In R software,does it have packages about confirmatory factor analysis in R
software? 3Q
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__
Dear Deepayan,
many thanks for answering.
Another thing I am wondering is the following: I know you can have (3d-like)
"crosses" in the wireframe plot. But are there any other 3d-like plot symbols?
Of course one can use different colors to distinguish between several points.
The problem is th
.libPaths() (on the same help page as .Library): and nothing has
changed since R 2.11.1.
However, to find a package, use .find.package, and your package should
almost certainly be using system.file(package='pkgname').
On Thu, 10 Mar 2011, Oxxx wrote:
I installed a custom package which needs
I having issues with interpreting the results of STL decomposition. The
following is the data used as well as the decompsed seasonality, trend and
the remainder components. It is a weekly data.
The original data doesn't appear to be seasonal. But there seems to be a
periodic peak in the seasonal c
On Wed, Mar 9, 2011 at 7:57 AM, Peter Ehlers wrote:
> On 2011-03-08 17:07, Jannis wrote:
>>
>> Searching the archive of the mailing list points to the following thread:
>>
>> https://stat.ethz.ch/pipermail/r-help/2009-March/193338.html
>>
>> Perhaps it helps you...
>>
>> On 03/08/2011 11:48 PM, Ma
Presumably the null hypothesis is that at least one of the differences
is larger in absolute magnitude than the chosen epsilon. I expect that
your procedure would be conservative: if it rejects the null
hypothesis, then you are ok, but presumably what you really want would
be based on a joint
Dear Aaron,
The problem is not with your function, but using apply(). Look at the
"Details" section of ?apply You will see that if the data is not an
array or matrix, apply will coerce it to one (or try). Now go over to
the "Details" section of ?matrix and you will see that matrices can
only co
Picking up an ancient thread (from Oct 2007), I have a somewhat more complex
problem than given in Simon Wood's example below. My full model has more than
two smooths as well as factor variables as in this simplified example:
b <- gam(y~fv1+s(x1)+s(x2)+s(x3))
Judging from Simon's example, my g
On Thu, Mar 10, 2011 at 9:56 PM, Oxxx wrote:
> I installed a custom package which needs to know its own location on the
> disk. (It runs an executable that is included in the package.) In version
> 2.11.1 the .Library command gave me the necessary information. It returned
> the path C:\Program F
Thank you so much for the advice. The R could not find function
"intersection". Do I need additional package to have this function active. I
tried "intersect" instead has no effect.
xyplot(p ~ xvar|chr, data=dataf,
panel=function(x, y, subscripts){
panel.xyplot(x, y)
ok= intersection(subscripts,
Hi,
I'm currently designing some global themes for use with lattice, and have
hit a snag. There doesn't appear to be (in xyplot at least) a way of setting
a lattice option for the 'scales' parameter at a global level - changes have
to be made in each function call. For example, consider the follow
Dear list,
I couldn't find a solution for this problem online, as simple as it seems.
Here's the problem:
#Construct test dataframe
tf <- data.frame(1:3,4:6,c("A","A","A"))
#Try the apply function I'm trying to use
test <- apply(tf,2,function(x) if(is.numeric(x)) mean(x) else unique(x)[1])
#Lo
I installed a custom package which needs to know its own location on the
disk. (It runs an executable that is included in the package.) In version
2.11.1 the .Library command gave me the necessary information. It returned
the path C:\Program Files\R\R-2.11.1\library. I have noticed that in the
2
On 2011-03-10 13:37, Marcos Prunello wrote:
Hello!
I want to display four xyplots in the same window (sorry if I don't use the
propers words, but my English is rude)
I know that if I use plot(y~x) then
x=(1:100)
y=(1:100)
par(mfrow=c(2,2))
plot(y~x)
plo
Yes, indeed I wanted them stored as characters instead of factor
levels. Thanks a lot for the help, Jim and Phil!
Gang
On Thu, Mar 10, 2011 at 5:19 PM, Phil Spector wrote:
> Gang -
> It sounds like you want your character variables to
> be stored as character values, not factor values. If tha
wang cheng yahoo.com.cn> writes:
>
> Hello:I run a gamm with following call
> :mode<-gamm(A~B,random=list(ID=~1),family=gaussian,na.action=na.omit,data=rs)
an error
> happened:ERROR names(object$sp) <- names(G$sp) : attempt to set an
attribute on NULL
with mgcv version 1.7-3
What so? How
On Mar 10, 2011, at 11:17 AM, Daniel Nüst wrote:
Hello!
I've been trying to get this right for quite a while now and fear
there is an easy solution I just don't see. I did not have this
problem in Linux, and I searched r-help and Google but did not find a
solution, but of course I am grateful
On 2011-03-10 11:37, Al Roark wrote:
Hi all:
I've been trying to figure out a way to move the tick mark labels
in my lattice plots closer to the the axes. I've tinkered with a
lot of the "buffer" parameters (those listed by
trellis.par.get()), but haven't had any luck. Any ideas?
If I underst
On Mar 10, 2011, at 11:08 AM, Paolo Cavatore wrote:
Hi David,
thanks for your comment...I managed to sort it out.
Below the final code...paolo
#
myExample <- data.frame(Ret=seq(-2, 2.5,
by=0.5),PE=seq(10,19),Sectors=rep(c("Financial","Industrial"),5))
myExa
G'day Maja,
On Thu, 10 Mar 2011 11:44:28 -0800 (PST)
maiya wrote:
> Aaah, it truly is wonderful, this technology!
> I guess I'm going to have to override it a bit though..
> Along the lines of
>
> tae <- ifesle(all.equal(obs, exp) == TRUE, 0, sum(abs(obs - exp)))
Please read the help page on
On 2011-03-09 12:14, chris20 wrote:
Hi,
I am trying to understand how the biplot.prcomp is constructed so I can
manipulate it to emphasise particular observations and reduce the number of
variables shown.
The prcomp model I have ran has cor=TRUE and scale=TRUE
I have worked out from looking at
I write about R every weekday at the Revolutions blog:
http://blog.revolutionanalytics.com
and every month (usually) I post a summary of articles from the
previous month of particular interest to readers of r-help. I somehow
missed this past January, though, so here are selected highlights from
Ja
Isn't reshape sun at first :).
The not obvious thing is that you don't need to cast anything. All you need is
the proper melt command (and a final sort if you want the table you have).
=
library(reshape)
xx = data.frame(country
I think we need a sample of the orginal data or reasonable facsimile to get an
idea of what you are doing.
Have a look at ?dput as a way to provide the data in a handy format.
--- On Thu, 3/10/11, Muzna Alvi wrote:
> From: Muzna Alvi
> Subject: [R] Reshape, melt and cast query
> To: r-help@r-
On Fri, Mar 11, 2011 at 11:07 AM, Arne Jonas Warnke
wrote:
> Dear all,
>
>
>
> First, I would like to thank you for your immense work. My question is
> about a frequent topic which I am not able to solve - even after hours
> of search in the mailing lisy.
>
> I would like to analyse random-effects
Hi:
Assuming that z constitutes the last two elements of your input vector,
try2 <- function(x) {
n <- length(x)
y <- x[1:(n - 2)]
z <- x[n - 1] + x[n]
sum(y) * z
}
x0 <- 1:4# result should be 21
x1 <- 1:8# z = 15, result = 21 * 15 = 315
try2(x0)
try2(x1)
HTH,
Dennis
On
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Arne Henningsen
> Sent: Thursday, March 10, 2011 12:09 PM
> To: Alex Olssen; r-help@r-project.org
> Subject: Re: [R] R beginner - Error in as.vector(x, mode)
>
> Dear Alex
>
> O
Try this:
> data$diff <- ave(as.numeric(data$date_obs), data$group, FUN=function(x)c(NA,
> diff(x)))
> data
group date_obs diff
1 IND1 1987-09-17 NA
2 IND1 1989-05-04 595
3 IND2 1997-04-30 NA
4 IND2 2008-11-03 4205
5 IND2 2009-05-08 186
6 IND3 1984-01-17 NA
7 IND4 1996-
Gang -
It sounds like you want your character variables to
be stored as character values, not factor values. If that's
the case, use
df = data.frame(n, s,stringsAsFactors=FALSE)
If you want them to be factors, but not to display as factors,
others have provided usable solutions.
try this:
> n = c(2, 3, 5)
> s = c("aa", "bb", "cc")
> df = data.frame(n, s, stringsAsFactors = FALSE)
> df
n s
1 2 aa
2 3 bb
3 5 cc
> str(df)
'data.frame': 3 obs. of 2 variables:
$ n: num 2 3 5
$ s: chr "aa" "bb" "cc"
>
On Thu, Mar 10, 2011 at 4:35 PM, Gang Chen wrote:
> A very simpl
Hi, I've got one silly question for evening.
I don't know is this reasonable, but can test with two the most extreme
proportions from the samples could be good enough evidence for testing
equivalence, or should I have to look for something else...?
--
Mi³ego dnia
[[alternative HTML vers
Dear all,
First, I would like to thank you for your immense work. My question is
about a frequent topic which I am not able to solve - even after hours
of search in the mailing lisy.
I would like to analyse random-effects (and fixed-effects)models of
longitudinal / panel data with sampling wei
Hi Calla,
On Thu, Mar 10, 2011 at 3:38 PM, Calla Carbone <6...@queensu.ca> wrote:
> Hi,
>
> I am trying to create 3 mds plots: one with axis 1 vs axis 2, one with axis 2
> vs axis 3, and one with axis 1 vs axis 3. When inputting my code, I only end
> up with one diagram and when inputting
> mdsg
Dear list,
I am trying to predict species volume from bioclimatic data, I have various
sites and I have a data frame with species volume and
the corresponding bioclimatic data for each site.
I read on a discussion forum that you can use ordination to predict species
abundance (in my case volume)
Dear R helpers
I have a table and i need to make new table
table1:
sire snp1 snp2 snp3 snp4 snp5 snp6 snp7 snp8 snp9 snp10
snp11 snp12 snp13 snp14 snp15 8877 -1 -1 -1 -1 0 0 -1 -1 -1 0 1 1 1 -1 -1
7765 1 1 1 0 0 0 -1 1 1 1 0 0 0 1 0 8766 1 1 -1 0 -1 -1 0 -1 0 -1 -1 -1 0 1
0 6756
Quite fascinating, if annoying. Nice example Petr!
Turns out my expected values are causing even more trouble because of this!
I've even gotten negative chi square values (calculated using Cressie and
Read's formula)!
So instead of kludging the error measurement code, I think I'm going to have
t
Hi,
I am trying to create 3 mds plots: one with axis 1 vs axis 2, one with axis 2
vs axis 3, and one with axis 1 vs axis 3. When inputting my code, I only end up
with one diagram and when inputting
mdsg.mds$dims, the program returns 2 for 2 dimensions. How can I create the
other two plots?
An
Thanks! You mean something like:
> print(df$s[1], max.levels=0)
It seems I could also do
> as.character(df$s[1])
Any other/better solutions?
On Thu, Mar 10, 2011 at 4:40 PM, Rob Tirrell wrote:
> See the max.levels argument in ?print. I think this is what you're looking
> for.
> --
> Robert Ti
On Thu, Mar 10, 2011 at 11:27 AM, mathijsdevaan wrote:
> Hi,
>
> I have a data.frame of the following type:
>
> F = data.frame(read.table(textConnection(" A B
> 1 1 4
> 2 1 3
> 3 1 1
> 4 1 4
> 5 1 2
> 6 1 2
> 7 1 2
> 8 2 1
> 9 2 1
> 10 2 1
> 11 2 1
> 12 3 2
> 13 3 4
> 14 3 1
> 15
See the max.levels argument in ?print. I think this is what you're looking
for.
--
Robert Tirrell | r...@stanford.edu | (607) 437-6532
Program in Biomedical Informatics | Butte Lab | Stanford University
On Thu, Mar 10, 2011 at 13:35, Gang Chen wrote:
> n = c(2, 3, 5)
> > s = c("aa", "bb", "cc"
Hello!
I want to display four xyplots in the same window (sorry if I don't use the
propers words, but my English is rude)
I know that if I use plot(y~x) then
x=(1:100)
y=(1:100)
par(mfrow=c(2,2))
plot(y~x)
plot(y~x)
plot(y~x)
plot(y~x)
wo
A very simple question. With a data frame like this:
> n = c(2, 3, 5)
> s = c("aa", "bb", "cc")
> df = data.frame(n, s)
I want df$s[1] or df[1,2], but how can I get rid of the extra line in
the output about the factor levels:
> df$s[1]
[1] aa
Levels: aa bb cc
Thanks,
Gang
_
On Fri, Mar 11, 2011 at 8:25 AM, Brian McLoone wrote:
> This is a follow-up to a query that was posted regarding some problems that
> emerge when running anova analyses for cox models, posted by Mathias Gondan:
>
> Matthias Gondan wrote:
>
>>* Dear List,*>**>* I have tried a stratified Cox Regress
Aaah, it truly is wonderful, this technology!
I guess I'm going to have to override it a bit though..
Along the lines of
tae <- ifesle(all.equal(obs, exp) == TRUE, 0, sum(abs(obs - exp)))
Do I like doing this? No. But short of reading the vast literature that
exists on calculation precision - wh
Dear Alex
On 10 March 2011 20:31, Alex Olssen wrote:
> I find it hard to provide a reproducible version of this error.
> When I use the exact same procedure but get data from data()
> everything works fine.
> I.e., I do not think the startvals are the problem - in fact I copied
> the syntax for t
This is a follow-up to a query that was posted regarding some problems that
emerge when running anova analyses for cox models, posted by Mathias Gondan:
Matthias Gondan wrote:
>* Dear List,*>**>* I have tried a stratified Cox Regression, it is working
>fine, except for*>* the "Anova"-Tests:*>**>
On Thu, Mar 10, 2011 at 10:34:21AM -0800, maiya wrote:
> Thanks Josh and Dan!
>
> I did figure it had something to do with the machine epsilon...
>
> But so what do I do now? I'm calculating the total absolute error over
> thousands of tables e.g.:
> tae<-sum(abs(obs-exp))
> Is there any easy way
On Fri, Mar 11, 2011 at 2:46 AM, Mike Marchywka wrote:
>
>> Date: Wed, 9 Mar 2011 10:55:46 +1300
>> From: darcy.web...@gmail.com
>> To: r-help@r-project.org
>> Subject: [R] minimum distance between line segments
>>
>> Dear R helpers,
>>
>> I think that thi
On Thu, 10 Mar 2011, Jen wrote:
Hi,
I'm trying to fit a tobit regression model to some data. When fitting the
exact same data in Stata, I have no problems at all, however R won't
converge. Its not a maxiters thing, since I've tried increasing this
already. I need to be able to fit the model in R
Hi all:
I've been trying to figure out a way to move the tick mark labels in my lattice
plots closer to the the axes. I've tinkered with a lot of the "buffer"
parameters (those listed by trellis.par.get()), but haven't had any luck. Any
ideas?
Cheers.
On Fri, Mar 11, 2011 at 6:08 AM, Yuelin Li wrote:
> Is there a function that calculates the power of a
> repeated-measure ANOVA design, e.g., 2 groups, 4
> within-subject factors, an average 0.40 correlation between
> the 4 within factors, etc.
>
> I don't think I can use power.anova.test() becaus
Maja,
Isn't modern technology wonderful? With computers we are able to do
calculations that we could never do by hand, and we get to complain about the
results not being exact. :-)
More comments below
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@
The function all.equal also might be helpful.
See the help page.
>help(all.equal)
-David Johnston
--
View this message in context:
http://r.789695.n4.nabble.com/identical-values-not-so-identical-newbie-help-please-tp3346078p3346581.html
Sent from the R help mailing list archive at Nabble.com.
On Thu, Mar 10, 2011 at 10:34 AM, maiya wrote:
> Thanks Josh and Dan!
>
> I did figure it had something to do with the machine epsilon...
>
> But so what do I do now? I'm calculating the total absolute error over
> thousands of tables e.g.:
> tae<-sum(abs(obs-exp))
> Is there any easy way to I kee
Good afternoon,
I am trying to create a plot where the bottom and top axes have the same
scale but different tick marks. I tried user-defined xscale.component
function but it does not produce desired results. Can anybody suggest
where my use of xscale.component function is incorrect?
For exampl
Did you scroll down the window to see the rules?
Regards,
Graham
On 10 March 2011 02:07, Xiaobo Gu wrote:
> Hi,
> I am using Rattle 2.6.4 with R 2.12.2 on win64, is this a bug ?
>
> Following is the content after execute the associate analysis process:
>
> Summary of the Apriori Association R
Thanks Josh and Dan!
I did figure it had something to do with the machine epsilon...
But so what do I do now? I'm calculating the total absolute error over
thousands of tables e.g.:
tae<-sum(abs(obs-exp))
Is there any easy way to I keep these ignorable errors from showing up?
And furthermore, wh
How about:
x <- rnorm(1000)
fn <- function(k) sum(x[1:k]*(x[k+1] + x[k+2]))
vals <- sapply(1:100, fn)
vals
--
Jonathan P. Daily
Technician - USGS Leetown Science Center
11649 Leetown Road
Kearneysville WV, 25430
(304) 724-4480
"Is the room still a room when it
Hi Mike,
You can use the ?axis function to add custom axes in. To prevent
overwriting, you will want to suppress the default y axis in your call
to plot. You can find details on the arguments and parameters in ?par
(I recommend pulling it up in a way that you can search because the
par documenta
Dear Natalie,
I am sure there are other ways, but one way you can do this is by
applying diff() to each group using tapply() or by(). Because those
return lists, if you want to add it back into your data frame, you can
wrap the whole call in unlist(). Here is an example:
dat <- structure(list(g
I want to increase the number of labels on my y-axis. Here is my code:
plot(fish$species, fish$fl, ylim=c(5,25), xlab="Species", ylab="Fork Length
(in)")
The plot is great but it only has numerical y-labes every five values (i.e.
labels at 5, 10, 15, and 25). I want R to give me the sa
Hello,
I'm trying to create some pdf-File via R containing hyperlink, which has in
the adress the "minus"-character used.
Example:
pdf(file="C:/R/test.pdf",width=11.69,height=8.27, onefile = TRUE)
plot.new()
mtext("http://www.sz-online.de";, side=1, line=-2, cex=0.7)
dev.off()
pdf-file will be cr
Dear R users,
Probably, this is quite a simpe question, but I do not find the proper way
to obtain want I need. To explain the problem, I constructed a simple
example.
Suppose I have the following function:
try1<-function(x){
y<-x[1:2]
z<-x[3:4]
y[1]*(z[1]+z[2])+y[2]*(z[1]+z[2])
}
This functio
Hi Maja,
Values that are theoretically the same may not be exactly identical
when using floating point arithmetic. Please read this FAQ page:
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
If you need to test that they are "equal", you may add
Try this:
rowsum(a$sales, gsub("(\\w*\\s\\w*\\s\\w*).*", "\\1", a$product))
On Thu, Mar 10, 2011 at 12:20 PM, Hui Du wrote:
>
>
> Thank you for your reply. I didn't state my problem very clearly in my
> previous post. The data could look like
>
> a = data.frame(date = c(20081201, 20081202, 2008
Hi David,
thanks for your comment...I managed to sort it out.
Below the final code...paolo
#
myExample <- data.frame(Ret=seq(-2, 2.5,
by=0.5),PE=seq(10,19),Sectors=rep(c("Financial","Industrial"),5))
myExample <- na.omit(myExample)
myecdf2 <- function(x, column, s
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of maiya
> Sent: Thursday, March 10, 2011 7:19 AM
> To: r-help@r-project.org
> Subject: [R] identical values not so identical? newbie help please!
>
> Hi there!
> I'm not sure I can
Hi,
I have a data.frame of the following type:
F = data.frame(read.table(textConnection("A B
1 1 4
2 1 3
3 1 1
4 1 4
5 1 2
6 1 2
7 1 2
8 2 1
9 2 1
10 2 1
11 2 1
12 3 2
13 3 4
14 3 1
15 3 1
16 3 1"),head=TRUE,stringsAsFactors=FALSE))
F
A B
1 1 4
2 1 3
3 1 1
Hi Everyone,
I would like to do sequential subtractions within a group so that I know the
time between separate observations for a group of individuals.
My data:
data <- structure(list(group = c("IND1", "IND1", "IND2",
"IND2", "IND2", "IND3", "IND4", "IND5",
"IND6", "IND6"), date_obs = struc
> Date: Wed, 9 Mar 2011 10:55:46 +1300
> From: darcy.web...@gmail.com
> To: r-help@r-project.org
> Subject: [R] minimum distance between line segments
>
> Dear R helpers,
>
> I think that this may be a bit of a math question as the more I
> consid
Benjamin,
A more elegant "R-style" solution would be to use one of R's
"apply"/aggregation routines, of which there are many. For example, the "by"
function can split a data.frame by some factor/categorical variable(s), and
then apply a function to each "slice". The result can then be pieced back
Hi there!
I'm not sure I can create a minimal example of my problem, so I'm linking to
a minimal .RData file that has only two objects: obs and exp, each is a 6x9
matrix. http://dl.dropbox.com/u/10364753/test.RData link to dropbox file
(I hope this is acceptable mailing list etiquette!)
Here's
Thank you for your reply. I didn't state my problem very clearly in my previous
post. The data could look like
a = data.frame(date = c(20081201, 20081202, 20081201), product = c("a b c d e",
"a bdfd c g h t", "def e h a c e h g"), sales = c(1, 2, 3)). The first three
items in "product" are th
Hello:I run a gamm with following call
:mode<-gamm(A~B,random=list(ID=~1),family=gaussian,na.action=na.omit,data=rs)an
error happened:ERROR names(object$sp) <- names(G$sp) : attempt to set an
attribute on NULLwith mgcv version 1.7-3What so? How can I correct the Error?
Thanks very much for any
Hi,
I am working on a population of an invasive clam. The data are the size of each
clam per station (2mm on average). Each station is found at a different
distance from a power nuclear station, so at different water temperatures. The
fist step I want to do is to identify cohort size at each st
To add to William's remarks, another advantage of the
apply family of functions is that they avoid growing an
object inside a loop, which is very inefficient in R.
In other words, without the *apply functions, users
might do something like this:
answer = NULL
for(i in 1:nrows)
answer = rbind
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of
> rex.dw...@syngenta.com
> Sent: Thursday, March 10, 2011 8:47 AM
> To: lig...@statistik.tu-dortmund.de; arun.kumar.s...@gmail.com
> Cc: r-help@r-project.org
> Subject: Re: [R] u
Is there a function that calculates the power of a
repeated-measure ANOVA design, e.g., 2 groups, 4
within-subject factors, an average 0.40 correlation between
the 4 within factors, etc.
I don't think I can use power.anova.test() because it does
not consider corr=0.40.
I am hoping that someone
But no one answered Kushan's question about performance implications of
for-loop vs lapply.
With apologies to George Orwell:
"for-loops BAAD, no loops GOOD."
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Uwe Ligges
Sent:
I did not know that. When reading the help topic for the first time
myself, I think I assumed that it returned no value since it had no Value
section, and I haven't used it in a way that it would return a value.
--
Jonathan P. Daily
Technician - USGS Leetown Sc
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Jonathan P Daily
> Sent: Thursday, March 10, 2011 7:49 AM
> To: Nipesh Bajaj
> Cc: r-help@r-project.org
> Subject: Re: [R] tryCatch - Continuing for/next loop after error
>
> Wow
Hello!
I've been trying to get this right for quite a while now and fear
there is an easy solution I just don't see. I did not have this
problem in Linux, and I searched r-help and Google but did not find a
solution, but of course I am grateful for and resources I might not
have found our not unde
Is there a package that does an EiV regression that uses the reliability of the
measures which are provided by the user? I've searched
RSiteSearch("error in variable regression")
As well as various permutations of these terms in google.
Though I say this with some trepidation, I am looking for
yes..thats what i was also confused about..but thats what my column is
named..
didnt know any other way of doing it.
On Thu, Mar 10, 2011 at 9:23 PM, jdnew...@gmail.com <
jdnew...@dcn.davis.ca.us> wrote:
> I can't see the value of your first melt, which appears to result in a
> column filled with
Hi list,
I have code which reads data from sql server using RODBC in R version
2.12.1. The code used to be ok, till today when I get the following error.
The data does not have any problem, but if I run the code in differnt pc, it
runs till different number of trial then it fails, for example on on
I can't see the value of your first melt, which appears to result in a column
filled with identical values "crop_group".
You really should read the posting guide and provide a reproducible example.
I suspect that you would benefit from learning how to use the plyr library.
--
Sent from my Andro
Wow, I had a major brain fart there. The function after error was supposed
to accept an argument, and next won't do anything there because inside the
function there is no loop. Also, the first argument of tryCatch is an
expression, so do the assignment there. Corrections inline below:
--
thanks for that ista
actually i want to break this into maybe three different data sets with each
of the 3 kinds of sugarcane...
it would be better if i could have this in one data set but if someone could
tell me how i can run descriplitive statistics on each of the groups
separately..
i would n
On Mar 10, 2011, at 12:54 , Duncan Murdoch wrote:
> On 11-03-10 5:54 AM, Ivan Calandra wrote:
>> Try with double "== "instead:
>> ifelse(val3 == "Monthly", val<- val1, val<- val2)
>
> That might work, but it is not how you should do it. (It should work if val3
> has a single entry, but will do
Hi Jonathan, I was also trying to understand this tryCatch function on
my own problem. Here is mine:
fn1 <- function(x) {
if(as.integer(x) == 5) {
stop("stop")
}
return(x+5)
}
res <- matrix(
Hi Muzna,
If I understand your question, it's not really about melting or
casting, but about re-ordering. Try this:
canefile <- canefile[order(canefile$value), ]
see ?"[" and ?order
HTH,
Ista
On Thu, Mar 10, 2011 at 2:20 PM, Muzna Alvi wrote:
> I have a dataset that is based on crop output for
Thanks to everybody who answered with suggestions (David Wolfskill,
Stephen Sefick, Jim Holtman, Gabor Grothendieck, and Phil Spector).
Beside the obvious end-of-the-day mixup ("int" in lieu of "integer"), I
was not aware of the existence of a "NULL" class - which proved pretty
useful in this cas
Hello,
I'm stuck with the following code.
p <- 0:100/100
a <- 0:200/100
beta <- matrix(a + p%x%(1-a), nrow=length(a), ncol=length(p))
lb <- c(expression(beta==0), expression(beta==0.2), expression(beta==0.4),
expression(beta==0.6), expression(beta==0.8), expression(beta==1),
expression(beta==1.2)
I have a dataset that is based on crop output for a single crop
*sugarcane*...but
each observation is further subdivided into 3 kinds of sugarcane
i have read the file and have also used melt it to sort it on the
basis of *crop
group* using
melt(sugarcane, m=c("Crop_group")) -> canefile
this is
Hi,
I have a newbie doubt. I try to use metaMDS in R, it is OK, but the
stress that I get is 16.32, in other programs stress is showed like
0.173. I don't find a good explanation about this calculations on books,
I think that R show stress em a percentage data (0-100) and other
software in a p
--
Jonathan P. Daily
Technician - USGS Leetown Science Center
11649 Leetown Road
Kearneysville WV, 25430
(304) 724-4480
"Is the room still a room when its empty? Does the room,
the thing itself have purpose? Or do we, what's the word... imbue it."
- Jubal E
On 10.03.2011 13:33, David Winsemius wrote:
On Mar 10, 2011, at 5:18 AM, wesley mathew wrote:
Dear All
I wanna install XML package, but it shows XML package is not available.
install.packages("XML")
*Warning: unable to access index for repository
http://www.stats.ox.ac.uk/pub/RWi
n/bin/win
Hi,
I'm trying to fit a tobit regression model to some data. When fitting the
exact same data in Stata, I have no problems at all, however R won't
converge. Its not a maxiters thing, since I've tried increasing this
already. I need to be able to fit the model in R since there are users of
the code
Hi:
To get the samples, here's one approach:
df <- data.frame(gender = rep(c('F', 'M'), c(165, 42)), y = rpois(207, 20))
# Sampling function to take 20 F and 20 M with replacement
# The sample function operates on the rows of df to get idx and then takes
# the y's corresponding to those rows
sfun
Hello list!
I have a data.frame which looks like this:
> serv
datum op.read op.write read write
1 2011-01-29 10:00:00 00 0 0
2 2011-01-29 10:00:01 00 0 0
3 2011-01-29 10:00:02 00 0 0
4 2011-01-29 10:00:03
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