Hello,
Sorry for not understanding your problem, but it really seemed like
homework.
Now, when I answered scale(x) I meant it, it transforms a matrix in (x -
mean)/sd, column by column.
If you're new to R, to use the on-line help the instruction is
help("scale")
?scale # shortcut
As for
On 06/03/2012 11:27 PM, stef salvez wrote:
Dear R users,
I have the following problem
I have a panel data across countries and individuals. For each country
I have a sequence of dates
For France for example
22/02/09
22/03/09
19/04/09
17/05/09
12/07/09
09/08/09
06/09/09
04/10/09
01/11/09
29/11/0
Hello R-users,
I'd like to load data from a CSV-file to a 3-dimensional array. However
the default seems to be 2-dimensional. It would also be ok to load the
data and then convert it to a 3-dimensional structure.
I've been trying:
dat = read.csv(filename)
myArr = as.array(dat, dim = c(12
Hi
Dear friend,
I encountered a problem when I read in one huge txt file which contains the
character *""*
when I run read.table("D:/test.txt",header=T,sep="\t",quote="")
the lines can only be read before "" appears.
Once "" appears, it will stop there.
Could you help me on this .
Many th
Hello dear R-users,
I have a problem in my code about ignoring NA values without removing them.
I'm working on a list of files. The aim is to fill one file from another
according to the highest correlation (correlation coeff between all my
files, so the file which looks like the most to the one I
Thanks for the replies,
1st, installing the BLAS library indeed worked out, and I was able to load
the packages.
2nd, I was trying to build from repository directly. But even worse, I
cannot tell you if i touched the BLAS linker flags, BUT there seem to be
more problems of a similar nature, ie.
I find that avoiding using the return() function at all makes my code easier to
follow. In your case it is simply incorrect, though, since ifelse is a vector
function and return is a control flow function.
Your code is not reproducible and your description isn't clear about how you
are handling
On 06/04/2012 12:14 AM, Prakash Thomas wrote:
> I am getting the following error while trying to install RPostgreSQL.
> RS-PostgreSQL.h:23:26: fatal error: libpq-fe.h: No such file or directory
> compilation terminated.
You need the PostgreSQL front end development headers installed. How you
On 6/3/2012 2:42 AM, Jinyan Huang wrote:
Dear list,
Some times I want to modify the pdf file which produced by R with
illustrator. But when I use Illustrator open the pdf file, it often
makes the pdf some little changed. Anyone have some suggestions? Is it
better to use other file type, not pdf?
Dear R-Users, I'd like to have some tips for a ternaryplot ("vcd").
I have this dataframe:
a<- c (0.1, 0.5, 0.5, 0.6, 0.2, 0, 0, 0.00417, 0.45) b<- c
(0.75,0.5,0,0.1,0.2,0.951612903,0.918103448,0.7875,0.45)c<- c
(0.15,0,0.5,0.3,0.6,0.048387097,0.081896552,0.20833,0.1)
Thanks for the links!
It is a very nice tutorial.
On Mon, Jun 4, 2012 at 8:47 PM, Michael Friendly wrote:
> On 6/3/2012 2:42 AM, Jinyan Huang wrote:
>>
>> Dear list,
>>
>> Some times I want to modify the pdf file which produced by R with
>> illustrator. But when I use Illustrator open the pdf fi
On 06/03/2012 07:38 PM, slipkid90515 wrote:
Hi All,
After a lengthy battle just to get the package installed, I am not able to
actually use Rgraphviz to generate any plots. I tried just using the sample
code in the documentation
(http://www2.warwick.ac.uk/fac/sci/moac/people/students/peter_cock
Hi all,
I have done a backward stepwise selection on a full binomial GLM where the
response variable is gender.
At the end of the selection I have found one model with only one explanatory
variable (cohort, factor variable with 10 levels).
I want to test the significance of the variable "cohort"
Hi all,
Like a lot of people I noticed that I get different results when I use nls
in R compared to the exponential fit in excel. A bit annoying because often
the R^2 is higher in excel but when I'm reading the different topics on this
forum I kind of understand that using R is better than excel?
On Jun 4, 2012, at 5:08 AM, David Zastrau wrote:
Hello R-users,
I'd like to load data from a CSV-file to a 3-dimensional array.
However
the default seems to be 2-dimensional.
No. that is not the problem.
It would also be ok to load the
data and then convert it to a 3-dimensional structu
Hello,
There's a ascii 26, hex 1a, between the quotes. Get rid of it and it
will probably work.
Hope this helps,
Rui Barradas
Em 04-06-2012 08:43, climmi escreveu:
Hi
Dear friend,
I encountered a problem when I read in one huge txt file which contains the
character *""*
when I run read
On Jun 4, 2012, at 7:00 AM, lincoln wrote:
Hi all,
I have done a backward stepwise selection on a full binomial GLM
where the
response variable is gender.
At the end of the selection I have found one model with only one
explanatory
variable (cohort, factor variable with 10 levels).
I wa
Hello,
'ifelse' is vectorized, what you want is the plain 'if'.
if(all(is.na(xx))) return(NA)
Hope this helps,
Rui Barradas
Em 04-06-2012 09:56, jeff6868 escreveu:
Hello dear R-users,
I have a problem in my code about ignoring NA values without removing them.
I'm working on a list of files.
On Mon, Jun 04, 2012 at 11:08:24AM +0200, David Zastrau wrote:
> Hello R-users,
>
> I'd like to load data from a CSV-file to a 3-dimensional array. However
> the default seems to be 2-dimensional. It would also be ok to load the
> data and then convert it to a 3-dimensional structure.
>
> I've
Thanks for answering Jeff.
Yes sorry it's not easy to explain my problem. I'll try to give you a
reproductible example (even if it'll not be exactly like my files), and I'll
try to explain my function and what I want to do more precisely.
Imagine for the example: df1, df2 and df3 are my files:
df1
Hello Rui,
Sorry I read your post after having answered to jeff.
If seems effectively to be better than ifelse, thanks. But I still have some
errors:
Error in x[1:8700, 1] : incorrect number of dimensions AND
In is.na(xx) : is.na() applied to non-(list or vector) of type 'NULL
It seems to have m
Hello again,
The complete function would be
na.fill <- function(x, y){
# do this immediatly, may save copying
if(all(is.na(y[1:8700,1]))) return(NA)
i <- is.na(x[1:8700,1])
xx <- y[1:8700,1]
new <- data.frame(xx=xx)
x[1:8700,1][i] <- predict(lm(x[1:8700,1]~xx, na.action=n
Thanks a lot, appreciate it.
On Sun, Jun 3, 2012 at 10:25 PM, David Winsemius wrote:
>
> On Jun 3, 2012, at 3:22 PM, Kai Mx wrote:
>
> Hi all,
>> probably really simple to solve, but having no background in programming I
>> haven't been able to figure this out: I have two dataframes like
>>
>> d
I'm trying to read in a tab separated table with read.delim().
I don't particularly care what the row names are.
My data file looks like this:
start stopSymbol Insert sequence Clone End Pair FISH
203048 67173930ABC8-43024000D23TI:993812543
TI:993834585
255176 87
Try help("read.delim") - always a good strategy before using a function for
the first time:
In it, you will find: "Using row.names = NULL forces row numbering. Missing
or NULL row.names generate row names that are considered to be 'automatic'
(and not preserved by as.matrix)."
---
On Jun 4, 2012, at 12:54 PM, arun wrote:
Hi Dave,
I tested the "as.array". It gives a different dimension when
compared to "array"
Interesting. `as.array basically throws away a 'dim' argument and
substitutes length(x). Thanks for the correction.
--
David.
set.seed(1)
datE<-data.fra
I did read that, and I still don't understand why I have a column
called row.names.
I used "row.names = NULL" in order to get numbered row names, which was
successful:
> row.names(BACS)
[1] "1" "2" "3" "4"
I don't see what this has to do with an extraneous column name. Can you be
more explicit a
Actually, I think it's ?data.frame that he should read.
The salient points are that:
1. All data frames must have unique row names. If not provided, they
are produced. Row numbers **are** row names.
2. The return value of read methods are data frames.
-- Bert
On Mon, Jun 4, 2012 at 11:05 AM, Da
Hello,
It must be something else. Mandatory row names have nothing to do with a
column header.
I've put the data example in a tab separated file and the strange
behavior was not reproduced.
> read.delim("test.txt", row.names=NULL, fill=TRUE)
start stop Symbol Insert.sequence
You will probably need to show us the first few lines of the .csv file.
Assuming that the lines look like this
start,stop,Symbol,Insert sequence,Clone End Pair,FISH
203048,67173930,ABC8-43024000D23,TI:993812543,TI:993834585
255176,87869359,ABC8-43034700N15,TI:995224581,TI:995237913
1022033,1060472
To jump into the fray, he really needs to read the Details section of
?read.table and arguably, the source code for read.table().
It is not that the resultant data frame has row names, but that an additional
first *column name* called 'row.names' is created, which does not exist in the
source d
Nerak hotmail.com> writes:
>
> Hi all,
>
> Like a lot of people I noticed that I get different results when I use nls
> in R compared to the exponential fit in excel. A bit annoying because often
> the R^2 is higher in excel but when I'm reading the different topics on this
> forum I kind of un
Hi All,
As you may already know, Vertica announced it now supports user defined R
function. Does anybody tried this alreay or maybe have more info than Vertica
site?
http://www.vertica.com/content/vertica-an-hp-company-enables-users-to-connect-access-analyze-and-manage-any-data-anywhere/
Best,
On 6/4/2012 12:12 PM, Marc Schwartz wrote:
To jump into the fray, he really needs to read the Details section of
?read.table and arguably, the source code for read.table().
It is not that the resultant data frame has row names, but that an
additional first *column name* called 'row.names' is cre
Dear Rui Barradas, Mackay and all R Users,
Thankyou
very much for your reply. You helped me a lot. I got what I wanted. I just want
one more favor from you, if you could.
Suppose
I have certain number of lists of data frame, say 50. Each list has yearly data
in it. Of-course, some lists have
Hi everybody!
I have a sample with n=2.000. This sample contains rare events (10, 20, 30
individuals with a specific illness).
Now I'd like to do a logistic regression in order to identify risk factors.
I have several independent variables on an interval
scale.
Does anyone know whether the number
Dear R People:
Are there any courses which describe how to use regular expressions in
R, please? Or any books, please?
I know a little bit (very little) but would like to know more.
Thanks,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University o
On Jun 4, 2012, at 3:48 PM, Erin Hodgess wrote:
> Dear R People:
>
> Are there any courses which describe how to use regular expressions in
> R, please? Or any books, please?
>
> I know a little bit (very little) but would like to know more.
>
> Thanks,
> Erin
Hi Erin,
The two places that
Thanks!
Jean
Uwe Ligges wrote on 05/26/2012 09:30:28
AM:
>
> On 23.05.2012 17:20, Jean V Adams wrote:
> > I'm trying to understand how a latent state matrix is updated by the
MCMC
> > iterations in a WinBUGS model, using the package R2WinBUGS and an
example
> > from Kery and Schaub's (2012)
hi
I have any problems with cspade; I convert my columns in factor, and then I
coerce my data frame in transaction; but when I want do the cspade exit this
message:
slot transactionInfo: missing 'sequenceID' or 'eventID
So I ask to everybody: "somebody help me?" because i tried everything
witho
Hello!
> I have a very simple set of data and I would like to analyze
> them with probit analysis.
> The data are: X Event Trial
> 1210 8 8
> 121 6 8
> 60.5 6 8
> I want to estimate the value of X that will give a 95% hit
> rate (Event/Trial) and the corresponding 95%
Hi, I run this code to get the power of the test for modified bartlett's
test..but I'm not really sure that my coding is right..
#normal distribution unequal variance
asim<-5000
pv<-rep(NA,asim)
for(i in 1:asim)
{print(i)
set.seed(i)
n1<-20
n2<-20
n3<-20
mu<-0
sd1<-sqrt(25)
sd2<-sqrt(50)
sd3<-sqrt
Hi, I run this code to get the power of the test for modified bartlett's
test..but I'm not really sure that my coding is right..
#normal distribution unequal variance
asim<-5000
pv<-rep(NA,asim)
for(i in 1:asim)
{print(i)
set.seed(i)
n1<-20
n2<-20
n3<-20
mu<-0
sd1<-sqrt(25)
sd2<-sqrt(50)
sd3<-sqrt
I am trying to run an R script to create a .png file containing a tree map on a
headless Linux server using Xvfb. When I try to run tmPlot, I get the
following errors and warnings:
Error in grid.Call(L_textBounds, as.graphicsAnnot(x$label), x$x, x$y, :
X11 font -adobe-helvetica-%s-%s-*-*-%d
(Just learning R)
I have this vector:
v <- c(1:10)
Now, I want to multiply each element of that vector with a scalar value
multiplied with its index:
vm <- v * scalar * indexOfCurrentElementOf_v
Is that possible without using a loop?
In a loop I would do this:
for (i in 1:length(a)) a[i] <-
So sorry,
My response variable is "site" (not "gender"!).
The selection process was:
> str(data)
'data.frame': 1003 obs. of 5 variables:
$ site : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
$ sex : Factor w/ 2 levels "0","1": NA NA NA NA 1 NA NA NA NA NA ...
$ age : Factor w/ 2
Hi list members,
I saw a couple lavaan posts here so I think I’m sending this to the
correct list.
I am trying to run a multigroup analysis with lavaan in order to
compare behavioural correlations across two populations. I’m following
the method suggested in the paper by Dingemanse et al. (2010)
Hi Dave,
I tested the "as.array". It gives a different dimension when compared to
"array"
set.seed(1)
datE<-data.frame(A=rnorm(30,0.5))
myArr=as.array(unlist(datE),dim=c(5,2,3))
> myArr
A1 A2 A3 A4 A5 A6
-0.12645381 0.68364332 -0.33562861
On Jun 4, 2012, at 3:47 PM, David Studer wrote:
> Hi everybody!
>
> I have a sample with n=2.000. This sample contains rare events (10, 20, 30
> individuals with a specific illness).
> Now I'd like to do a logistic regression in order to identify risk factors.
> I have several independent variabl
Hello,
I'm glad it helped.
To answer to this new question, we need to see what your data looks like.
When you say 'list' are you refering to the type of R data structure
'list'? Or to data.frames?
For us to see the best way is to use function dput. Example:
df1 <- data.frame(A=rnorm(10), B=r
On Sun, Jun 3, 2012 at 3:43 AM, peter dalgaard wrote:
>
> On Jun 2, 2012, at 15:43 , jalantho...@verizon.net wrote:
>
>> Why is igraph 0.5.5 not on the igraph sourceforgwe home page?
>
> You'll have to ask the igraph developers... Possibly, they just ran out of
> "round tuits".
Yes, kind of. :)
You could try the Cairo() device from the Cairo package. It's my
understanding (from a time when I was using Xvgb for the same reason) that
Cairo does not depend on X Windows.
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On Sat, Jun 2, 2012 at 9:58 AM, jalantho...@verizon.net
wrote:
> Could someone tell me the difference between igraph and igraph0?
>
> I searched the CRAN web site, but cannot find an explantion of the
> differences.
You are right that this should be in the package description.
There is no diffe
On Jun 4, 2012, at 4:25 PM, Erdal Karaca wrote:
(Just learning R)
I have this vector:
v <- c(1:10)
Now, I want to multiply each element of that vector with a scalar
value
multiplied with its index:
vm <- v * scalar * indexOfCurrentElementOf_v
Almost:
vm <- v * scalar * seq(v)
Is th
On Mon, Jun 4, 2012 at 4:48 PM, Erin Hodgess wrote:
> Dear R People:
>
> Are there any courses which describe how to use regular expressions in
> R, please? Or any books, please?
>
> I know a little bit (very little) but would like to know more.
>
You might want to go through the regular express
i realy appreciate your concern..
here is a small piece of my data. if you see the first and last part data, they
contain 366 entries but the middle one has 365 entries. i want to put first and
last entries is one matrix.
list(c(0.86, 0.86, 0.86, 0.86, 0.86, 1.08, 1.08, 1.08, 1.08,
1.08, 1.08,
Hello,
Just learning the alphabet? If yes, there's a difference between 'v' and
'a'.
Now more seriously. Your description and your loop don't do the same.
Description:
vm <- scalar * v * seq_along(v)
Loop:
a <- scalar * a
Also, seq_along is the way to do it, it works even if length(a) == 0.
Hello,
Try
# 'x' is your list
xlen <- sapply(x, length)
i1 <- which(xlen == 365)
i2 <- which(xlen == 366)
mat365 <- matrix(unlist(x[i1]), nrow=365)
mat366 <- matrix(unlist(x[i2]), nrow=366)
Hope this helps,
Rui Barradas
Em 04-06-2012 22:46, eliza botto escreveu:
i realy appreciate your con
On Jun 4, 2012, at 11:31 AM, lincoln wrote:
So sorry,
My response variable is "site" (not "gender"!).
The selection process was:
If there is a natural probability interpretation to "site"==1 being a
sort of event, (say perhaps a non-lymphatic site for the primary site
of a lymphoma) th
FOR what it's worth this dev.hold/dev/flush combination does not seem
to prevent flickering on the quartz device on Macs.
Best,
luke
On Sat, 2 Jun 2012, Duncan Murdoch wrote:
On 12-06-02 4:02 PM, Daniel Carr wrote:
Most of my animations that used to work
on windows() version 2.11.1 and earli
dear rui,
lots of hugs for you.
thnkyou very much 4 your support.
eliza
> Date: Mon, 4 Jun 2012 22:58:12 +0100
> From: ruipbarra...@sapo.pt
> To: eliza_bo...@hotmail.com
> CC: r-help@r-project.org
> Subject: Re: Spliting Lists into matrices
>
> Hello,
>
> Try
>
> # 'x' is your list
> xlen
Dear R users,
we generally apply approx() command to a list data. how can we apply this
command to a matrix, so that we can approximate 366 readings from certain
number of each column over 365 intervals??
hope i am clear in my statement.
eliza botto
> From: eliza_bo...@hotmail.com
> To: ruipba
On Sun, Jun 3, 2012 at 4:11 PM, Aziz, Muhammad Fayez wrote:
>
> Hmm interesting. To come to think of it there could be many disconnected
> components in the graph and thus there should be a generic way to either
> mitigate the disconnectedness in the dendrogram or in the original graph. I
> had
Dear R users,
we generally apply approx() command to a list data. how can we apply this
command to a matrix, so that we can approximate 366 readings from certain
number of each column over 365 intervals?? more precisely, i want to
interpolate 366 discharge readings, in each 8 columns of a matr
Dear R users,
we generally apply approx() command to a list data. how can we apply this
command to a matrix, so that we can approximate 366 readings from certain
number of each column over 365 intervals?? more precisely, i want to
interpolate 366 discharge readings, in each 8 columns of a mat
Dear R users,
I have data on 4 types of interest rates. These rates evolve over
time and across regions of countries . So for each type of interest
rates I want to run a regression of rates on some other variables.
So my regression for one type of interest rate will be I_{ij}_t= a
+regressors
Hi Rui
You got in before me - time zone differences and other commitments
I have done a base graphics of my lattice with your data - it might
give some more insight
Sites <- 1:92
windows(8,10)
par(mfrow = c(12,8),
mai = c(0,0,0,0)
)
invisible(
sapply(seq_len(nc), function(j)
Hi all,
Could you please point me to good materials on various
tricks/intuitions/techniques of regression, and hopefully in R?
For example, what does lm(y~ x * w - 1) mean vs. lm(y ~ x/w -1 ) vs. lm (y
~ x:w-1), etc...
I just found that even simple linear regression is not that simple and
there
Hi Michael,
This is far from exhaustive (I wrote it as an introduction some years
ago) but you may find it useful to start:
https://joshuawiley.com/R/formulae_in_R.aspx
Cheers,
Josh
On Mon, Jun 4, 2012 at 9:06 PM, Michael wrote:
> Hi all,
>
> Could you please point me to good materials on vari
Try Frank Harrell's RMS book.
Michael Weylandt
On Jun 5, 2012, at 12:06 AM, Michael wrote:
> Hi all,
>
> Could you please point me to good materials on various
> tricks/intuitions/techniques of regression, and hopefully in R?
>
> For example, what does lm(y~ x * w - 1) mean vs. lm(y ~ x/w -1
Stef:
1. Read and follow the posting guide. I could make no sense of your
post. This may be because I didn't work hard enough to decrypt it -
which I shouldn't have to do -- or because I'm too stupid -- which I
can't do anything about anyway.
2. What does this have to do with R anyway? Try postin
1. Please follow the posting guide and provide a small reproducible
example. See ?dput to provide data.
2. Please do not double post.
-- Bert
On Mon, Jun 4, 2012 at 4:47 PM, eliza botto wrote:
>
> Dear R users,
> we generally apply approx() command to a list data. how can we apply this
> com
Thanks all, that worked!
Yes, it should be
for (i in 1:length(a)) a[i]<- scalar * a[i] * i
And now is...
a <- a * scalar * seq_along(a)
That is almost as cool as the PERL programming language :-)
2012/6/4 Rui Barradas
> Hello,
>
> Just learning the alphabet? If yes, there's a difference betw
Dear R-listers,
I am giving part of my R code :
###
n=15
m=1
library("partitions")
library("gregmisc")
library("combinat")
x = t(restrictedparts(n-m,m))
l = length(x[,1])
for(u in 1:l){
A= unique(matrix( unlist(permn(x[u,])), ncol=m,
Hi Trinh,
Please check ?dose.p() from library MASS.
Since you wanted the rate, the code was a bit modified from the example.
> datPr
X Event Trial
1 1210.0 8 8
2 121.0 6 8
3 60.5 6 8
fm<-glm(Event/Trial~X,data=datPr,family=quasibinomial(link=probit))
XD50<-d
Thank you ! Rui Barradas .
Yes, the arrow character "" can not be recoginised by R.
the character will be treat as a stop command...
So I replace all the character "" with space, then it works..
Just want to know is there any way for R to handle it.
Not sure does eveybody able to view the cha
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