On 16/10/2012 02:33, veepsirtt wrote:
Currently download methods internal, wget, curl and lynx are
available.
If wget is not possible under windows,how to use 'curl'?.
But it is. You need the programs wget or curl installed and in your
path: pre-compiled versions for Windows are available if
Hi All,
I'm running into a problem with GAM (in the MGCV package). When I try
to estimate the model, I get the following error message:
1 fit -
gam(ndvi~s(rain)+s(temp)+s(rainl1)+s(rainl2)+s(rainxY)+s(rainl1xY)+s(rainl2xY)+s(tempxY),
data=dsub, weights=wvec)
Error in while
Hi Sachin,
You may find this tutorial useful:
http://goanna.cs.rmit.edu.au/~fscholer/anova.php
And you'll need the car package; but become yourself familiar with Type I, II
and III sums of squares models before running the Anova; the tutorial explains
these in detail.
Hope it helps.
José
Hi All,
I have a data frame where nearly 10K columns of data, where most of them
have standard deviation( of all rows) as zero.
I want to exclude all the columns from the data frame and proceed to further
processing.
I tried like blow.
*data - read.csv(data.CSV, header=T)
for(i in
hello,
I work on map anamporphosis, and I have managed to :
- import and view mif/mid in R
- work on coordinates in R.
But I don't get how I can transform a list of x, y, id in a sp object (or
other) to visualize my transformation without having to go back in mapinfo
(Actually I can see points,
Hello,
svyvar from the survey package computes variances (with standard errors)
from survey design objects. Is there any way to compute standard
deviations and their standard errors in a similar manner?
Thanks a lot,
Sebastian
__
How to resolve this below Error.I have tried my best but again i have got
error.
Error in plot.new() : figure margins too large.
Iam waiting for Reply
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Hello Prof Brian Ripley
I installed wget for windows from http://users.ugent.be/~bpuype/wget/
http://users.ugent.be/~bpuype/wget/ and got the results
very thanks
veepsirtt
hey
I'd like to divide my data into four seasons. for this I made a function:
Jahreszeit - function(x) {
if (x=02 || x==12) {return(Winter)
}else{
if (x=03 x=05) {return(Fruehling)
}else{
if (x=06 x=08) {return(Sommer)
}else{
if (x=09 x=11) {return(Herbst)
}
Now, I have some trouble to
On Tue, Oct 16, 2012 at 9:08 AM, siddu479 onlyfordigitalst...@gmail.com wrote:
Hi All,
I have a data frame where nearly 10K columns of data, where most of them
have standard deviation( of all rows) as zero.
I want to exclude all the columns from the data frame and proceed to further
Hello,
Your function works,
sapply(1:12, Jahreszeit)
[1] WinterWinterFruehling Fruehling Fruehling Sommer
[7] SommerSommerHerbstHerbstHerbst Winter
but not with your data? We don't know what your data looks like. Post an
example using ?dput
dput( head(dat, 30) )
Hello,
You're calling your dataset 'data' and 'df' in the same instruction,
hence the error. (Even if you were to call it different names in
different instructions...)
Also, both 'data' and 'df' are really bad names for objects, they're
already are R functions. Name your dataset something
Likely you've used layout() or similar to add lots of columns rows
to your plot, where lots is determined by the size of the device.
Simply make the device larger (either just drag it to be larger if its
interactive or change the default arguments if its a non-default
device you initialize
I want to test whether the MEAN of two different variables,
(and different number of observations) are the same. I am
trying to use the anova test but it doesn't seem to like that
the number of observations are different:
a=c(1:5)
b=c(1:3)
aov_test=aov(a~b)
Error in
On 16/10/12 07:32, Andrew Crane-Droesch wrote:
Hi All,
I'm running into a problem with GAM (in the MGCV package). When I try
to estimate the model, I get the following error message:
1 fit -
gam(ndvi~s(rain)+s(temp)+s(rainl1)+s(rainl2)+s(rainxY)+s(rainl1xY)+s(rainl2xY)+s(tempxY),
Hi Andrew,
Could you send me a bit more information (off list, as this is likely to
get into obscure details), please?
In particular can you let me know the mgcv and R version numbers, the
server operating system and, if possible, what BLAS it has installed?
best,
Simon
On 16/10/12 07:32,
Hi all,
I have a list of 2 data, and the list look like below. I wonder what is
the simplest way to extract 'kappa' value (or 'xi' or 'alpha' for the
matter) from each of the data. How can I simply code it without having to
change the list to a dataframe first? Many thanks!
$X19997
Hello,
Try the following.
lapply(lst, `[[`, 'kappa')
Also, please post a data example using ?dput, in case of lists it's
difficult for us to reproduce what you've posted.
Hope this helps,
Rui Barradas
Em 16-10-2012 12:17, Balqis escreveu:
Hi all,
I have a list of 2 data, and the list
ok thanks, but I just got it right. Although I don't know what was the
problem.
just thanks for your effort!!!
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Hi,
How can an xts object be subset using a date in a variable?
E.g. in below, what is the right syntax for Q to get the same value as P?
Obviously the syntax shown below doesnt work...
V is an xts object.
d = 2007-01-01
P - V['2007-01-01/']
Q - V['d/']
Thanks,
J
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View this message in
Hi,
I'm experiencing the same problems.. Installed the package DRM yestarday..
first couple of runs it worked..
I closed down R and now I want to do the same code again and I get the
error:
Error in drm(TotDmShoot ~ DAP, data = DATA, fct = gompGrowth.2()) :
Cluster not specified. Use e.g. y ~
Hi,
I am writing a .R script to plot some data from a file, all info given from
command line.
For example:
Rscript Rtest.R sas_summary.csv sas_score_pre sas_score_post
...will try to plot columns of data sas_score_pre and sas_score_post
from the file sas_summary.csv.
I can extract these
Maybe something along the following lines?
m - rep(list(rnorm(10)),3) # dummy data
data - c()
for (a in 1:length(m)) {
data[a] - m[[a]][1] }
Muhammad
On Tue, Oct 16, 2012 at 1:17 PM, Balqis aehan3...@gmail.com wrote:
Hi all,
I have a list of 2 data, and the list look like below. I
On Tue, Oct 16, 2012 at 5:04 AM, john.dempster78
john.dempste...@gmail.com wrote:
Hi,
How can an xts object be subset using a date in a variable?
E.g. in below, what is the right syntax for Q to get the same value as P?
Obviously the syntax shown below doesnt work...
V is an xts object.
R
HI,
Try this:
To extract kappa:
list1-list(X17=data.frame(xi=784.77,alpha=165.406,kappa=-0.2709),X19998=data.frame(xi=860.783,alpha=215.114,kappa=-0.1769))
sapply(list1,`[`,3)
#$X17.kappa
#[1] -0.2709
#$X19998.kappa
[1] -0.1769
#To extract alpha:
sapply(list1,`[`,2)
#$X17.alpha
#[1]
Whoops... tried to get all descriptive with the argument names before
I posted but after I tested. Try this
toDaySecond - function(date.string, format.string='%d%b%Y:%T', tz='') {
d - as.POSIXct(date.string, format=format.string, tz=tz)
sum(mapply(function(f, l) as.numeric(format(d,
Ok here's the lines of codes:
sIr=replicate(5,ts(arima.sim(list(ar=0),n=150,innov=rgev(150,xi=0.2,
mu=inP[1,5],sigma=inP[1,4]),start.innov=rgev(150,xi=0.2,
mu=inP[1,5],sigma=inP[1,4]
sIrr=data.frame(cbind(sIr))
EsIpr=sapply(sIrr[1:100,1:2],function(m) pargevn(lmom.ub(m)))
HI,
May be this helps.
set.seed(1)
dat1-data.frame(col1=rep(5,10),col2=rnorm(10,15),col3=rep(25,10),col4=rnorm(10,25))
dat1[sapply(dat1,function(x) sd(x)!=0)]
# col2 col4
#1 14.37355 26.51178
#2 15.18364 25.38984
#3 14.16437 24.37876
#4 16.59528 22.78530
#5 15.32951 26.12493
Hi Roland (and Terry),
Is this the model you want to fit?
--a separate 'baseline' hazard for each stratum defined by cov1
--a coefficient for cov2 that is different for each stratum defined by cov1
Then you can have a separate call to coxph for each stratum.
sCox0 - coxph(Surv(time, status) ~
Hi Simon,
I tried your code, but I am getting some error message:
toDaySecond('04MAY2011:08:19:00')
#Error in as.POSIXct(d, format = format.string, tz = tz) :
#object 'd' not found
A.K.
- Original Message -
From: Simon Knapp sleepingw...@gmail.com
To: Jeff Newmiller
Hello everybody,
I've got a problem concerning the function unique. I have got a
data.frame shopdata with 1000 shop which were evaluated at different
points in time.
With function subset I chose those shops with more then 10 employee and
store it in data.frame bigshopdata with 700 shops.
Le mardi 16 octobre 2012 à 14:45 +0200, paladini a écrit :
Hello everybody,
I've got a problem concerning the function unique. I have got a
data.frame shopdata with 1000 shop which were evaluated at different
points in time.
With function subset I chose those shops with more then 10
Véronique,
in addition to Bert's comments, I would like to bring to your attention
that there are several packages that perform
threshold/breakpoint/changepoint estimation in R, including
cumSeg, segmented, strucchange, and bcp for a Bayesian approach
Moreover some packages, such as
Hello,
Your code doesn't work, which libraries are you using? There are several
with function rgev.
And pargevn() also doesn't exist. Nor does the matrix/df inP.
(And, for reproducible example there's no need to replicate 50K times...)
Rui Barradas
Em 16-10-2012 13:41, Balqis escreveu:
Ok
For some reason this seems to work:
d - 04MAY2011:08:19:00
toDaySecond(d)
John Kane
Kingston ON Canada
-Original Message-
From: smartpink...@yahoo.com
Sent: Tue, 16 Oct 2012 05:22:29 -0700 (PDT)
To: sleepingw...@gmail.com
Subject: Re: [R] time extraction and normalization
Hi
Please supply some sample data and code.
The easiest way to supply data is to use the dput() function. Example with
your file named testfile:
dput(testfile)
Then copy the output and paste into your email. For large data sets, you can
just supply a representative sample. Usually,
If I understand your problem correctly (as Milan has pointed out, sample data
and code would help enormously) this should get you where you want:
unique( shopdata$name[ shopdata$employee 10 ] )
If not, something is wrong from the outset (or with my understanding, but then
... see above)!
I need an analogue of uniq -c for a data frame.
xtabs(), although dog slow, would have footed the bill nicely:
--8---cut here---start-8---
x - data.frame(a=1:32,b=1:32,c=1:32,d=1:32,e=1:32)
system.time(subset(as.data.frame(xtabs( ~. , x )), Freq != 0 ))
Perhaps this but your results example did not include Char1.
ibrary(reshape2)
md - structure(list(Coutry = structure(c(3L, 3L, 3L, 3L, 2L, 2L, 1L,
1L), .Label = c(J, M, U), class = factor), State = structure(c(1L,
1L, 4L, 2L, 5L, 5L, 3L, 6L), .Label = c(A, C, K, O, S,
T), class = factor), City =
Thanks for the information, Simon.
I had both the 64-bit version of R and the 64-bit Windows 7, but not the
64-bit version of Java.
Now that I've downloaded the proper version of Java, the problem has been
solved.
Thanks -
SR
Steven H. Ranney
On Mon, Oct 15, 2012 at 5:12 PM, Simon Knapp
Hi,
I want to get a clean succinct list of all levels for all my factor variables.
I have a dataframe that's something like #1 below. This is just an example
subset of my data and my actual dataset has 70 variables. I know how to narrow
down my list of variables to just my factor variables by
Have you looked at using table() directly? If I understand what you
want correctly something like:
table(do.call(paste, x))
Also, if you take a look at the development version of R, changes are
being put in place to allow much larger data sets.
Cheers,
Micael
On Tue, Oct 16, 2012 at 4:03 PM,
On Tue, Oct 16, 2012 at 4:19 PM, Lopez, Dan lopez...@llnl.gov wrote:
Hi,
I want to get a clean succinct list of all levels for all my factor variables.
I have a dataframe that's something like #1 below. This is just an example
subset of my data and my actual dataset has 70 variables. I know
Using unlist() did not produce the result I wanted. I have a dataframe. I tried
playing with the parameters of unlist but each time it just tried to return
each observation.
unlist(x, recursive = TRUE, use.names = TRUE)
Dan
-Original Message-
From: R. Michael Weylandt
Hello,
Inline.
Em 16-10-2012 16:22, Balqis escreveu:
well that function is the one I created, manipulating pargev from evir
package. maybe this is where I have to check I presume.
Maybe yes, we can't tell.
for the 50k,the more the robust perhaps?
Yes, the more the merrier but if you want
Hello,
The problem is with clean?
dat - data.frame(X = sample(letters[1:4], 100, TRUE),
Y = sample(LETTERS[1:6], 100, TRUE),
Z = factor(rep(1:5, 4)))
levs - lapply(dat, levels)
clean - lapply(seq_along(levs), function(i)
paste(names(levs)[i], :, paste(levs[[i]],
On Oct 15, 2012, at 10:13 PM, Nicola Cesca wrote:
I'm trying to get significance of coefficients as for lm() but I
news help.
In order for there to be significance tests for the coefficients, the
problem would needs to be embedded in some sort of a probability
model. Such may be
Perfect!
Thank you!
Dan
-Original Message-
From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
Sent: Tuesday, October 16, 2012 9:03 AM
To: Lopez, Dan
Cc: R. Michael Weylandt; R help (r-help@r-project.org)
Subject: Re: [R] List of Levels for all Factor variables
Hello,
The problem is with
On Oct 13, 2012, at 5:38 PM, Bhupendrasinh Thakre wrote:
HI Team,
I am currently working on problem and stumped on for loop.
Data:
structure(list(Coutry = structure(c(3L, 3L, 3L, 3L, 2L, 2L, 1L,
1L), .Label = c(J, M, U), class = factor), State =
structure(c(1L,
1L, 4L, 2L, 5L, 5L, 3L,
* R. Michael Weylandt zvpunry.jrlyn...@tznvy.pbz [2012-10-16 16:19:27
+0100]:
Have you looked at using table() directly? If I understand what you
want correctly something like:
table(do.call(paste, x))
I wished to avoid paste (I will have to re-split later, so it will be a
performance
Have you tried downloading it manually using a web browser or similar?
If so, compare that downloaded file to that you get from
download.file() and see if they're the same. If the same, the problem
is elsewhere (which is not unlikely to be the case). /H
On Mon, Oct 15, 2012 at 6:15 PM,
Hi,
Using R 2.15.1 on Windows 7. Have installed both versions 32 and 64bit.
In both of them among others I have installed a package rgenoud
When I open R gui of 32bit and write library(rgenoud) it responds by
showing a functional rgenoud version 5.7-8. The same it does on Rgui
64bit.
Now I
On 16/10/2012 12:29 PM, Sam Steingold wrote:
* R. Michael Weylandt zvpunry.jrlyn...@tznvy.pbz [2012-10-16 16:19:27
+0100]:
Have you looked at using table() directly? If I understand what you
want correctly something like:
table(do.call(paste, x))
I wished to avoid paste (I will have to
On 16/10/2012 12:41 PM, Aldi Kraja wrote:
Hi,
Using R 2.15.1 on Windows 7. Have installed both versions 32 and 64bit.
In both of them among others I have installed a package rgenoud
When I open R gui of 32bit and write library(rgenoud) it responds by
showing a functional rgenoud version 5.7-8.
Dear R usuer,
I need to fit logistic regression with binomial response. The
objective is to compare treatment groups controlling other categorical
and continuous predictors. The GLM procedure with
family=binomial(Logit) gives me parameters estimates as well as odd
ratios. But objective is to
HI,
You can also try this:
set.seed(1)
dat1-data.frame(col1=factor(sample(1:25,10,replace=TRUE)),col2=sample(letters[1:10],10,replace=TRUE),col3=factor(rep(1:5,each=2)))
sapply(lapply(mapply(c,lapply(names(sapply(dat1,levels)),function(x)
x),sapply(dat1,levels)),function(x)
Thank you Duncan,
No I did not install R from cygwin. R is installed with windows 7.
I am calling R with a symbolic link from /usr/bin part of cygwin paths,
but my symbolic link is pointing to /usr/bin/R -
/cygdrive/d/RHome/bin/R.exe
Is it possible R is lost in forward paths recognized by
Probably because when you run it from Cygwin the R_LIBS variable does not point
to the user and install library directories. I don't know how Rgui knows where
they are (registry?) but you can look in the .Library and .Library.site
variables to see the results.
In a case like this, posting
I would guess that you installed rgenoud as user, not as administrator.
That would put the
file inside
c:/Users/YourName/AppData/Local/VirtualStore/Program Files/R
instead of where you think it is. I can imagine that could easily cause
confusion.
Rich
On Tue, Oct 16, 2012 at 1:14 PM, Aldi
I wonder if there is a simple way of doing this?
My data is very simple, a right censored outcome (T,delta) and the
predictor is simply Z*t, i.e., the cox model is like
h(t)=h0(t)exp(beta*Z*t)
can this be done with coxph?
__
R-help@r-project.org
On 16/10/2012 1:14 PM, Aldi Kraja wrote:
Thank you Duncan,
No I did not install R from cygwin. R is installed with windows 7.
I am calling R with a symbolic link from /usr/bin part of cygwin paths,
but my symbolic link is pointing to /usr/bin/R -
/cygdrive/d/RHome/bin/R.exe
Is it possible R is
Thank you Richard and Jeff,
There is a difference in the reporting of extra packages in 32bit 64bit,
see following:
32bit although it does not report the extra package when I call it with
Rgui it has the rgenoud.
Instead 64bit Rgui it reports that extra package of rgenoud.
More info follows:
* Duncan Murdoch zheqbpu.qha...@tznvy.pbz [2012-10-16 12:47:36 -0400]:
On 16/10/2012 12:29 PM, Sam Steingold wrote:
x is sorted.
sparseby(data=x, INDICES=x, FUN=nrow)
this takes forever; apparently, it does not use the fact that x is
sorted (even then - it should not take more than a few
why?
rle
Run Length Encoding
lengths: int [1:1650061] 2 2 8 2 4 5 6 3 26 46 ...
values : chr [1:1650061] 4bbf9e94cbceb70c BG bg 4fbbf2c67e0fb867 SK sk ...
as.data.frame(rle)
Error in as.data.frame.default(vertices.rle) :
cannot coerce class 'rle' into a data.frame
it seems that
rle.df
Hi,
I keep getting the below error regarding rJava which is required by package
xlsx (I have used it in the past to directly import data from Excel 2010).
I was on R version 2.15.0 when I was getting this error this morning. So I
upgraded to 2.15.1 but still the same problem. I tried
Hi,
vec1-c(1,2,3,4,4,5,6)
str(rle(vec1))
#List of 2
# $ lengths: int [1:6] 1 1 1 2 1 1
# $ values : num [1:6] 1 2 3 4 5 6
#- attr(*, class)= chr rle
do.call(data.frame,rle(vec1))
# lengths values
#1 1 1
#2 1 2
#3 1 3
#4 2 4
#5 1 5
#6
Here follows also the Sys.getenv():
R Windwos 32bit Rgui run:
==
R_HOME
D:/RHome
R_LIBS_USER
C:\\Users\\aldi\\Documents/R/win-library/2.15
R_USER
C:\\Users\\aldi\\Documents
R cygwin run:
===
Sys.getenv()
/usr/bin/R
BIBINPUTS
* Duncan Murdoch zheqbpu.qha...@tznvy.pbz [2012-10-16 12:47:36 -0400]:
sparseby(data=x, INDICES=x, FUN=nrow)
Error in `[-.data.frame`(`*tmp*`, index, , value = list(user = c(2L, :
missing values are not allowed in subscripted assignments of data frames
--
Sam Steingold
On Oct 16, 2012, at 12:54 PM, Sam Steingold s...@gnu.org wrote:
why?
rle
Run Length Encoding
lengths: int [1:1650061] 2 2 8 2 4 5 6 3 26 46 ...
values : chr [1:1650061] 4bbf9e94cbceb70c BG bg 4fbbf2c67e0fb867 SK sk
...
as.data.frame(rle)
Error in as.data.frame.default(vertices.rle) :
Hello,
I cannot answer directly to your question, but I had problems with
package xlsx and for what I've seen on R-Help so had others. Maybe you
are using the wrong architecture of R (32/64 bits).
Package XLConnect is more flexible and a bit more complex, but it comes
with a vignette that
I do not think that this has anything to do with the issue at hand,
but just to clarify ...
On Tue, Oct 16, 2012 at 10:18 AM, Jeff Newmiller
jdnew...@dcn.davis.ca.us wrote:
Probably because when you run it from Cygwin the R_LIBS variable does not
point to the user and install library
On 16/10/2012 1:46 PM, Sam Steingold wrote:
* Duncan Murdoch zheqbpu.qha...@tznvy.pbz [2012-10-16 12:47:36 -0400]:
On 16/10/2012 12:29 PM, Sam Steingold wrote:
x is sorted.
sparseby(data=x, INDICES=x, FUN=nrow)
this takes forever; apparently, it does not use the fact that x is
sorted (even
* Duncan Murdoch zheqbpu.qha...@tznvy.pbz [2012-10-16 14:22:51 -0400]:
On 16/10/2012 1:46 PM, Sam Steingold wrote:
* Duncan Murdoch zheqbpu.qha...@tznvy.pbz [2012-10-16 12:47:36 -0400]:
On 16/10/2012 12:29 PM, Sam Steingold wrote:
x is sorted.
sparseby(data=x, INDICES=x, FUN=nrow)
Thank you all including Bert,
In my opinion the library() when is working in cygwin env., does not
check the additional libraries, where R installed the additional packages.
So the solution for this problem was to define manually in my programs
where is located the library of additional
On 16/10/2012 3:02 PM, Aldi Kraja wrote:
Thank you all including Bert,
In my opinion the library() when is working in cygwin env., does not
check the additional libraries, where R installed the additional packages.
So the solution for this problem was to define manually in my programs
where is
Veronique,
Can you write down the likelihood function in R and send it to me (this is very
easy to do, but I don't want to do your work)? Also send me the code for
simulating the data. I will show you how to fit such models using optimization
tools.
Best,
Ravi
hi all
i'm trying to open gplots library but i keep on getting this error:
/Loading required package: gdata
Error in readRDS(mapfile) : embedded nul in string: 'ref\0\0\002\0\0\0\0'
Error : unable to load R code in package ‘gdata’
Error: package ‘gdata’ could not be loaded/
if i run session
Hello. I apologize in advance for the VERY lengthy e-mail. I endeavor to
include enough detail.
I have a question about survival curves I have been battling off and on for
a few months. No one local seems to be able to help, so I turn here. The
issue seems to either be how R calculates
Sam Steingold s...@gnu.org writes:
I need an analogue of uniq -c for a data frame.
The count.rows() function is the R analogue.
See
http://orgmode.org/worg/org-contrib/babel/examples/Rpackage.html#sec-6-1
No need to install the package - just copy and paste the function into an
R session.
Hi All,
I am trying to use optim() to minimize a function with a penalty function
term. This is a simple model bioeconomic model of a fishery. The penalty
function constrains the amount of effort (f) at 9. This works fine. The code
is:
**
nfleets-2
M-1
M-array(M,dim=c(nfleets))
N-1000
Hi, everyone
I need to create a 429497 x 429497 matrix.
When I use
*matrix(0,429497,429497)*
I got the error information : Error in matrix(0, 429497, 429497) : too many
elements specified
Then I use ff package, try to store this matrix on disk
* x-ff(0,dim=c(429497,429497)*
And I got the error
You are right! But that was my bad, I was trying to simplify the part of the
data I used in the example. The real data have exactly the same IDs as both
rows and columns. Thanks for catching that. :)
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Thank you both for the suggestion. So this means that when the vector is one
of characters the string inside it is interpreted as itself, rather than as
an index. That makes so much sense! Thanks again :)
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Okay, I've now tried to the predict function and get the SE, although it seem
to calculate SE for each observation from the line (I assume), while I want
the CI-interval and SE for each line fitted line for the treatment. I do not
really understand what parameter mean these SEs are calculated
On 16.10.2012 17:13, Seb wrote:
hi all
i'm trying to open gplots library but i keep on getting this error:
/Loading required package: gdata
Error in readRDS(mapfile) : embedded nul in string: 'ref\0\0\002\0\0\0\0'
Error : unable to load R code in package ‘gdata’
Error: package ‘gdata’ could
You said you wanted the equivalent of the Unix 'uniq -c' but said
that xtab's results were roughly right and the rle might be what
you want. rle() is the equivalent of 'uniq -c', they both output the
lengths of runs of identical elements. if the data is sorted they
are equivalent to using
Duncan wrote
You can override this by setting the R_LIBS_USER variable in Cygwin to
C:/Users/aldi/Documents/R/win-library/2.15 before you start R; then
things should be fine.
If you use R_LIBS_USER then you have to be careful when later calling .libPaths
to add more libraries:
% env
On Tue, Oct 16, 2012 at 10:40 PM, Sebastian Weirich
sebastian.weir...@iqb.hu-berlin.de wrote:
Hello,
svyvar from the survey package computes variances (with standard errors)
from survey design objects. Is there any way to compute standard deviations
and their standard errors in a similar
On Tue, Oct 16, 2012 at 8:46 PM, lrl liurl1...@gmail.com wrote:
Hi, everyone
I need to create a 429497 x 429497 matrix.
When I use
*matrix(0,429497,429497)*
I got the error information : Error in matrix(0, 429497, 429497) : too many
elements specified
Then I use ff package, try to store
On Tue, Oct 16, 2012 at 7:07 PM, Marc Schwartz marc_schwa...@me.com wrote:
On Oct 16, 2012, at 12:54 PM, Sam Steingold s...@gnu.org wrote:
why?
rle
Run Length Encoding
lengths: int [1:1650061] 2 2 8 2 4 5 6 3 26 46 ...
values : chr [1:1650061] 4bbf9e94cbceb70c BG bg 4fbbf2c67e0fb867 SK
I was looking at rank() and I came across:
...
first = sort.list(sort.list(xx)), ...
line 32 of rank.r [1]
sort.list(x) returns the indices of the values of x in ascending (by
default) order. So sort.list(sort.list(x)) returns the same list.
So, what am I missing here?
-Tyler
[1]
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of Tyler Ritchie
Sent: Tuesday, October 16, 2012 2:23 PM
To: r-help@r-project.org
Subject: [R] Question about use of sort.list(sort.list(x)) in rank.r
I was looking at rank()
Hello,
If you want confidence intervals for the beta coefficients of the model,
try the following.
ci_lm - function(object, level = 0.95){
sfit - summary(object)
beta - sfit$coefficients[, 1]
se - sfit$coefficients[, 2]
df - sfit$df[1]
alpha - 1 - level
lower - beta +
Dear R-Users,
I'd like to have your help on this problem:
I have two vectors:x- c(0,1,0,0,0,0,0,0,0,0,0,1,0)y-
c(0,0,-10,0,0,-10,0,-10,0,0,0,0,0)
And I want to know where the value -500 in y have a correspondence value 1 in
x.Considering a buffer of one position before and after in x.i.e. in
Your question does not seem to make sense - there is no value of -500
in Y (did you mean -10?). Anyway, I think this might work:
which(y==-10 (x==1 | c(0, x[-length(x)]) == 1 | c(x[-1], 0) == 1))
... though one would think there is a more elegant way
On Wed, Oct 17, 2012 at 10:07 AM,
Dear List members
I want to do the sliding window analysis of some specific values. Here is my
code:
require(zoo)
dat - read.table(chr1.txt, header = TRUE, sep=\t)
dat2 - cbind(dat[1,3]) #The first column is also important. It represents the
position of the site on the chromosome.
TS -
Should it be .Machine$integer.max- 429497 ^2?
The matrix is not a sparse matrix, but a symmetric matrix.
It's a process of feature selection. I am choosing the most important
variables from 439497 variables. Now I am considering divide the whole
dataset into several part and process a small
I came seeking the answer to the same question.
Looking over my work, the equation I had to deliver the times variable (in
rep, the integer vector giving the (non-negative) number of times to repeat
each element , and seeking to replicate a number NaN times too.
Thanks!
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I would like to create a frequency table with custom bands.
seq1 = seq(0, 100, by = 5)
seq2 = seq(100, 1000, by = 100)
Bands = c(seq1, seq2)
Prices = sample(1:1000, 200, replace=F)
How would I go about find the frequency of prices within each band?
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Good afternoon,
In the code below, I have a set of functions (m1,m2,m3,s1,s2, and s3) which
represent response surface designs for the mean and variance for three response
variables, followed by an objective function that uses the Big M method to
minimize variance (that is, push s1, s2, and
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