okay, I should have remembered... Gabor suggested a pretty nice solution,
see:
http://r.789695.n4.nabble.com/How-to-create-an-array-of-lists-of-multiple-components-td3167576.html
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Duncan Murdoch gmail.com> writes:
>
> On 18/05/2011 10:02 PM, Nick Matzke wrote:
> > Hi,
> >
> > If I load 2 packages that have a function with the same
> > name, how do I tell R to run one or the other?
> >
>
> If you are using a package without a namespace, expect problems. Having
> the na
> --
>
> Message: 15
> Date: Tue, 17 May 2011 13:03:57 +0100
> From: Timothy Bates
> To: R list
> Subject: Re: [R] Box Plot under GUI (R Commander/RKward)
> Message-ID: <9ca34005-9076-439f-aec2-1931e1472...@gmail.com>
> Content-Type: text/plain; charset=windows-1252
>
Thank you so much for this reply, Peter. It helps.
I know this is one way to adjust for covariates. However, if what I want is
to get the 'remaining values' after adjustment. For example, say, 'gene
expression' value is denoted as 'ge', and for each gene,
ge=a*age+b*sex+c*per_se
My question is:
I'm pretty new in R... try to learn ASAP for a project, I wonder if anyone
can provide some advice on which package to use when fit Mixture regression
model with both categorical(more than two levels) and continuous variables.
I tried to use Flexmix... and transposed the multi-level variables as
s
On Thu, May 19, 2011 at 7:21 PM, karena wrote:
> Hi, I have a question about how to do covariate adjustment.
>
> I have two sets of 'gene expression' data. They are from two different
> tissue types, 'liver' and 'brain', respectively.
> The purpose of my analysis is to compare the pattern of the w
For that kind of operation (unusual as it is) work with numeric matrices. When
you are finished, if you still want a data frame, make it then, not before.
If your data starts off as data frame to begin with, turn it into a matrix
first. E.g.
myMatrix <- data.frame(myData)
myMatrix2 <- myMatr
Hi,
Does the following work for you?
set.seed(123)
d <- data.frame(x = rpois(10, 4), y = rnorm(10))
d
d + 2
See ?within for one more option.
HTH,
Jorge
On Thu, May 19, 2011 at 11:54 PM, Ramya <> wrote:
> Hi there
>
> I just want to add 2 to all the values in dataframe.
>
> I tried using sapp
Thanks for your help. How would I extract each of the 3 values in the vector
individually?
Thanks again
On Thu, May 19, 2011 at 10:40 PM, Rolf Turner wrote:
> On 20/05/11 13:46, Cheryl Johnson wrote:
>
>> Hello,
>>
>> I am randomly generating values and then using an ANOVA table to find the
>> m
Hi there
I just want to add 2 to all the values in dataframe.
I tried using sapply but it seem to die all the time.
Any help would be appreciated.
Thanks
Ramya
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Hi, I have a question about how to do covariate adjustment.
I have two sets of 'gene expression' data. They are from two different
tissue types, 'liver' and 'brain', respectively.
The purpose of my analysis is to compare the pattern of the whole genome
'gene expression' between the two tissue typ
Thanks Barry. I have now got as far as creating the png (or gif or tiff) images
(as per your suggestions). But I am unable to work out how to convert each to a
2 dimensionional numeric matrix showing the spatial distribution of
occupied/unoccupied pixels in R. Any suggestions anybody?
Thanks
T
That only applies if you have the same factors a and b each time. If this is
the case you can do things in a much more slick way.
u <- matrix(rnorm(5000), nrow = 10) ## NB, nrow
AB <- expand.grid(a = letters[1:2], b = letters[1:5])
M <- lm(u ~ a+b, AB)
rmsq <- colSums(resid(M)^2)/M$df.resid
an
Hi:
It's easier to use an apply family function than a loop for this type
of problem, as illustrated below:
# Generate 30 random samples of size 10 from a standard
# normal distribution and put them into a matrix
u <- matrix(rnorm(300), ncol = 10)
a <- factor(rep(1:5, each = 2))
b <- factor(rep(1
On 20/05/11 14:51, Cheryl Johnson wrote:
Thanks for your help. How would I extract each of the 3 values in the
vector individually?
If you are going to use R, it would pay you to learn a little bit about it.
Like, e.g., vector indexing.
In your example
anova(mylm)[["Mean Sq"]][3]
would g
On 20/05/11 13:46, Cheryl Johnson wrote:
Hello,
I am randomly generating values and then using an ANOVA table to find the
mean square value. I would like to form a loop that extracts the mean square
value from ANOVA in each iteration. Below is an example of what I am doing.
a<-rnorm(10)
b<-fact
Hello,
I am randomly generating values and then using an ANOVA table to find the
mean square value. I would like to form a loop that extracts the mean square
value from ANOVA in each iteration. Below is an example of what I am doing.
a<-rnorm(10)
b<-factor(c(1,1,2,2,3,3,4,4,5,5))
c<-factor(c(1,2,
On Thu, May 19, 2011 at 4:28 PM, Michael Conklin
wrote:
> I have been trying to get Rgraphviz to work (I know it is from Bioconductor)
> unsuccessfully. Since I have no experience with Bioconductor I thought I
> would ask here if anyone has advice. I have installed Graphviz 2.20.3 as is
> recom
Hi:
Could you post a small, reproducible data set that illustrates what
you want to do? It sounds like you're creating 'spaghetti plots',
which can be done with a minimal amount of pain in ggplot2.
Dennis
On Thu, May 19, 2011 at 11:29 AM, 1Rnwb wrote:
> Hello gurus,
>
> I have a dataframe conta
Hi
In addition to getting the graphviz version EXACTLY right (as per the
README file), and ensuring that the right paths are set up, I also
seemed to need to install graphviz in a path WITHOUT spaces (but that
may just have been superstition)
Paul
On 20/05/2011 8:28 a.m., Michael Conklin wr
Hi Tao,
For you situation (and even MUCH larger number of events), multivariable
modeling will be unreliable unless you use shrinkage, variable selection
will select the wrong variables, and univariable screening leads to massive
bias in later stages.
Terry converted me from SAS to S-Plus in 1991
Thank you, Frank and Terry, for all your answers! I'll upgrade my "survival"
package for sure!
It seems to me that you two are pointing to two different issues: 1) Is
stepwise
model selection a good approach (for any data)? 2) Whether the data I have has
enough information that even worth to
What you should and can do depends on the expectations you have regarding the
correlation structure of the data and is limited by your degrees of freedom.
For example, what is wrong about:
reg<-lmer(response~femaleset+treatment+(1|group))
?
This assumes that there are constant group effects on
>>> Martin Maechler writes:
>
> Well, then you don't know *THE ONE* case where modern users of
> R should use attach() ... as I have been teaching for a while,
> but seem not have got enought students listening ;-) ...
>
> --- Use it instead of load() {for save()d R objects} ---
>
> Th
On May 19, 2011, at 5:20 PM, hsu ya-hui wrote:
Dear R-user,
I would like to get the survival probability (surv) for each subject
id.
That is, I want to have additional column surv as follows:
The survival probability for each subject is 1 until death and 0
thereafter.
id OS OS_DUR
Hi Bert,
thanks for your advice on lapply which I am still not very familair with.
With regard from the issue of the names I realised that I can simply call
names on the data.frame and then join them with other strings through paste
and add them to the data.frame by using the [] syntax rather than
Hmmm... I am sorry Patrick, it should have been
*
a <- codboot[c(4)]
round(a$bca[4:5], 2) # note I replaced "," by ":"
Best,
Jorge
*
>
> On Thu, May 19, 2011 at 3:08 PM, Patrick Santoso <> wrote:
>
>> Thanks for the suggestion Jorge!
>>
>> - that gives me subscript out of bounds error.
Perhaps this is useful:
x=c(-2,0,2)
sign(x)*abs(x)
[1] -2 0 2
--
Robert W. Baer, Ph.D.
Professor of Physiology
Kirksville College of Osteopathic Medicine
A. T. Still University of Health Sciences
800 W. Jefferson St.
Kirksville, MO 63501
660-626-2322
F
(1) What has this to do with recursion?
(2) You probably need to use ifelse(). I believe that this is
(in effect) an FAQ.
cheers,
Rolf Turner
On 20/05/11 07:42, Tremblay, Pierre-Olivier wrote:
Hi,
I created a function for obtaining the normal cumulative distribution (I
know all
Hi Yighua,
Try
> m <- 2.35343
> m
[1] 2.35343
> cat(m)
2.35343
HTH,
Jorge
On Thu, May 19, 2011 at 2:35 PM, Hu, Yinghua <> wrote:
> Hi,
>
> I am running some function in Ubuntu command line and get some problem. I
> used some command like below
>
> $ R --slave -vanilla < my_infile > my_outfile
Hi,
I am running some function in Ubuntu command line and get some problem. I used
some command like below
$ R --slave -vanilla < my_infile > my_outfile
The return should be one number in the my_outfile. In the my_infile, I call one
extra function and it has code like "return x".
However, in
You can see the whole function by printing its name in console.
Example:
> plot
function (x, y, ...)
{
if (is.function(x) && is.null(attr(x, "class"))) {
if (missing(y))
y <- NULL
hasylab <- function(...) !all(is.na(pmatch(names(list(...)),
"ylab")))
Dear R-user,
I would like to get the survival probability (surv) for each subject id.
That is, I want to have additional column surv as follows:
id OS OS_DUR surv
eg:
fitKM<-survfit(Surv(OS_DUR,OS)~1,data=data)
fitKM$surv # this will only give me survival probability and
Thank you all for the help and solutions presented!
Luisa
On Thu, May 19, 2011 at 9:33 PM, Dennis Murphy wrote:
> Oops, didn't see Marc's reply. His solution is much more compact. For
> R 2.11.0 and above, aggregate() now has a formula interface that
> usually works nicely:
>
> aggregate(Var3 ~
Hello gurus,
I have a dataframe containing two groups viz., 'control' and 'case', each of
these groups contains longitudinal data for 100 subjects. I have to plot all
these subjects on a single chart and then put a regression line for each of
the group for all the subjects. I have written a functi
Hi,
I created a function for obtaining the normal cumulative distribution (I
know all this already exists in R, I just wanted to verify my
understanding of it). below is the code I came up with.
cdf<-function(x) {
erf<-function(x) {
# approximation to the error function
Thanks for the suggestion Jorge!
- that gives me subscript out of bounds error. perhaps I can tweak the
parameters? I've never worked with that command.
Pat
On Thu, May 19, 2011 at 2:12 PM, Jorge Ivan Velez
wrote:
> Hi Patrick,
>
> How about this (untested)?
>
> a <- codboot[c(4)]
> round(a$bca
Dear all,
I am analysing a data set based on 6 groups of individuals. Each group is
observed for 10 days. 5 days with one manipulation 5 days with another
manipulation. I therefore have 6 replicate groups (n=6) each with one mean
measurement for manipulation A and manipulation B. Each group con
On Thu, May 19, 2011 at 11:41 PM, Manderscheid Katharina
wrote:
> hi,
>
> i am trying to work with the survey package in order to apply survey design
> weights. the data set i am using - ess - contains missing values.
> my question: when using svytable(~variable1+variable2,
> design=my.svydesign
John Chambers is also very widely quoted describing the aim of S as "to turn
ideas into software, quickly and faithfully." Perhaps S3 is for people
wanting to do things "quickly" and S4 is for those wanting to do things
"faithfully".
Kevin
On Wed, May 18, 2011 at 7:39 PM, wrote:
> I used to th
Dear list,
I came up with a two functions that flatten arbitrary deeply nested
lists (as long as they're named; not tested for unnamed) and
environments (see attachment; 'flatten_examples.txt' contains some
examples).
The paradigm is somewhat similar to that implemented in 'unlist()', yet
e
Thank you all for the quick answers!
I googled first instead of having the idea to search this forum...
I'm using R 2.12.2 on a 32-bit Computer with windows installed.
Up to this point I was trying to get the image the way I would like to have
it, but didn't get fully satisfactory results.
I trie
I have been trying to get Rgraphviz to work (I know it is from Bioconductor)
unsuccessfully. Since I have no experience with Bioconductor I thought I would
ask here if anyone has advice. I have installed Graphviz 2.20.3 as is
recommended on the Bioconductor site but basically R cannot seem to fi
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Dennis Murphy
> Sent: Thursday, May 19, 2011 11:38 AM
> To: infoc...@gmx.net
> Cc: r-help@r-project.org
> Subject: Re: [R] vectorial search for elements with certain attributes
>
I took a look at sendmailR when I was trying to figure out how to send
email a few years ago. I ended up using my SQL Server dbmail facility,
which I can easily access from R using the RODBC package. In case my
scenario applies to anyone else, I will paste the simple function I
wrote to serve as
Oops, didn't see Marc's reply. His solution is much more compact. For
R 2.11.0 and above, aggregate() now has a formula interface that
usually works nicely:
aggregate(Var3 ~ Var1 + Var2, data = d, FUN = table)
Var1 Var2 Var3.D Var3.I
1 S1 T1 2 2
2 S2 T1 2 2
3 S1 T
Hi:
The dummy column really isn't necessary. Here's another way to get the
result you want. Let d be the name of your example data frame.
d <- d[, 1:3]
(dtable <- as.data.frame(ftable(d, row.vars = c(1, 2
Var1 Var2 Var3 Freq
1 S1 T1D2
2 S2 T1D2
3 S1 T2D2
Hi!
I try to find out how bootstrap mediation work! How I coud see the
whole function or algorithm? (rather whole bootstrap).
t.Mete
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting gui
On May 19, 2011, at 9:43 AM, guy33 wrote:
Hey all,
I can't seem to get the princurve package to produce correct
results, even
in the simplest cases. For example, if you just generate a 1 period
noiseless sine wave, and ask for the principal curve and plot, the
returned
curve is clearly w
I wrote this goody a while back, and thought I'd post it in case anyone
can find a use for it. I'm mentioning it here because I'm not convinced
it warrants a full package release.
The function is named "ktsolve," and is available at
http://witthoft.com/rtools.html or directly at
http://witt
Hi:
Here's one possible approach:
A <- c(1,3,7,8,10)
B <- c(5,8)
apply(outer(A, B, '-'), 2, function(x) min(which(x >= 0)))
[1] 3 4
HTH,
Dennis
On Thu, May 19, 2011 at 2:58 AM, wrote:
> I have 2 vectors A and B. For each element of B I'd like to find the index of
> the next higher or equal e
Another approach, using aggregate(), presuming that the data is in a data frame
called 'DF':
> with(DF, aggregate(Var3, list(Var1 = Var1, Var2 = Var2), table))
Var1 Var2 x.D x.I
1 S1 T1 2 2
2 S2 T1 2 2
3 S1 T2 2 2
4 S2 T2 0 4
HTH,
Marc Schwartz
On May 19, 20
Hi Patrick,
How about this (untested)?
a <- codboot[c(4)]
round(a$bca[4, 5], 2)
HTH,
Jorge
On Thu, May 19, 2011 at 7:20 AM, Patrick Santoso <> wrote:
> Good Morning,
>
> I'm having what I hope to be a simple problem. I am generating bootstrap
> confidence intervals using package (boot) - whic
Luma -
If I understand you correctly, I think the easiest way
to get what you want is to use the reshape function on
the output from aggregate:
reshape(Count.Cells,idvar=c('Var1','Var2'),timevar='Var3',direction='wide')
Var1 Var2 dummy.D dummy.I
1 S1 T1 2 2
2 S2 T1
As I (thought I) understood from the sendmailR manual, the package does
currently not support server authentication, or does it?
Daniel
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Hello R-Community
I did a lot of research, but i found no answer for my issue.
I try to plot multiple Plots. I do this with the layout() function.
The following two pictures show the defined layout:
http://www.stuber.info/layout_1.jpg
http://www.stuber.info/layout_2.jpg
The final plots looks lik
From: wolley.chir...@hotmail.com
To: biomathjda...@gmail.com
Subject: RE: [R] problem with optim()
Date: Thu, 19 May 2011 15:45:15 +0200
First, thank you for ur response...
Actually I didn't write the entire code ...X, Y and p_tilde are of course all
defined at the beginning of my code.
I made a pie chart and the names of the levels are outside
the circle. How do I put the names of the levels within each
sector?
names(tab13) = paste(c('Regular', 'Bom', 'Excelente'),
round(100*prop.table(tab13), dig=1), "%")
pie(tab13, col=c("LightYellow", "lightgreen", 'lightblue',
'white'),
Dear all,
I am having trouble creating summary tables using aggregate function.
given the following table:
Var1 Var2Var3 dummy
S1 T1 I 1
S1 T1 I 1
S1 T1 D1
S1 T1 D1
S1 T2 I 1
S1
Dear all,
Can some please help me to estimate heteroskedastic regression model in R. I
know that in Stata reghv function (regh package) can be used. Waht are their
equivalents in R?
Thanks in advance.
Best,
Tomas
[[alternative HTML version deleted]]
Good Morning,
I'm having what I hope to be a simple problem. I am generating bootstrap
confidence intervals using package (boot) - which works perfectly. The issue
I am having is getting the results into a format which I can write out to a
database. To be clear I am having no problems generating t
> From: dwinsem...@comcast.net
> To: julio.flo...@spss.com.mx
> Date: Thu, 19 May 2011 10:40:08 -0400
> CC: r-help@r-project.org
> Subject: Re: [R] Help, please
>
>
> On May 18, 2011, at 6:29 PM, Julio César Flores Castro wrote:
>
> > Hi,
> >
> > I am
I have 2 vectors A and B. For each element of B I'd like to find the index of
the next higher or equal element in A. And I'd like to store it effectiv. E.g.:
A <- c(1,3,7,8,10)
B <- c(5,8)
result: 3, 4
I have a possibility but for long vectors it works not very effectiv:
ans <- sapply(B, functi
Hey all,
I can't seem to get the princurve package to produce correct results, even
in the simplest cases. For example, if you just generate a 1 period
noiseless sine wave, and ask for the principal curve and plot, the returned
curve is clearly wrong (doesn't follow the sine wave). Here's my cod
Hi,
have the same question. did you manage to make x11 your default ploting
device?
- Chriss
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_
Dear Help List:
read.csv() seems to have changed in R version 2.13.0 as compared to version
2.12.2 when reading in simple CSV files.
In reading a 2-column CSV file ("test.csv"), say
1, a
2, b
If file is encoded as UTF-8 (on Windows 7), then under R 2.13.0
read.csv("test.csv",fileEncoding="UTF-8
You can try something like this, at the command line:
gs -sDEVICE=pdfwrite -dCompatibilityLevel=1.5 -dPDFSETTINGS=/screen
-dNOPAUSE -dQUIET -dBATCH -sOutputFile=output.pdf input.pdf
evidently, the new compactPDF() function in R 2.13 does something very similar.
-Aaron
On Thu, May 19, 2011 at
krusty the klown libero.it> writes:
> I found this useful package for generalized beta, yet the function that
> calculates its density leaves me puzzled, especially when I plotted it:
>
> plot(function(y) dbetagen(y,4,1.2,min=0,max=40),xlim=c(0,40))
>
> The area between the density curve and th
Is this what you are looking for:
> mdat3
sp.1 sp.2 sp.3 sp.4 sp.5
T110010
T210010
T311100
T410111
>
> # create a matrix of when species first appeared
> first <- apply(mdat3, 2, function(x) (cumsum(x == 1) > 0) + 0
Thanks a lot, Jim.
This is exactly what I was looking for!
Dimitri
On Thu, May 19, 2011 at 12:11 PM, jim holtman wrote:
> try this:
>
> mydf<-data.frame(mydate=seq(as.Date("2011-01-01"), length = 92, by = "day"))
> # add 'day' to the dataframe
> mydf$day <- format(mydf$mydate, "%d")
> mydf$newfac
Hi all,
I found this useful package for generalized beta, yet the function that
calculates its density leaves me puzzled, especially when I plotted it:
plot(function(y) dbetagen(y,4,1.2,min=0,max=40),xlim=c(0,40))
The area between the density curve and the x axis does not seem to measure
1!!!
--
try this:
mydf<-data.frame(mydate=seq(as.Date("2011-01-01"), length = 92, by = "day"))
# add 'day' to the dataframe
mydf$day <- format(mydf$mydate, "%d")
mydf$newfactor <- cumsum(mydf$day == '20')
mydf
On Thu, May 19, 2011 at 10:51 AM, Dimitri Liakhovitski
wrote:
> Hello!
> I have a data frame w
1. This won't work. The lagged variables have length one less than the
originals.
2. How about:
lagged_Q <- data.frame( lapply( QuarterlyData,diff))
You can then change the names in lagged_Q to something like
lagged_originalName via paste() if you like.
3. I strongly suspect that none of this i
On Thu, 19 May 2011, Liaw, Andy wrote:
From: Prof Brian Ripley
Hmm, load() does have an 'envir' argument. So you could simply use
that and with() (which is pretty much what attach() does internally).
If people really wanted a lazy approach, with() could be extended to
allow file names (as at
On 19/05/2011 11:14 AM, Layman123 wrote:
Hi everyone,
My data consists of a system of nearly 75000 roads, available as a
shapefile. When I plot the road system, by adding the individual roads with
'lines' and store it as a pdf-file with 'pdf' I get a file of size 13 MB.
This is way too large to
This was answered on this list a few days ago. See
https://stat.ethz.ch/pipermail/r-help/2011-May/278029.html
On Thu, 19 May 2011, Layman123 wrote:
Hi everyone,
My data consists of a system of nearly 75000 roads, available as a
shapefile. When I plot the road system, by adding the individual
On 19/05/2011 10:34 AM, Liaw, Andy wrote:
From: Prof Brian Ripley
>
> Hmm, load() does have an 'envir' argument. So you could simply use
> that and with() (which is pretty much what attach() does internally).
>
> If people really wanted a lazy approach, with() could be extended to
> allow fi
On May 19, 2011, at 9:32 AM, Timothy Bates wrote:
Likewise it would help reduce confusion when
plot(~Source, data=Df) # works
# but
boxplot(~Source, data=Df)
# Error in boxplot.formula(~Source, data = Df) :
# 'formula' missing or incorrect
The formula isn’t missing or illformed, it’s that bo
Hi everyone,
My data consists of a system of nearly 75000 roads, available as a
shapefile. When I plot the road system, by adding the individual roads with
'lines' and store it as a pdf-file with 'pdf' I get a file of size 13 MB.
This is way too large to add it in my LaTeX-document, because there
Is a list like Henrik's (below) available someplace on r-project.org or
on your local CRAN mirror? If no, where do you think it might most
conveniently fit? What about "www.r-project.org/other-docs.html"?
Perhaps add a bullet right under the link to "DevCheatSheet has a
collection of R Refere
Hello,
I would like to create lagged and delta variables from a set of variables
and then add them to a dataframe
Suppose that GDPPcSa is a variable. I would like to be able to do this
QuarterlyData$D1GdpPcSa = diff(GDPPcSa , 1)
in an automated fashion so that I loop over Quartely data to compu
Hello!
I have a data frame with dates. I need to create a new "month" that
starts on the 20th of each month - because I'll need to aggregate my
data later by that "shifted" month.
I wrote the code below and it works. However, I was wondering if there
is some ready-made function in some package - th
On May 18, 2011, at 6:29 PM, Julio César Flores Castro wrote:
Hi,
I am using R 2.10.1 and I have a doubt. Do you know how many cases
can R
handle?
I was able to handle (meaning do Cox proportional hazards work with
the 'rms' package which adds extra memory overhead with a datadist
obj
On 19/05/2011 9:32 AM, Timothy Bates wrote:
The most interesting thing in this thread for me was "within()" - that is what
I want 99% of the time, but I had only ever heard of with()
A real boon! Thanks Bill V!
The help for within gives this example for with(), which seems unnecessary, as
glm
From: Prof Brian Ripley
>
> Hmm, load() does have an 'envir' argument. So you could simply use
> that and with() (which is pretty much what attach() does internally).
>
> If people really wanted a lazy approach, with() could be extended to
> allow file names (as attach does).
I'm not sure if
Check data.table ... It benefits a lot from indexing data in data frames
On Thu, May 19, 2011 at 7:41 PM, xiagao1982 wrote:
> Hi all,
>
> I wonder if I can create indexes for columns in a data frame to speed up data
> selection by matching column values, just like indexes for large tables in a
Hi all,
I wonder if I can create indexes for columns in a data frame to speed up data
selection by matching column values, just like indexes for large tables in a
relational database? Thank you!
xiagao1982
2011-05-19
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Thanks Martin,
Your points are, of course, well taken. Nevertheless, I still think it
might be useful to put a link or links to one or more style guides in
the FAQ with a comment to the effect that these are various
recommended ways to help write better, more readable code. Something
like:
--
Thought I'd share with the community: A very interesting (free) report
by McKinsey Global Institute about the importance of Big Data for the
present/future economy:
http://www.mckinsey.com/mgi/publications/big_data/index.asp
Hoping R will be an important player in this area.
--
Dimitri Liakhovi
I cc'd r-help to get this message back on the board.
Again, you haven't really answered my questions. Comments inline.
On Thu, May 19, 2011 at 9:45 AM, chirine wolley
wrote:
> First, thank you for ur response...
> Actually I didn't write the entire code ...X, Y and p_tilde are of
> course all de
It works!
Thank you.
CecÃlia
De: Scott Chamberlain [mailto:scttchamberla...@gmail.com]
Enviada: quinta-feira, 19 de Maio de 2011 13:40
Para: Cecilia Carmo
Cc: r-help@r-project.org
Assunto: Re: [R] balanced panel data
# If you know how many years are needed you could do this
maken
On Wed, May 18, 2011 at 9:49 PM, jim holtman wrote:
> Is this what you were after:
>
>> mdat <- matrix(c(1,0,1,1,1,0), nrow = 2, ncol=3, byrow=TRUE,
> + dimnames = list(c("T1", "T2"),
> + c("sp.1", "sp.2", "sp.3")))
>>
>> mdat
> sp.1 sp.2 sp.3
> T1
The most interesting thing in this thread for me was "within()" - that is what
I want 99% of the time, but I had only ever heard of with()
A real boon! Thanks Bill V!
The help for within gives this example for with(), which seems unnecessary, as
glm supports data=
with(anorexia, {
anorex.1
What do you mean when you say "wrong results"? What do you expect for
the output? Your code doesn't work for me because it references X in
places and X is not defined.
Have you tested your functions to make sure they return reasonable values?
On Thu, May 19, 2011 at 9:17 AM, chirine wolley
wrote
Dear R-users,
I would like to maximize the function g above which depends on 4 parameters (2
vectors, 1 real number, and 1 matrix) using optim() and BFGS method. Here is
my code:
# fonction to maximize
g=function(x)
{
x1 = x[1:ncol(X)]
x2 = x[(ncol(X)+1)]
x3 =
matrix(x[(ncol(X)+2):(ncol(
I am using the package rpart to explore various classification structures.
The call looks like:
seekhi1<-rpart(pvol~spec+a1+psize+eppres+numpt+icds+bivalcrt+stents+ppshare+
nhosp+nyrs,data=dat,method="class",
control=rpart.control(minsplit=30,xval=10))
The output is
1)
You generally aren't allowed to have duplicate row names in a data frame in R.
If you want to keep your rownames you can make them a column in the data frame
before unsplitting.
Scott
On Thursday, May 19, 2011 at 5:27 AM, Cecilia Carmo wrote:
> Hi everyone,
>
>
>
> I have already used split
# If you know how many years are needed you could do this
makenewtable <- function(x, years) {
xlist <- split(x, x$firm)
new <- list()
dat <- lapply(xlist, function(z) if(length(unique(z$year)) == years) {new <- z}
)
dat_ <- do.call(rbind, dat)
return(dat_)
}
makenewtable(finaldata, 5)
Scott
On
On 18/05/2011 10:02 PM, Nick Matzke wrote:
Hi,
If I load 2 packages that have a function with the same
name, how do I tell R to run one or the other?
(Instead of having R automatically use the first- or
last-loaded one, whichever it is. (Which is it, by the way.))
Cheers!
Nick
It is the m
Hi.
No tired.
I need the Quasi-Newton and Nelder-Mead algorithm for estimate the
3parameters function.
I need really. please help me.
Thanks a lot.
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