On Sat, 23 Jan 2010, Nai-Wei Chen wrote:
Hi,
How can I find and download??a function in R to do the LU decompostion for
finding the upper and lower triangular matrix.?? Thank you so much.
Joe,
The 'Do your homework' advice in the _posting guide_ should get you there.
Specifically,
Update on above. I sampled my data to create a 10,000 observation data set.
I then tried lme with a correlation = corSpher and only one predictor, as a
test. I set my memory.limit to the max allowable. It ran for a while then
returned
Error: cannot allocate vector of size 64.0 Mb.
I can see
Hi,
I have many files which look like this:
"
2009/02/07 12:30:10.0 5.0161 13.208
2009/02/07 12:45:10.0 5.0102 13.350
2009/02/07 13:00:10.0 5.0044 13.473
2009/02/07 16:30:10.0 4.9366 13.788
2009/02/07 16:45:10.0 4.9397 13.798
end d
Hi,
How can I find and download a function in R to do the LU decompostion for
finding the upper and lower triangular matrix. Thank you so much.
Joe
___
æ¨ççæ´»å³æé ï¼ æºéãå¨æ¨ãçæ´»ãå·¥ä½ä¸æ¬¡æå®ï¼
[[
Hi to all. I make a little movie with myself, is it good?
http://tin.bz/tiffany
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Dear R-Help Group:
R will not start on my imac running 10.6.2. I installed both R2.10.1 and
2.10.0 from pkg. It crashed on both. I tried both the R and R64 apps and
both versions 2.10.1 and 2.10.0. I googled and found nothing on this.
Below is part of report:
Process: R [767]
Path:
Sorry...
also, how do I remove the last line which says "end data"
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Why am I getting a wrong result for quartiles?
here is my code:
> cbiomass = c(910, 1058, 929, 1103, 1056, 1022, 1255, 1121, , 1192,
> 1074, 1415)
> summary(cbiomass)
> IQR(cbiomass)
The result R gives me is:
For the summary
> Min. 1st Qu. MedianMean 3rd Qu.Max.
9101048
Hi,
I have a spatial data set with many observations (~50,000) and would like to
keep as much data as possible. There is spatial dependence, so I am
attempting a mixed model in R with a spherical variogram defining the
correlation as a function of distance between points. I have tried nlme,
lme
It's looks like you think that type=2 are the 'true' quantiles, but the
default method in R is type=7
You might want to look at ?stats::quantile
hth
david freedman
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Hi:
# Start with some fictitious data:
> firm <- rep(1:10, each = 5)
> year <- rep(2002:2006, 10)
> resp <- rnorm(50)
> dat.long <- data.frame(firm = firm, year = year, resp = resp)
> head(dat.long)
firm yearresp
11 2002 -0.77367688
21 2003 -0.79069791
31 2004 0.69257133
4
sqldf, authored by Gabor, if I read your intention correctly.
D.
On Fri, Jan 22, 2010 at 5:00 PM, Don MacQueen wrote:
> Perhaps using the R merge() function, possibly twice in succession, will do
> the job. (merge() does a one to many relational join, but with only two
> dataframes at at time).
Perhaps using the R merge() function, possibly twice in succession,
will do the job. (merge() does a one to many relational join, but
with only two dataframes at at time).
Or, there is an R package that lets you use the SQL language on
dataframes. I don't recall its name, but a search on R pac
Hi,
how to compile with tcl/tk support ? command line options?
thank you:)
2010/1/22 Uwe Ligges
> You need to reinstall R and compile with tcl/tk support (best is to use
> R-2.10.1 nowadays since yours is already outdated).
>
> Best,
> Uwe Ligges
>
>
>
>
> On 22.01.2010 08:13, æé丰 wrote:
On Jan 22, 2010, at 6:49 PM, Marlin Keith Cox wrote:
I can plot this just fine:
t<-seq(0,4, by=.1)
y<- t^3-6*t^2+5*t+30
plot(t,y ,xlab="t-values", ylab="f(t)", type="l")
This is the first derivative, how I I make a similar plot?
t<-seq(0,4, by=.1)
y<- t^3-6*t^2+5*t+30
y1<-D(expression(t^3-6*t^2
On Fri, 22 Jan 2010, Jason Morgan wrote:
Hello Jean-Baptiste,
On 2010.01.22 16:32:53, Jean-Baptiste Combes wrote:
Hello,
I am learning R and I am fluent in Stata and I try to translate part of my
Stata code to R to check the reliability of the data under R. I have a
proportion variable as a d
I can plot this just fine:
t<-seq(0,4, by=.1)
y<- t^3-6*t^2+5*t+30
plot(t,y ,xlab="t-values", ylab="f(t)", type="l")
This is the first derivative, how I I make a similar plot?
t<-seq(0,4, by=.1)
y<- t^3-6*t^2+5*t+30
y1<-D(expression(t^3-6*t^2+5*t+30), 't')
Thanks ahead of time.
kc
On Fri, J
Dear R Community,
I am using the package DRC ( to fit a boltzman model to my data. I
can fit the model and extract the lower limit, upper limit, and ED50
(aka V50), but I cannot figure out how to get the slope of the curve
at ED50. Is there a simple way to do this? I've searched the mailing
lis
I have plotted the dataset below into a stripchart, which gives me the correct
graph. However, my x-axis labels are to long to display hortizonal. I would
like to display the labels at a 45 degree angle so that all of the labels are
displayed. I have used the following text+par code on other gra
Hello Jean-Baptiste,
On 2010.01.22 16:32:53, Jean-Baptiste Combes wrote:
> Hello,
>
> I am learning R and I am fluent in Stata and I try to translate part of my
> Stata code to R to check the reliability of the data under R. I have a
> proportion variable as a dependent variable pQSfteHT . Indepe
My edtdbg debugging tool is now on CRAN, at
http://cran.r-project.org/web/packages/edtdbg/index.html
I've added a few new commands since the last time I announced the
package here. I'll enclose command list at the end of this message.
Currently the package is implemented for Vim. I have a vol
dear R wizards: I am wrestling with reshape. I have a long data set
that I want to convert into a wide data set, in which rows are firms
and columns are years.
> summary(rin)
firm fyear sim1
Min. :1004.00 Min. :1964.0 Min. : -1.0
1st Qu.:1010.00 1s
On Tue, Jan 5, 2010 at 5:25 PM, Carl Witthoft wrote:
> quote:
> > There are certainly formulas for solving polynomials numerically up to
> 4th degree non-iteratively, but you will almost certainly get better results
> using iterative methods.
>
> I must be missing something here. Why not use t
One approach is to do a permutation test on the covariances. You need a
measure of how different they are (maximum difference, mean difference).
Calculate that for the original data. Then randomly permute observations as to
which group they are in and recalculate your difference measure. Rep
> D(expression(t^3-6*t^2+5*t + 30), 't')
3 * t^2 - 6 * (2 * t) + 5
> D(D(expression(t^3-6*t^2+5*t + 30), 't'), 't')
3 * (2 * t) - 6 * 2
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Marlin Keith Cox
Sent: Friday, January 22, 2010
I would like to calculate a first and second derivative and am having
problems finding a simple solution. My syntax may be off as I am not a
mathematician, so pardon ahead of time.
data:
t<-seq(0,4, by=.1)
The function is:
H(t) = t^3-6*t^2+5*t + 30
from here I plot the curve:
plot(x,y ,xlab="x-va
Hi Michael,
It looks like you're having fun, I hope I don't spoil it for you:
title <- "some variety of words that are descriptive"
paste(strwrap(title, width=16), collapse="\n")
-Ista
On Fri, Jan 22, 2010 at 9:07 PM, Michael Pearmain wrote:
> Hi All,
>
> I'm trying to write a function to autom
Hi All,
I'm trying to write a function to automatically split long strings so they
will appear nicely in a chart i'm trying to create,
Say i have a string
title <- "some variety of words that are descriptive"
In this instance i want to place carriage return where there is a space just
prior to
Hello dear R help group,
I learned recently that one can change the rotation of labels in the axis,
when using a lattice plot, for example:
library(lattice)
barchart(yield ~ variety , data = barley,
groups = year,
ylab = "Barley Yield (bushels/acre)",
scales = list(rot
Hi! I have managed to sort this problem out by adapting script from a
published 'instruction'. This gives:
< m1 <- mantelkdist(kdist(data gene, data geo),999)
< plot (m1).
The script produced two sub-plots: one showing an xy plot of gene(x) end
geo(y) data and another showing a histogram of gene-
On Jan 22, 2010, at 2:07 PM, Fabrice DELENTE wrote:
Here's another R-way:
lets<-factor(c( 'A', 'B', 'A', 'C', 'B', 'D', 'B'))
# you did say they were factors, right?
nums <- factor(c('1', '2', '2', '3', '2', '2', '3'))
lets=="B"
[1] FALSE TRUE FALSE FALSE TRUE FALSE TRUE
sum(lets=="B"
A note about xts:
The time class layer is abstracted from the implementation. This
allows for some very time specific manipulation without having to
manage the underlying details.
Take a look at the xts docs and vignette.
Also, the functions .indexmon .indexyear etc correspond to the POSIX
> Here's another R-way:
>
> > lets<-factor(c( 'A', 'B', 'A', 'C', 'B', 'D', 'B'))
> # you did say they were factors, right?
> > nums <- factor(c('1', '2', '2', '3', '2', '2', '3'))
> > lets=="B"
> [1] FALSE TRUE FALSE FALSE TRUE FALSE TRUE
> > sum(lets=="B" & nums=="2")
> [1] 2
Thanks very muc
On Jan 22, 2010, at 1:58 PM, Fabrice DELENTE wrote:
Try this;
f <- c( 'A', 'B', 'A', 'C', 'B', 'D', 'B')
n <- c('1', '2', '2', '3', '2', '2', '3')
table(paste(f, n))
Thanks for the incredibly fast answer! I'll give this a shot!
Here's another R-way:
> lets<-factor(c( 'A', 'B', 'A', 'C',
> Try this;
>
> > f <- c( 'A', 'B', 'A', 'C', 'B', 'D', 'B')
> > n <- c('1', '2', '2', '3', '2', '2', '3')
> > table(paste(f, n))
Thanks for the incredibly fast answer! I'll give this a shot!
--
Fabrice DELENTE
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ht
Try this;
> f <- c( 'A', 'B', 'A', 'C', 'B', 'D', 'B')
> n <- c('1', '2', '2', '3', '2', '2', '3')
> table(paste(f, n))
On Fri, Jan 22, 2010 at 4:51 PM, Fabrice DELENTE wrote:
> Hello.
>
> I'm trying to count string data that correspond to a given
> condition in two factors of the same length.
>
I need to merge three datasets and don't know how. If I were using SQL, I
would use df3, look up the characteristics of each date in df1 and the value
for each observation in df2.
df1 - unique list of Dates and characteristics of those dates
Date, MM, WW, DOW
df2 - the raw data
Date,
Hello.
I'm trying to count string data that correspond to a given
condition in two factors of the same length.
For example, I have one factor
[ 'A', 'B', 'A', 'C', 'B', 'D', 'B' ]
and another is
[ '1', '2', '2', '3', '2', '2', '3' ]
I'd like to count the occurences of 'B' and '2' (so in my e
Normally one wants to store time indexes as a single column so its
likely that this is not what you really want to do. You may wish to
explain what your final objective is and why you want to do this.
However, if you must there are many ways and here is one The first
two lines load chron and set
On 22/01/2010 12:52 PM, Christophe Genolini wrote:
Thanks both of you.
>
> > Inf - Inf
> [1] NaN
So isn't the line 9 useless ? If either x[i] or y[i] are NA, then dev
will be NA and !ISNAN(dev) will detect it...
Sothe loop cool be
8.for(i = 0 ; i < taille ; i++) {
10.dev = (x[i] -
I would suggest that you use a 'list' for the data. Something like
this should work:
objs <- grep("^mrunoff_", ls(), values=TRUE) # get the names of the objects
result <- lapply(objs, function(.obj){
.obj <- get(.obj) # get the value
lapply(1:12, function(. iter) .obj[,, .iter])
})
A side note: The NA vs NaN does not seem to play a role here, because:
#define both_non_NA(a,b) (!ISNAN(a) && !ISNAN(b))
So, it is the same type of test used in line 9 and in line 11.
/Henrik
On Fri, Jan 22, 2010 at 9:52 AM, Christophe Genolini
wrote:
> Thanks both of you.
>
>>
>> > Inf - Inf
On 22/01/2010 8:52 AM, mlcarte3 wrote:
Which line do you want? There's more than one choice. For example, you
could regress y and z on x, or you could find the principle axis of the
x, y, z point cloud.
Duncan Murdoch
Thanks for the response. I need to find the principal axis for the point
Hi all,
I have a quick question about lm on group, say I have:
x <- 1:10
y <- x*3
buckets <- seq(0, 10, by=5)
g <- cut(x, buckets)
summary(lm(y ~ g - 1))
Coefficients:
Estimate Std. Error t value Pr(>|t|)
g(0,5] 9.000 2.121 4.243 0.00283 **
g(5,1
> x <- as.POSIXct('4/17/2008 16:01', format='%m/%d/%Y %H:%M')
> x
[1] "2008-04-17 16:01:00 EDT"
> # now create a four element vector of your desired values
> y <- strsplit(format(x, "%d %m %Y %H:%M"), ' ')
> y
[[1]]
[1] "17""04""2008" "16:01"
>
On Fri, Jan 22, 2010 at 1:09 PM, FMH wrot
Try this:
strDate <- c('4/17/2008 16:01', '4/18/2008 17:13')
do.call(rbind, strsplit(strDate, "[[:punct:]]|\\s"))[,c(2, 1, 3:5)]
On Fri, Jan 22, 2010 at 4:09 PM, FMH wrote:
> Dear All,
>
> I have a series of data in which the first column consist of a combination of
> date and time, for instan
Hello everyone, I am trying to use the as.timeSeries.ts()/as.timeSeries()
method on a ts object. Here is the error I get:
as.timeSeries.ts(mes)
Error in .local(.Object, ...) :
unused argument(s) (Data = c(NA, NA, 14.2,15.5,
I tried the same removing the NA values it gave me same error so I don'
FMH wrote:
Dear All,
I have a series of data in which the first column consist of a combination of
date and time, for instance 17 April 2008 at 4.01pm, such data is recorded as:
4/17/2008 16:01
I'd like to seperate it into four different columns which consist of Day,
Month,Year and Time, r
On Jan 22, 2010, at 12:01 PM, Na'im R. Tyson wrote:
Thank you for your very prompt response. The authors of the ROCR
package informed me the package works as stated in the documentation
as long as you use R version 2.9.0--and indeed, it does! I do not
mind using a slightly older version
Dear All,
I have a series of data in which the first column consist of a combination of
date and time, for instance 17 April 2008 at 4.01pm, such data is recorded as:
4/17/2008 16:01
I'd like to seperate it into four different columns which consist of Day,
Month,Year and Time, respectively.
(i) you EITHER correct the p value (by multiplying by 8 in your case) OR
you use the Bonferroni-threshold of 0.05/8, not both. If you correct the
p values, your threshold remains 0.05. If you use 0.05/8, you use the
original p values.
(ii) Yes, if the p value is 0.15, then the corrected one for 8 t
Thanks both of you.
> Inf - Inf
[1] NaN
So isn't the line 9 useless ? If either x[i] or y[i] are NA, then dev
will be NA and !ISNAN(dev) will detect it...
Sothe loop cool be
8.for(i = 0 ; i < taille ; i++) {
10.dev = (x[i] - y[i]);
11.if(!ISNAN(dev)) {
12. dist +
data.table is the package name too. Make sure you find ?"[.data.table" which
is linked from ?data.table.
You could just do a mean of one variable first, and then build it up from
there e.g. dataset[, mean(epLsar), by="SPECSHOR,BONE"].
To get multiple columns of output, wrap with DT() like this
?"[" should give you enough information. In short, "[" is an operator
to extract elements, you can think of it as a function with special
semantics. For a simple vector,
v = c("one", "two")
v[2] selects the second element of the vector, and is equivalent to,
`[`(v, 2)
# "two"
as you can see fro
I have conducted a discriminant function analysis with lda() in the MASS
Package, and I am interested in testing that the covariance matrices of the
groups are equal.
Does anybody have any suggestions on how I could test for equality between
covariance matrices?
Any help would be great. Thank yo
William thanks a lot that is exactly what I wanted! :working:
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The problem I see here is not so much the 27136 observations, but the fact that
the first two factors have up to 101 different levels and the third factor up
to 1001 different levels. That means that lm() is essentially creating
100+100+1000 dummies for those factors, leading to a large (and spa
If I understand your problem correctly, then one option is to use a permutation
test. This tests the null hypothesis that set A and set B are identical and
the only reason you are seeing a difference is due to random chance. The
procedure is to randomly shuffle the observations between the set
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of anna
> Sent: Friday, January 22, 2010 8:56 AM
> To: r-help@r-project.org
> Subject: [R] Counting Na values on a time serie only on the past datas
>
>
> Hello everyone, I have a
Hi,
It sounds to me like you need to learn to use R's help functions. Try
help.search("logistic regression")
help.search("spearman")
etc.
If you don't find what you need with help.search() you can broaden
your search filed by switching to RSiteSearch().
-Ista
On Fri, Jan 22, 2010 at 4:53 PM, Di
Thank you for your very prompt response. The authors of the ROCR
package informed me the package works as stated in the documentation
as long as you use R version 2.9.0--and indeed, it does! I do not
mind using a slightly older version of R to get the results I need.
It is useful to have
Thanks for your advice, I will work on it then!
Just one last question. In which package can I find the function
data.table?
Ivan
Le 1/22/2010 17:18, Matthew Dowle a écrit :
Great.
If you mean the crantastic r package, sorry I wasn't clear, I meant the
crantastic website http://crantastic.or
Hello everyone, I have a time serie of n values, some are na's. I want to get
a vector of size n whose elements represent on the time t < n the number of
missing values between o and t. I know I can do it with a loop but I wanted
to know if there was a function or a special syntax to do this. Than
Hi Jim,
You might also want to see ?list.files() for a quick way to get the
filenames in the first place, and then you can manipulate them using
the functions Gabor suggested.
-Ista
On Fri, Jan 22, 2010 at 4:48 PM, Gabor Grothendieck
wrote:
> Check out the various help pages on string manipulati
anyone?
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Fantastic. You're much more likely to get a response now. Best of luck.
"werner w" wrote in message
news:1264175935970-1100164.p...@n4.nabble.com...
>
> Thanks Matthew, you are absolutely right.
>
> I am working on Windows XP SP2 32bit with R versions 2.9.1.
>
> Here is an example:
> d <- as.d
Check out the various help pages on string manipulation:
help.search(keyword = "character", package = "base")
and particularly ?sub, ?substr and ?regex
On Fri, Jan 22, 2010 at 11:33 AM, Jim Bouldin wrote:
>
> Is there a way to capure all, or part, of a filename and assign it to an
> object. Sa
Thanks Baptiste, it does help.
However, I don't really understand what "[" means. Could you please tell
me more about it? I didn't find anything helpful on that in the help.
Thanks in advance
Ivan
Le 1/22/2010 17:19, baptiste auguie a écrit :
Hi,
Try this,
a = replicate(3, data.frame(x=1
Hello,
I am learning R and I am fluent in Stata and I try to translate part of my
Stata code to R to check the reliability of the data under R. I have a
proportion variable as a dependent variable pQSfteHT . Independent variables
are dummies for two categorical variables called dQSvacrateHTQuali3
Is there a way to capure all, or part, of a filename and assign it to an
object. Say I wanted to read in a file tiled "example.txt" and then assign
the character string "example" (or "exa" or any other substring of
"example" for that matter), to object a. Is there a simple way to do so?
Thanks
Jean-Baptiste
The most immediate difference I see is that you use a logit link in the R
code but a probit link function
in the stata code.
Joe
On Fri, Jan 22, 2010 at 8:25 AM, Jean-Baptiste Combes
wrote:
> Hello people,
>
> I am in the process of migrating from Stata to R and I would like to chec
Dear all,
I have 30 arrays, each with dimensions 720,360,12. The naming format for each
of these 30 objects is: mrunoff_5221, mrunoff_5222... mrunoff_5250.
For example:
> str(mrunoff_5221)
num [1:720, 1:360, 1:12] NA NA NA NA NA NA NA NA NA NA ... (the initial NA's
are nothing to worry abo
Jean-Baptiste,
You are not doing the same thing in R as in Stata. In stata you used the
probit link, in R the logit link.
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek
team Biometrie & Kwa
Hello people,
I am in the process of migrating from Stata to R and I would like to check
if my results are similar under the two softwares:
Here is my GLM command under R
nurse.model<-glm(pQSfteHT~dQSvacrateHTQuali3_2 + dQSvacrateHTQuali3_3 +
dQSvacrateHTQuali3_4 + dQSvacrateHTQuali3_5 + cluster_
Thanks Matthew, you are absolutely right.
I am working on Windows XP SP2 32bit with R versions 2.9.1.
Here is an example:
d <- as.data.frame(matrix(trunc(rnorm(6*27136, 1, 100)),ncol=6))
d[,4:5] <- trunc(100*runif(2*27136, 0, 1))
d[,6] <- trunc(1000*runif(27136, 0, 1))
for (i in 4:6)
Thank you, Mario.
Biostudent asked how one could perform repetitive tasks, e.g., plotting,
with subsets of data. I originally provided a flexible example based on
lapply. Mario suggested a variation that permits flexible control of
options. This reply shows how Mario's objective and naming of
Hi,
Try this,
a = replicate(3, data.frame(x=1:10, y=rnorm(10)), simplify=FALSE)
lapply(a, "[", "y")
HTH,
baptiste
2010/1/22 Ivan Calandra :
> Hi everybody!
>
> I have a (stupid) question but I cannot find a way to do it!
>
> I have a list like:
>> SPECSHOR_tx_Asfc
> $cotau
> SPECSHOR Asfc.m
Which line do you want? There's more than one choice. For example, you
could regress y and z on x, or you could find the principle axis of the
x, y, z point cloud.
Duncan Murdoch
Thanks for the response. I need to find the principal axis for the point
cloud. How can I plot this in R?
Thank
Great.
If you mean the crantastic r package, sorry I wasn't clear, I meant the
crantastic website http://crantastic.org/.
If you meant the description of plyr then if the description looks useful
then click the link taking you to the package documentation and read it.
Same for any of the othe
Thank you Don for the code,
but I get the following error massage:
> by(AlexETF,AlexETF$Industry,function(a) {filename = paste("C:/ab/",gsub("
> ","",a$Industry[1]),".txt",sep="")
+ print(filename)
+ write.table(a[,3,drop=FALSE],quote=FALSE,col.names=FALSE,row.names=FALSE)
+ }
Hi everybody!
I have a (stupid) question but I cannot find a way to do it!
I have a list like:
> SPECSHOR_tx_Asfc
$cotau
SPECSHOR Asfc.median
38cotau381.0247
39cotau154.6280
40cotau303.3219
41cotau351.2933
42cotau156.5327
$eqgre
SPECSHOR Asfc.median
netrunner wrote:
> Dear Peter,
> thank you for your explanations!
>
> Originally, my idea was to apply pairwise.wilcox.test, but sincerely I do
> not know how have to use my data to do that. I read the documentation for
> the function, but I did not understand how I have to arrange my data set.
>
Hi Werner,
On Thu, Jan 21, 2010 at 9:23 PM, Werner W. wrote:
> Hi,
>
> I have browsed the help list and looked at the FAQ but I don't find
> conclusive evidence if this is normal or I am doing something wrong.
> I am running a lm() on a data.frame with 27136 observations of 6 variables (3
> num
helloo
let me try rephrase my previous post this way. I have a distribution of
points ( with cords) and I want to select only or two points that fall
within a buffer distance from each point. I guess there should be some way
to do this in R. Can someone help or suggest where I might find a solutio
On Fri, Jan 22, 2010 at 3:33 PM, Dennis Murphy wrote:
> Hi:
>
> On Thu, Jan 21, 2010 at 1:06 PM, Peng Yu wrote:
>>
>> On Thu, Jan 21, 2010 at 2:41 PM, Dennis Murphy wrote:
>> > Hi:
>> >
>> > On Thu, Jan 21, 2010 at 12:29 PM, Peng Yu wrote:
>> >>
>> >> On Thu, Jan 21, 2010 at 2:16 PM, Dennis Mur
Access the information in
help(predict)
help(fitted)
With your reproducible code, run the following
predict(model1)
predict(model1, type="response")
predict(model1, type="response", se.fit=TRUE)
require(lattice)
xyplot(value+fitted(model1)~treatments, data=mydata1,
distribute.type=TRUE
Without reading all the details of your question, it looks like maybe
split() is what you want.
split( dataset, paste(dataset$SPECSHOR,dataset$BONE) )
or
split( dataset[,3], paste(dataset$SPECSHOR,dataset$BONE) )
-Don
At 5:12 PM +0100 1/21/10, Ivan Calandra wrote:
Hi everybody!
To use s
Trafim Vanishek wrote:
> Dear all,
>
> I cannot find to explicitly get the R-squared or adjusted R-squared from
> summary(lm())
>
> names(summary(RegModel.1))
[1] "call" "terms" "residuals" "coefficients"
[5] "aliased" "sigma" "df""r.squared"
[9]
On 1/22/2010 10:10 AM, Trafim Vanishek wrote:
> Dear all,
>
> I cannot find to explicitly get the R-squared or adjusted R-squared from
> summary(lm())
fm1 <- lm(Sepal.Length ~ Sepal.Width, data=iris)
summary(fm1)
str(summary(fm1))
summary(fm1)$r.squared
?str
> Thanks a lot!
>
> [[alte
On 10/01/2010 1:51 PM, Duncan Murdoch wrote:
On 10/01/2010 1:42 PM, Gabor Grothendieck wrote:
> Perhaps you can add it to the bug list before you leave it including
> what you tried in case its a symptom of some larger underlying
> problem.
This list is archived, so there's a record. But I thin
Does this example help?
a <- matrix(letters[1:12], ncol=3)
a
[,1] [,2] [,3]
[1,] "a" "e" "i"
[2,] "b" "f" "j"
[3,] "c" "g" "k"
[4,] "d" "h" "l"
write.table(a[,3,drop=FALSE],quote=FALSE,col.names=FALSE,row.names=FALSE)
i
j
k
l
At 4:11 PM -0800 1/21/10, Peter Rote wrote:
T
Hi Trafim,
Take a look at the following example:
set.seed(123)
x <- runif(100)
y <- 3 + 1.5*x + rnorm(100)
fit <- lm(y ~ x)
summary(fit)
str(summary(fit))
summary(fit)$r.squared
HTH,
Jorge
On Fri, Jan 22, 2010 at 10:10 AM, Trafim Vanishek <> wrote:
> Dear all,
>
> I cannot find to explicitly
Great example Glen!
I want to add simply a small thing that could be useful to someone.
Suppose in your last step you want to change the line color for each chart.
Using a for loop it is simple to use the integer index to access the df.lst
elements and
set the color:
for(i in 1:length(df.lst)
Trafim Vanishek wrote:
Dear all,
I cannot find to explicitly get the R-squared or adjusted R-squared from
summary(lm())
Thanks a lot!
[[alternative HTML version deleted]]
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Dear Peter,
thank you for your explanations!
Originally, my idea was to apply pairwise.wilcox.test, but sincerely I do
not know how have to use my data to do that. I read the documentation for
the function, but I did not understand how I have to arrange my data set.
Can you help me?
I have two g
Dear all,
I cannot find to explicitly get the R-squared or adjusted R-squared from
summary(lm())
Thanks a lot!
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do rea
yes I am using the rollMean function with trim =false and na.rm = true, it
works but it removes the na's from the results...
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I didn't know about crantastic actually.
I've looked what it is exactly and it indeed looks interesting, but I
don't really see how I would know that it would help me for the task.
There's a description of what it was built for, but how can I then know
which function from this package can help
netrunner wrote:
> Dear Michael,
> thank you very much for your help.
>
> I perfomed the wilcox.exact function on each of the 8 items for the two
> groups that I am analysing (that is, I performed 8 times the wilcox test).
> Here an example for the values (ratings from a questionnaire) of one of t
Dear useRs,
How could I obtain the confidence intervals for the means of my treatments,
when my data was fitted to a GLM?
I need the CI's for the Poisson and Negative Binomial distributions.
Here's what I have:
mydata1 <- data.frame('treatments'=gl(4,20), 'value'=rpois(80, 1))
model1 <- glm(val
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