Hi ,
I am working in Eclipse IDE , I want to use rscript to produce statistical
analysis , I tested a sample rcode in the script its working fine in my
Eclipse IDE , but I don't know how to pass my java values into rscript . I
need some guidance ,Please help me .
Thanks ,
Janarthanan .M
Hi David,
Apologies again and thankyou for your help, I've edited my original post to
clarify what I was asking. What I meant was that the factor had only 1
degrees of freedom when it should have had 2 (14 in total), so you're right
there were 14 but not in the right place.
In SPSS you select one
You should try http://math.tutorvista.com/statistics.html Tutorvista.com
the site offers http://math.tutorvista.com/statistics.html online
statistics help to anyone that is in need. The site is very interactive and
helps you get the ace in your desired subject.
--
View this message in conte
Without really knowing this code, I can guess that it may be the
"triangular" prior at work. Bayes Factors are notorious for being sensitive
to the prior. Presumably, the prior somehow prefers to see the rarer allele
as the "BB", and not the "AA" homozygous genotype (this is a common
assumption:
Hi Ivo,
I'll just be brief, but regarding data.table's syntax: one person's
"strange" is another's "intuitive" :-)
Note also that data.table also provides a `merge.data.table` function
which works pretty much as you expect merge.data.frame to. The
exception being that its `by` argument will, by d
hi patrick. thanks. I think you are right..
combined <- merge( main, aggregate.data, by="day", all.x=TRUE, all.y=FALSE )
lm( stockreturn ~ marketreturn, data=combined )
becomes something like
main <- as.data.table(main)
setkey(main, "mmdd")
aggregate.data <- as.data.table(aggregate.data)
s
shardman hotmail.com> writes:
>
> Hi,
>
> I'm trying to analyse some data I have imported into R from a .csv file but
> when I carry out the aov command the results show only one degree of freedom
> when there should be 14. Does anyone know why? I'd really appreciate some
> help, the data is pa
Thanks for both comments. Indeed the sep = "" is needed.
On Sun, Oct 9, 2011 at 9:43 PM, Rolf Turner wrote:
> On 10/10/11 04:53, R. Michael Weylandt wrote:
>
>
>
> Try this:
>>
>> for (i in 1990:2009) {
>> varName = paste("pci", i, collapse = "")
>> assign(varName, log(get(varName))
>>
On 10/10/11 04:53, R. Michael Weylandt wrote:
Try this:
for (i in 1990:2009) {
varName = paste("pci", i, collapse = "")
assign(varName, log(get(varName))
}
I believe that ``sep= " '' is needed here rather than collapse.
Try:
paste("junk",42,collapse="")
You get
[1]
Dear colleagues,
Do you know any R code implemented for the Mittag-Leffler function
(generalisation of the exponential function, fractional calculus) ?
Thanks in advance,
--
Dr Eric B Ferreira
Exact Sciences Department
Federal University of Alfenas
Brazil
[[alternative HTML version del
Note that exprs returns a matrix, so we can manipulate that just as we
would for any other type of matrix. There is also a Bioconductor
mailing list, which may be helpful.
On Thu, Oct 6, 2011 at 4:56 AM, Clayton K Collings wrote:
> Hello R people,
>
>>dim(exprs(estrogenrma)
>
> I have an expressi
On Sat, Oct 8, 2011 at 5:59 AM, Allan Sikk wrote:
> Hello,
>
> I'm trying to plot connected time series of two variables in a lattice plot:
> xyplot(y1 + y2 ~ t, data=size, type="b")
>
> y2 has missing data for some of the observations and some points are
> therefore not connected. It would make
You could also check this function I implemented awhile back:
http://www.fernandohrosa.com.br/en/P/sphericity-test-for-covariance-matrices-in-r-sphericity-test/
On Fri, Jun 17, 2011 at 4:43 PM, thibault grava wrote:
> Hello Dear R user,
>
> I want to conduct a Principal components analysis and I
Hi,
Use function ltext() instead, also available in lattice package.
Regards,
Carlos Ortega
www.qualityexcellence.es
On Sun, Oct 9, 2011 at 6:30 PM, Richard O. Legendi <
richard.lege...@gmail.com> wrote:
> Hi all,
>
> I'm new to R and to the mailing list, so please bear with me :-)
>
> I would
On Oct 9, 2011, at 2:08 PM, shardman wrote:
Hi David,
Thanks for your message.
I can assure you this is not homework. I'm working on an ecology
project and
am trying to analyse the results from the fieldwork. I don't want
other
people to do the work for me I was just hoping someone might
Hi,
Another way to do that is with function barchart() in package lattice.
Barchart requires a function which relates your variables with the option to
specify groups.
Check the examples (are under xyplot help) to apply them to your case.
Regards,
Carlos Ortega
www.qualityexcellence.es
On Thu,
Hi David,
Thanks for your message.
I can assure you this is not homework. I'm working on an ecology project and
am trying to analyse the results from the fieldwork. I don't want other
people to do the work for me I was just hoping someone might be able to spot
where I have made a mistake, I'm sti
On Oct 9, 2011, at 2:24 PM, Carlos Ortega wrote:
Hello,
In R you just need to take the log of the whole whole data.frame
where you
have your pci* and store in a new variable.
You do not need to use a "for" loop:
log.df <- log(your_data_frame)
Possibly with a selection for the column name
On Oct 9, 2011, at 1:46 PM, beaulieu.j...@epamail.epa.gov wrote:
The reference below describes the utility of the two-dimensional
Kolmogorov-Smirnow (2DKS) test for detecting relationships in
bivariate
data. If this test has been implemented in R I would love to know
about
it!
I have n
Hello,
In R you just need to take the log of the whole whole data.frame where you
have your pci* and store in a new variable.
You do not need to use a "for" loop:
log.df <- log(your_data_frame)
Regards,
Carlos Ortega
www.qualityexcellence.es
On Sun, Oct 9, 2011 at 5:34 PM, deepankar wrote:
>
On Oct 9, 2011, at 1:04 PM, Maas James Dr (MED) wrote:
If possible I'd like to produce a function that applies a formula to
a column in a matrix (essentially calculating the mse) and then
inserts it between values of a an array ...
confusing I know, here is an example of what I'm trying to
The reference below describes the utility of the two-dimensional
Kolmogorov-Smirnow (2DKS) test for detecting relationships in bivariate
data. If this test has been implemented in R I would love to know about
it!
Thanks,
jake
Garvey, J. E., E.A. Marschall, and R.A. Wright (1998). "From star
I think you are looking for the 'data.table'
package.
On 09/10/2011 17:31, ivo welch wrote:
Dear R experts---I am struggling with memory and speed issues. Advice
would be appreciated.
I have a long data set (of financial stock returns, with stock name
and trading day). All three variables, st
Hey Guys,
I just started fooling around with the twitteR package in order to get a record
of all tweets from a single public account. When I run userTimeline, I get the
default 20 most recent tweets just fine. However, when I specify an arbitrary
number of tweets (as described in the documentat
On Oct 9, 2011, at 11:10 AM, shardman wrote:
Hi,
I'm trying to analyse some data I have imported into R from a .csv
file but
when I carry out the aov command the results show only one degree of
freedom
when there should be 14. Does anyone know why? I'd really appreciate
some
help, the d
If possible I'd like to produce a function that applies a formula to a column
in a matrix (essentially calculating the mse) and then inserts it between
values of a an array ...
confusing I know, here is an example of what I'm trying to accomplish:
## create a matrix
(a <- matrix(c(3,6,4,8,5,9,1
Thanks a lot for the help
From: Jeff Newmiller
Sent: Sunday, October 9, 2011 6:35 PM
Subject: Re: [R] Substract "previous" element
Vectors are not matrices in R, though matrices are special cases of vectors.
See ?diff for a solution that works with vectors.
Dear Peter and Tim,
Thank you very much for taking the time to explain this to me! It is much
more clear now.
And sorry for using the space here maybe inappropriately, I really hope this
is OK and gets posted, I think it is really important that non-statisticians
like myself get a good idea of the
Have you taken a look at the provided introductory materials? type
help.start() to get "An Introduction to R" which I believe is provided
as a part of every pre-packaged version of R so it's most likely
already on your machine.
Read that and then we can sort through how to correctly implement your
Hi All,
This is surely an easy question but somehow I am not being able to get it.
I am using R 2.13.2 and have a data set where variable names like this
appear:
pci1990, pci1991, ... , pci2009.
"pci1990" has data on per capita income for 1990, "pci1991" has data on
per capita income for 19
I'll concur with Peter Dalgaard that
* a permutation test is the right thing to do - your problem is equivalent
to a two-sample test,
* don't bootstrap, and
* don't bother with t-statistics
but I'll elaborate a bit on on why, including
* two approaches to the whole problem - and how your approa
Hi,
I'm trying to analyse some data I have imported into R from a .csv file but
when I carry out the aov command the results show only one degree of freedom
when there should be 14. Does anyone know why? I'd really appreciate some
help, the data is pasted below.
/The imported table looks ike thi
Hi all,
I'm new to R and to the mailing list, so please bear with me :-)
I would like to create multiple levelplots on the same chart with a nice
main title with something like this:
print(levelplot(matrix(c(1,2,3,4), 2, 2)), split=c(1, 1, 2, 1))
print(levelplot(matrix(c(1,2,3,4), 2, 2)),
Hi Alex,
If "data" is a matrix, probably the easiest option would be:
tips <- as.data.frame(data)
mclapply(tips, foo)
By the way, I would recommend not using 'data' (which is also a
function) as the name of the object storing your data. If your data
set has many columns and performance is an i
Dear R experts---I am struggling with memory and speed issues. Advice
would be appreciated.
I have a long data set (of financial stock returns, with stock name
and trading day). All three variables, stock return, id and day, are
irregular. About 1.3GB in object.size (200MB on disk). now, I nee
"This is surely an easy question but somehow I am not being able to get it."
get() is the key -- it takes a string and returns the object with that
string as its name. Assign() goes the other way
Try this:
for (i in 1990:2009) {
varName = paste("pci", i, collapse = "")
assign(varName, lo
Dear all I want to convert a apply to lapply. The reason for that is that there
is a function mclappy that uses exact the same format as the lapply function.
My code looks like that
mean_power_per_tip <- function(data) {
return((apply(data[,],2,MeanTip)));
}
where data is a [m,n] matrix
Hi All,
This is surely an easy question but somehow I am not being able to get it.
I am using R 2.13.2 and have a data set where variable names like this
appear:
pci1990, pci1991, ... , pci2009.
"pci1990" has data on per capita income for 1990, "pci1991" has data on per
capita income for 1991,
Vectors are not matrices in R, though matrices are special cases of vectors.
See ?diff for a solution that works with vectors.
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead:
Take a look at diff() though offhand I dont know what it does to matrices
M
On Oct 9, 2011, at 11:00 AM, Alaios wrote:
> Dear all,
> I have a matrix with data and I want to substract from every value the
> previous element.
>
> Let's assume that my vector(matrix) is c<-(1,2,3,4,5)
> I want to
Excellent! Thank you!
Ben
On Sat, Oct 8, 2011 at 3:36 PM, Duncan Murdoch wrote:
> On 11-10-08 11:04 AM, Ben qant wrote:
>
>> Thank you!
>>
>> Sorry, I have a couple more questions:
>> 1) How to I turn off the box shading completely? I figured out how to
>> lighten it up to a grey color with col=
In many cases a flexible parametric fit, using regression splines, will
result in a fit that is as good as a gam, with similar regression shapes.
The rms package has a latex method that will represent such fits in
interpretable algebraic form. latex(fit) does that, and print(fit,
latex=TRUE) will
Dear all,
I have a matrix with data and I want to substract from every value the previous
element.
Let's assume that my vector(matrix) is c<-(1,2,3,4,5)
I want to get remove_previous c<-(0,1,2,3,4).
How I can do that efficiently in R?
I would like to thank you in advance for your help
B.R
Alex
On Sun, 9 Oct 2011, buehlerman wrote:
I want to apply Nyblom-Hansen test with the strucchange package, but I
don't know how is the correct way and what is the difference between the
following two approaches (leeding to different results):
The difference is that sctest(formula, type = "Nyblom-
I may be speaking out of turn here, but I would prefer not to see R-help
turn into a tutorial site for basic statistics.Such sites already exist
(e.g. http://stats.stackexchange.com/).
I realize that there is occasionally reason to venture down this path a way
within legitimate R contexts, but thi
Hi there,
I am reading the 2004 paper "Smoothing with mixed model software" in
Journal of Statistical Software, by Ngo and Wand. I tried to run
their first example in Section 2.1 using R but I had some problems.
Here is the code:
library(nlme)
fossil <- read.table("fossil.dat",header=T)
x <-
R is not the right tool for all things. This looks like a job for a computer
algebra system.
That said, R **does** have at least one interface to such a system. See the
Ryacas package (check my capitalization, which may be wrong). HelpeRs may
provide you with others.
-- Bert
On Sun, Oct 9, 2011
Thank you very much to both Ken and Peter for the very helpful
explanations.
Just to understand this better (sorry for repeating but I am also new in
statistics
so please correct me where I am wrong):
Ken' method:
Random sampling of the mean, and then using these means to construct a
distribution
I want to apply Nyblom-Hansen test with the strucchange package, but I don't
know how is the correct way and what is the difference between the following
two approaches (leeding to different results):
data("longley")
# 1. Approach:
sctest(Employed ~ Year + GNP.deflator + GNP + Armed.Forces, data
On Oct 9, 2011, at 12:00 , francesca casalino wrote:
> Thank you very much to both Ken and Peter for the very helpful explanations.
>
> Just to understand this better (sorry for repeating but I am also new in
> statistics…so please correct me where I am wrong):
>
> Ken' method:
> Random sampl
On Oct 9, 2011, at 14:02 , Bogaso Christofer wrote:
> Dear all, I have a system of simultaneous equations with 2 unknowns as
> follows:
>
>
>
> x*y + (1-x) = 0.05
>
> x*(y - .5)^2 + (1-x)*0.6 = 0.56^2
>
>
>
> Ofcourse I can do it manually however wondering whether there is any direct
> way
Dear all, I have a system of simultaneous equations with 2 unknowns as
follows:
x*y + (1-x) = 0.05
x*(y - .5)^2 + (1-x)*0.6 = 0.56^2
Ofcourse I can do it manually however wondering whether there is any direct
way in R available to get the solution of this system?
Thanks and regards,
Hi,
I'm a newbie to R. My knowledge of statistics is mostly self-taught. My
problem is how to measure the effect of users in groups. I can calculate a
particular attribute for a user in a group. But my hypothesis is that the
user's attribute is not independent of each other and that the user's
att
You may "build" your customized matrix merging the components of the objects
before calling the xtable function:
my.matrix <- rbind(model$coefficients, [vector containing errors])
xtable(my.matrix)
(I'm sorry I don't know exactly where the standard errors are stored / how
to compute them)
You can
On Sat, 8 Oct 2011, Duncan Murdoch wrote:
On 11-10-07 5:26 PM, Carl Witthoft wrote:
Just wondering here -- I tested and found to my delight that
% round(325.4,-2)
[1] 300
gave me exactly what I would have expected (and wanted). Since it's not
explicitly mentioned in the documentation that n
On 11-10-09 4:00 AM, (Ted Harding) wrote:
On 09-Oct-11 00:46:58, Carl Witthoft wrote:
On 10/8/11 6:11 PM, (Ted Harding) wrote:
Carl Witthoft's serendipitous discovery is a nice example
of how secrets can be guessed by wondering "what if ... ?".
So probably you don;t need to tell the secrets.
On Oct 8, 2011, at 8:43 PM, maspitze wrote:
Hi,
I have a series of id's with multiple visits and questionnaire
scores. This
is a clinical trial that will be analyzed using the last observation
carried
forward method. In other words, in order to comply with intent to
treat
analysis when
Hi,
It is probably more confusing with several steps combined, but you are
correct that it is because there are NAs. It is fairly common for R
functions to return NA if there are any NA values unless you
explicitly set an argument on what to do with missing values. A quick
look at ?cor clearly s
Hello. I am wondering why I am getting NA for all
in cors=sapply(pred,cor,y=resp). I suppose that each column in pred has NAs in
them. Is there some way to fix this? Thanks
> str(pred)
'data.frame': 200 obs. of 13 variables:
$ mnO2: num 9.8 8 11.4 4.8 9 13.1 10.3 10.6 3.4 9.9 ...
$ Cl :
On 09-Oct-11 00:46:58, Carl Witthoft wrote:
>
> On 10/8/11 6:11 PM, (Ted Harding) wrote:
>
>> Carl Witthoft's serendipitous discovery is a nice example
>> of how secrets can be guessed by wondering "what if ... ?".
>> So probably you don;t need to tell the secrets.
>>
>> Taking the "negative digi
On Oct 8, 2011, at 16:04 , francy wrote:
> Hi,
>
> I am having trouble understanding how to approach a simulation:
>
> I have a sample of n=250 from a population of N=2,000 individuals, and I
> would like to use either permutation test or bootstrap to test whether this
> particular sample is s
61 matches
Mail list logo
