Hi I want to import around 75 files. Each file has a name and a time and is
comma separated.
For example some of my file names are
Asheville_Dec.txt
Asheville_Jan.txt
Asheville_Feb.txt
Charlotte_Dec.txt
Chapelhill_Jan.txt
The time months are only Dec Jan and Feb. The locations are different. I
Hello everyone,
Im relatively new to R and new to the randomForest package and have scoured
the archives for help with no luck. I am trying to perform a regression on a
set of predictors and response variables to determine the most important
predictors. I have 100 response variables collected
Hi
Best way is to put those files in one directory and start R from this directory
or set this directory as working one.
?set.wd
Then you can use list.files() to get file names.
?list.files
myfiles-list.files()
Than strip txt. E.g. by
filenames-strsplit(list.files(), \\.))
After that you
Hm. We also do not know what will be done with those files. Save/load is an
option but in that case nothing prevents you to use original file and read.*
function interactively.
I understood that OP wanted to load all files as 75 objects into one R session
environment. In that case having one
On 03/20/2014 06:00 AM, r-help-requ...@r-project.org wrote:
My question is related to a cox model with time-dependent variable.
When I think about it more, I get a little confused about
non-increasing assumption for survival probability for an individual.
For example, for a time-dependent ,say
I noticed that dnearneigh::spdep shows diverging behaviour with matrix and
SpatialPoints objects respectively:
reproducible example:
library(spdep)
set.seed(5)
spdf-SpatialPointsDataFrame(cbind(lon=runif(1000,2,8),lat=runif(1000,53,56)),
data=data.frame(par=runif(1000,0,1)),coords.nrs =
Dear Sirs
I would like to make you aware of the following training taking place in
Chicago next month.
Analysis of experiments using ASReml-R, Chicago, 24/25th April
This workshop is aimed at scientist/practitioners that are interested in
analyzing complex datasets by fitting linear mixed
# Dear all,
# I simulate individual paths in a landscape (let's say coordinates in x
and y range
# from -100 to 100 both) and would like to replace each individual going
outside the
# landscape by a new simulation... until none of them goes outside the
landscape.
# For example, I simulate
But I don't think a list can contain that much data - we dont know how big
each txt file is.
So I'd like to do:
city.set - c('Asheville','Charlotte')
month.set - c('Dec', 'Jan', 'Feb')
comb.set - expand.grid(city.set,month.set)
comb.set$filename = paste(comb.set[,1],comb.set[,2],sep = '_')
for
Hi all,
I've been using the randomForest package and I'm trying to make the switch
over to party. My problem is that I have an extremely unbalanced outcome
(only 1% of the data has a positive outcome) which makes resampling methods
necessary.
randomForest has a very useful argument that is
Hi
I have a strong suspiction that you want to reinvent wheel.
does functions
?chull (grDevices) and ?pip (splancs) do what you want?
Regards
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Gwennaël Bataille
Sent:
Hi
It seems to me that it is task for while loop with break but I never used it.
see
?for
something like
counter-0
is.outside - TRUE
while(is.outside) {
counter-counter+1
{do what you want here and set some.test/another.test}
is.outside - some.test another.test
if (counter100) break # just to
Dear R family,
I am trying to read a real large dataset in R (~ 2Gb). Its in binary format.
When i tried to read it by using following command
readBin(DAT.dat.nc, numeric(), n=9e8, size=4, signed=TRUE, endian='little')
I got the following error
Error: cannot allocate vector of size 5.2 Gb
I have
Hi friends
Iwould like to smooth my plot. I have a plot which is wiggly and I want to
smooth it. here is my data
539, 532, 531, 538, 544, 554, 575, 571, 543, 559, 511, 525, 512, 540, 535,
514, 524, 527, 532, 547, 564, 548, 572, 564, 549, 532, 519, 520, 520, 543,
550, 542, 528, 523, 531, 548,
5.2 won't go into 4 but there may be more problems.
32-bit or 64 bit operating system?
RAM is cheap but will your motherboard support more than 4 GB?
And don't forget there are other processes that need to run while you are
using R.
Clint BowmanINTERNET:
Perhaps you will understand the meaning better if you try the following
sequence:
plot(datalist, ylab='Value')
lines(lowess(datalist),col=blue, lwd=2)
lines(lowess(datalist,f=1/3),col=red, lwd=2)
lines(lowess(datalist,f=1/10),col=green, lwd=2)
-Don
--
Don MacQueen
Lawrence Livermore National
thanks you so much
On Thu, Mar 20, 2014 at 10:25 AM, MacQueen, Don macque...@llnl.gov wrote:
Perhaps you will understand the meaning better if you try the following
sequence:
plot(datalist, ylab='Value')
lines(lowess(datalist),col=blue, lwd=2)
lines(lowess(datalist,f=1/3),col=red, lwd=2)
Dear Petr,
Thank you for your answer. This will probably save me three lines of
script, which is appreciable.
Unfortunately, this does not solve my main issue, that is:
If my condition is not fulfilled (some points are outside the
landscape), I have to go back in the script to the first step,
Hi,
Another way would be:
dat11 - transform(dat[rep(1:nrow(dat),each=12),1:2],
weekdatesunday=rep(0:11,3), RevenueWeekN00=as.vector(t(dat[,-c(1:2)])))
row.names(dat11) - 1:nrow(dat11)
dat22 - unsplit(lapply(split(dat11,
with(dat11,list(customer_id,CountryName)),drop=TRUE),function(x) {m1 -
Hi,
Not sure if this helps or not:
vec1 - c(MaxUpdated_row, as.Date(c(2013-06-28,2013-06-29), %Y-%m-%d)) +
as.Date(1970-01-01)
vec1
#[1] 2013-06-28 2013-06-29
str(vec1)
#Date[1:2], format: 2013-06-28 2013-06-29
rbind(MaxUpdated_row, as.Date(c(2013-06-28,2013-06-29), %Y-%m-%d)) +
Hi,
It is better to use ?dput() to show the data.dput(dataset)
dat -
structure(list(customer_id = c(8L, 33L, 12L), CountryName = c(US,
CA, UK), RevenueWeekN00 = c(2.28, 0, 30.18), RevenueWeekN01 = c(9.57,
14.69, 43.9), RevenueWeekN02 = c(7.54, 3.31, 90.4), RevenueWeekN03 = c(8.99,
5.21, 45),
Dear Group,
I am bit new in R. I am looking for a code repository of few projects
developed in R.
I want to see them to learn more.
If any one of the esteemed members of the group may kindly help.
Regards,
Subhabrata Banerjee.
[[alternative HTML version deleted]]
Please read the posting guide (there is a link at the bottom of every
post) and post in plain text, not HTML, and also avoid all caps.
The biggest repository of R code is probably CRAN which is linked from
the main R page. There are also R-forge and Bioconductor and many R
packages on Github.
Look into the while() statement. E.g.,
f - function(n, absLimit)
{
x - rep(absLimit+1, n) # start with all x,y outside limits
y - x
while(any(isOutside - abs(x)absLimit | abs(y)absLimit)) {
x[isOutside] - rnorm(sum(isOutside), 0, 100)
y[isOutside] -
Hello,
Try
res[order(res[,1]), ]
Hope this helps,
Rui Barradas
Em 20-03-2014 17:48, Luigi Marongiu escreveu:
Dear Rui,
Thnak you very much: this works exactly as I requested, but the merged
vetor (res) order the variables by alphabetical order rather than by the
factors I indicated during
For time series plots using zoo I want the layout to be horizontal
rectangles rather than squares. When I specify par(pin=5,2.5) an error is
returned:
Error in par(pin = 5,2.5) :
graphical parameter pin has the wrong length
In addition: Warning message:
In par(pin = 5,2.5) : NAs introduced
Hello,
Use c(), not
par(pin = c(5, 2.5))
Hope this helps,
Rui Barradas
Em 20-03-2014 19:30, Rich Shepard escreveu:
For time series plots using zoo I want the layout to be horizontal
rectangles rather than squares. When I specify par(pin=5,2.5) an error is
returned:
Error in
You're specifying a string rather than two numbers.
Try c(5, 2.5) instead.
Sarah
On Thu, Mar 20, 2014 at 3:30 PM, Rich Shepard rshep...@appl-ecosys.com wrote:
For time series plots using zoo I want the layout to be horizontal
rectangles rather than squares. When I specify par(pin=5,2.5) an
Hi R User,
I was trying to convert a decimal value into integral (whole number). I used
round function but some of the cells have the value less than 0.5 and it
converted into 0. But I wanted these cell to have 1 instead of 0. Another way,
I could multiply by 10. But l did not want it because
What about:
ceiling(data[,2:3])
A B
1 0 1
2 3 1
3 3 3
4 2 3
Note that ceiling is referenced in ?round
Thanks for the clear reproducible example.
Sarah
On Thu, Mar 20, 2014 at 3:42 PM, Kristi Glover
kristi.glo...@hotmail.com wrote:
Hi R User,
I was trying to convert a decimal value into
On 20-Mar-2014 19:42:35 Kristi Glover wrote:
Hi R User,
I was trying to convert a decimal value into integral (whole number). I used
round function but some of the cells have the value less than 0.5 and it
converted into 0. But I wanted these cell to have 1 instead of 0. Another
way, I could
On Thu, 20 Mar 2014, Rui Barradas wrote:
Use c(), not
par(pin = c(5, 2.5))
Rui/Sarah,
Thank you both. I did not pick up on that even though that's how plt is
described.
Rich
__
R-help@r-project.org mailing list
Excellent! Thank you for your help.
-Original Message-
From: Therneau, Terry M., Ph.D. [mailto:thern...@mayo.edu]
Sent: Monday, 17 March 2014 11:15 PM
To: Lucy Leigh; r-help@R-project.org
Cc: David Winsemius
Subject: Re: survfit question - Q1 and Q3 survival time?
Try
On Thu, 20 Mar 2014, Rich Shepard wrote:
For time series plots using zoo I want the layout to be horizontal
rectangles rather than squares. When I specify par(pin = c(5,2.5)) nothing
happens; the plot frame is square regardless of specifying pin within the
plot command or prior to issuing the
Hello,
It is not the right way to read a NetCDF file (according to the
extension) in R. Please have a look at the ncdf4 package. The
raster package is also able to read this kind of files.
Regards,
Pascal
On Fri, Mar 21, 2014 at 1:25 AM, eliza botto eliza_bo...@hotmail.com wrote:
Dear R
Although there is a great deal of value to be gained by understanding how
package code is written, most of my R projects consist of a file with
10-100 lines of code working with data in the global environment that call on
packages. Such code is not typically posted online.
I do find it
On Thu, Mar 20, 2014 at 7:13 PM, Rich Shepard rshep...@appl-ecosys.com wrote:
On Thu, 20 Mar 2014, Rich Shepard wrote:
For time series plots using zoo I want the layout to be horizontal
rectangles rather than squares. When I specify par(pin = c(5,2.5)) nothing
happens; the plot frame is
Hi,You could do:
m1 - as.matrix(read.table(text=3 0.9 1
5 0.5 0.7
7 0.1 0.3,sep=,header=FALSE))
dimnames(m1) - NULL
d2 - data.frame(V1=1:10)
d1 - as.data.frame(m1)
library(zoo)
res - na.locf(merge(d1,d2,all=TRUE))
res[is.na(res)] - 1
res
#or
vec1 - 1:10
indx - vec1[!vec1 %in% m1[,1]]
res2 -
This is really a question of programming logic - not of R per se.
If I understand your problem correctly, the preferred approach is not to create
your paths all at once, but individually, and check that they fulfil your
requirements as you generate them. Keep on doing this until you have enough
You can check the function read_folder() in the package phenology.
read_folder(folder = try(file.choose(), silent = TRUE), wildcard = *.*,
read = read.delim, ...)
Arguments
folder
Where to search for files; can be or a file path or a folder path
wildcard
Define which files are to be read
Try ?ceiling
ceiling(0.4)
[1] 1
B.
On 2014-03-20, at 3:42 PM, Kristi Glover wrote:
Hi R User,
I was trying to convert a decimal value into integral (whole number). I used
round function but some of the cells have the value less than 0.5 and it
converted into 0. But I wanted these
I'm trying to figure out how to import data into a dataframe in R.
The table I'm trying to import is:
HistoricalTradingReportServlet.csv
http://r.789695.n4.nabble.com/file/n4687246/HistoricalTradingReportServlet.csv
I would like to add two more column for each block of data to have the
Hi,
May be this helps:
strsplit(sub((.*)(\\|.*),\\1 \\2,test), )[[1]]
#[1] comp99810_c0_seq1 |m.8409
sub((.*)\\|.*,\\1,test)
#[1] comp99810_c0_seq1
sub(.*(\\|.*),\\1,test)
#[1] |m.8409
A.K.
Hi Could some one help me regular expression for this. I am really struggling.
Basically i
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