Hi,
I see no need to construct the vector, try this instead,
belong = function(x=4, y=c(1,10)) x <= y[2] && x >= y[1]
see also ?findInterval
HTH,
baptiste
On 13 August 2010 01:10, fishkbob wrote:
>
> So basically I want to do this -
>
> 4 %in% 1:10
> should return true
>
> Would there be an
Hi,
Here is one way:
> is.there <- function(mynumber, a, b) mynumber %in% a:b
> is.there(4, 1, 10)
[1] TRUE
> is.there(459124, 103000, 983000)
[1] TRUE
> is.there(4, 103000, 983000)
[1] FALSE
HTH,
Jorge
On Thu, Aug 12, 2010 at 7:10 PM, fishkbob <> wrote:
>
> So basically I want to do this -
Hi,
I'm trying to fit a multinomial logistic regression to my data which
consists of 5 discrete variables (scales 1:10) and 1000 observations.
I get the following error:
Error in `row.names<-.data.frame`(`*tmp*`, value = c("NA.NA", "NA.NA", :
duplicate 'row.names' are not allowed
In addition:
So basically I want to do this -
4 %in% 1:10
should return true
Would there be another way of doing this without having to do the 1:10 part?
I am using a very large data set and trying to do
459124 %in% 103000:983000
multiple times for many values, and it is taking quite a long time
Also, I
R CMD build (lower case) indeed works. I was confused with older versions of
Rtools where it didn't matter if you used lower or upper case (as well as that
I thought under DOS upper and lower cases don't make a difference). I was wrong
however.
Thanks for the help
__
I need a code to export my output to excel 2007.I am dealing with
observations of 15,000.Thank you
--
View this message in context:
http://r.789695.n4.nabble.com/help-tp2323542p2323542.html
Sent from the Export many data to Excel 2007 mailing list archive at Nabble.com.
Dear R-list
Suppose I have a data set stored in hmet, for which I did get confidence
limit on a linear regression as shown below.
My question is how I can subset only data points which are within the
confidence limit.
Thank you.
Keun-Hyung
---
Hi David,
Thanks for your advice.
The remaining commands will be;
### Create a side-by-side boxplot of the test01
boxplot(test01$DO ~ test01$Stream)
### Create an ANOVA table
dotest01.aov <-(test01$DO ~ test01$Stream)
summary(dotest01.aov)
### Conduct a Tukey's multiple comparison procedure
Tu
On Aug 12, 2010, at 11:34 PM, Stephen Liu wrote:
Hi folks,
File to be used is on;
data(InsectSprays)
I can't figure out where to insert it on following command;
test01 <- read.csv(fil.choose(), header=TRUE)
Don't think that would work, even if you did spell file.choose()
correctly.
?d
Hi folks,
File to be used is on;
data(InsectSprays)
I can't figure out where to insert it on following command;
test01 <- read.csv(fil.choose(), header=TRUE)
Please help. TIA
B.R.
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Thank you both for response!
Enh, I never think about function though. Good.
My way is using 'aggregate' and 'merge' but it is a little bit trivial.
On Wed, Aug 11, 2010 at 9:13 PM, Dennis Murphy wrote:
> Hi:
>
> Try this from package plyr:
>
> library(plyr)
> # function to compute sum of sq
The magic is not in setting options(digits=10). I did that just to show you
the accuracy of the answer. The magic is in setting a more stringent error
tolerance, rel.tol=1.e-10.
The divergent error is due to the fact that the default tolerance of (roughly)
1.e-04 is too large.
Ravi.
__
Hi folks,
Ubuntu 10.04 64 bit
Where can I find 64 bit RSQLite?
It seems not there;
RSQLite: SQLite interface for R
http://cran.r-project.org/web/packages/RSQLite/index.html
TIA
B.R.
Stephen L
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https://stat.eth
On Thu, Aug 12, 2010 at 4:12 PM, Lars Bishop wrote:
> The company where I work is considering getting a license for Revolution
> Enterprise - Windows 64-bit. I'll appreciate for those familiar with the
> product if can share your experiences with it? In particular, how does it
> compare to the "fr
On Aug 12, 2010, at 9:00 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)
There is a space between your "<" and your "-". R is interppreting
that a
Hi Ravi - Thank you. I'm wondering how the magic happens here with
options(digits=10)? And the most important point, I do not quite
understand why divergent error could occur. In this case, the function
is analytically integrable. Any insight would be greatly appreciated
so that I learn something o
On Aug 12, 2010, at 7:54 PM, Greg Snow wrote:
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Thursday, August 12, 2010 3:13 PM
To: Greg Snow
Cc: Joshua Wiley; r-help@r-project.org
Subject: Re: [R] storing the results of an apply call
On Aug 12, 2010, at
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
hi
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
[[alternative HTML version deleted]]
__
R-help@r-p
> -Original Message-
> From: David Winsemius [mailto:dwinsem...@comcast.net]
> Sent: Thursday, August 12, 2010 3:13 PM
> To: Greg Snow
> Cc: Joshua Wiley; r-help@r-project.org
> Subject: Re: [R] storing the results of an apply call
>
>
> On Aug 12, 2010, at 5:07 PM, Greg Snow wrote:
> >
> -Original Message-
> From: Marc Schwartz [mailto:marc_schwa...@me.com]
> Sent: Thursday, August 12, 2010 3:32 PM
> To: Greg Snow; Joshua Wiley
> Cc: r-help@r-project.org Forum
> Subject: Re: [R] storing the results of an apply call
>
> P.S. Pardon me while I go visit with the Emacs Psy
I downloaded their Academic version and installed it on a Windows virtual
machine (as there is not a Mac version available).
I played around with it a little bit and wasn't overly impressed. I still like
my current configuration: textmate with R bundle on Mac OSX.
With textmate data entry is a
Dear users,
The company where I work is considering getting a license for Revolution
Enterprise - Windows 64-bit. I'll appreciate for those familiar with the
product if can share your experiences with it? In particular, how does it
compare to the "free" version of R 64-bit?
Thanks in advance.
Re
On Thu, Aug 12, 2010 at 2:31 PM, Marc Schwartz wrote:
> On Aug 12, 2010, at 4:07 PM, Greg Snow wrote:
>
>>> -Original Message-
>>> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
>>> project.org] On Behalf Of Joshua Wiley
>>> Sent: Wednesday, August 11, 2010 6:31 PM
>>> To: Lo
I write about R every weekday at the Revolutions blog:
http://blog.revolutionanalytics.com
and every month I post a summary of articles from the previous month
of particular interest to readers of r-help.
In case you missed them, here are some articles related to R from the
month of July:
http:/
On Aug 12, 2010, at 6:15 PM, David Winsemius wrote:
On Aug 12, 2010, at 5:20 PM, Toby Gass wrote:
Hi,
I do want to look only at slope.
If there is one negative slope measurement for a given day and a
given chamber, I would like to remove all other slope measurements
for that day and that c
On Aug 12, 2010, at 5:20 PM, Toby Gass wrote:
Hi,
I do want to look only at slope.
If there is one negative slope measurement for a given day and a
given chamber, I would like to remove all other slope measurements
for that day and that chamber, even if they are positive.
On one day, I will
OK, looks sensible, although I said either abs() OR squared
differences. I don't think it makes sense to use both the L1 and the
L2 metric at the same time.
--
David.
On Aug 12, 2010, at 4:58 PM, TGS wrote:
# just to clean it up for my own understanding, the "difference"
approach as you ha
Use nls().
But you should also take the time to learn a bit about data fits in
general. I'm far from expert, but it sure looks like that P(t) is
overconstrained. Since cos(wt) = sin(wt+pi/2), why do you have all
those different terms?
Dear all,
I have a time series and I like to fit S
On Aug 12, 2010, at 4:07 PM, Greg Snow wrote:
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
>> project.org] On Behalf Of Joshua Wiley
>> Sent: Wednesday, August 11, 2010 6:31 PM
>> To: Lorenzo Cattarino
>> Cc: r-help@r-project.org
>> Subject: Re: [R]
Hi,
I do want to look only at slope.
If there is one negative slope measurement for a given day and a
given chamber, I would like to remove all other slope measurements
for that day and that chamber, even if they are positive.
On one day, I will have 20 slope measurements for each chamber.
If you want a grid of hexagons, look at the examples in the my.symbols function
(TeachingDemos package), one of those gives a way to create the grid of
hexagons (not as efficient as the hexbin package, but allows you to set your
own colors).
For choosing colors you may want to look at the RColo
On Aug 12, 2010, at 5:07 PM, Greg Snow wrote:
[1] "metacarpal sit-down unfriendliness chapel ME"
My wife gives "ME" the "middle phalangeal unfriendliness" message
about once a week.
--
David Winsemius, MD
West Hartford, CT
__
R-help@r-proje
On Aug 12, 2010, at 3:06 PM, Toby Gass wrote:
> Thank you all for the quick responses. So far as I've checked,
> Marc's solution works perfectly and is quite speedy. I'm still
> trying to figure out what it is doing. :)
>
> Henrique's solution seems to need some columns somewhere. David's
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Joshua Wiley
> Sent: Wednesday, August 11, 2010 6:31 PM
> To: Lorenzo Cattarino
> Cc: r-help@r-project.org
> Subject: Re: [R] storing the results of an apply call
>
> What foll
Hi all,
I'm having difficulties passing data from the master to the children. I
haven't seen any complete tutorial, all the tutorial broadcast data where
I would like to send parts of my big tablechip list.
Everytime I run this, the last line always jams. I even tried to reduce
the number of r
On Thu, 12 Aug 2010, Tim Gruene wrote:
I don't know if it's elegant enough for you, but you could split the file into
two files with 'grep "^3" file > file_3' and 'grep "^4" file > file_4'
and then read them in separately.
along the same lines, but all in R (untested)
original.lines <- readL
# just to clean it up for my own understanding, the "difference" approach as
you had suggested would be
x <- seq(.2, .3, by = .1)
f1 <- function(x){
x*cos(x)-2*x**2+3*x-1
}
plot(x,f1(x), type = "l")
abline(h = -.1)
abline(v = x[which.min(abs(diff((f1(x) - (-.1))**2)))], lty = 'dotted'
Hi Marcio,
Your friend has given the answer.
x <- rnorm(10)
y <- rnorm(10)
ind <- c(3,0,1,0,3,0,2,2,0,0)
plot(x, y, col = grey(0:max(ind)/max(ind))[ind], pch = 16)
Mestat wrote:
>
> I am working o a scatterplot where I would like to plot the variables
> according with another frequency var
On Aug 12, 2010, at 4:06 PM, Toby Gass wrote:
Thank you all for the quick responses. So far as I've checked,
Marc's solution works perfectly and is quite speedy. I'm still
trying to figure out what it is doing. :)
Henrique's solution seems to need some columns somewhere. David's
solution do
Try this:
setdiff(list, s)
On Thu, Aug 12, 2010 at 5:32 PM, André de Boer wrote:
> Hi,
>
> I want to eliminate an element of a list:
>
> list <- seq(1,5,1)
> s <- sample(list,1)
>
> lets say s=3
> Now I want to remove 3 from the list: list2 = {1,2,4,5}
>
> Can someone give me a tip?
>
> Thanks,
Hi,
I want to eliminate an element of a list:
list <- seq(1,5,1)
s <- sample(list,1)
lets say s=3
Now I want to remove 3 from the list: list2 = {1,2,4,5}
Can someone give me a tip?
Thanks,
André
[[alternative HTML version deleted]]
__
R-hel
Hi listers...
I am working o a scatterplot where I would like to plot the variables
according with another frequency variable.
Another friend here proposed this code...
x <- rnorm(10)
y <- rnorm(10)
ind <- c(1,0,1,0,1,0,1,1,0,0)
plot(x, y, col = ind + 1, pch = 16) # 1 is black, 2 is red
But in
What did you try?
R CMD build (build all lower case) does work for me
Uwe Ligges
On 09.08.2010 22:30, Hintzen, Niels wrote:
Dear all,
As I couldn't find any thread on the internet I hope the help-list might help
me out.
I've tried to update Rtools from R210 used in combination with R2.9.1
Thank you all for the quick responses. So far as I've checked,
Marc's solution works perfectly and is quite speedy. I'm still
trying to figure out what it is doing. :)
Henrique's solution seems to need some columns somewhere. David's
solution does not find all the other measurements, possibl
I was meaning something like the following:
x <- seq(.2, .3, by = .01)
f <- function(x){
x*cos(x)-2*x**2+3*x-1
}
plot(x,f(x), type = "l")
abline(h = -.1)
But I'm guessing "uniroot" will do this?---I haven't looked far into the
uniroot function to see if it will solve this.
On Aug 12, 20
I don't know if it's elegant enough for you, but you could split the file into
two files with 'grep "^3" file > file_3' and 'grep "^4" file > file_4'
and then read them in separately.
Tim
On Thu, Aug 12, 2010 at 01:57:19PM -0400, Denis Chabot wrote:
> Hi,
>
> I know how to read fixed width forma
On Aug 12, 2010, at 3:54 PM, TGS wrote:
Actually I spoke too soon David.
I'm looking for a function that will either tell me which point is
the intersection so that I'd be able to plot a point there.
Or, if I have to solve for the roots in the ways which were
demonstrated yesterday, then
I did. Did not work. Did you try your code? The matrix did not result into
integer numbers as expected. MY approach resulted in a correct scan
result, at least.
M.
> Martin Tomko wrote:
>> Hi Peter,
>> apologies, too fast copying and pasting.
>> So, here is the explanation:
>> f<-"C:/test/mytab.tx
Actually I spoke too soon David.
I'm looking for a function that will either tell me which point is the
intersection so that I'd be able to plot a point there.
Or, if I have to solve for the roots in the ways which were demonstrated
yesterday, then would I be able to specify what the horizontal
Yes, I'm playing around with other things but the "points()" function is what I
was looking for. Thanks
On Aug 12, 2010, at 12:47 PM, David Winsemius wrote:
On Aug 12, 2010, at 3:43 PM, TGS wrote:
> I'd like to plot a point at the intersection of these two curves. Thanks
>
> x <- seq(.2, .3,
I searched with "print x-axis label in 45 degree" which didn't return useful
links. Apparently I used poor search keywords.
- Original Message
From: David Winsemius
To: Marc Schwartz
Cc: array chip ; r-help@r-project.org
Sent: Thu, August 12, 2010 12:34:16 PM
Subject: Re: [R] x-ax
Than you Marc.
John
- Original Message
From: Marc Schwartz
To: array chip
Cc: r-help@r-project.org
Sent: Thu, August 12, 2010 12:17:12 PM
Subject: Re: [R] x-axis label print in 45 degree
On Aug 12, 2010, at 2:14 PM, array chip wrote:
> Hi how can print x-axis labels in 45 degree in
I'd like to plot a point at the intersection of these two curves. Thanks
x <- seq(.2, .3, by = .01)
f <- function(x){
x*cos(x)-2*x**2+3*x-1
}
plot(x,f(x), type = "l")
abline(h = 0)
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Hi Olga,
not directly related to your question. We have also a server
installation and subsequently our IT department determines which version
and packages I can use on R.
A few days ago I have switched to R-portable. Works without any problems
from my USB stick on any locked-for-installation W
On Aug 12, 2010, at 3:17 PM, Marc Schwartz wrote:
On Aug 12, 2010, at 2:14 PM, array chip wrote:
Hi how can print x-axis labels in 45 degree in boxplot() (or plot
in general)? I
can use las=2 to print in 90 degree, but it looks ugly. Is there a
simple option
to do 45 degree easily?
Thank
On Aug 12, 2010, at 2:24 PM, Marc Schwartz wrote:
> On Aug 12, 2010, at 2:11 PM, Toby Gass wrote:
>
>> Dear helpeRs,
>>
>> I have a dataframe (14947 x 27) containing measurements collected
>> every 5 seconds at several different sampling locations. If one
>> measurement at a given location is
On Aug 12, 2010, at 2:11 PM, Toby Gass wrote:
> Dear helpeRs,
>
> I have a dataframe (14947 x 27) containing measurements collected
> every 5 seconds at several different sampling locations. If one
> measurement at a given location is less than zero on a given day, I
> would like to delete al
Try this:
subset(toy, !rowSums(mapply(is.element, toy[c('CH', 'DAY')], subset(toy,
SLOPE < 0, CH:DAY))) > 1 | SLOPE < 0)
On Thu, Aug 12, 2010 at 4:11 PM, Toby Gass wrote:
> Dear helpeRs,
>
> I have a dataframe (14947 x 27) containing measurements collected
> every 5 seconds at several differen
On Aug 12, 2010, at 3:11 PM, Toby Gass wrote:
Dear helpeRs,
I have a dataframe (14947 x 27) containing measurements collected
every 5 seconds at several different sampling locations. If one
measurement at a given location is less than zero on a given day, I
would like to delete all measuremen
On Aug 12, 2010, at 2:14 PM, array chip wrote:
> Hi how can print x-axis labels in 45 degree in boxplot() (or plot in
> general)? I
> can use las=2 to print in 90 degree, but it looks ugly. Is there a simple
> option
> to do 45 degree easily?
>
> Thanks
>
> John
John,
See R FAQ 7.27 How c
Hi how can print x-axis labels in 45 degree in boxplot() (or plot in general)?
I
can use las=2 to print in 90 degree, but it looks ugly. Is there a simple
option
to do 45 degree easily?
Thanks
John
__
R-help@r-project.org mailing list
https://stat.
Dear helpeRs,
I have a dataframe (14947 x 27) containing measurements collected
every 5 seconds at several different sampling locations. If one
measurement at a given location is less than zero on a given day, I
would like to delete all measurements from that location on that day.
Here is a t
hi Peter,
There's no single function for ICC with variable number of judges. To
estimate the variances in that case you need hierarchical linear modeling. I
posted code for this at Stackoverflow in answer to your question there:
http://stackoverflow.com/questions/3205176/
Stats questions get
Another suggestion: compare
-1:4
with
-(1:4)
-Peter Ehlers
On 2010-08-12 12:28, 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
Yes, please do as Erik said in the future but here's one way to do it.
(A <- matrix(data = rnorm(n = 9, mean = 0, sd = 1), nrow = 3, ncol = 3, byrow =
FALSE, dimnames = NULL))
matrix(rowSums(A))
On Aug 12, 2010, at 11:28 AM, Amit Patel wrote:
Hi
I am trying to calculate the row sums of a matr
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)
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)
When I carry out the
On Wed, Aug 11, 2010 at 10:14 PM, Brian Tsai wrote:
> Hi all,
>
> I'm interested in doing a dot plot where *both* the size and color (more
> specifically, shade of grey) change with the associated value.
>
> I've found examples online for ggplot2 where you can scale the size of the
> dot with a va
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?
Ah. Indeed, this is from the glmulti. I had not realized there would be
problems using Java. Is there a way around this to still use a multicore
approach? For what other packages that use multiple cores will this not be a
problem?
-Jarrett
On Aug 12, 2010, at 11:02 AM, Thomas Lumley wrote:
You can use following scriptI think
#create a vector of random numbers on which to test script
v<-sample(1:3,size=90,replace=TRUE)
#creates two matrixes out of vector v which can be assigned to M to test
script
M2<-matrix(v,ncol=2)
M3<-matrix(v,ncol=3)
M<- #Assign you're matrix or a te
I think your code will work but only for the two columns I gave. I used those
as an example but my actual data is 200 in length with two columns and I
need code that will give a label to each unique pair but still have the
original length for instance, one that will turn something such as
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?
Thank you,
Olga
[
On Thu, 12 Aug 2010, Jarrett Byrnes wrote:
I'm running r 2. on a mac running 10.6.4 and a dual-core macbook pro. I'm
having a funny time with multicore. When I run it with 2 cores, mclapply, R
borks with the following error.
The process has forked and you cannot use this CoreFoundation func
Hi,
I know how to read fixed width format data with read.fwf, but suddenly I need
to read in a large number of old fwf files with 2 types of lines. Lines that
begin with "3" in first column carry one set of variables, and lines that begin
with "4" carry another set, like this:
…
3A00206546L070
I'm running r 2. on a mac running 10.6.4 and a dual-core macbook pro. I'm
having a funny time with multicore. When I run it with 2 cores, mclapply, R
borks with the following error.
The process has forked and you cannot use this CoreFoundation functionality
safely. You MUST exec().
Break on
Thanks everyone for your help and advice. For the R-help archives, here is
what I ended up doing.
First creating a separate function to handle one day at a time -
byrow.gen2 <- function(genmat,rownum,use1,use2,num,ortho_obs_used){
prev = rownum-1
ran = runif(length(rownum),0,1)
if(genmat[rownum,
Martin Tomko wrote:
> Hi Peter,
> apologies, too fast copying and pasting.
> So, here is the explanation:
> f<-"C:/test/mytab.txt";
> R<-readLines(con=f);
>
> where mytab.txt is a table formatted as noted in previous post (space
> delimited, with header, rownames, containing integers).
>
> Now,
On 12/08/2010 12:10 PM, Tim Gruene wrote:
Hello,
when I ran "R CMD INSTALL circular_0.4.tar.gz" on a machine with Debian stable,
the command also created the documentation in various fomrats (latex, html,
online).
The same command on Debian testing only provides the online documentation.
How c
On 12/08/2010 10:35 AM, szisziszilvi wrote:
I've tried lm, but something is wrong.
I've made a test dataset of 599 data points, my original equation is
zz = 1 +0.5*xx -3.2*xx*xx -1*yy +4.2*yy*yy
but the R gives this result:
---
> mp <- read.cs
Hello,
when I ran "R CMD INSTALL circular_0.4.tar.gz" on a machine with Debian stable,
the command also created the documentation in various fomrats (latex, html,
online).
The same command on Debian testing only provides the online documentation.
How can I persuade R to create the different form
@JLucke:
As for the africa variable: I took it out of the model, so that we can
exclude this variable itself and collinearity between the africa and the
litrate variable as causes for the litrate-problem. This also removed the
singularity remark at the top. However, the problem with litrate-varia
On Thu, Aug 12, 2010 at 4:51 PM, Stephen Liu wrote:
> Hi Barry,
>
>
> Following 2 commands are useful to me;
>> row.names(subset(file.info(list.files(getwd(),full.name=TRUE)),isdir))
> showing directories.
>
>> row.names(subset(file.info(list.files(getwd(),full.name=TRUE)),!isdir))
> showing files
On Aug 12, 2010, at 11:51 AM, Stephen Liu wrote:
Hi Barry,
Following 2 commands are useful to me;
row
.names(subset(file.info(list.files(getwd(),full.name=TRUE)),isdir))
showing directories.
row.names(subset(file.info(list.files(getwd(),full.name=TRUE)),!
isdir))
showing files
What is
thanks !!!
On 12/08/10 17:49, William Dunlap wrote:
segments(x0=ix-w, x1=ix+w, y0=mediansByGroup, col=ix)
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PLEASE do read the posting guide http://www.R-project.org/post
I would like to build an XML file from scratch using the XML package.
I would like to save the following vector in it:
1:10
Could someone help me by briefly outlining how I go about it ?
And maybe provide a few lines of code?
Thanks!
Mark
Mark Heckmann
Di
Hi Barry,
Following 2 commands are useful to me;
> row.names(subset(file.info(list.files(getwd(),full.name=TRUE)),isdir))
showing directories.
> row.names(subset(file.info(list.files(getwd(),full.name=TRUE)),!isdir))
showing files
What is "!" for? TIA
B.R.
Stephen L
- Original Messa
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of David martin
> Sent: Thursday, August 12, 2010 7:42 AM
> To: r-h...@stat.math.ethz.ch
> Subject: [R] Median abline how-to ?
>
> Hi,
> I'm newbie with R and don't really know how to add a median
> line to ea
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Learn how to use R's search resources!
--
Bert Gunter
Genentech Nonclinical Statistics
On Wed, Aug 11, 2010 at 8:21 PM, Geoffrey Smith wrote:
> Hello, does anyone know how to compute the following two normality tests
> using R:
>
> (1) the Kiefer-Salmon (1
I think I understand your question and the following would produce the result
you've posted.
(x <- matrix(c(1, 2, 2, 3, 1, 2, 1, 2, 3, 4), nrow=5, byrow=TRUE))
On Aug 12, 2010, at 5:41 AM, clips10 wrote:
Thanks for the help,
I tried to apply this to a vector with two columns, well I suppose i
> You're not seeing the .Rdata file containing the data objects. Try:
> list.files(getwd(),full.name=TRUE, all.files=TRUE)
Hi Keith,
Thanks for your advice
On R console running
> list.files(getwd(),full.name=TRUE,all.files=TRUE)
The output is similar to running following command on Ubuntu
On Aug 12, 2010, at 10:35 AM, asdir wrote:
This command
cdmoutcome<- glm(log(value)~factor(year)
+log(gdppcpppconst)+log(gdppcpppconstAII)
+log(co2eemisspc)+log(co2eemisspcAII)
+log(dist)
+fdiboth
+odapartnertohost
Hi,
I'm newbie with R and don't really know how to add a median line to each
of the groups that is not all the plot long.
Here is a small working code that i have adapted for my purpose. If
somebody could tell me how to draw median lines on each group and not
all plot long.
ctl <- c(4.17,5
I have a simple dataset of a numerical dependent Y, a numerical independent X
and a categorial variable Z with three levels. I want to do linear
regression Y~X for each level of Z. How can I do this in a single command
that is without using lm() applied three isolated times?
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right. How does it come that if I devide the result vector with
10*interception, I get a much better result?
> zz2 <- 25.86 -2239.86*mp$xx -595.01*mp$xx*mp$xx + 2875.54*mp$yy +
> 776.84*mp$yy*mp$yy
> mp$zz2 <- zz2
> library(lattice)
> cloud(zz2/258.6 + zz ~ xx * yy, data=mp)
looks quite pretty.
Hi guys,
I have a code in R and it was work well but when I decrease the epsilon
value (indicated in the code) , then I am getting this error
Error: evaluation nested too deeply: infinite recursion /
options(expressions=)?
any help please
y = 6.8;
w = 7.4;
z = 5.7;
muy = 7;
muw = 7;
muz =
I've tried lm, but something is wrong.
I've made a test dataset of 599 data points, my original equation is
zz = 1 +0.5*xx -3.2*xx*xx -1*yy +4.2*yy*yy
but the R gives this result:
---
> mp <- read.csv(file="sample.csv",sep=";",header=TRUE)
> lm
Thanks for the help,
I tried to apply this to a vector with two columns, well I suppose it is not
a vector but for instance like this:
[,1] [,2]
[1,]1 2
[2,]2 3
[3,]1 2
[4,]1 2
[5,]3 4
and return a vector :
1,2,1,1,3, so that it recognises b
baptiste,
I have two more questions. How can I get the category labels right
justified? It seems that I need to change the size of the text too. Since in
my real data the text size are big and they appears to be even bigger than
my plot area. So, my second question is how can I change the text siz
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