?.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
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
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),
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
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.
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
?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 lplo...@usc.edu wrote:
Hi,
I am trying to generate all possible permutations (choose 2)
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 :).
From
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(d$z)
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
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
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.df, 20)
Erik Iverson wrote
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
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(location=c(loc
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 to
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
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
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
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
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
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
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.
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
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 er...@ccbr.umn.edu:
What class of object / structure do you exactly want
in the end? A matrix
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 er...@ccbr.umn.edu:
So what do you want the matrix to
look like, since the number
?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
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
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
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
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,
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
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,
Your code is not reproducible. Can you come up with a small example
showing the crux of your data structures/problem, that we can all run in
our R sessions? You're likely get much higher quality responses this way.
Ted Byers wrote:
From the documentation I have found, it seems that one of
On 07/12/2010 07:16 PM, Roger Deangelis wrote:
Hi,
I am new to R.
I am trying to create an R function to do a SAS proc means/summary
proc.means ( data=bsebal;
class team year;
var ab h;
output out=BseBalAvg mean=;
The high-level concept you need is called Regular Expressions. R
supports these through several functions, see ?regex .
Ralf B wrote:
Hi all,
I would like to detect all strings in the vector 'content' that
contain the strings from the vector 'search'. Here a code example:
content -
Is it possible to download data from password-protected ftp sites? I saw
another thread with instructions for uploading files using RCurl, but I
could not find information for downloading them in the RCurl documentation.
Did you try the ?getURL function in RCurl? See the `Test the
Using this dataframe with quite long column headers, how can I wrap the
text so that the columns are narrower. I was trying to use strwrap without
success. Thanks
reportDF - structure(list(IDDate = c(3/12/2010, 3/13/2010, 3/14/2010,
3/15/2010), FirstRunoftheYear = c(33 (119 ? 119), n (0 ?
Raubertas, Richard wrote:
I agree that 'list' is a terrible package name, but only secondarily
because it is a data type. The primary problem is that it is so generic
as to be almost totally uninformative about what the package does.
For some reason package writers seem to prefer
Suppose I want create variables with indexes in their names, e.g., X_1_1,
X_1_2, X_1_3, ..., X_1_10, X_2_1, X_2_2, X_2_3, .. X_2_10,..., X_10_1,
X_10_2, ... X_10_10. It looks like I need to use 2 indexes I and J so I is
looped from 1 to 10, and J is looped from 1 to 10. But I don't know how to
Do you care to share your sessionInfo() , as the Posting Guide asks?
I cannot reproduce on:
R version 2.11.1 (2010-05-31)
i486-pc-linux-gnu
On 07/13/2010 09:07 PM, Ian Seow wrote:
Hi, I'm encountering a strange error in POSIXlt... anyone got a clue on
this?
as.POSIXlt(1982-01-01)
Error in
syrvn wrote:
Thanks again for your quick reply.
I understood your procedure but now it is even more strange why my
conversion does not work.
In your example, the NA values are in brackets NA and what your
procedure does is to
convert these NA values into NA and then it seems to be possible to
Bogaso Christofer wrote:
Hi, there is a function Skewness() under fBasics package. If I type
skewness, I get followings:
Case matters in R, so please be precise.
skewness
function (x, ...)
{
UseMethod(skewness)
}
environment: namespace:timeDate
Would be great if
On 07/15/2010 08:02 AM, Jannis wrote:
Dear R community,
is there any way to include a backslash in a charakter string without meaning
some escape sequence?
E.g. i need a string like:
a- '\hline'
Error: '\h' is an unrecognized escape in character string starting \h
to include some latex
David Bickel wrote:
Sage and Marc, thank you for your helpful replies.
Since RLink enables R calls from Mathematica, I wonder if it would make
the Mathematica Front End useful for organizing R work even if no
Mathematica functions are needed. It would be nice R had something like
the Front
btc1 wrote:
Hello, I have a vector, dates, as a series of 3 digit elements, i.e. date
[1] 528 528 528 528 528 528 528 528 528 528 528 528 708 708 708 708 708
708
[19] 708 708 708 708 529 529 529 529 529 529 529 529 529 529 529 529 529
529
[37] 529 624
I need to convert them into julian,
louisey wrote:
Hello everyone!!
I have written a function in R and need to call it more than once say M
times then take an average of my output. How would I tell R to automatically
call my function many times instead of just the once?
The natural way will depend on what you're trying to
It's high time for a small, reproducible example. Don't say, for
example, *give* us an example in R.
That being said, something like
combined[!names(combined) %in% names(miceSample)]
might be a start...
Addi Wei wrote:
But miceSample has multiple columns... For example if nvars=4, I have
Doran, Harold wrote:
I just need to confirm something with pattern matching folks. I have
a factor with the following levels in a very large data set:
levels(all$Classical.Statistic)
[1] AB;ABD
CollapsedSteps CR_PCR_Prop;CR_P;AB
[6] NMK
Doran, Harold wrote:
Thanks. Yes, I did that on a toy data set and with my real data. It
*seems* to have worked. I just work with grep so rarely that I didn't
want to miss something.
Yes, I see no reason why it won't work. You can do another check
'manually'. You know how many of each you
On 07/15/2010 07:10 PM, James wrote:
I'm completely new to R, and I'd like to do something like this:
x=c(1,2,3)
plot(x,x)
At this point, R creates a file Rplots.pdf, since the default device is
PDF and the default filename is Rplots.pdf.
Since you're completely new to R, I
snip
So instead of:
mypng()
x=c(1,2,3)
plot(x,x)
I can just do:
x=c(1,2,3)
plot(x,x)
Isn't that what you can do after you define mypng and set
options(device = mypng)
?
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On 07/15/2010 08:32 PM, Axel Urbiz wrote:
Dear users,
My apologies for the simple question. I'd like to create a function where
the number of arguments is as big as the size of my data set. Supose I have
n observations in my data, how can I write a function like
fun- function (x1,x2,,xn)
We need to know how you're doing this, with a minimal example that we
can run. Most graphics devices accept a file or filename argument, so
that's one way. If you're using the pdf device, multiple plots will
create multiple pages in the final output..
linda.s wrote:
I made a plot, but
Anyway, I have often wished that something like
new.mt.sample - miceTrainSample[, -pID50]
would return miceTrainSample without the pID50 column. Here are three
alternative ways to do it.
# Method 1: Assign NULL to the column
new.mt.sample - miceTrainsSample
new.mt.sample$pID50 - NULL
# Method
This is not reproducible, and does not look minimal. You'll get better
answers, and probably solve many issues on your own, if you construct
small examples that illustrate the same problem you're having with your
real data.
Addi Wei wrote:
iterations - 100
nvars - 4
combined -
snip
I think I have identified the problem such that when I identify the
structure of some of the files that I am reading in, columns are labeled as
Factors. In other files, the same columns are labeled as numeric
values. Is there a way to assign the data structure to these columns in the
On 07/16/2010 06:59 PM, Joshua Wiley wrote:
Hello,
Is there an way to easy comment of sections of code? I was thinking
something along the lines of
\dontrun{
codeline 1
codeline k
}
but that could be used in regular script files. When I am still
working on a script, I often want to
On 07/17/2010 01:05 AM, Joshua Wiley wrote:
I use XEmacs + ESS, and looking for ways to add text to a region of
code, I see it is quite easy
with C-x r t. Thanks for your great advice.
I use GNU Emacs. With a region of code actve, M-; will add the comment line (or
remove it if it's already
Hello,
What I would like to do is have a data.frame with column names and have
these column names stored as strings in another vector. Then I would like
to be able to access the data.fram columns via referencing the vector of
names. The code below shows the last few executions that failed to
jd6688 wrote:
myDF =
data.frame(id=c(A10,A20),d1=c(.3,.3),d2=c(.4,.4),d3=c(-.2,.5),d4=c(-.3,.6),d5=c(.5,-.2),d6=c(.6,-.4),d7=c(-.9,-.5),d8=c(-.8,-.6))
doit=function(x)c(x[1],sum_LK_positive=sum(x[-1][x[-1]0]),sum_LK_negative=sum(x[-1][x[-1]0]))
myDF
id d1 d2 d3 d4 d5 d6
Josh B wrote:
Hi all,
I am trying to concatenate words together to create new column names, using a
loop.
Please consider the following toy example:
x - matrix(nrow = 1, ncol = 3)
colnames(x) - c(a, b, c)
x[1,1] - 1
x[1,2] - 2
x[1,3] - 3
I would like to create a new matrix with column
On 07/30/2010 06:04 PM, putri wikie wrote:
Hi,
I have problem with my R program. Actually, it just a simple program, but I do
not know the error is. Here is part of my program:
if (tau2ca==0) {MVvc- 0.01} else {MVvc- tau2ca}
ri_vc- si2/MVvc
vi_vc- ri_vc + 1
Y_bar_vc-
On 08/01/2010 08:48 PM, thmsfuller...@gmail.com wrote:
Hi,
The following two 'df's should be the same, although their
constructions are different.
But they aren't the same.
df1 - data.frame(X=c(1, 2, 3), Y=c(4, 5, 6))
df2 - data.frame(X=1:3, Y=4:6)
identical(df1, df2)
yields FALSE
See
Michael Lachmann wrote:
On 08/01/2010 08:48 PM, thmsfuller...@gmail.com wrote:
Hi,
The following two 'df's should be the same, although their
constructions are different.
But they aren't the same.
df1 - data.frame(X=c(1, 2, 3), Y=c(4, 5, 6))
df2 - data.frame(X=1:3, Y=4:6)
identical(df1,
I would like to apply a function to two vectors
For example,
A-c(NA,1,2,3,NA)
B-6:10
I would like C to be equal to A but if any A element is NA that
corresponding c element is equal to B, i.e.
C = c(6,1,2,3,10)
#untested...
C - ifelse(is.na(A), B, A)
Alastair wrote:
Hi,
I've got some boolean data in a data.frame in the form:
XYZA B C
[1] T TFT F F
[2] F TTF F F
.
.
.
What I want to do is create a new column which is the logical disjunction of
several of the columns.
Just like:
Trying To learn again wrote:
Hi all,
I wanto to run a plot about the levels of a variable parting on an ols
regression. The regression in done on the rate of return of the variable.
Imagine R_{t}=a+b*R_{t-1}
So If P, the estimated price would be P_{t}=P_{t-1}*R_{t}
Imagine that I obtain
GL wrote:
If I have a column with 2 levels, but one level has no remaining
observations. Can I remove the level?
What is a 'column'? An element of a data.frame?
Does the following help?
f1 - factor(L1, levels = c(L1, L2))
levels(f1)
f1 - factor(f1)
levels(f1)
In absence of a
Cristian Montes wrote:
Even shorter is
x-c(a,b,c,d)
paste(x, sep=+)
x-c(a,b,c,d)
paste(x, sep=+)
[1] a b c d
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PLEASE do read the posting guide
Christy Denckla wrote:
Can anyone help me with the necessary code to relevel a numeric*factor
interaction term in a linear model? I would like to report the estimate,
std. error and t-value for the reference factor.
First, I estimated a linear model with dummy variables and was able to
j.delashe...@ed.ac.uk wrote:
Quoting Prof Brian Ripley rip...@stats.ox.ac.uk:
See also Omegahat package Rcompression (a copy of which for Windows is
on CRANextras).
Thank you, I will look into that.
But I would do this via unzip, modify, zip
once or even 10 times, yes. After that it
Cristian Montes wrote:
Ok, sorry, I misinterpreted the question! Here is the right solution
x - c(a, b, c, d, e)
z - x[1]
for (i in 2:length(x))
{
z- paste(z, x[i], sep = +)
}
print(z)
[1] a+b+c+d+e
Or
paste(x, collapse = '+')
which was the solution that was in the
Dimitri Liakhovitski wrote:
I am sorry, I'd like to split my column (names) such that all the
beginning of a string (X..) is gone and only the rest of the text is
left.
x-data.frame(names=c(X..aba,X..abb,X..abc,X..abd))
x$names-as.character(x$names)
(x)
str(x)
Can't figure out how to apply
Hello,
I have a problem which has bitten me occasionally. I often need to
prepare graphs for many variables in a data set, but seldom for all.
or for any large number of sequential or sequentially named variables.
Often I need several graphs for different subsets of the dataset
for a given
Jennifer Hou wrote:
Hello,
I have got a linear model that looks like this:
lm(criterion ~ variable.A*variable.a + variable.B*variable.b + variable.C
*variable.c)
The output computed with stdCoeff() seems to be all right, but it does not show the
coefficients of the interaction of the first
What OS are you specifically referring to?
Even the most basic web search surely would have turned up the R
Installation and Administration Manual
http://cran.r-project.org/doc/manuals/R-admin.html#Installing-R-under-Unix_002dalikes
There may be pre-built packages available for your
Hello,
Please read the posting guide found at the bottom of every post to
this list. We need to be able to see a small, reproducible example
of code that illustrates your question.
It sounds like you might be looking for ?replicate.
On 08/05/2010 03:47 AM, Turn Fall wrote:
I have some code
On 08/05/2010 05:50 AM, Giuseppe Amatulli wrote:
Hi,
can somebody tell me why R is not able to calculate a linear model
written in this way?
lm (seq(1:100)~seq(1:100))
Call:
lm(formula = seq(1:100) ~ seq(1:100))
Coefficients:
(Intercept)
50.5
Warning messages:
1: In
Just FYI, the Hmisc package has had an implementation of %nin% for some
time now.
Ken Williams wrote:
Sometimes I write code like this:
qf.a - subset(qf, pubid %in% c(104, 106, 107, 108))
qf.b - subset(qf, !pubid %in% c(104, 106, 107, 108))
and I get a little worried that maybe I've
Jennifer Hou wrote:
Thank you very much for your kind reply, I have found that the error
was in the stdCoeff function and not in the linear model. summary()
works pretty well on my model, I will simply use another function to
compute my coefficients. Best regards, Jennifer
See ?coef ,
On 08/06/2010 08:03 AM, Werner W. wrote:
Hi,
I know ways to do this but they all seem awkward and I somehow believe that
there is a convenient shortcut.
Since you don't show us what you tried, I don't know what you consider
'awkward'.
If I have a data.frame with many columns, how can I
?Rprof
or
?system.time
TGS wrote:
Is there a way to find out what the computation duration time was to complete
executing a code chunk?
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PLEASE do read the posting
Raghuraman Ramachandran wrote:
guRus
I have say a dataframe, d and I wish to do the following:
1) For each row, I want to take one particular value of the row and multiply
it by 2. How do I do it. Say the data frame is as below:
OPEN HIGH LOW CLOSE 1931.2 1931.2 1931.2 1931.2 0 0 0 999.05
On 08/07/2010 10:29 AM, Stephen Liu wrote:
Hi steve,
After starting R which command shall I use to exit it rather than close the
console. exit/quit did not work. Thanks
What does did not work mean?
help.search(quit)
leads you to:
?quit
quit()
or
q()
should terminate the R session.
On 08/07/2010 01:00 PM, GMail (KU) wrote:
Hello,
While learning how to manipulate data with R, I found one example that I
could not understand.
In the following example, the function definition of maxcor has an argument
named i and I don't understand why. Could someone explain why the maxcor
On 08/07/2010 03:08 PM, josef.kar...@phila.gov wrote:
I have read the R manual and help archives, sorry but I'm still stuck.
How would I do a t-test with an adjusted p-value?
Please be more specific about what you mean by 'adjusted p-value'. See below...
Suppose that I use t.test ( ) ,
On 08/08/2010 08:36 PM, meetsiddu1 wrote:
Hi folks,
I am new to the R software. I have been going through different materials to
know more about R.
I have the R software installed on my windows machine.I would like to know
the R source code for the following problems on iris flower data set.
On 08/08/2010 08:47 PM, Stephen Liu wrote:
Hi folks,
x
[1] 1 2 3 4 5
y
[1] 6 7 8 9 10
plot(x,y, type = 1 )
Error in plot.xy(xy, type, ...) : invalid plot type '1'
I looked on man plot but unable to resolve. Please help. TIA
You never say what you expect type = 1 (that's the
On 08/08/2010 09:14 PM, Stephen Liu wrote:
Hi Erik,
I followed following video;
Statistics with R (part 3: plot and history tutorial)
http://www.youtube.com/watch?v=NfH5peM1RtIfeature=related
It is type = 1.
A graph was generated there. But I got an error msg
No, it was type = l (the
On 08/09/2010 01:16 AM, Alexander Eggel wrote:
Using R, I would like to find out which Samples (S1, S2, S3, S4, S5) fulfill
the following criteria:contain minimally one value (x, y or z) bigger than
4. Any ideas? Thanks, Alex.
data
Sample xy z
1S1 -0.35.3
Stephen Liu wrote:
The Introduction to R is rather terse, but a reasonable starting
point. The other manuals listed on that page are special-purpose
manuals for installation administration, development, etc.. You
can try the manuals listed in the contributed documentation section
too.
David,
I was working on a project involving a linear model, and wanted to
extract the standard error of a predictor. I am able to do so, but not
in the way I would expect.
I would have expected that if a created a model such as Model1 -
lm(y~x,z,d), the object Model1 would contain that
GL wrote:
Know that if I have List_1 and List_2 that I can check to see if the
intersect via the code below:
List _1:
a, b, c, d, e, f, g
List_2:
z, y, x, w, v, u, b
length(intersect(List_1, List_2)) 0
return = true
If instead I wanted to check a dataframe that is a list of lists, how
On 08/10/2010 09:04 PM, Alexander Eggel wrote:
How can I extract the samples (S1-S5) containing a TRUE value in their row?
Solution should apply to a much bigger data frame.
a
Samples A B C D . . .
1 S1 FALSE FALSE FALSE FALSE
2 S2 FALSE FALSE NA TRUE
3 S3 FALSE FALSE FALSE FALSE
4 S4 FALSE
Are you for some reason against writing your function to accept a single
argument, a formula, that you simply pass on to coxph?
Mendolia, Franco wrote:
Hello!
I have something like this:
test1 - data.frame(intx=c(4,3,1,1,2,2,3),
status=c(1,1,1,0,1,1,0),
Gene Leynes wrote:
*What I want to do:
*Create a windows shortcut that will start the R gui **and**
simultaneously source a file
*What I have already tried:
*This almost works, but it's not the interactive R GUI:
R --no-save --sdi -file=C:\SomePath\example.R
These open the R GUI, but
?match, look at the %in% operator.
Mestat wrote:
Hi listers,
I made some search, but i didn`t find in the forum.
I have a data set.
I would like to make a search (conditon) on my data set.
x-c(1,2,3,4,5,6,7,8,9,10)
count-0
if (CONDITON){count-1}else{count-0}
My CONDITION would be: is there
Olga Shaganova wrote:
Hi,
I am a brand new user and may be my question is too simple. I have R on
our (not Unix) server. I am trying to build a decision tree and the error
message says couldn't find function rpart. Does it mean I have to ask our
server guy to install an additional package?
Amit Patel wrote:
Hi
I am trying to calculate the row sums of a matrix i have created
The matrix ( FeaturePresenceMatrix) has been created by
1) Read csv
2) Removing unnecesarry data using [-1:4,] command
3) replacing all the NA values with as.numeric(0) and all others with as.numeric
(1)
On 08/12/2010 08:00 PM, elaine kuo wrote:
Dear list,
factor function was tried but failed.
Pls kindly help and thank you.
E.
code
rm(list=ls())
library(faraway)
data (pima)
pima$test - factor (pima$test)
[1] NA NA NA NA NA NA
The unreported warning message should have given you a
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