But ...
On Mon, Nov 19, 2012 at 9:23 PM, Jeff Newmiller
wrote:
> Try using the lm function:
... Better yet, post to a statistics list, like
stats.stackexchange.com, as the questions appear primarily
statistical, and not R related.
Incidentally, loess() is specifically designed to deal with
"out
You can use ave(), like this:
ke$maxa <- ave(ke$a, as.factor(ke$patid), FUN=max)
greetings,
remko
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
No.
---
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Try using the lm function:
?lm
---
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On 11/19/2012 04:21 PM, Andrea Spano wrote:
Hello the list,
As a simple example:
rm(list = ls())> setClass("tre", representation(x="numeric"))> setMethod("show", "tre", def = function(object) cat(object@x))[1] "show">
setMethod("summary", "tre", def = function(object) cat("This is a tre of v
In your case j$y are the original Y values (response), not the fitted
values. Try
plot(j$x,fitted(j))
On 19-Nov-12, at 7:20 PM, eric wrote:
Not sure what I'm doing wrong. Can't seem to get loess values. It
looks like
loess is returning the same values as the input.
j <-loess(x1$total~a
Hello,
Your problem is not reproducible.
Regards,
Pascal
Le 20/11/2012 12:20, eric a écrit :
Not sure what I'm doing wrong. Can't seem to get loess values. It looks like
loess is returning the same values as the input.
j <-loess(x1$total~as.numeric(index(x1)
plot(x1$total,type='l', ylab='M
You might want to check out the bootstrap package.
Also consider clarifying what you want to bootstrap ...mec or vec or what
Lastly, it is not clear what you mean when you say ...
and I have the next errors:
ro 12 = ro (mec,vec)
ro 34 = ro (alg,ana)
ro 35 = ro (alg,sta)
ro 45 = ro (ana,sta
Not sure what I'm doing wrong. Can't seem to get loess values. It looks like
loess is returning the same values as the input.
j <-loess(x1$total~as.numeric(index(x1)
plot(x1$total,type='l', ylab='M coms/y global',xlab='')
lines(loess(total~as.numeric(index(x1)),x1))
The plot statement works fine
Hi Rui, how are you?
I want to thank you for your assistance again.
I'm sorry, but the code that you provided for me did not work this time.
The code and the warning message are both below:
> NSErr <- t(matrix(NSEr)) # which is a 1x1000 double matrix
> Vsim[] <- Vsim[ , NSErr > 0.6]
Warning me
Why don't you try clicking on the "Help" link at the top of their site?
You can even google "github for dummies" to great success as well ...
-steve
On Mon, Nov 19, 2012 at 6:07 PM, Muhuri, Pradip (SAMHSA/CBHSQ)
wrote:
>
> Hello,
>
> I would like to learn how to set up Github/repository and upl
Just curious, once you have a model that works well, does it make sense to then
tune it against 100% of the dataset (with known outcomes)
so you can apply it to data you wish to predict for or is that a bad approach?
I have done like is explained in this thread many times, taken a sample,
learn
Hi!
It seems the data file wasn't transmit. Please find a copy in attachment.
Best,
ge
On 11/19/2012 09:02 AM, Terry Therneau-2 [via R] wrote:
> I can't reproduce the problem.
>
> Tell us what version of R and what version of the survival package.
> Create a reproducable example. I don't know
Hello the list,
As a simple example:
> rm(list = ls())> setClass("tre", representation(x="numeric"))>
> setMethod("show", "tre", def = function(object) cat(object@x))[1] "show">
> setMethod("summary", "tre", def = function(object) cat("This is a tre of
> value ", object@x, "\n"))Creating a ge
On Nov 19, 2012, at 5:33 PM, Georges Dupret wrote:
> Hi David,
>
> Sorry for the signature files... this is automatic. I should disable that.
>
> Please find in attachment a copy of small.csv.gz
I found it but I suspect nobody else will. I think Terry Therneau already got a
copy. when you att
On Nov 19, 2012, at 6:07 PM, Muhuri, Pradip (SAMHSA/CBHSQ) wrote:
>
> Hello,
>
> I would like to learn how to set up Github/repository and upload/update files
> and am looking for "Github for Dummies". Any help will be appreciated.
Wrong list for this question.
--
David Winsemius, MD
Alam
Hello,
Why don't you search on Internet?
Regards,
Pascal
Le 20/11/2012 10:57, Muhuri, Pradip (SAMHSA/CBHSQ) a écrit :
Hello,
I am an Intro-level R and ggplot2 user and looking for resources to self teach
dynamic report generation in R using knitr. Any advice would be highly
appreciated.
T
Hello,
I would like to learn how to set up Github/repository and upload/update files
and am looking for "Github for Dummies". Any help will be appreciated.
Thanks,
Pradip
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Hello,
I am an Intro-level R and ggplot2 user and looking for resources to self teach
dynamic report generation in R using knitr. Any advice would be highly
appreciated.
Thanks,
Pradip
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Hello,
Just a note, you can (should?) have an argument na.rm in your function
definition with a small modification, like this:
co.var <- function(x,na.rm=TRUE) 100*(sd(x,na.rm=na.rm)/mean(x,na.rm=na.rm))
Then you can choose to use the default TRUE or not.
Hope this helps,
Rui Barradas
Em 19
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
RMW
On Mon, Nov 19, 2012 at 5:27 PM, swiss_guy
wrote:
> *Hi
>
> I have problem with plot.xts .
> I try to subset some data in a xts time series.*
>
>
>
> *subseting works fore more that one event*
>
> *But I re
I believe that you need to consult a local statistician, as there are
likely way too many statistical issues here that you do not fully
understand. Alternatively, try posting to a statistical list like
stats.stackexchange.com, as I think most of your issues are primarily
statistical, not R related.
HI,
No problem.
You got two NA in the previous example.
According to the coefficient of variaion documentation in R
(http://hosho.ees.hokudai.ac.jp/~kubo/Rdoc/library/raster/html/cv.html)
Compute the coefficient of variation (expressed as a percentage). If there is
only a single value, sd is NA
Fantastic, thank you!
On Mon, Nov 19, 2012 at 3:44 PM, arun wrote:
> HI,
>
> Your example dataset is in unreadable format. You could use dput().
> set.seed(5)
> mat1<-matrix(sample(c(1:9,NA),20,replace=TRUE),ncol=5)
> rowleyi<-data.frame(mat1)
> co.var<-function(x) 100*(sd(x,na.rm=TRUE)/mean
HI,
Your example dataset is in unreadable format. You could use dput().
set.seed(5)
mat1<-matrix(sample(c(1:9,NA),20,replace=TRUE),ncol=5)
rowleyi<-data.frame(mat1)
co.var<-function(x) 100*(sd(x,na.rm=TRUE)/mean(x,na.rm=TRUE))
rowleyi<-data.frame(subspecies=rep(LETTERS[1:2],2),rowleyi)
wit
The key is to not change the margins, set them once and stick with those
margins. The next question then becomes "how do I leave area at the
top/bottom for the title and common axis?" to which the answer is "Set
outer margins at the beginning".
Modifying your code:
y<-rnorm(1:100)
x<-rnorm(1:100
genome.wustl.edu> writes:
>
> I created a 3d scatter plot and am trying to change the color of outer box
> lines with box3d.
>
> Anybody can help me to figure out how to do this?
>
> My example is:
>
> library(scatterplot3d)
> x=seq(1:6)
> y=seq(7:12)
> z=x*2
> scatterplot3d(x, y,z)
This
On Nov 19, 2012, at 1:25 PM, Sam Steingold wrote:
> Thanks Steve,
> what is the analogue of .N for min and max?
?seq
> i.e., what is the data.table's version of
> aggregate(infl$delay,by=list(infl$share.id),FUN=min)
> aggregate(infl$delay,by=list(infl$share.id),FUN=max)
> DT[, list( max(v)),
On 12-11-19 12:20 PM, x...@genome.wustl.edu wrote:
I created a 3d scatter plot and am trying to change the color of outer box
lines with box3d.
Anybody can help me to figure out how to do this?
My example is:
library(scatterplot3d)
x=seq(1:6)
y=seq(7:12)
z=x*2
scatterplot3d(x, y,z)
See ?scat
HI,
For the first part, may be this helps:
set.seed(5)
mat1<-matrix(sample(c(1:9,NA),20,replace=TRUE),ncol=5)
rowleyi<-data.frame(mat1)
co.var<-function(x) 100*(sd(x,na.rm=TRUE)/mean(x,na.rm=TRUE))
apply(rowleyi,2,function(x) co.var(x))
# X1 X2 X3 X4 X5
#53.29387 49.
arun4 gmail.com> writes:
> I am using rbetabinom ( to generate beta binomial random variables) function
> available in the "emdbook"package written by Professor. Ben Bolker for my
> research study.
> I have no questions with this function. However, I am looking for the
> theoretical method/algori
Christophe Genolini u-paris10.fr> writes:
>
> Hi the list,
>
> I am a member of the organizing comity of the French Statistics
> Association (SFdS)'s conference. We
> are looking for sponsors. Some software (SAS, RITME, ...)
> are represented. Do you know if there is
> any possibility to be
Hi Cyril,
please let me know the following details:
- sessionInfo() output from R
- Version of XLConnect you are using
- Version of rJava you are using
- Version of Java you are using (complete output of "java -version" on the
command line)
- Value of the JAVA_HOME environment variable if it is se
Hi everyone,
I'm fairly new to R, and I don't have a background in statistics, so
please bear with me. ;-)
I'm dealing with 2^k factorial designs, and I was just wondering if
there's any way to analyze more than two factors of a gage R&R study in
R. For example, Minitab has an "expanded gage R&R"
Trying to do a meta-analysis of correlations in R using the meta package; have
tried several things and keep getting a similar error. Can anyone help explain
the error?
> cor<-c(-0.3018, 0.667, -3.8002, -0.607, -0.4885, -3.8002, -0.0701, 0.1348,
> -0.9505, -0.5709, -0.6127, -1.2419, -0.1511, -0
Hello helpers,
I have a two part issue. FIRSTLY, I am attempting to write a function
for coefficient of variation, using
> co.var <- function(rowleyi) ( 100*sd(rowleyi)/mean(rowleyi) ) #where rowleyi
> is my data set, which has multiple columns and rows of data.
This is not working because som
I created a 3d scatter plot and am trying to change the color of outer box
lines with box3d.
Anybody can help me to figure out how to do this?
My example is:
library(scatterplot3d)
x=seq(1:6)
y=seq(7:12)
z=x*2
scatterplot3d(x, y,z)
Thanks.
Xin
__
R-
You can also use layout() with base graphics. This
example sets up a column of 14 strips and allocates
3 strips to the top and bottom graphs and 2 strips
To the four middle graphs. Using Richard's tmp dataframe:
layout(matrix(c(1, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5,
6, 6, 6), 14, 1))
layout.show(6)
p
Hi the list,
I am a member of the organizing comity of the French Statistics Association
(SFdS)'s conference. We
are looking for sponsors. Some software (SAS, RITME, ...) are represented. Do
you know if there is
any possibility to be sponsored by R (or by an association close to R)? Do you
th
Dear all,
I have two vectors x and y, both of which are of length 1000.
I created a factor as.factor(rep(1:100,each=10)).
Now I want to count, for each factor level, the number of intersections
between x and y. In other words, I would like to count the number of
intersections between x[1:10] a
Hi,
On Mon, Nov 19, 2012 at 1:25 PM, Sam Steingold wrote:
> Thanks Steve,
> what is the analogue of .N for min and max?
> i.e., what is the data.table's version of
> aggregate(infl$delay,by=list(infl$share.id),FUN=min)
> aggregate(infl$delay,by=list(infl$share.id),FUN=max)
> thanks!
It would be
Thanks Steve,
what is the analogue of .N for min and max?
i.e., what is the data.table's version of
aggregate(infl$delay,by=list(infl$share.id),FUN=min)
aggregate(infl$delay,by=list(infl$share.id),FUN=max)
thanks!
Sam.
On Fri, Sep 14, 2012 at 3:40 PM, Steve Lianoglou
wrote:
> Hi,
>
> On Fri, Sep
On 19/11/2012 3:27 PM, Bush, Daniel P. DPI wrote:
No, that did not resolve the issue, but thanks for the suggestion.
Here's another possibility: your Sweave session may not be setting the
same option defaults in startup code as your regular session. This one
just bit me: I normally work wit
No, that did not resolve the issue, but thanks for the suggestion.
Daniel Bush | School Finance Consultant
School Financial Services | Wis. Dept. of Public Instruction
daniel.bush -at- dpi.wi.gov | 608-267-9212
-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
I think this task would be easier in lattice
library(lattice)
xyplot(y + y + y + y + y + y ~ x, outer=TRUE, layout=c(1,6),
strip=FALSE, strip.left=TRUE,
ylab="6 copies of the Y variable",
main="put an interesting title here")
Six different y variables instead of six
On 12-11-19 10:18 AM, S Ellison wrote:
>
>> -Original Message-
>>> Can I use simple linear regression when I have proportion data
>>> for both dependent and independent variables? Or, should I use
>>> beta regression analysis? Or any suggestion?
>>>
>>
>> The distribution of the indepe
Dear Max,
first: Thanks a lot for your suggestion and the open words about methods in
real life. I guess: Thats my problem.
Regarding my analysis: Yes, thats the problem and I have to coerce to do this
analysis regarding lack of time to start something/other methods.
So you suggest Linear Discr
On Nov 19, 2012, at 10:46 AM, William Dunlap wrote:
If you have a list and want to add a new (or replace a)
named component use
myList[[compName]] <- compValue
as in
myList <- list()
compName <- "Incr"
compValue <- function(x) x + 1
myList[[compName]] <- compValue
If you want to make
*Hi
I have problem with plot.xts .
I try to subset some data in a xts time series.*
*subseting works fore more that one event*
*But I receive nothing, If I try to get one event *
I'm happy for every hint!
Thanks!
--
View this message in context:
http://r.789695.n4.nabble.com/xts-plot-
Hi!
In answer to:
I noticed that you were using what might be called an "externally
created Surv object". I have a memory that Terry Therneau has
criticized that practice. I cannot remember if it was in exactly this
situation but I might ask if setting up the model as:
cox = coxph
Please take the advice of Berend if you ever want to get help here. Also,
you will need to do some basic and initial work yourself; read what he
suggested, use Google to search keywords, and see great help like this:
http://cran.r-project.org/doc/contrib/Short-refcard.pdf
also, use the built in
Hi - I've seen a similar issue going on with survfit when using strata in
the model, although I get a different error message from ge. If it helps
to track down the problem (rather than confusing things further) here is
some code that should reproduce the issue I've seen. I'm running R 2.15.2,
wi
Dear colleagues,
I wish to create a figure with 6 plots arranged vertically with no spacing
between them as they all have a common x-axis.
However, using the code below I'm unable to get the plot area the same size
for each plot.
The bottom plot with the x-axis label is smaller than the others
I missed the last snipet; just saw the first. So you have your
solution. If you want a function, try:
> f.newList <- function(x,name){.x <- list(x);names(.x) <- name;.x}
> f.newList(10, paste('f', 'oo', sep = ''))
$foo
[1] 10
On Mon, Nov 19, 2012 at 1:32 PM, Sam Steingold wrote:
>> * jim holt
Hello,
I have code that will generate data for a 5 category IRT graded response
model. However, the code could be improved through vectorizing. Here is
the code below:
###Inputs
N <- 100 #Number of people taking test
n <- 10 #Number of items
nCat <- 5 #number of categories
###Generate Item pa
On Mon, Nov 19, 2012 at 5:42 AM, AnjaM wrote:
> Using the mtcars dataset, how to define the grouping variable to be valid
> only for the upper or lower panel?
>
> The following doesn't work:
>
> # Code start
>
>
Almost :
splom(~data.frame(mpg, disp, hp, drat, wt, qsec),
> data=mtcars, psca
Thanks a lot! I got some ideas from all the replies and here is the final one.
newdata
select <- sample(nrow(newdata), nrow(newdata) * .7)
data70 <- newdata[select,] # select
write.csv(data70, "data70.csv", row.names=FALSE)
data30 <- newdata[-select,] # testing
write.csv(data30, "data30.csv",
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hi Terry,
I attached a small data set to this email. This is what I get (I
restricted the formula to avoid NA's):
> surv = with(small, Surv(time=absence, event=(censored==FALSE)))
> (cox.s = coxph(surv ~ bucket*(today) + strata(activity), data =
>
If you have a list and want to add a new (or replace a)
named component use
myList[[compName]] <- compValue
as in
myList <- list()
compName <- "Incr"
compValue <- function(x) x + 1
myList[[compName]] <- compValue
If you want to make a new list-with-names from scratch try
structur
> * jim holtman [2012-11-19 13:14:05 -0500]:
>
> How about this (if you don't like writing two lines, encapsulate it in
> a function):
>
>> x <- list(10)
>> names(x) <- paste('f', 'oo', sep = '')
>> str(x)
> List of 1
> $ foo: num 10
>>
I am sorry, how is this different from my second snippet (e
On Nov 19, 2012, at 1:57 AM, wouterjohannes wrote:
No, I never got any response. I used an article by Knol to solve the
issue. Ik hope this is useful for you too.
Best regards,
Wouter
Knol, M.J., van, d.T., Grobbee, D.E., Numans, M.E., Geerlings, M.I.,
2007. Estimating interaction on a
How about this (if you don't like writing two lines, encapsulate it in
a function):
> x <- list(10)
> names(x) <- paste('f', 'oo', sep = '')
> str(x)
List of 1
$ foo: num 10
>
On Mon, Nov 19, 2012 at 1:07 PM, Sam Steingold wrote:
> How can I create lists with element names created on the fly?
How can I create lists with element names created on the fly?
--8<---cut here---start->8---
> list (foo = 10)
$foo
[1] 10
> list ("foo" = 10)
$foo
[1] 10
> list (paste("f","oo",sep="") = 10)
Error: unexpected '=' in "list (paste("f","oo",sep="") ="
--8<---
HI,
May be this helps:
dat1<-read.table(text="
V1 V2
1 5 10
2 6 3
3 8 4
4 9 20
5 15 30
6 25 40
7 2 4
8 3 1
9 1 5
10 8 10
",header=TRUE)
dat2<-dat1[sample(NROW(dat1),NROW(dat1)*(1-0.3)),] #70% of data
dat2$newcol<-TRUE
dat1$newcol1<-TRUE
dat4<-merge(dat1,dat2,by=c("V1","V2"),all=TRUE)
dat5
Hello all, could you explain me, how to get bootstrap estimates if i have
the next data:
>scor
mec vec alg ana sta
1 77 82 67 67 81
2 63 78 80 70 81
3 75 73 71 66 81
4 55 72 63 70 68
5 63 63 65 70 63
6 53 61 72 64 73
7 51 67 65 65 68
8 59 70 68 62
I'm not sure what you mean by "balance", but you can use sample() to
randomly order the values 1:1000, then use the first 700 as row
indices for the first set, and the last 300 as the test set.
Sarah
On Mon, Nov 19, 2012 at 12:16 PM, Eddie Smith wrote:
> Hi guys,
>
> I have 1000 rows of a datase
Hello,
See the following example.
x <- matrix(rnorm(2000), ncol = 2)
idx <- sample(nrow(x), 0.7*nrow(x))
x2 <- x[idx, ]
nrow(x2) # 700
x3 <- x[-idx, ]
nrow(x3) # 300
Hope this helps,
Rui Barradas
Em 19-11-2012 17:16, Eddie Smith escreveu:
Hi guys,
I have 1000 rows of a dataset. In my ana
Hi guys,
I have 1000 rows of a dataset. In my analysis, I need 70% of the data,
run my analysis and then use the remaining 30% to test my model.
Could anybody kindly help me on this?
Cheers
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mai
I'd like to expand the following data to perform a daily time series.
It should cover from '2012-07-01' to '2012-10-06' with the values I
actually have being the mean from one point measurement to another. Does
anyone has a clue to perform this task.
structure(list(Date.beg = structure(c(15635, 1
I can't reproduce the problem.
Tell us what version of R and what version of the survival package.
Create a reproducable example. I don't know if some variables are numeric and some are
factors, how/where the "surv" object was defined, etc.
Terry Therneau
On 11/17/2012 05:00 AM, r-help-req
At 17:13 18/11/2012, Jeff Newmiller wrote:
Michael, this comment doesn't seem appropriate to the question,
since the sample data is a ragged array that requires the addition
of NAs to fit into a wide format.
It does now but we do not know whether that was how it started life.
My remarks were
Sorry, the comma is in the wrong place, it should be
Vsim[] <- Vsim[ , NSErr > 0.6]
Rui Barradas
Em 19-11-2012 16:18, Rui Barradas escreveu:
Hello,
Try
Vsim[] <- Vsim[NSErr > 0.6, ]
Hope this helps,
Rui Barradas
Em 19-11-2012 14:30, iembry escreveu:
Hi everyone, now I am trying to finish
On 19-11-2012, at 15:48, Anna23 wrote:
> . Q>>>Use the built in dataset called iris in this task.
> (a) Calculate the result of following formula separately in every species
> for all of the
> numerical variables: log(x)/x.
> (b) Calculate trimmed mean for each of the numerical variables using
>
On 19-11-2012, at 15:32, andrew wrote:
> I am a Biologist and a beginner, please help me to solve this: please
> anyone..its my homework and I dont have a clue abt R yet..
> Write a function which does the following tasks:
>
> (a) Calculates minimum and maximum value of a given argument x.
> (
Hello,
Try
Vsim[] <- Vsim[NSErr > 0.6, ]
Hope this helps,
Rui Barradas
Em 19-11-2012 14:30, iembry escreveu:
Hi everyone, now I am trying to finish writing the code (I had asked for
assistance on subtracting arrays)
This is what I what I am running in R:
source("/home/ie/Documents/TTU/GA_R
Sorry for not taking care of this... If anyone would like to take over
maintainership of RMySQL I'm sure the R community would greatly appreciate
it. I just don't have the time these days.
Jeff
[[alternative HTML version deleted]]
__
R-help@r-p
Hello,
I am using rbetabinom ( to generate beta binomial random variables) function
available in the "emdbook"package written by Professor. Ben Bolker for my
research study.
I have no questions with this function. However, I am looking for the
theoretical method/algorithm of the function "rbetabi
Dear All,
I am Ph.D student at Chulalongkorn University in Thailand, I want to use
Package 'sampleSection' to estimate missing data which generate under IRT
model(3-PL);
n<-500 ## number of examinee
I<-20 ## number of items
num.imp<-5 ##number of imputations
p.missing<-c(0.09, 0.01) #prob of
David, thanks for the feedback!
Steve, thanks for the direction! I have heard and read some about Dr. Harrell's
work but somehow had missed the term "penalized logistic regression." That was
helpful for finding more specific sources to follow Dr. Harrell's (and other's)
suggestions. I may have
1 down vote favorite
my problem is the following:
I am using the R SNA package for social network analysis. Lets say, my
starting point is an edgelist with the following characteristics. Every row
contains a firm name, the ID of a project they are involved and further
characteristics, l
. Q>>>Use the built in dataset called iris in this task.
(a) Calculate the result of following formula separately in every species
for all of the
numerical variables: log(x)/x.
(b) Calculate trimmed mean for each of the numerical variables using
apply–function. Choose
your own trimming percentage
Hi everyone, now I am trying to finish writing the code (I had asked for
assistance on subtracting arrays)
This is what I what I am running in R:
> source("/home/ie/Documents/TTU/GA_Research/GLUE/R-Project/R_GLUE_Example/NSEr.R")
NSEr <- function (obs, sim)
{
{jjh <- (as.vector(obs) - sim)^2
Xjjhs
HI,
May be this helps:
Vobsr<-read.table(text="
81.071
73.187
66.991
62.482
59.662
58.529
59.085
61.328
65.259
70.878
",sep="",header=FALSE)
Vsimr=read.table(text="
81.07 81.07
73.19 73.19
65.81 67.16
58.93 63
52.55 60.7
46.68 60.25
41.31 61.67
36.44 64.95
32.08 70.08
28.22 77.08
",sep="",header=F
I am a Biologist and a beginner, please help me to solve this: please
anyone..its my homework and I dont have a clue abt R yet..
Write a function which does the following tasks:
(a) Calculates minimum and maximum value of a given argument x.
(b) If x is positive, some new vector gets the value
Using the mtcars dataset, how to define the grouping variable to be valid
only for the upper or lower panel?
The following doesn't work:
# Code start
splom(~data.frame(mpg, disp, hp, drat, wt, qsec),
data=mtcars, pscales=0,
auto.key=list(columns=3),
upper.panel = function(...){
Hi Peter,
why you are involving -1 with this concept? Can you explain more please
Cheers
Date: Sun, 18 Nov 2012 23:28:26 -0800
From: ml-node+s789695n4650012...@n4.nabble.com
To: frespi...@hotmail.com
Subject: Re: R-Square in WLS
On Nov 18, 2012, at 21:32 , Thomas Lumley wrote:
>
Hi everyone,
I’m working on my Master’s Degree thesis about the pricing of C.B. trying to
do that with “R”.
I read the paper “RQuantLib: Interfacing QuantLib from R” and now I’m
matching several market price (taken from Bloomberg or Deutsche Bank
database) with “R” output.
Could you help me to unde
Thank you for the quick reply.
Two more questions:
1. For example, if this is my code:
>RegModel =
lm(glucose~sex+BMI+height+weight+education+ses,weight=w_without_non_response)
>summary(RegModel)
>step(RegModel, direction ="backward",scope=list(lower=?,upper=?))
and I want the sex and height var
Hi,
No, I never got any response. I used an article by Knol to solve the issue. Ik
hope this is useful for you too.
Best regards,
Wouter
Knol, M.J., van, d.T., Grobbee, D.E., Numans, M.E., Geerlings, M.I., 2007.
Estimating interaction on an additive scale between continuous determinants in
I want to plot the results of multiple paired comparison in a side-by-side, in
opposite direction with the standard error, graph. Would you please give me
some help of howto. Thanks. Charlie.
lbp
[[alternative HTML version deleted]]
__
R
Dear all,
i searched for some classification methods and I have no glue if i took the
right once.
My problem: I have a matrix with 17000 rows and 33 colums (genes and patients).
The patients are grouped into 3 diseases.
No I want to classify the patients and for sure i want to know which rows are
> -Original Message-
> > Can I use simple linear regression when I have proportion data for
> > both dependent and independent variables? Or, should I use beta
> > regression analysis? Or any suggestion?
> >
>
> The distribution of the independent variable is irrelevant
> (in so
> I have a data matrix with 570 columns containing 95 (samples)
> with 6 replicates each.
> How can I calculate the mean of the replicates for 95 samples?
Write a function that calculates the sample means for a vector of 95
observations and then use apply() to apply that function to the whole
Hi Arun, thanks for your assistance. That worked as well.
Irucka
<-Original Message->
>From: arun [smartpink...@yahoo.com]
>Sent: 11/19/2012 7:22:09 AM
>To: iruc...@mail2world.com
>Cc: r-help@r-project.org
>Subject: Re: [R] loop to subtract arrays / error
>
>HI,
>May be this helps:
>
>Vo
thank you all for the great help, in particular to dennis murphy
in order to close the thread I'm posting here the final solution to my
question
new.ex<-structure(list(TEC = c(0.21, 0.077, 0.06, 0.033, 0.014, 0.007,
0.21, 0.077, 0.01, 0.033, 0.05, 0.014),
Thanks, Rui and Jim, for your replies. I tried to post this question to
r.devel but its admin told me that the question rather belongs into
r-help. Thanks, Jim, for your suggestion. I have already constructed
something similar. I posted my question to suggest modifying the
function so that not
Dear useRs,
I wanted to know something about the Index.G2 and Index.G3 which Calculate G2
and G2 internal cluster quality indexes.
i tried to find material from internet but it seems that the file have been
removed. Is it good to have higher values of these indexes or lower?
thanks in advan
Hi Rui, thank-you.
That was simple and worked great.
Irucka
<-Original Message->
>From: Rui Barradas [ruipbarra...@sapo.pt]
>Sent: 11/19/2012 4:13:24 AM
>To: iruc...@mail2world.com
>Cc: r-help@r-project.org
>Subject: Re: [R] loop to subtract arrays / error
>
>Hello,
>
>Or simpler, sin
Hi Rui Barradas, how are you?
Thank-you very much. That worked perfectly.
Irucka Embry
<-Original Message->
>From: Rui Barradas [ruipbarra...@sapo.pt]
>Sent: 11/19/2012 4:05:11 AM
>To: iruc...@mail2world.com
>Cc: r-help@r-project.org
>Subject: Re: [R] loop to subtract arrays / error
>
Thank you very much, It works!
--
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