Ultimately, I would like for this to be 1 conponent in a larger function
that will produce PROC CONTENTS style output. Something like...
data1.contents<-data.frame(Variable=names(data1),
Class=sapply(data1,class),
n.valid=sapply(data1,sum(!is.na)),
n.miss=sapply(data1,sum(is.na)))
data1.cont
Dan,
I am attempting to write a function to count the number of non-missing
values of each column in a data frame using the sapply function. I have the
following code which is receiving the error message below.
n.valid<-sapply(data1,sum(!is.na))
Error in !is.na : invalid argument type
Tha
On 04/07/2011 07:55 PM, Rolf Turner wrote:
On 08/04/11 12:40, Erik Iverson wrote:
On 04/07/2011 07:32 PM, Rolf Turner wrote:
On 08/04/11 12:22, Erik Iverson wrote:
Rolf,
What does
capabilities("tcltk")
return?
FALSE
I believe R needs to be compiled with tcltk support for the
On 04/07/2011 07:32 PM, Rolf Turner wrote:
On 08/04/11 12:22, Erik Iverson wrote:
Rolf,
What does
> capabilities("tcltk")
return?
FALSE
I believe R needs to be compiled with tcltk support for the package to work. Do
you compile your own version or use a package manager (e.
Rolf,
What does
> capabilities("tcltk")
return?
On 04/07/2011 07:15 PM, Rolf Turner wrote:
Perhaps I'm being even thicker than usual, but I can't find
the tcltk package on CRAN. There is a tcltk2 package, which says
that it is a collection of supplements to tcltk, but I cannot
see a just-pla
jim holtman wrote:
They are probably the same. It isjust that summary is printing out 4
significant digits. Try:
options(digits = 20)
FYI, the default summary method also has its own digits argument.
On Tue, Apr 5, 2011 at 12:38 PM, Daniel Malter wrote:
Hi,
I summary() a variable w
Holly,
try
> length(unique(x)) == 1
where x is your vector of interest. But think about
how you want NA values to be treated, and also think about
R FAQ 7.31 if dealing with floating point numbers.
--Erik
Beale, Holly (NIH/NHGRI) [F] wrote:
Is there a less cryptic way to compare three or mo
See:
http://en.wikipedia.org/wiki/Coefficient_of_determination#Adjusted_R2
and the implementation in summary.lm :
ans$adj.r.squared <- 1 - (1 - ans$r.squared) * ((n -
df.int)/rdf)
Brian Smith wrote:
Hi,
Sorry for the naive question, but what exactly does the 'Adjusted R
Jonas,
Try looking at the tikzDevice package, and/or
the pgfSweave package.
--Erik
Jonas Stein wrote:
Hi,
i want to make my plots look uniform in LaTeX documents.
- usage of the same font on axes and in legend like LaTeX uses
(for example "Computer Modern")
- put real LaTeX formulas on th
John,
as.data.frame is a generic function that will call different methods
depending on what class of object you pass to it. The different
methods may have different arguments that they expect or honor.
The stringsAsFactors parameter is only used in certain methods of
as.data.frame. When you p
Gene, it's described in ?NumericConstants
HTH, Erik
Gene Leynes wrote:
I've been wondering what L means in the R computing context, and was
wondering if someone could point me to a reference where I could read about
it, or tell me what it's called so that I can search for it myself. (L by
itse
Is this what you mean?
z[which(z[,"x"] == 5) - 1, "y"]
?which is probably what you're looking for...
Hongwei Dong wrote:
Hi, R users,
I'm wondering if I can identify an element in a column by an element in
another column. For example:
x<-1:10
y<-11:20
z<-cbind(x,y)
z
x y
[1,] 1 11
[
complete.cases useful.
Also try:
sum(!is.na(o$m))
?tapply may also be useful in general: e.g.,
tapply(o$m, o$n, mean, na.rm = TRUE)
None of this is tested...
hope this explains, what I need to know.
Thanks,
S.
Am 22.02.2011 16:50, schrieb Erik Iverson:
Sandra,
Please provide a small
Sandra,
Please provide a small, reproducible example of this issue.
You probably want to use ?is.nan and not the inequality
operator.
Similar example, contrast:
x <- NA
is.na(x)
x == NA
Sandra Stankowski wrote:
Hey there,
I tried to count the number of rows, where my data isn't NaN in a
cer
Gene,
?rapply is a recursive version of ?lapply, and should work.
rapply(masterlist, function(x) x*NA, how = "replace")
--Erik
Gene Leynes wrote:
Hello all,
Maybe I'm being thick, but I was trying to figure out a simple way to create
a list with the same dimension of another list, but popul
Julia,
While not a direct answer to your question, you may find the
pgfSweave driver/package to be interesting.
http://cran.r-project.org/web/packages/pgfSweave/
Among other things, it uses the tikzDevice package for figures, which
will cause the labels and text to use the same font as in your
On 02/09/2011 09:21 PM, Benjamin Caldwell wrote:
Am trying to merge about 15 .csv tables - tried a test run but came up with
0 rows (no data for each variable/column head)
CAHSEE.EA.feb.2009<-read.csv("2009 CAHSEE EA feb 2009.csv", header=TRUE)
CAHSEE.IM.MATH.2009<-read.csv("2009 CAHSEE Impact
Mark Knecht wrote:
Title asks it all.
Thanks in advance,
Mark
a = 1:5
b1 = 2:6
Z = data.frame(a,b1)
Z
Z$b1
count = 1
MyName = paste("b",count,sep="")
MyName
Z$MyName
Z[[MyName]]
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koooee wrote:
is this the expected behavior of as.character ?
resultset is a data.frame from a sqlQuery() using RODBC
a = as.character(as.vector(resultset[1]))
a
[1] "c(-1, 1, 2, 3, 4, 5, 6, 7, 8, 9)"
I would expect the statement above to return similar to the result below, am
I missing so
H Roark wrote:
I have a data set covering a large number of cities with values for
characteristics such as land area, population, and employment. The problem I
have is that some cities lack observations for some of the characteristics and
I'd like a quick way to determine which cities have m
H Roark wrote:
I'm wondering about the behavior of the merge function when using factors as by
variables. I know that when you combine two factors using c() the results can
be odd, as in:
c(factor(1:5),factor(6:10))
which prints: [1] 1 2 3 4 5 1 2 3 4 5
I presume this is because factors ar
?tapply would be one way to get the max for all
conditions at once.
Your example is not reproducible, so I cannot give you a
reproducible answer, but adapt the following:
tapply(df$responce, df$condition, max)
A. Ramzan wrote:
Hello
I am trying to find a way to find the max value, for only
I am trying to create a function that is able to calculate this sum:
a<-c(2,3,5)
b<-(8,7)
with "a" meaning 235 and "b" 87. So the result of this sum would be 235 + 87
= 322.
a <- c(2,3,5)
b <- c(8,7)
vectorToScalar <- function(x) {
as.numeric(paste(x, collapse = ""))
}
vectorToScalar(a)
Suppose I have a function, like list, that takes a variable number of
arguments, and I have those arguments in some vector x. How can execute the
function with the *contents* of x as its arguments? I.e., I don't want
list(x), but rather list(x[[1]], x[[2]], ..., x[[n]]), but I don't want to
spel
http://crantastic.org/
On 01/24/2011 09:08 PM, zubin wrote:
We should really have an R package rating site, comments, reviews or
such, like folks do for apps or movie reviews. Does anyone know of a
site trying to do this. If i remember correctly a few R user
conferences ago this was talked abo
Roy Mathew wrote:
Thanks for the reply Erik, As you mentioned, grouping consecutive
elements of 'a' was my idea.
I am unaware of any R'ish way to do it. It would be nice if someone in
the community knows this.
Is this the idea you're trying to execute? It uses ?rle and ?mapply.
a <- c(2,3
Roy,
I have no idea what you're actually trying to do here, but
it looks like there would be a more natural R'ish way if
you're concerned about grouping consecutive elements of 'a'.
At any rate, within your while loop, you're incrementing n by
1, and eventually n will be 10, which will be transf
Joe P King wrote:
So many matrices are square symmetrical (i.e. variance-covariance matrices),
is there any way to get R to split the matrix on its diagonal and just
return one diagonal?
mat<-matrix(c(1,4,3,4,1,2,3,2,1), nrow = 3, ncol=3, byrow=TRUE)
is there anyway to get the lower rig
Silvano wrote:
Hi,
I used subset command, like this:
grupoP = subset(dados, grupos=='P', select=c(mortos, vivos, doses,
percevejos, p))
and the variables in select option are numeric.
They may *look* numeric, but are they really? You don't give us
enough information to determine that.
Does the prtest argument help when you actually use the 'print' function
around your summary.formula object? I think that's how I
solve it.
I.e.,
sf1 <- summary(trt~sex+ascites,data=ex,test=T,method="reverse",catTest=u)
print(sf1, prtest = "P")
Descriptive Statistics by trt
+---+---+
The closest I get is
u<-function(a,b){
j<-fisher.test(a)
p<-list(P=j$p.value,stat=1,df=1,testname=j$method,statname="")
return(p)
}
However then I manually have to edit the output. Is there a smart way of doing
this?
You're not explaining
Please provide a reproducible example!
E.g., use ?dput to dump a minimal data.frame that
exhibits this issue on the newest version of R.
Justin Fincher wrote:
Howdy,
I have written a small function to generate a simple plot and my
colleague is having an error when attempting to run it. Esse
all object that the
parser would not make).
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Erik Iverson
Sent: Monday, December 13, 2010 12:17 PM
To: R-help
Subject: [R] Does
Hello,
Does anyone know of a function that will determine whether
or not a formula object has a left hand side?
I.e., can differentiate between
y ~ x + z
and
~ x + z
Perhaps I'm overlooking the obvious...
Thanks!
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On 12/13/2010 07:14 AM, Özgür Asar wrote:
Dear Researchers,
I am looking for to read a SAS macro in R. Although I searched it on web, I
couldn’t find anything.
Are you hoping just to read it in, or to actually have it
execute the macro as SAS would?
What gives you the idea the latter is ev
On 12/11/2010 04:48 PM, Tal Galili wrote:
Hello dear R-help mailing list,
My question is *not* about how factors are implemented in R (which is, if I
understand correctly, that factors keeps numbers and assign levels to them).
My question *is* about why so many functions that work on factors don
Does the following help?
A = c("A\\nB", "C\\nD")
test <-data.frame(A)
#access levels directly to change names
levels(test$A) <- sub("n", "\n", levels(test$A))
#re-order levels of the factor
test$A <- relevel(test$A, "C\nD")
Rob James wrote:
I am trying to reorder a factor variable that ha
One way:
as.vector(t(cbind(ColA, ColB)))
Ross, Stephanie wrote:
Hello,
I have a general formatting question. I have two columns of data:
ColA <- c("m", "m", "m", "m")
ColB<- c("d","d","d","d")
And I would like to reorder them into a new column that looks like this:
ColC<- c("m","d","m","d",
On 12/07/2010 05:29 PM, Paul Miller wrote:
Hello Everyone,
Been learning R over the past several months. Read several books and have
learned a great deal about data manipulation, statistical analysis, and
graphics.
Now I want to learn how to make nice looking documents and about "literate
progr
Phil Spector wrote:
Jahan -
Try
dat$target = toupper(dat$target)
Note that in this case, the above *will* coerce dat$target
to a character vector, which may or may not be what
is intended.
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Hello,
To do what you want, see ?toupper :
levels(dat$target) <- toupper(levels(dat$target))
However, for clarity,
dat$target is not a "string variable", it is a factor,
which you can verify with
> str(dat)
Factors are enumerated types, and have a discrete
set of 'levels' associated with the
On 12/02/2010 09:35 PM, Jinsong Zhao wrote:
Hi there,
In function, it's usually using ``='' to assign default value for function
argument. For newbie, it's possible to using ``<- '' to assign value for
function argument. Although it's not a correct way, R don't give any warning
message.
> matr
Edwin -
I think the usual way to do this would be to use a function
like lapply or mapply to call your function multiple times
with varying arguments.
For example, with one varying argument:
lapply(list(3,5), test, z = 4)
With multiple varying arguments:
mapply(test, y = list(3,5), z = list(4,
Chris,
Well, the 'answer' could be:
which(sapply(a, function(x) all(x == c(1,2
But I wonder how these elements of 'a' in your
actual application are coming to be? If you're
constructing them, you can give the elements of
the list names, and then it doesn't matter what
numerical index they
Is the algorithm converging? Is there separation (i.e.,
perfect predictor) in the model?
Are you getting a warning about fitted probabilities of
0 or 1?, etc.
We would need much more information (preferably a reproducible
example) before we can help.
Benjamin Godlove wrote:
Dear R developers,
Well, the error message seems relatively straightforward.
When you run str(x) (you did not provide the data)
you should see 1 or more components are factors that have more than 32
levels. Apparently you can't include those predictors in a call
to randomForest.
You might find the following line
On 11/09/2010 09:16 PM, vicho wrote:
I want to make a function for geometric seqeunce
since
testing=function(x){i=1;ans=1;while(true){ans=ans+(1/x)^i ; i=i+1}
;return(ans)}
doesn't work... the program is freeze...
What exactly are you trying to do? Where does "true" get set?
Did you mean T
This type of object has the "matrix" class in R.
So just use ?matrix to create it.
matrix(1:25, ncol = 5)
for example.
On 11/09/2010 08:55 PM, sachinthaka.abeyward...@allianz.com.au wrote:
Hi All,
I want to have an array/ matrix that looks this
x<- 0 0 1 1
1
?set.seed is what you're looking for
Xiaoxi Gao wrote:
Hello R users,
Here is my question about generating random sample. How to set the random seed to
recreate the same random numbers? For example, 10 random numbers is generated from
N(0,1), then "runif(10)" is used.What if I want to get the
Then one solution is to use
rbind.fill from the plyr package.
johannes rara wrote:
This is the ideal result (data.frame):
result
names X1 X2 X3
1 a 1 2 NA
2 b 1 2 3
2010/11/8 Erik Iverson :
So what do you want the matrix to
look like, since the number of columns
will be
So what do you want the matrix to
look like, since the number of columns
will be different between the two rows?
johannes rara wrote:
Thanks, data.frame or matrix.
-J
2010/11/8 Erik Iverson :
What class of object / structure do you exactly want
in the end? A matrix, a data.frame, a vector
What class of object / structure do you exactly want
in the end? A matrix, a data.frame, a vector?
johannes rara wrote:
Hi,
How to rbind these vectors from a list?:
l <- list(a = c(1, 2), b = c(1, 2, 3))
l
$a
[1] 1 2
$b
[1] 1 2 3
do.call(rbind, l)
[,1] [,2] [,3]
a121
b
On 11/06/2010 11:36 AM, Robert Ruser wrote:
Hello R Users,
I'm wondering if there exists any elegant and simple way to do the
following: I have a data.frame X fills in numbers. I have a vector y with
numbers as well. Every value in X that is equal to any values in y should
be replaced by e.g. 1.
Could you give a small reproducible example please?
It is not clear to me what your looping structure is
doing, or what your goal here is.
There may be a much simpler method than introducing
subscripts.
--Erik
Wade Wall wrote:
Hi all,
I have a dataframe (df1) that I am trying to select values
Well, %in% returns a logical vector...
So
subset(dat, ! ID %in% someID)
Also, from ?subset:
Note
that ‘subset’ will be evaluated in the data frame, so columns can
be referred to (by name) as variables in the expression
Thus, you don't need 'dat$ID', bur just 'ID' in the subset argum
The learning curve of R is rather steep at start.
Yes, it can be. I think the general advice would be
to get a good Intro to R book if you're just starting
out.
That's certainly my advice.
Or, get a book on some methods you're interested
in that uses R (the Springer useR! series is
really us
DomDom wrote:
Hi there,
i´ve got a problem with how to create a vector with three variables out of
three seperate ascii files.
These three ascii files contain pixel information of the same image but
different bands and i need a matrix of
vectors, with each vector containing the corresponding
Anand Bambhania wrote:
Hi all,
I am processing 24 samples data and combine them in single table called
CombinedSamples using following:
CombinedSamples<-rbind(Sample1,Sample2,Sample3)
Please use reproducible examples.
Now variables Sample1, Sample2 and Sample3 have many different columns
Hello,
The best way to get help from people on the list is
for you to give us *reproducible* examples of exactly
what is you want.
Usually, you can come up with some sample data and code
that corresponds to your situation, and that we can run
directly by cutting and pasting from the email.
You
Just read the help page :).
This is under "Note" in the ?pdf.
On some systems the default plotting character ‘pch = 1’ is
displayed in some PDF viewers incorrectly as a ‘"q"’ character.
(These seem to be viewers based on the ‘poppler’ PDF rendering
library). This may be due
Just
merge(df1, df2, all = TRUE)
does it, yes?
Dimitri Liakhovitski wrote:
Hello!
I have 2 data frames like this (well, actually, I have 200 of them):
df1<-data.frame(location=c("loc 1","loc 2","loc
3"),date=c("1/1/2010","1/1/2010","1/1/2010"), a=1:3,b=11:13,c=111:113)
df2<-data.frame(locati
Hadley's reshape package (google for it) can
do this. There's a nice intro on the site.
> library(reshape)
> cast(melt(mydf, measure.vars = "value"), city ~ brand,
fun.aggregate = sum)
city x y z
1a 3 23 450
2b 12 42 231
Although the numbers differ slightly?
I've heard of
Note that these methods don't 'delete' observations.
They all create brand new objects that are
subsets of the test.df object. You can effectively
'delete' the observations by replacing the original
data.frame with the returned object...
so
test.df <- head(test.
It depends on which 20 you want.
If you have a data.frame called 'test.df', you can do:
#first 20
test.df[20, ]
-or-
head(test.df, 20)
#random 20
test.df[sample(nrow(test.df), 20), ]
None of this was tested, but it should be a start.
--Erik
Matevž Pavlič wrote:
Hi,
I am sure that can
alas, should R not come with an is.defined() function?
?exists
a variable may
never have been created, and this is different from a variable
existing but holding a NULL. this can be the case in the global
environment or in a data frame.
> is.null(never.before.seen)
Error: objected 'ne
ivo welch wrote:
quick programming questions. I want to "turn on" more errors. there
are two traps I occasionally fall into.
* I wonder why R thinks that a variable is always defined in a data frame.
> is.defined(d)
[1] FALSE
> d= data.frame( x=1:5, y=1:5 )
> is.defined(
lord12 wrote:
s= "Hey"
a = "Hello"
table = rbind(s,a)
write.table(table,paste("blah",".PROPERTIES",sep = ""),row.names =
FALSE,col.names = FALSE)
In my table, how do I output only the words and not the words with the
quotations?
You read the help page for the function you're using :).
Fro
?head or just
df[1:20, ]
Louis Plough wrote:
Hi,
I am simply looking for the function that will allow you to look at the top
20 lines of a long dataset?
LP
On Mon, Nov 1, 2010 at 10:46 AM, Louis Plough wrote:
Hi,
I am trying to generate all possible permutations (choose 2) of a vector,
b,
Simply read the ?library help page, where you'll find under Details:
To suppress messages during the loading of packages use
‘suppressPackageStartupMessages’: this will suppress all messages
from R itself but not necessarily all those from package authors.
Christofer Bogaso wrote
I suggest using the most recent version of R (2.12.0 I believe)
and providing a reproducible example, showing us exactly how you
are creating this data.frame (assuming it still exhibits the
behavior).
--Erik
ANJAN PURKAYASTHA wrote:
Hi,
I'm creating a data frame of 24 columns and 45101 rows.
He
claudia tebaldi wrote:
Hi all
Just this morning I upgraded to R 2.12.0 (for Mac OS X 10.6.4).
All went well until I needed to run a help() or help.search() in my session,
which I'm running within Emacs (ESS 5.3.7).
That's very old version of ESS, I have no problems with 2.12.0 with ESS
5.11
Hello,
How would you go about handling the following situation?
This is on R 2.12.0 on Ubuntu 32-bit.
I have a wrapper function to lm. I want to pass in a
subset argument. First, I just thought I'd use "...".
## make example reproducible
set.seed(123)
df1 <- data.frame(age = rnorm(100, 50, 10
On 10/24/2010 04:57 PM, Jason Kwok wrote:
I'm trying to import a CSV file into R and when it gets imported, the
entries get numbered down the left side. How do I get rid of that?
When you imported the CSV file into R, an object of class data.frame
was created, and since you did not assign it t
?.libPaths
clee wrote:
hi all,
How can I change the library path in R? I don't have permission to write to
the default R library on the computer I am running R on.
I have searched the forum and have not found anything that I understand, so
I apologize if this has been asked before.
Thanks ve
Hello,
There are probably many ways to do this, but I think
it's easier if you use a data.frame as your object.
The easy solution for the matrix you provide is escaping
me at the moment.
One solution, using the plyr package:
library(plyr)
A <- data.frame(a = rnorm(100),b = runif(100), c = rep
Julia,
Can you provide a reproducible example? Your code calls the
'rq' function which is not found on my system.
Any paring down of the code to make it more readable would
help us help you better, too.
Julia Lira wrote:
Dear all,
I am trying to run a loop in my codes, but the software r
Hello,
Jon Zadra wrote:
Hi,
I've looked all over for a solution to this, but haven't had much look
in specifying what I want to do with appropriate search terms. Thus I'm
turning to R-help.
In the process of trying to write a simple function to rename individual
column names in a data fr
Bart,
I'm hardly one of the lists regex gurus: but this can
get you started...
tests <- c("pH", "Assay (%)", "Impurity A(%)", "content (mg/ml)")
x <- regexpr("\\((.*)\\)", tests)
substr(tests, x + 1, x + attr(x, "match.length") - 2)
Bart Joosen wrote:
Hi,
this should be an easy one, but I c
Shiv wrote:
I am trying a simple toin coss simulation, of say 200 coin tosses. The idea
is to have a data frame like so:
Experiment#Number_Of_Heads
1 104
296
3101
So I do:
d <-data.frame(exp_num=c(1,2,3)); /* J
See the R Data Import/Export manual for the first step:
http://cran.r-project.org/doc/manuals/R-data.html
?read.table should help you out. You might use
?lapply along with read.table to read in multiple files.
Then, use ?merge, possibly in tandem with the ?Reduce
function, depending on how man
Hello,
You can use ddply from the very useful plyr package to do this.
There must be a way using "base R" functions, but plyr is
worth looking into in my opinion.
> install.packages("plyr")
> library(plyr)
> ddply(myData, .(class, group, name), function(x) mean(x$height))
class group name V
What output medium? On a graphic? In LaTeX ouput?
Or do you mean in the R console?
Joe P King wrote:
I wasn't sure about the subject so I am sorry about the vagueness, but if I
have a vector of values, how do I get a certain type of vectors to be bold
or italics?
So let x be a vector from
1) Please use an informative subject line
2) The answer may depend on the package, and can be easy
to difficult. Which package are you trying to install?
wenyue sun wrote:
*Dear All, *
**
*I want to install tar.gz in R on a Windows operating system. I know that
R in Win accepts packages in
Another alternative is to use Geany [1]. It would save you the trouble
of learning Emacs,
/s/save/deprive
/s/trouble/thrill
:)
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PLEASE do read the posting guide http:
Forgot to mention you should be able to install from a CRAN
mirror.
Felipe Carrillo wrote:
Hi:
I get a couple of warnings when trying to download gridExtra:
install.packages("gridExtra",repos=http://R-Forge.R-project.org)
Warning: unable to access index for repository
http://R-Forge.R-proj
Well, a quick check of
http://r-forge.r-project.org/bin/windows/contrib/
shows that there is no 2.10 directory, possibly because
it's not very new, 2.12 is scheduled for release in 2 weeks.
Felipe Carrillo wrote:
Hi:
I get a couple of warnings when trying to download gridExtra:
install.pack
Mike,
Small, reproducible examples are always useful for the rest of the us.
x <- c(0, NA, NaN, 1 , 10, 20, 21, Inf)
x[!is.na(x) & x >=1 & x<= 20]
Is that what you're looking for?
mlar...@rsmas.miami.edu wrote:
I calculated a large vector. Unfortunately, I have some measurement error
in my d
Hello,
Although on the surface a simple request,
I think you need to be more specific.
?paste returns a character vector. The first question
is: do you want a character vector of length 1 or 2?
It sounds like you're trying to format text for display
on screen or on a graphics device. Perhaps y
On 09/22/2010 07:24 PM, sisxy wrote:
Hello, i am trying to import the csv file into R .
i have a file saved as csv in my desktop.
My laptop is Window vista, version R is 2.10.1.
then i used the code
Q<-read.csv("Q.csv",header=TRUE)
R will search in its working directory for Q.csv.
What is t
Ozlem,
Just read ?formula, where it says:
There are two special interpretations of ‘.’ in a formula. The
usual one is in the context of a ‘data’ argument of model fitting
functions and means ‘all columns not otherwise in the formula’:
see ‘terms.formula’. In the context of
Everyone is free to create as they please. My opinion is that between
this list and stackexchange (as Tal pointed out), that there isn't
a market for such a thing.
If you want a web front-end to this list, such things exist, like
nabble.
Vojtěch Zeisek wrote:
Hello,
this might be little off-to
Erik Iverson wrote:
Brian J Mingus wrote:
Hi all,
I'm having trouble getting access to the pwr. This is on Ubuntu Lucid
Lynx,
64 bit.
I'm installing pwr via packages.install('pwr'), and loading it via
library(pwr), both of which appear successful.
Perhaps the actu
Brian J Mingus wrote:
Hi all,
I'm having trouble getting access to the pwr. This is on Ubuntu Lucid Lynx,
64 bit.
I'm installing pwr via packages.install('pwr'), and loading it via
library(pwr), both of which appear successful.
Perhaps the actual output would help here.
Strangely, I neve
What are you trying to do?
Did you look at the arguments for
the first argument of ?available.packages ?
What did you expect "emacs" to do?
On 09/17/2010 10:20 PM, Stephen Liu wrote:
Hi folks,
Debian 504 64-bit
What is the correct syntax to check the available of a package on R repo?
avai
On 09/17/2010 09:52 PM, Stephen Liu wrote:
Hi folks,
Debian 504 64-bit
Emacs Version 22.1.1
I have Emacs+ESS running on the box. R can work on ESS. But the fonts on the
menu bar (top) of Emacs are NOT clear, difficult to read, grey foreground. I
have been
googling around for solution withou
Sarah Goslee wrote:
Leaving aside the question of whether this is a good thing to do,
Let's not leave that aside. Surely there is a much more
straight-forward way to accomplish what you want. Why not store
all the X's in a vector, and then you can avoid this for/assign/get
hack.
you
can
Diogo B. Provete wrote:
I have a data set and I want to procedure to model fitting (e.g., Poisson,
Gausian, binomial, quasipoisson etc.). I'd like to know if there is an
easier way to do this in R.
Easier than what ?
There is no shortage of R functions and packages to fit almost any type
of
David Winsemius wrote:
On Sep 15, 2010, at 1:44 PM, Sarah Jilani wrote:
Hi,
I need to call an R program from Sas. I have tried using the following
code
in Sas using the x command but it just calls up dos and says
I went searching for a worked example and found this:
http://www.nesug.or
Sarah,
This is a SAS question, not R. However, it seems clear that it
has something to do with the fact that there are spaces in the
command that you're sending to Windows.
Maybe try calling with the 'short directory name' notation,
I forget what that's called in Windows.
Or else follow-up on
See ?list.files and ?file.info.
E.g.,
file.info(list.files("~/tmp", full.names = TRUE))
Michael D wrote:
I have a bunch of files named
"[identifier1].[identifier2].[mmdd].[hhmmss].txt" and im having
trouble updating my read script for the timestamp.
I've been using file.exists and a for l
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