
> I haven't tried bigglm() personally (BTW, it's in the biglm package, which
> wasn't mentioned), but this syntax works in the standard glm() fu
On 9/02/2011 6:25 a.m., David Winsemius wrote:
On Feb 8, 2011, at 11:43 AM, kateF87 wrote:
I have a relatively simple question.
I am trying to post a title to a plot using a symbol and multiple
lines.
Right now I have:
title(main = c('Hazard Ratio for women with score', expression('>='),
'A
Yep, that finally does it. Many thanks, Rex.
Cheers,
Marius
On 2011-02-08, at 21:11 ,
wrote:
> Dear Marius,
> Try this:
>
> plot.list = lapply(1:10,
>
>function(i) xyplot(i~i,type="p",xlim=c(0,11),panel=function(...) {
>panel.xyplot(...); panel.abline(v=i)})
> )
> plot.list[[3]]
>
Hello,
This is mostly to developers, but in case I missed something in my
literature search, I am sending this to the broader audience.
- Are there any plans in the works to make "time" classes a bit more
friendly to the rest of the "R" world? I am not suggesting to allow for
fancy
On Feb 8, 2011, at 2:35 PM, Mauro Ferreira wrote:
Dear members,
I would like a help for extracting the values from a step function
(stepfun).
From help(stepfun) we have the following example:
Y0<-c(1.,2.,4.,3.)
y0<-c(1.,2.,3.,4.)
sfun<-stepfun(1:3,y0,f=0)
plot(sfun)
Now, suppose instead I
Look at ?par and specifically the ask option.
Or, you can use the pdf function to send a set of graphs directly to a pdf
file, then open the pdf file and step through (and go back if you want) the
graphs.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s..
Hi:
Did you try putting the offset in the model formula, as in
bigglm( y ~ offset(z) + x, ...) ?
I haven't tried bigglm() personally (BTW, it's in the biglm package, which
wasn't mentioned), but this syntax works in the standard glm() function, so
perhaps it maps to bigglm() as well... (?)
maxsilva uc.cl> writes:
> I generated a glm logit model using a sample of a bigger set, and now I want
> to test the model with the bigger set. How can I do this? predict makes
> predictions only for a number of rows equal to the sample used. The question
> is, how can I use the coefficients of
On 2011-02-08 11:17, wang peter wrote:
my program run well on my Pc
but when run one the server(linux),it meet the error
Error in match.fun(FUN) : could not find function "Negate"
Is Negate your own function or is it the function in
the base package? If the latter, check your R versions
(which
Another approach (still using the theory of sufficient statistics) is to
generate data from a normal distribution that matches exactly the sizes, means,
and standard deviations that you have, then analyze the simulated data.
The mvrnorm function in the MASS package can generate data with a given
Hi,
I have found Pinheiro and Bates an absolute godsend for mixed modelling,
http://www.amazon.com/Mixed-Effects-Models-S-S-Plus/dp/0387989579
alternatively, there is a REALLY good chapter in the 3rd edition of DAAG:
http://www.amazon.com/Data-Analysis-Graphics-Using-Example-Based/dp/05217629
On 11-02-07 19:21, Sascha Vieweg wrote:
Hello, knowing that some index y, with y(341)=2, SE=3, is t-distributed, I
(think I) can find an appropriate (left/one-sided) p-value with
R: y <- 2
R: yse <- 3
R: (p <- 1-pt(y/yse, df=341))
Now, some simulation resulted in the non-parametric distrib
On Feb 8, 2011, at 1:17 PM, wang peter wrote:
> my program run well on my Pc
> but when run one the server(linux),it meet the error
> Error in match.fun(FUN) : could not find function "Negate"
> thx
> peter
The Negate() function was introduced in base R version 2.7.0, which was
released about t
Dear R people,
I generated a glm logit model using a sample of a bigger set, and now I want
to test the model with the bigger set. How can I do this? predict makes
predictions only for a number of rows equal to the sample used. The question
is, how can I use the coefficients of a glm with other s
Hi all, thank you for your patience.
I am dealing with a large dataset detailing patients and medications
Medications are hard to code, as they are (usually) meaningless unless
matched with doses.
I have a dataframe with vectors (Drug1, Drug2. Drug 16) and individual
patients are represente
Hi,
This was a terrific suggestion. I'm no expert in R, but I managed to find and
read the axis command to do the following.
x=c(rep(2,14),rep(4,13),rep(6,11),rep(8,4),rep(10,3),rep(12,2),14,16,18)
stripchart(x,method="stack",pch=21,at=0,bg="lightblue",col="blue",offset=.5,cex=1.5,xlab="Bag
B"
my program run well on my Pc
but when run one the server(linux),it meet the error
Error in match.fun(FUN) : could not find function "Negate"
thx
peter
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__
R-help@r-project.org mailing list
https://stat.ethz.ch
Dear members,
I would like a help for extracting the values from a step function
(stepfun).
>From help(stepfun) we have the following example:
Y0<-c(1.,2.,4.,3.)
y0<-c(1.,2.,3.,4.)
sfun<-stepfun(1:3,y0,f=0)
plot(sfun)
Now, suppose instead I was given the object (*sfun*, say) from which I
wanted
On Tue, Feb 8, 2011 at 11:29 AM, Phil Spector wrote:
> Mark -
> Here's a few possibilites:
>
>> dts = c('6/10/2009 10:04:00 AM','6/15/2009 9:47:00 AM','6/15/2009 9:47:00
>> AM')
>> as.Date(sapply(strsplit(dts,' '),'[',1),'%m/%d/%Y')
>
> [1] "2009-06-10" "2009-06-15" "2009-06-15"
>>
>> as.Date(su
On Tue, Feb 8, 2011 at 11:30 AM, Prof Brian Ripley
wrote:
> Try as.Date() with a suitable format (it only knows about internationally
> standard formats), e.g. maybe you mean
>
>> as.Date("6/10/2009 10:04:00 AM", format="%m/%d/%Y")
>
> [1] "2009-06-10"
>
Thank you!
- Mark
__
Try as.Date() with a suitable format (it only knows about
internationally standard formats), e.g. maybe you mean
as.Date("6/10/2009 10:04:00 AM", format="%m/%d/%Y")
[1] "2009-06-10"
On Tue, 8 Feb 2011, Mark Knecht wrote:
I have hundreds of CSV files coming in from another program that have
Hi,
either upscale the circles (increase cex) or narrow your plot, as in
x11(3,5)
stripchart(x,method="stack",pch=21,at=0,bg="lightblue",col="blue",offset=.5,cex=1.5,xlab="Bag
B",frame.plot=FALSE,axes=FALSE)
there's a trade off between 'closing the gap' and having all your ticks
readable, so just
Mark -
Here's a few possibilites:
dts = c('6/10/2009 10:04:00 AM','6/15/2009 9:47:00 AM','6/15/2009 9:47:00 AM')
as.Date(sapply(strsplit(dts,' '),'[',1),'%m/%d/%Y')
[1] "2009-06-10" "2009-06-15" "2009-06-15"
as.Date(sub('(\\d+/\\d+/\\d+) .*','\\1',dts),'%m/%d/%Y')
[1] "2009-06-10" "2009-06
On Tue, 8 Feb 2011, apepe wrote:
Hi,
I have read some of the documentation relative to character encodings and
non-standard fonts (including previous answered questions and the 2006-2 R
issue), but I am still struggling with very basic plotting of Chinese text.
I am using R for Mac OS X GUI 1
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