On Aug 28, 2012, at 9:54 PM, arun wrote:
HI,
If I understand your question correctly, this should give the result:
dat1<-read.table(text="
A B C
a b a
x y z
",sep="",header=TRUE,stringsAsFactors=FALSE)
within(dat1,{new_column<-ifelse(A==C|B==C,"y","n")})
# A B C new_column
#1 a b a y
Have a look at the documentation for ifelse()
http://stat.ethz.ch/R-manual/R-devel/library/base/html/ifelse.html that should
do what you need.
Rhydwyn
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Sapana Lohani
Sent: Wednesda
Hi,
Try this:
dat1<-read.table(text="
id price distance
1 2 4
1 3 5
2 4 8
2 5 9
3 6 3
3 4 8
",sep="",header=TRUE)
dat2<-split(dat1,dat1$id)
lapply(dat2,function(x) cor(x[2],x[3],method="spearman"))
A.K.
- Original Message -
From: Yi
To: R-help@r-project.org
Cc:
Sen
Hi
I am trying to export dataframes to Stata with stata's labels, I have looked
though the documentation for the foreign library and code for write.dta, (but
not the complied code) and I can't find anything. Ideally what I want is
something like write.dta(myDataFrame, labels = listOfLabels, fi
HI,
If I understand your question correctly, this should give the result:
dat1<-read.table(text="
A B C
a b a
x y z
",sep="",header=TRUE,stringsAsFactors=FALSE)
within(dat1,{new_column<-ifelse(A==C|B==C,"y","n")})
# A B C new_column
#1 a b a y
#2 x y z n
A.K.
- Original M
On Aug 28, 2012, at 9:20 PM, R. Michael Weylandt wrote:
On Tue, Aug 28, 2012 at 9:01 PM, Yi wrote:
Dear all,
Suppose my data frame is as follows:
id price distance
1 2 4
1 35
...
2 4 8
2 5 9
...
n 3 7
n 8 9
I would like to calculate the rank-order correlation betw
> * Duncan Murdoch [2012-08-28 21:06:33 -0400]:
>
> On 12-08-28 5:55 PM, Sam Steingold wrote:
>>> * R. Michael Weylandt [2012-08-28 13:45:35
>>> -0500]:
>>>
always you shouldn't need manual garbage collection.
>>
>> my observation is that gc in R sucks.
>> (it cannot release small objects).
On Tue, Aug 28, 2012 at 11:39 PM, Sam Steingold wrote:
>> * R. Michael Weylandt [2012-08-28 17:16:35
>> -0500]:
>>
>> install.packages by default goes to CRAN (the stable repository):
>
> how often are stable released made?
> weekly? monthly?
>
At the discretion of the package maintainers and t
> * Jeff Newmiller [2012-08-28 15:21:39 -0700]:
>
> Sam Steingold wrote:
>
>>> * R. Michael Weylandt [2012-08-28
>>13:45:35 -0500]:
>>>
always you shouldn't need manual garbage collection.
>>
>>my observation is that gc in R sucks.
>>(it cannot release small objects).
>>this is not specific
> * R. Michael Weylandt [2012-08-28 17:30:06
> -0500]:
>
>> In my experience, the one point I've needed it was after freeing
>> multiple very large objects when hitting memory limits.
that's what I am doing.
>> Rewriting that code to use functions rather than as one long
>> imperative slog was
> * R. Michael Weylandt [2012-08-28 17:16:35
> -0500]:
>
> install.packages by default goes to CRAN (the stable repository):
how often are stable released made?
weekly? monthly?
> if you want the bleeding edge, install e1071 from R-forge (note you
> might need a source install if the build bot
On Tue, Aug 28, 2012 at 9:58 PM, Salma Wafi wrote:
> Hi, I am trying to get the maximum likelihood estimator for lognormal
> distribution with censored data;when we have left, interval and right
> censord. I built my code in R, by writing the deriving of log likelihood
> function and using newt
On Tue, Aug 28, 2012 at 9:01 PM, Yi wrote:
> Dear all,
>
> Suppose my data frame is as follows:
>
> id price distance
> 1 2 4
> 1 35
> ...
> 2 4 8
> 2 5 9
> ...
> n 3 7
> n 8 9
>
> I would like to calculate the rank-order correlation between price and
> distance for each
Suppose your data is called dfs:
dfs$NEW_COLUMN <- ifelse(dfs$A == dfs$C | dfs$B == dfs$C, "y", "n")
which says
if dfs column A equals dfs column C or dfs column B equals dfs column
C, put "y" else put "n"; then put the resulting vector into NEW_COLUMN
of dfs.
Note that if your data are factors
On Tue, Aug 28, 2012 at 8:19 AM, Rantony wrote:
> Here i have a variable
>
> MyVar <- data.frame(read.csv("D:\\Doc.csv"))
read.csv() returns a data.frame so the outer call to data.frame() is
superfluous.
>
> And now i am storing this variable name into a list.
>
> MyList <- list()
> MyList [len
On Aug 28, 2012, at 11:52 AM, Mauricio Cornejo wrote:
Peter,
You're right ... and I do think I've finally understood the help
file on this. I somehow missed the connection between my variable's
values, which include NAs, and the fact that those returned values
are in turn indices for '[
Dear all,
Suppose my data frame is as follows:
id price distance
1 2 4
1 35
...
2 4 8
2 5 9
...
n 3 7
n 8 9
I would like to calculate the rank-order correlation between price and
distance for each id.
cor(price,distance,method = "spearman") calculate a correlation for
HI,
Replacing seq_along() with which() slightly improved CPU time.
system.time({
set.seed(1)
A<-sample(1:5,1e6,replace=TRUE)
which(!duplicated(A))
A[which(!duplicated(A))]
})
# user system elapsed
#0.040 0.012 0.052
A.K.
- Original Message -
From: Bronwyn Rayfield
T
Hi, I am trying to get the maximum likelihood estimator for lognormal
distribution with censored data;when we have left, interval and right censord.
I built my code in R, by writing the deriving of log likelihood function and
using newton raphson method but my estimators were too high " overest
Here i have a variable
MyVar <- data.frame(read.csv("D:\\Doc.csv"))
And now i am storing this variable name into a list.
MyList <- list()
MyList [length(MyList )+1]<- "MyVar"
Now what is the requirement is,
i need to call the variable name "MyVar" from the list "MyList " and get
the data.
---
Hi, My data frame looks like
A B C
a b a
x y z
I want to add a new column which says "y" if either A or B matches with C and
"n" if there is no match. How can I do that?
Thanks
[[alternative HTML version deleted]]
__
R-help@r-project.o
HI,
I was thinking about duplicated(). But, Bert already posted the solution. The
solution below is not very efficient.
A<-c(9,2,9,5)
unik<-as.numeric(names(table(A)))
match(unik,A)
#[1] 2 4 1
#Bert's solution wins here.
system.time({
set.seed(1)
A<-sample(1:5,1e6,replace=TRUE)
unik <- !duplic
HI,
I just answered to one with similar contents but with a different title. I
guess both are the same.
A.K.
- Original Message -
From: Sapana Lohani
To: R help
Cc:
Sent: Tuesday, August 28, 2012 5:41 PM
Subject: [R] Error in ops.factor: check and verify
Hi, may be i was not clear
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Sola
Hello all,
How do I actually use the output of predict.lm(..., type="terms") to
predict new term values from new response values?
I'm a chromatographer trying to use R (2.15.1) for one of the most
common calculations in that business:
- Given several chromatographic peak areas measured for
Here are two methods:
> A<-c(9,2,9,5)
> f1 <- function(x) { d <- !duplicated(x) ; data.frame(uniqueValue=x[d],
> firstIndex=which(d)) }
> f2 <- function(x) { u <- unique(x) ; data.frame(uniqueValue=u,
> firstIndex=match(u, x))}
> f1(A)
uniqueValue firstIndex
1 9 1
2
Tom Roche Tue, Aug 28, 2012 at 4:23 PM
>> Can ncdf4 be made to fail more helpfully? E.g., to fail immediately
>> on nc_open without assignment?
David W. Pierce Tue, 28 Aug 2012 13:23:53 -0700
> I'll see if there exists some way to do a better check for this.
TIA.
> One of the many great things
your specification was to have the data centered at 15 minute intervals and
extend 7.5 minutes on either side. on my output you see the time of the lower
value of the interval. 11:52:30 is the lower value of the 12:00:00 time. so
you can just add back the 7.5 minutes to the time value to get
On 12-08-28 5:45 PM, Rui Barradas wrote:
Hello,
Try the following.
occur1 <- function(pat, vec){
m <- length(pat)
n <- length(vec)
candidate <- seq.int(length=n-m+1)
for (i in seq.int(length=m))
candidate <- candidate[pat[i] == vec[candidate + i - 1]]
can
On 12-08-28 5:55 PM, Sam Steingold wrote:
* R. Michael Weylandt [2012-08-28 13:45:35 -0500]:
always you shouldn't need manual garbage collection.
my observation is that gc in R sucks.
(it cannot release small objects).
this is not specific to R; ocaml suffers too.
Sorry, I didn't realize y
On 12-08-28 5:32 PM, Marc Schwartz wrote:
On Aug 28, 2012, at 4:05 PM, Duncan Murdoch wrote:
Is there a function to efficiently search for a subsequence within a vector?
For example, with
x <- 1:100
I'd like to search for the sequence c(49,50,51), and be told that it occurs
exactly once, s
De rien. You're welcome.
Pascal
Le 12/08/29 5:23, Tom Roche a écrit :
Pascal Oettli: MERCI BEAUCOUP! (though I would have thanked you
earlier if I hadn't had to dig through the r-help digest first :-)
Tom Roche Mon, 27 Aug 2012 23:31:23 -0400
summary: I can successfully ncvar_put(...) data
On 2012-08-27 15:49, iwaite wrote:
I have read multiple books and looked at many posts online and can't seem to
figure out how to add a 1:1 line in lattice xyplot. I've tried multiple
versions of getting this to work, including trying "panel.abline(h=0,v=1),
panel=function and others.
Second que
On 8/28/2012 5:52 PM, Bert Gunter wrote:
Sheesh!
I would have thought that someone would have noticed that on the
?unique Help page there is a link to ?duplicated, which gives a
_logical_ vector of the duplicates. From this, everything else can be
quickly derived -- and packaged in a simple Matl
Sheesh!
I would have thought that someone would have noticed that on the
?unique Help page there is a link to ?duplicated, which gives a
_logical_ vector of the duplicates. From this, everything else can be
quickly derived -- and packaged in a simple Matlab like function, if
you insist on that. e.
Hi,
Try this:
order(A)[!duplicated(sort(A))]
--
Noia Raindrops
noia.raindr...@gmail.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and
It all looks up to date to me. The only one we can't see is lme4
(because it won't load) so could you also give us
installed.packages()[c("lme4","nlme","Matrix"),c(2,3,12)]
Beyond that, can you give your linux distro and installation details?
It might be necessary to punt this to the R-Debian or
On Tue, Aug 28, 2012 at 2:58 PM, Bronwyn Rayfield
wrote:
> I would like to efficiently find the first index of each unique value in a
> very large vector.
>
> For example, if I have a vector
>
> A<-c(9,2,9,5)
>
> I would like to return not only the unique values (2,5,9) but also their
> first indi
On Tue, Aug 28, 2012 at 5:13 PM, R. Michael Weylandt
wrote:
> On Tue, Aug 28, 2012 at 4:55 PM, Sam Steingold wrote:
>>> * R. Michael Weylandt [2012-08-28 13:45:35
>>> -0500]:
>>>
always you shouldn't need manual garbage collection.
>>
>> my observation is that gc in R sucks.
>> (it cannot
On Tue, Aug 28, 2012 at 4:55 PM, Sam Steingold wrote:
>> * R. Michael Weylandt [2012-08-28 13:45:35
>> -0500]:
>>
>>> always you shouldn't need manual garbage collection.
>
> my observation is that gc in R sucks.
> (it cannot release small objects).
> this is not specific to R; ocaml suffers too
install.packages by default goes to CRAN (the stable repository): if
you want the bleeding edge, install e1071 from R-forge (note you might
need a source install if the build bot hasn't gotten to it since the
update)
https://r-forge.r-project.org/R/?group_id=786
Cheers,
Michael
On Tue, Aug 28, 2
HI,
If you wish to keep it as factors (A-D), then try this:
test<-read.table(text="
A B C D E F
a b c d 40 30
a f a b 20 10
x m y m 50 30
",sep="",header=TRUE)
str(test)
#'data.frame': 3 obs. of 6 variables:
# $ A: Factor w/ 2 levels "a","x": 1 1 2
# $ B: Factor w/ 3 levels "b","f","m":
Hi, may be i was not clear when I made this query before.
I have 6 columns in my dataframe (test)
A B C D E F
a b c d 40 30
a f a b 20 10
x m y m 50 30
I want a new column which gets value in E if character value in column A
matches with that in column C. One more column comparing B and D. It
Peter,
You're right ... and I do think I've finally understood the help file on this.
I somehow missed the connection between my variable's values, which include
NAs, and the fact that those returned values are in turn indices for '['.
OK ... I'm all straight on this now. (I lost count of how
The listserver is very good about passing on attachments IF AND ONLY IF they
are text or an acceptable image format.
--
David
Sent from my iPhone
On Aug 28, 2012, at 11:17 AM, John Kane wrote:
> No data is attached. The R-help list usually strips out attachments to
> prevent viruses or malw
I was wondering if there was an R equivalent to the two phased approach that
MATLAB uses in performing the Kmeans algorithm. If not is there away that I
can determine if the kmeans in R and the kmeans in MATLAB are essentially
giving me the same clustering information within a small amount of erro
HI,
Try this:
test<-read.table(text="
A B C D E F
a b c d 40 30
a f a b 20 10
x m y m 50 30
",sep="",header=TRUE,stringsAsFactors=FALSE)
#Here, stringsAsFactors=FALSE is the key here. If you did not add that (or
=TRUE) , by default, A-D columns will be factors and you will get the error
mess
I would like to efficiently find the first index of each unique value in a
very large vector.
For example, if I have a vector
A<-c(9,2,9,5)
I would like to return not only the unique values (2,5,9) but also their
first indices (2,4,1).
I tried using a for loop with which(A==unique(A)[i])[1] to
Here's the sessinoInfo(). I didn't do the upgrade and I don't
know how to interpret the output.
> sessionInfo()
R version 2.15.1 (2012-06-22)
Platform: x86_64-pc-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8 LC_COL
HI,
You can also try this:
Students1<-data.frame(ID=c(101,201,303,304),Name=c("Andrew","John","Julie","Monica"),Fav_Place=c("Phoenix
AZ","San Francisco","California/New York","New York"))
gsubfun<-function(pattern,replacement,x, ...){
for(i in seq_along(pattern))
x<-gsub(pattern[i],replacement
Roy Mendelssohn Tue, 28 Aug 2012 09:07:32 -0700
> here is the relevant section from your code:
> > netcdf.file <- nc_create(
> > filename=netcdf.fn,
> > # vars=list(emis.var),
> > # verbose=TRUE)
> > vars=list(emis.var))
> > # Write data to data variable: gotta have file first.
> > # Gotta
Thanks for all help so far!
And I seems as you are correct Peter (and Jean too).
And I have now investigated and found how it is connected with the standard
errors:
If use the following code (taking from Jeans example code), where we have
one manova and two individual models (continue read comment
Thanks, when will I be able to use the new code?
I just did install.packages("e1071") + library(e1071). no good.
do I need to restart R?
thanks!
> * David Meyer [2012-08-27 22:57:17 +0200]:
>
> done, thanks for the suggestion.
>
> On 2012-08-27 21:15, Sam Steingold wrote:
>>
>> David, could you p
> * R. Michael Weylandt [2012-08-28 13:45:35
> -0500]:
>
>> always you shouldn't need manual garbage collection.
my observation is that gc in R sucks.
(it cannot release small objects).
this is not specific to R; ocaml suffers too.
> since a loop doesn't define its own scope like some languages
Hello,
Try the following.
occur1 <- function(pat, vec){
m <- length(pat)
n <- length(vec)
candidate <- seq.int(length=n-m+1)
for (i in seq.int(length=m))
candidate <- candidate[pat[i] == vec[candidate + i - 1]]
candidate
}
patrn <- c(1,2,3,4)
exmpl <- c(3,3,4,2,3,1,2
On Aug 28, 2012, at 4:05 PM, Duncan Murdoch wrote:
> Is there a function to efficiently search for a subsequence within a vector?
>
> For example, with
>
> x <- 1:100
>
> I'd like to search for the sequence c(49,50,51), and be told that it occurs
> exactly once, starting at location 49. (The
Is there a function to efficiently search for a subsequence within a vector?
For example, with
x <- 1:100
I'd like to search for the sequence c(49,50,51), and be told that it
occurs exactly once, starting at location 49. (The items in the vectors
might be numeric or character, and there migh
>
> - Nothing about the error text (IMHO) would lead one to the fix.
>
> (Note also that neither ncvar_put or nc_close appear to require
> assignment, which is probably what made me think I could nc_open
> without assignment.)
>
> Can ncdf4 be made to fail more helpfully? E.g., to fail immediately
Pascal Oettli: MERCI BEAUCOUP! (though I would have thanked you
earlier if I hadn't had to dig through the r-help digest first :-)
Tom Roche Mon, 27 Aug 2012 23:31:23 -0400
>> summary: I can successfully ncvar_put(...) data to a file, but when
>> I try to ncvar_get(...) the same data I get
>> >
On Aug 28, 2012, at 12:24 PM, Tom Roche wrote:
>
> Roy Mendelssohn Tue, 28 Aug 2012 09:07:32 -0700
>> here is the relevant section from your code:
>
>>> netcdf.file <- nc_create(
>>> filename=netcdf.fn,
>>> # vars=list(emis.var),
>>> # verbose=TRUE)
>>> vars=list(emis.var))
>
>>> # Write d
You could try running the dependency checker function directly and
then using the usual R debugging tools to start tracking down the problem.
E.g., I have a package in ~/packages/foo with a bogus entry in
the Depends line of DESCRIPTION:
> tools:::.check_package_depends(dir="~/packages/foo")
Packa
Sample data? Some representative sample data really is needed to see what you
are doing. Have a look at dput() for a handy way to present data.
Also where did theme_complete_bw come from? Is it a special formattng
function?
John Kane
Kingston ON Canada
> -Original Message-
> F
Hi,
I have revised my package cwhmisc and want to re-submit it (meeting a
deadline soon). One big stumbling stone is the 'sudden' appearance of
> sessionInfo()
R version 2.15.1 (2012-06-22)
Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit)
locale:
[1] C
attached base packages:
[1] tools
Hello,
You should say which package you are getting 'corrgram' from. I'll
assume it's from package corrgram.
The answer to your question is not in the help page but part of it is,
it says it calls function 'col.corrgram'.
And R is open source so a quick look at the package *.R sources (only
On 28/08/2012 2:16 PM, Liviu Andronic wrote:
On Tue, Aug 28, 2012 at 7:34 PM, Duncan Murdoch
wrote:
> Assign a class to the object, and write a print method for it.
>
> For example, this doesn't quite do what you want, but it's a start:
>
> print.noattributes <- function(x, ...) {
>attribute
On 28/08/2012 2:29 PM, Sam Steingold wrote:
At the end of a for loop its variables are still present:
for (i in 1:10) {
x <- vector(length=1)
}
ls()
will print "i" and "x".
this means that at the end of the for loop body I have to write
rm(x)
gc()
is there a more elegant way
On Aug 28, 2012, at 1:29 PM, Sam Steingold wrote:
> At the end of a for loop its variables are still present:
>
> for (i in 1:10) {
> x <- vector(length=1)
> }
> ls()
>
> will print "i" and "x".
> this means that at the end of the for loop body I have to write
>
> rm(x)
> gc()
>
>
On Tue, Aug 28, 2012 at 1:37 PM, R. Michael Weylandt
wrote:
> On Tue, Aug 28, 2012 at 1:29 PM, Sam Steingold wrote:
>> At the end of a for loop its variables are still present:
>>
>> for (i in 1:10) {
>> x <- vector(length=1)
>> }
>> ls()
>>
>> will print "i" and "x".
>> this means that
Perhaps I'm dense, but huh*?
-- Bert
*e.g. What are you trying to do? R does it's own garbage collection --
why do you think you need it?
And, as a general comment which may or may not be applicable, if you
create variables in a function they are local only to the function --
they disappear once t
Hello,
Maybe local().
Continue your example with
#?local
local(for (i in 1:10) {
x <- vector(length=1)
})
ls() # not 'i' nor 'x'
Hope this helps,
Rui Barradas
Em 28-08-2012 19:29, Sam Steingold escreveu:
At the end of a for loop its variables are still present:
for (i in 1:10) {
On Tue, Aug 28, 2012 at 1:29 PM, Sam Steingold wrote:
> At the end of a for loop its variables are still present:
>
> for (i in 1:10) {
> x <- vector(length=1)
> }
> ls()
>
> will print "i" and "x".
> this means that at the end of the for loop body I have to write
>
> rm(x)
> gc()
>
At the end of a for loop its variables are still present:
for (i in 1:10) {
x <- vector(length=1)
}
ls()
will print "i" and "x".
this means that at the end of the for loop body I have to write
rm(x)
gc()
is there a more elegant way to handle this?
Thanks.
--
Sam Steingold (http
On Tue, Aug 28, 2012 at 1:16 PM, Liviu Andronic wrote:
> On Tue, Aug 28, 2012 at 7:34 PM, Duncan Murdoch
> wrote:
>> Assign a class to the object, and write a print method for it.
>>
>> For example, this doesn't quite do what you want, but it's a start:
>>
>> print.noattributes <- function(x, ...
On Tue, Aug 28, 2012 at 7:34 PM, Duncan Murdoch
wrote:
> Assign a class to the object, and write a print method for it.
>
> For example, this doesn't quite do what you want, but it's a start:
>
> print.noattributes <- function(x, ...) {
>attributes(x) <- NULL
>print(x)
> }
>
> class(x) <-
Hello,
Inline.
Em 28-08-2012 18:23, Nordlund, Dan (DSHS/RDA) escreveu:
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Norbert Skalski
Sent: Tuesday, August 28, 2012 9:49 AM
To: r-help@r-project.org
Subject: [R] date in plot, can'
On 2012-08-28 07:44, Mauricio Cornejo wrote:
William, David, and Peter,
Thank you all so much for your help on this. Though I had read the help files
on 'subset' and '[', I had not been able to discern from that text what the
problem was. I could not have solved it without your help.
The
First of all, a practical way to supply data is to use the function dput()
Just do dput(mydata) and copy and paste the results into your email. The
reader can copy and paste into R and have an identical data set.
I am not sure I have followed exactly what you are doing but here is something
th
On 2012-08-28 10:34, Duncan Murdoch wrote:
On 28/08/2012 1:12 PM, Liviu Andronic wrote:
Dear all
Suppose the object below:
require(Hmisc)
require(plyr)
x <- dlply(iris, .(Species), describe)
How can I print the object without displaying the attributes? I
inspected ?print and ?print.default wi
On 28/08/2012 1:12 PM, Liviu Andronic wrote:
Dear all
Suppose the object below:
> require(Hmisc)
> require(plyr)
> x <- dlply(iris, .(Species), describe)
How can I print the object without displaying the attributes? I
inspected ?print and ?print.default with no luck.
Assign a class to the obje
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Norbert Skalski
> Sent: Tuesday, August 28, 2012 9:49 AM
> To: r-help@r-project.org
> Subject: [R] date in plot, can't add regression line
>
> Hello all,
>
> I have been using R
No data is attached. The R-help list usually strips out attachments to prevent
viruses or malware spreading.
Use dput to include the data . Just do : dput(mydata) and copy the results into
your email.
Also it is a good idea to include the code that you are using.
John Kane
Kingston ON Canad
Dear all
Suppose the object below:
> require(Hmisc)
> require(plyr)
> x <- dlply(iris, .(Species), describe)
How can I print the object without displaying the attributes? I
inspected ?print and ?print.default with no luck.
> x
$setosa
x[, "Sepal.Length"]
n missing uniqueMean .05
On Aug 27, 2012, at 19:15 , BrutishFruit wrote:
> Hi David.
> I mean that I want to get the *standard error of the predicted means* (which
> is a type standard deviation, if I have understand everything right), which
> the se.fit switch mentioned above should require from the "predict()"
> functi
On 28/08/2012 9:42 AM, Pedro Henrique Lamarão Souza wrote:
Hello,
I am a student of Materials Engineering and I want to minister an
introductory course of R at the university I study here in Brazil. I know R is
a free software, but I just want to know if I do need a special authorization
for
HI,
Try this:
Variable_1<-"MyDataFrame"
set.seed(1)
assign(Variable_1,sample(1:5,replace=TRUE))
assign(Variable_1,rbind(get(Variable_1),1:5))
get(Variable_1)
# [,1] [,2] [,3] [,4] [,5]
#[1,] 2 2 3 5 2
#[2,] 1 2 3 4 5
A.K.
- Original Message -
From
Though, for the record, it is perfectly legal and rather common to
charge for _instruction_ in R, just not for R itself.[1] This is done,
inter alia, at the UserR conferences. Similarly, one could charge for
books on R (physical or digital), for code deliverables in R, etc.
Cheers,
Michael
[1] Ac
Hi @ all,
I have a problem concerning the barplot (barchart lattice) of a dataframe. I
work with the attached dataframe.
When I try to plot this dataframe I only get two rows plottet. When I try
to plot the whole dataframe, there is message, that it is 'height' must be a
vector or a matrix.
On the
Others have mentioned assign and get, but generally when the answer to
a question is "assign" it means that you are asking the wrong question
(see fortune(236)).
This is actually FAQ 7.21, the most useful part of the answer in the
FAQ is the last few lines.
If you tell us more about what you are
Hello all,
I have been using R for about 3 weeks and I am frustrated by a problem. I have
read R in a nutshell, scoured the internet for help but I either am not
understanding examples or am missing something completely basic. Here is the
problem:
I want to plot data that contains dates on t
Ahhh that makes sense now.
I figured that the name would be that of the file you were downloading.
Feel a bit stupid now, many thanks!
--
View this message in context:
http://r.789695.n4.nabble.com/R-Download-Permission-Denied-tp4641568p4641582.html
Sent from the R help mailing list archiv
Hi Tom:
Sorry for top-posting. My bad etiquette for the day. The order that you do
things when you create nercdf files matters a lot. Here is a section from the
help for ncdf4:
> If you want to WRITE data to a new netCDF file, the procedure is to first
> define the dimensions your data arra
> > Now, " MyDataFrame " is a variable and containing some
> values in that.
> > And Now, the problem what is I need to do "rbind" into the
> variable "
> > MyDataFrame ".
> >
> > I tried to do,
> > rbind(as.character(Variable_1),
> > data.frame(read.csv("c:\\My2ndData.csv")))
> >
The above
I am not a lawyer, and you should not ask for such advice on the Web. However,
if you read the GNU license, it does not mention such activities. Money is not
traditionally the issue with free software... the obligation to pass the source
along if you change it is. I have personally taken a cours
On 28/08/2012 10:31 AM, Scott Raynaud wrote:
I suddenly started getting the error message below.
Not sure why. If I type intalled.packages() it
shows Matrix and lme4 installed. Can someone tell
what's going on and what I need to do to remedy the
problem? I'm running on a Linux box.
Dear Pedro
in your R session, enter the commands
license()
RShowDoc("COPYING")
"R is free software and comes with ABSOLUTELY NO WARRANTY.
You are welcome to redistribute it under certain conditions."
Those imply no restriction on charging a fee for presenting courses in R.
albyn
On 28-08-2012, at 16:57, TwistedSkies wrote:
> Hello,
>
> I am receiving a 'Permission Denied' error when trying to use the R Download
> funtion.
>
> I am wondering if this is a:
>
> - Error in the code
> - Permissions issue at source
> - Permissions issue at destination.
>
> If you could she
Hello,
I am a student of Materials Engineering and I want to minister an
introductory course of R at the university I study here in Brazil. I know R is
a free software, but I just want to know if I do need a special authorization
for doing it. The course will be totaly free and I also will not
googling "R unzip files" gives this as the first hit:
http://stackoverflow.com/questions/3053833/using-r-to-download-zipped-data-file-extract-and-import-data
Dirk's answer should get you started, but note that there are
different zip protocols, so you might need to modify for your data.
Cheers,
Hi,
Try this:
dat1<-readLines(textConnection(
"ID Name Fav_Place
101 Andrew Phoenix,AZ
201 John SanFrancisco
303 Julie California/New York
304 Monica New York"))
gsub("Phoenix","Tucson",gsub("New York","New York City",dat1))
#[1] "ID Name Fav_Place" "101 Andrew Tucson,AZ
Hi,
Try this:
set.seed(1)
dat1<-data.frame(A=sample(letters[20:25],replace=TRUE),B=sample(letters[1:6],replace=TRUE),C=c(letters[1:3],letters[3:1]),D=sample(letters[2:7],replace=TRUE),E=sample(letters[21:26],replace=TRUE))
newdat<-list()
for(i in 1:ncol(dat1)){
newdat[[i]]<-list()
newdat[[i]]<
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