Hi,
So I have everything written out. I used R-studio (same as R console
but with an IDE) and I had no trouble calling upon and opening the csv
file for computation.
However, when I try to run my code in plain old R, it says that the
file could not be located. Heres my code:
func <- function(fil
Like Denis said, you are asking ts to do things that don't make sense;
in particular, some of your statements suggest you don't fully
understand what window does or what its frequency argument does.
Specifically, when you set frequency = 1 in window, that doesn't mean
take a window and treat it as
Dear David,
Thanks a lot for your kind reply.
Unfortunately, distm() can not give me what I want. distm(), output the
distance matrix (but not in dist class).
Fortunately, I just realized that as.dist() from base R can convert matrix
to dist class.
Thanks again.
Best
Jian-Feng,
2011/11/8 Davi
Thank you very much DJ MuseR!
Curtis K
~ Sent from a PDA. I apologize in advance for any typing errors.
On Nov 8, 2011, at 4:36 PM, "djmuseR [via R]"
wrote:
> Here's another approach using the plyr and data.table packages, where
> df is the name I gave to your example data:
>
> # plyr
Hi:
Here are a couple of ways using the plyr and ggplot2 packages:
library('plyr')
library('ggplot2')
# Create a summary data frame that computes the mean
# and standard deviation at each time/group combination
xsumm <- ddply(x, .(Time, Group), summarise, m = mean(Score), s = sd(Score))
# Creat
On Tue, Nov 8, 2011 at 1:04 PM, Ben quant wrote:
> Because I don't know anything about sqldf. :)
>
> Here is what happens, but I"m sure it is happening because I didn't read
> the manual yet:
>
>> s <- sqldf('create table r.dat("id" int primary key,"val" int)')
> Error in ls(envir = envir, all.nam
The problem is when I use the window function an try to extract a subset of
the time series an specify the frequency as 1 (not only will ets not take a
time series with a frequency greater than 24, now that I am taking a subset
there is no frequency so I would like to set it to 1 (which is one of t
On Tue, Nov 8, 2011 at 11:43 AM, Ben quant wrote:
> Hello,
>
> I'm using package RpgSQL. Is there a better way to create a multi-line
> query/character string? I'm looking for less to type and readability.
>
> This is not very readable for large queries:
> s <- 'create table r.BOD("id" int primar
See downloadFile() of the R.utils package (on CRAN);
it allows you to download to a file to a given path (without have to
"replicate" the filename), and it also have an option (the default) to
drop zero-size files that are (sometimes) created when download.file()
tries to download a non-existing f
On Tue, Nov 8, 2011 at 12:16 PM, Ben quant wrote:
> Hello,
>
> When I do:
>
> dbWriteTable(con, "r.BOD", cbind(row_names = rownames(BOD), BOD))
>
> ...can I specify the data types such as varchar(12), float, double
> precision, etc. for each of the fields/columns?
>
> If not, what is the best way
Here's another approach using the plyr and data.table packages, where
df is the name I gave to your example data:
# plyr
library('plyr')
ddply(df, .(TIME), mutate, L1 = sort(S1))
# Another way with the data.table package:
library('data.table')
dt <- data.table(df, key = 'TIME')
dt[, list(X1, S1,
As I mentioned earlier, if you use different labels/breaks/titles for
the legends, you'll get separate legends in a ggplot. I'm assuming you
want to keep your original labels in scale_shape_manual(); there's no
problem if you do that. The code I provided showed you how to align
all three legends; i
Hi,
I often use plotmeans() from the gplots package to quickly visualize a
pattern of change. I would like to be able to plot separate lines for
different groups, but the function gives an error when a grouping
variable is included in the formula argument.
For instance,
> require(gplots)
>
The ets() function in the forecast package requires either a numeric
vector or a Time-Series object (produced from ts()). The frequency
argument in ts() refers to the time duration between observations;
e.g., frequency = 7 means that the data are weekly; frequency = 12
means that the data are month
On Nov 8, 2011, at 6:32 PM, Mary Kindall wrote:
I am downloading say 100 files from ucsc website and storing it into
dest
folder.
download.file function create a file in destination folder even if
the file
is not present which is something I dont want.
So I wrote if condition to remove the
I am downloading say 100 files from ucsc website and storing it into dest
folder.
download.file function create a file in destination folder even if the file
is not present which is something I dont want.
So I wrote if condition to remove the file if the download function has non
zero value.
Now i
On Nov 8, 2011, at 2:58 PM, economicurtis wrote:
I have panel data in the following form:
TIME X1 S1
1 1 0.99
1 2 0.50
1 3 0.01
2 3 0.99
2 1 0.99
2 2 0.25
3 3 0.75
3 2 0.50
3 1 0.25
...
Hi Dennis,
Thank you, it looks much better now the three characteristics are combined
in the label. However, I would like to keep two different sets of labels on
the side as I want to describe lines for each facet (high and low). Do you
think this is possible, without breaking up the characteristi
Dear John,
Thank you for your reply. My data is actually simulated under the model X
= Lambda*F + E.
Since my post, I've simplified the simulation of my data and I still get
the error. This is what I've done since my last post.
I constructed Lambda apriori (so I know exactly which observed var
Dan
Nordlund, Dan (DSHS/RDA) wrote:
>
>> -Original Message-
>> From: r-help-bounces@ [mailto:r-help-bounces@r-
>> project.org] On Behalf Of SarahJoyes
>> Sent: Tuesday, November 08, 2011 5:57 AM
>> To: r-help@
>> Subject: Re: [R] Sampling with conditions
>>
>> That is exactly what I want
I have panel data in the following form:
TIME X1 S1
1 1 0.99
1 2 0.50
1 3 0.01
2 3 0.99
2 1 0.99
2 2 0.25
3 3 0.75
3 2 0.50
3 1 0.25
... ... ..
And desire a new vector o
I expect the frequency to be set to what I set it at and the window to
return all of the data in the window from the original time series. The
error is not because it is prime. I can generate a time series with just 52
values (or 10) and it still occurs. I am building these objects for use with
the
Dear Lisa,
As I said, I don't have a whole lot of time today. At a quick glance, you
simulation seems straightfoward except for
> F_i ~ N(mu_i, Phi).
>
> mu_i is fixed to Phi*z_i, where z_i is a 5x1 vector.
Why don't the factors have 0 means? I assumethat Phi is a correlation matrix.
John
O
Thanks! Makes sense now.
On Nov 8, 2011, at 1:32 PM, jim holtman wrote:
> Try this:
>
>> cbind(scores[,1,drop = FALSE], scores[,2:3])
> name round1 round2
> 1 Bob 40 5
> 2 Ron 30 6
> 3 Bud 20 4
>>
>
> Then do ?'[' to learn about 'drop'
>
> On Tue, Nov 8, 201
Hi:
The problem is that you use different sets of labels in each
scale_*_manual. To get all of the scales to merge into one, you need
to have the same title, same breaks and same labels for the legend.
This gets you a single legend:
ggplot(data = df, aes(x = year, y = total, colour = treatment,
Gyanendra Pokharel wrote on 11/08/2011 10:57:38 AM:
>
> Hi R community!
> I am analyzing the data set "motorins" in the package "faraway" by using
> the generalized additive model. it shows the following error. Can some
one
> suggest me the right way?
>
> library(faraway)
> data(motorins)
> moto
Feel free to post again if you need help. If you are comfortable with
slightly-heavy-lifting in programming, but are finding "the R way" of
getting things done a little counterintuitive, you can do much worse
than reading Patrick Burns' two books: S Poetry and the R Inferno,
available here: http://
Dear Lisa,
There doesn't seem to be anything logically wrong with your model.
I don't have much time today to look into it, but trying different optimizers
in version 2.0-0 of sem, using the correlation matrix in place of the
covariance matrix, and setting the par.size parameter, I was unable t
In addition to Dan's quite valid concern, the final sample is not
truly 'random' - the first k - 1 elements are randomly chosen, but the
last is determined so that the constraint is met.
Dennis
On Tue, Nov 8, 2011 at 9:59 AM, Nordlund, Dan (DSHS/RDA)
wrote:
>> -Original Message-
>> From
I am not going through all of your code to understand what you are trying to
demonstrate. R uses the IEEE Standard 754 for Floating Point Numbers. There
is a sign bit, 11 bits for the exponent, and 52 bits for the mantissa. Because
the standard normalizes the mantissa you get an extra bit of
I'm not entirely sure that your request makes sense: what do you
expect the frequency to be? It makes sense to me as is...Might your
troubles be because 53 is prime?
More generally, most people don't like working with the raw ts class
and prefer the zoo or xts packages because they are much more p
Hi:
Here's another way of doing this on the simplified version of your example:
L <- vector('list', 3) # initialize a list of three components
## populate it
for(i in seq_along(L)) L[[i]] <- rnorm(20, 10, 3)
## name the components
names(L) <- c('Monday', 'Tuesday', 'Wednesday')
## replace va
This doesn't seem to work:
> d <- rnorm(2*53)
> ds <- ts(d, frequency=53, start=c(2000,10))
> dswin <- window(ds, start=c(2001,1), end=c(2001,10), frequency=1)
> dswin
Time Series:
Start = 2001
End = 2001
Frequency = 1
[1] 1.779409
> dswin <- window(ds, start=c(2001,1), end=c(200
Dear all,
I am having a data stucture that contains Products and Time Stamps,
I have made also two lists
ProductList=list(c('Example1','Example2'...)
TimeStamp=list(c("1990-02-03 12:57:60"),c("1990-02-03 12:57:60"),
then I have made few functions that call each other
do_analysis_for_all the d
On Nov 08, 2011 at 11:16am Colin Aitken wrote:
> An unresolved problem is: what does R do when the explanatory factors
> are not defined as factors when it obtains a different value for the
> intercept but the correct value for the fitted value?
Colin,
I don't think that happens (that the fit
I found
names(y)<-y
before the end is working ;)
some more ideas ?
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Dear all,
I think that every number x in R can be represented in floating point
arithmetic as:
x = (-1)^s (1+f) 2^(e-1023)
where s is coded on 1 bit, e (positive integer) is coded on 11 bits, and
f (real in [0,1)) is coded on 52 bits.
Am I right?
We have f=\sum_{i=1}^{52} k_i 2^{-i} for some
Hi there,
I am having a little problem with combining three scale_manual commands in a
facet plot. I am not able to combine the three different characteristics,
instead ending up with three different descriptions next to the graph for
the same geom. I would like to see two separate labels (not th
It might make the discussion easier to follow if you used
a smaller dataset that anyone can make and did some experiments
with contrasts. E.g.,
> D <- data.frame(expand.grid(X1=LETTERS[1:3], X2=letters[24:26])[-1,],
> Y=2^(1:8))
> D
X1 X2 Y
2 B x 2
3 C x 4
4 A y 8
5 B y 16
6 C
It's polite to include the whole list in your replies so the threads
get archived properly.
Yes, underflow is occasionally a problem: hence the common use of
log-likelihood in MLE and other applications. Might that help you out
here? You can get log-likelihoods directly from pnorm and dnorm with
t
Hi:
(1) Here is one way to reorganize the levels of a factor:
plotData[['infection']] <- factor(plotData[['infection']],
levels = c('InfA', 'InfC', 'InfB', 'InfD'))
Do this ahead of the call to ggplot(), preferably after plotData is defined.
relevel() resets the ba
Hi Dennis, Thanks very much for the details. All those explanations about
non-estimable mu_{12} when it has no data make sense to me!
Regarding my specific example data where mu_{12} should NOT be estimable in a
linear model with interaction because it has no data, yet the linear model I
creat
On Nov 8, 2011, at 1:20 PM, katiab81 wrote:
up
down ... for a variety of reasons.
And...
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
It's very possible that you are not communicating with the package
authors by posting to rhelp (by way of Nabble) , an
Try this:
> cbind(scores[,1,drop = FALSE], scores[,2:3])
name round1 round2
1 Bob 40 5
2 Ron 30 6
3 Bud 20 4
>
Then do ?'[' to learn about 'drop'
On Tue, Nov 8, 2011 at 1:06 PM, Eric Rupley wrote:
>
>
>
> Hi all ---
>
> I note that the column name of the first co
up
--
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__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLE
Dear Co-Forumeees
Does anybody have experience with using rpanel..or how to retrieve data from
created panel.
For example my panel draws some interactive graph and computes something
inside the panel.
Question : is there a way to retrieve those data ?
For illustration:
if (interactive()) {
The cell mean mu_{12} is non-estimable because it has no data in the
cell. How can you estimate something that's not there (at least
without imputation :)? Every parametric function that involves mu_{12}
will also be non-estimable - in particular, the interaction term and
the population marginal m
On Nov 8, 2011, at 12:36 PM, array chip wrote:
Sure it does, but still struggling with what is going on...
Have you considered redefining the implicit base level for "treat" so
it does not create the missing crossed-category?
> test$treat2_ <- factor(test$treat, levels=c(2:7, 1) )
> lm(y~
Hi all ---
I note that the column name of the first column in a dataframe does not
necessarily get passed on when using cbind (example below)…
I'm looking for help in clarifying why this behavior occurs, and how I can get
all col names, including the first, passed on to the result…while I su
Because I don't know anything about sqldf. :)
Here is what happens, but I"m sure it is happening because I didn't read
the manual yet:
> s <- sqldf('create table r.dat("id" int primary key,"val" int)')
Error in ls(envir = envir, all.names = private) :
invalid 'envir' argument
Error in !dbPreExi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of SarahJoyes
> Sent: Tuesday, November 08, 2011 5:57 AM
> To: r-help@r-project.org
> Subject: Re: [R] Sampling with conditions
>
> That is exactly what I want, and it's so simple!
Why not just send it in as is. I use SQLite (via sqldf) and here is
the way I write my SQL statements:
inRange <- sqldf('
select t.*
, r.start
, r.end
from total t, commRange r
where t.comm = r.comm and
t.loc betw
true, why it has to omit "treat 7-group 2"
Thanks again
From: David Winsemius
Cc: "r-help@r-project.org"
Sent: Monday, November 7, 2011 10:19 PM
Subject: Re: [R] why NA coefficients
On Nov 7, 2011, at 10:07 PM, array chip wrote:
> Thanks David. The only
Sure it does, but still struggling with what is going on...
Thanks
John
From:David Winsemius
To:David Winsemius
oject.org>
Sent:Monday, November 7, 2011 10:27 PM
Subject:Re: [R] why NA coefficients
But this output suggests there may be alligators in the swam
On 08.11.2011 17:59, Ana wrote:
Hi
Can someone help me with this?
How can I apply a function to a list of variables.
something like this
listvar=list("Monday","Tuesday","Wednesday")
This is a list of length one character vectors rather than a "list of
variables".
func=function(x){x
Hello,
When I do:
dbWriteTable(con, "r.BOD", cbind(row_names = rownames(BOD), BOD))
...can I specify the data types such as varchar(12), float, double
precision, etc. for each of the fields/columns?
If not, what is the best way to create a table with specified field data
types (with the RpgSQL
I suspect you are lost. This is almost certainly more appropriate on
the BioConductor list. There are relatively few people on this list
who will know what (un-named) packages you might be using. Read the
Posting Guide, note the link to the Bioconductor mailing list page,
configure your
On Nov 8, 2011, at 11:31 AM, flokke wrote:
Thanks for the replies!
Indeed, I could use order() instead of sample() but that it wouldnt
be
random anymore, as it sorts data points in increasing(!) order.
Second, I
have to use 1000 different samples.
Please re-read the help page for `orde
Hi R community!
I am analyzing the data set "motorins" in the package "faraway" by using
the generalized additive model. it shows the following error. Can some one
suggest me the right way?
library(faraway)
data(motorins)
motori <- motorins[motorins$Zone==1,]
library(mgcv)
>amgam <- gam(log(Paymen
Dear R-community,
I have a fitted bivariate polynomial, i.e:
fit = lm(cbind(x, y)~poly(t, 15))
and I would like to determine the length of the line in the interval t =
[a, b]. Obviously, I could use predict and go through all the points, i.e.
for (t in a:(b-1)) {
length = length + sqrt((x.pred[
Hi
Can someone help me with this?
How can I apply a function to a list of variables.
something like this
listvar=list("Monday","Tuesday","Wednesday")
func=function(x){x[which(x<=10)]=NA}
lapply(listvar, func)
were
Monday=[213,56,345,33,34,678,444]
Tuesday=[213,56,345,33,34,678,444]
...
in m
Hi
My name is Rocky and I am trying to use the org.Dm.eg.db library.
When I am using the org.Dm.egFLYBASE2EG[fb_ids] it is stopping at a point where
it cannot find any value for a given ID such as the following:
Error in .checkKeys(value, Rkeys(x), x@ifnotfound) :
value for "FBgn0004461" not fou
Hello.
I started using the sem package in R and after a lot of searching and trying
things I am still having difficulty. I get the following error message when
I use the sem() function:
Warning message:
In sem.default(ram = ram, S = S, N = N, param.names = pars, var.names =
vars, :
Could no
That is exactly what I want, and it's so simple!
Thanks so much!
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__
R-help@r-projec
Hello,
I recently downloaded an updated version of the {spatstat} package 1.24-1. But
now, when I do >help() to look up a certain function in the spatstat library,
I get an error message. It only seems to be for spatstat fns. The following error:
Error in fetch(key) : internal error -3 in R_d
Sorry about being confusing, I have so many loops in loops and ifelses that I
get mixed up sometimes, it was just a typo, it was supposed to be "for(i in
1:5)" Sorry,
Thanks for you help!
SJ
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Thanks for the replies!
Indeed, I could use order() instead of sample() but that it wouldnt be
random anymore, as it sorts data points in increasing(!) order. Second, I
have to use 1000 different samples.
I got the hint that I have to do somethin with indeces, but still cant
figure out what thi
Hello,
I'm using package RpgSQL. Is there a better way to create a multi-line
query/character string? I'm looking for less to type and readability.
This is not very readable for large queries:
s <- 'create table r.BOD("id" int primary key,"name" varchar(12))'
I write a lot of code, so I'm looki
Am 08.11.2011 13:01, schrieb R. Michael Weylandt:
See the replies given to you two days ago when you asked the same
Hi Michael,
thank you for your second answer.
I did not get my first question and I did not get your answer via mail -
strange
Knut
_
On Nov 8, 2011, at 9:43 AM, R. Michael Weylandt wrote:
Have you tried wrapping it in a function and using integrate()? R is
pretty good at handling numerical integration. If integrate() isn't
good for you, can you say more as to why?
Michael
On Nov 6, 2011, at 4:15 PM, JeffND wrote:
H
Thanks Jim,
I didn't think it worked on directories but I've got it working now.
Cheers,
Gavin.
-Original Message-
From: jim holtman [mailto:jholt...@gmail.com]
Sent: 08 November 2011 14:18
To: Gavin Blackburn
Cc: r-help@r-project.org
Subject: Re: [R] Rename a directory in R
apropos('
On Nov 8, 2011, at 8:57 AM, Jean V Adams wrote:
SML wrote on 11/07/2011 09:10:30 PM:
I'm trying to rekey values which denote there is no values, i.e.,
'-999' in a dataset which contains both '-999' and NA entries.
When I try the following command I get the following error:
data.frame[data.f
This is great, thanks!
I have another unrelated question. I'll create a new email for that one.
ben
On Mon, Nov 7, 2011 at 4:16 PM, Gabor Grothendieck
wrote:
> On Mon, Nov 7, 2011 at 5:34 PM, Ben quant wrote:
> > Hello,
> >
> > Using the RpgSQL package, there must be a way to get the row names
On 08.11.2011 16:26, Kevin Burton wrote:
Can someone enlighten me on why the following doesn't work?
setwd('C:/Temp/R')
d<- rep(1:53,2)
(s<- ts(d, frequency=53, start=c(2000,10)))
n<- length(s)
k<- n%/%3
for(i in (n-k):n)
{
st<- c(start(s)[1] + (start(s)[2] + i)%/%frequency
Can someone enlighten me on why the following doesn't work?
setwd('C:/Temp/R')
d <- rep(1:53,2)
(s <- ts(d, frequency=53, start=c(2000,10)))
n <- length(s)
k <- n%/%3
for(i in (n-k):n)
{
st <- c(start(s)[1] + (start(s)[2] + i)%/%frequency(s), (start(s)[2] +
i) %% frequency(s
Hi Alex,
For the R part, I would abstract it a bit:
mydir <- paste("./", abs(threshold), sep = "")
if (!file.exists(mydir)) dir.create(mydir)
save(var, file = paste(mydir, DataSource[[4]], sep = "/"))
if you use file.exists('threshold') you are testing for the existence
of threshold, not the val
On 08.11.2011 16:09, Alaios wrote:
Hmm I will try something like that
if (file.exists('threshold')==FALSE)
dir.create(paste('./',abs(threshold)))
save(var,file=paste('./',abs(threshold),'/',DataSource[[4]],sep="")
I just need a bit of confirmation If I am doing soemthing ter
Hmm I will try something like that
if (file.exists('threshold')==FALSE)
dir.create(paste('./',abs(threshold)))
save(var,file=paste('./',abs(threshold),'/',DataSource[[4]],sep="")
I just need a bit of confirmation If I am doing soemthing terribly wrong that
might harm my filesystem
R CMD check is *not* 'building a package'. Nor is making a Windows
binary package. 'Building a package' is creating a source tarball
from a source directory.
On Tue, 8 Nov 2011, Joshua Wiley wrote:
Hi Ed,
If the only error is in examples then this should work:
R CMD check --no-examples fo
Just because your model specification is valid does not mean your data can be
analyzed using that particular model specification.
---
Jeff NewmillerThe . . Go Live...
DCN:Basi
On 08.11.2011 13:50, Joshua Wiley wrote:
Hi Ed,
If the only error is in examples then this should work:
R CMD check --no-examples foopkg
Disabling the example checks is not the solution - well, it is the one
to hide the errors, of course.
should not have anything to do with vignettes
Thank you. I was confused because the output of the line "summary(lm(formula
=..." reads:
"Coefficients: (1 not defined because of singularities)",
which did not look like a normal message (which can safely be ignored)
to me.
--Holger
On Tue, 8 Nov 2011 15:06:28 +0100
Peter Konings wrote:
> O
Have you tried wrapping it in a function and using integrate()? R is pretty
good at handling numerical integration. If integrate() isn't good for you, can
you say more as to why?
Michael
On Nov 6, 2011, at 4:15 PM, JeffND wrote:
> Hello to all,
>
> I am having trouble with intregrating a co
Look at distm() in package geosphere.
--
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77843-4352
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Beha
apropos('rename')
you will find 'file.rename' which also works on directories.
On Tue, Nov 8, 2011 at 7:45 AM, Gavin Blackburn
wrote:
> Hi,
>
> I want to be able to rename a folder using R, similar to file.rename. I want
> to paste " - done" onto the folder name. The reason for this is we run a
SML wrote on 11/07/2011 09:10:30 PM:
>
> I'm trying to rekey values which denote there is no values, i.e.,
> '-999' in a dataset which contains both '-999' and NA entries.
> When I try the following command I get the following error:
>
> > data.frame[data.frame$MAR <= -9,"MAR"] <- NA
>
> "mi
Dear R-listers,
Here, I would like to hearing helps from you.
I have GPS data (multiple points in the geographic scale) in
longitude/latitude. I intend to calculate
distance (in kilometer) among such points and output the distance matrix in
dist class.
I have gotten some progress, but I still ca
On 11-11-08 7:24 AM, Matthew Young wrote:
Hi,
I need to do some simple coordinate transforms between cartesian,
cylindrical and spherical coordinates. I can't find any built in
functions or packages to do this, is this because they do not exist?
Obviously I can write my own code, but don't want
Yes, adding "-1" or "+0" both return a lm without an intercept term.
Michael
On Tue, Nov 8, 2011 at 8:52 AM, Holger Taschenberger
wrote:
> Hi,
>
> I'm trying to compare two linear regressions. I'm using the
> following approach:
> ##
> xx<-1:100
> df1 <- data.frame(x = xx,
Hi,
I want to be able to rename a folder using R, similar to file.rename. I want to
paste " - done" onto the folder name. The reason for this is we run a loop on a
large number of folders on a server and it would be nice for people to be able
to log in and instantly see if their data has been p
The 'user' + 'system' will give you how much CPU is required which is
an indication of how many of the CPU cycles you are using. The
elapsed time is just how long it spent. If the script is CPU
intensive, and there is no paging going on, you should see the CPU
time close to the elapsed time. Lon
Dear R-Helpers,
this is my first post ever to a mailing list, so please feel free to point out
any missunderstandings on my side regarding the conventions of this mailing
list.
My problem:
Assuming the following character vector is given:
names <- c("filia Maria", "vidua Joh Dirck Kleve (oo 0
Hi,
I'm trying to compare two linear regressions. I'm using the
following approach:
##
xx<-1:100
df1 <- data.frame(x = xx, y = xx * 2 + 30 + rnorm(n=length(xx),sd=10), g = 1)
df2 <- data.frame(x = xx, y = xx * 4 + 9 + rnorm(n=length(xx),sd=10), g = 2)
dta <- rbind(df1, df2
Hi Alex,
Look at some of these functions:
apropos("dir")
apropos("exists")
Cheers,
Josh
On Tue, Nov 8, 2011 at 5:36 AM, Alaios wrote:
> Dear all,
> I have a variable called thres and before I run a script I set it to a value
> like
> thres<- -10
> at the end of the execution I am issuing a s
Dear all,
I have a variable called thres and before I run a script I set it to a value
like
thres<- -10
at the end of the execution I am issuing a save(variablename,file='Results')
which will end up with a file saved at the current directory with the name
Results
I would like though to use thres
Actually I want to have a rough approximation. A process that takes one day and
a half it is good to send me how many hours it gets. It is not a problem to
convert the values of system.error the major is that I am not sure If I should
use the user or elapsed time for getting an estimation of how
Celine wrote on 11/08/2011 02:28:32 AM:
>
> Thanks for your help, I still have a little problem with this function
> because I don't have the same number of line in my two datarame so when
I
> try to apply the dataframe function, I obtain this response ; that I
have a
> different number of lines
Hi
I am falling in love with iplots, especially the linked graphs, i.e.
selecting points in one will highlight the same points in the other graph
as well.
But I would like to be able to "export" the graphs, so that I can give them
to somebody else to be able to look at the graphs and explore the d
Dear Iain,
RSiteSearch("ggplot2 reorder", restrict = c("Rhelp10", "Rhelp08")) gives you
the solution.
Best regards,
Thierry
> -Oorspronkelijk bericht-
> Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> Namens Iain Gallagher
> Verzonden: dinsdag 8 november 2011 1
Hi, Iain-
You might want to have a look at
?relevel
i.e. plotData$infection<-relevel(plotData$infection, ref = 'InfC')
Ken
On 11/08/11, Iain Gallagher wrote:
>
>
> Dear List
>
> I am trying to draw a heatmap using ggplot2. In this heatmap I have faceted
> my data by 'infection' o
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