Thanks for the codes. Maybe I elaborate more here.
In my actual dataset, the dim(E) = (200, 20, 1000) and dim(C) = (200,15,
965) in which I don't know which 35 matrix are missing. I want to ask how
merge the two arrays by matching the first 5 columns of values. Thanks you.
Best Regards,
Ray
On
On Fri, 8 Feb 2013, Richard M. Heiberger wrote:
require(vcd)
mosaic(matrix(1:6, 2, 3), gp=gpar(fill=c("red","gray","blue")), main="three
colors in each row")
mosaic(matrix(1:3, 1, 3), gp=gpar(fill=c("red","gray","blue")), main="only
one color in each row")
I anticipated three colors in the secon
Hi,
You can write "biomaRt::getSequence" instead of "getSequence".
Hope this helps,
Pascal
Le 08/02/2013 15:07, Mohammad Tanvir Ahamed a écrit :
Hi !
Facing problem with " getSequence" commend .
when only biomaRt package loaded the following example working well
mart <- useMart("ensembl",
Hi !
Facing problem with " getSequence" commend .
when only biomaRt package loaded the following example working well
>mart <- useMart("ensembl",dataset="hsapiens_gene_ensembl")
>seq = getSequence(id="BRCA1", type="hgnc_symbol", seqType="peptide", mart =
>mart)
show(seq)
but when i have lo
require(vcd)
mosaic(matrix(1:6, 2, 3), gp=gpar(fill=c("red","gray","blue")), main="three
colors in each row")
mosaic(matrix(1:3, 1, 3), gp=gpar(fill=c("red","gray","blue")), main="only
one color in each row")
I anticipated three colors in the second plot. My guess is that there is
a matrix subsc
Thanks, Bert! Never used it, but reading it now.
Tao
- Original Message -
> From: Bert Gunter
> To: "Shi, Tao"
> Cc: R. Michael Weylandt ; "r-help@r-project.org"
>
> Sent: Thursday, February 7, 2013 2:05 PM
> Subject: Re: [R] compare objects in two different workspaces
>
> Ummm.
Thanks.
On Thu, 07 Feb 2013 11:46:36 +
Rui Barradas wrote:
> Hello,
>
> Try the following.
>
> > maintainer("spatstat")
> [1] "Adrian Baddeley "
>
>
> Hope this helps,
>
> Rui Barradas
>
> Em 07-02-2013 02:05, Hiroshi Saito escreveu:
> > Dear sir,
> >
> > Which mailing list is approp
Please see in-line below.
On 02/08/2013 05:20 AM, Giuseppe Amatulli wrote:
Hi Rolf,
sorry for this late answer and thanks for your kind explanation and
relative R code. I really appreciate.
In reality the concept that I'm trying to address is a bit more complex.
I'm fitting a model y vs 6 predi
It is also acceptable (and often useful) to ask on the R-Sig-Geo list.
Bug reports should of course be sent to Prof. Baddeley.
cheers,
Rolf Turner
On 02/08/2013 12:46 AM, Rui Barradas wrote:
Hello,
Try the following.
> maintainer("spatstat")
[1] "Adrian Baddeley "
Hope this h
I am trying to calculate adjusted means according to a glm model with a
Poisson distribution using effect() in the 'effects' package. I've done
this many times before with great success, but this time I'm getting an
error. My model contains two factors and two covariates/blocks and I would
like the
In fact, they have objects with exact same names. They were created by the
same code at different time points. The code is not version-controlled, so I
only have the most recent version, but I want to know what's changed.
Tao
- Original Message -
> From: Bert Gunter
> To: R. Micha
On Feb 7, 2013, at 6:21 PM, Julio Sergio wrote:
> Julio Sergio gmail.com> writes:
>
>>
>> I'm trying to produce a series of powers of a number as follows:
>>
>> |> 0.05^0:5
>
>
> I'm sorry for the question. The answer is simple: the result is due to
> operator precedence not to coercing:
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Julio Sergio
> Sent: Thursday, February 07, 2013 6:13 PM
> To: r-h...@stat.math.ethz.ch
> Subject: [R] Coercing of types when raising a number to a series of
> powers
>
> I'm tr
Julio Sergio gmail.com> writes:
>
> I'm trying to produce a series of powers of a number as follows:
>
> |> 0.05^0:5
I'm sorry for the question. The answer is simple: the result is due to
operator precedence not to coercing:
|> 0.05^(0:5)
[1] 1.000e+00 5.000e-02 2.500e-03 1.250e-04 6.2
I'm trying to produce a series of powers of a number as follows:
|> 0.05^0:5
[1] 1 2 3 4 5
This is not the result I expected. I guess some kind of coercion happened,
since,
|> class(0.05^0:5)
[1] "integer"
Could anyone explain me what is happening here?
Thanks,
-Sergio.
_
On 8 Feb 2013, at 01:24, "Antonio Silva" wrote:
> # Two-sided p-value should be estimated with
> dt(t,summary(fit1)$df[2]) # isn't it?
No, on two counts.
>
First, dt is the density; you need pt, the cumulative probability, not dt.
Second, pt(t, ...) would calculate the one sided lower tail pro
OK, I found the answer:
2*pt(t,summary(fit1)$df[2])
Thanks anyway ...
2013/2/7 Antonio Silva
> Dear list members
>
> I have a doubt on how p-values for t-statistics are calculated in the
> summary of Linear Models.
>
> Here goes an example:
>
> x <- rnorm(100,50,10)
> y <- rnorm(100,0,5)
> fi
On 02/07/2013 08:22 PM, Antonio Silva wrote:
Dear list members
I have a doubt on how p-values for t-statistics are calculated in the
summary of Linear Models.
Here goes an example:
x <- rnorm(100,50,10)
y <- rnorm(100,0,5)
fit1<-lm(y~x)
summary(fit1)
summary(fit1)$coef[2] # b
summary(fit1)$coe
Thanks, Ellison. Another question is if this p-value is a good parameter to
test if the fitting is good,
Absolutely not.
***
This email and any attachments are confidential. Any use, copying or
disclosure other than by the intende
Dear list members
I have a doubt on how p-values for t-statistics are calculated in the
summary of Linear Models.
Here goes an example:
x <- rnorm(100,50,10)
y <- rnorm(100,0,5)
fit1<-lm(y~x)
summary(fit1)
summary(fit1)$coef[2] # b
summary(fit1)$coef[4] # Std. Error
summary(fit1)$coef[6] # t-sta
Thanks, Bill.
Andrew
On Thu, Feb 07, 2013 at 09:21:53PM +, William Dunlap wrote:
> >> Notice that the '[[' function is superior in every way to the '$'
> >> function.
> > I'm curious to know whether you can point to some consolidated comparison?
>
> Two problems are that List$name allows a
There are 4 ways, other than generating a core dump :-), that could
fit your vague desctiption:
Call q("no"); this is the only way to avoid 'finally' expressions
in tryCatch if that is really what you want.
You can invoke an "abort" restart; there will be one established
by the i
P. S. I just noticed that you referred to the envelopes produced as
"confidence" envelopes.
They are ***NOT*** confidence envelopes!!! They are "critical" envelopes.
This is a very different concept. Repeat after me, 50 times:
critical envelopes, critical envelopes, critical envelopes, .
No, it does not. If you donât use print.eval=TRUE, you would need to wrap
the dotplot() call in a print() inside your script file, i.e.
print(dotplot(resamps, metric = "Accuracy"))
Dan
Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Acc
Thanks Dan. The ... is an accidental typo in the email. Would you mind
verifying that dotplot opens a plot *without* print.eval=TRUE?
Thanks,
J.
On Thu, Feb 7, 2013 at 5:08 PM, Nordlund, Dan (DSHS/RDA) <
nord...@dshs.wa.gov> wrote:
> James,
>
> ** **
>
> Your code seems to have
sitti
Forgot to send to R-help
From: Nordlund, Dan (DSHS/RDA)
Sent: Thursday, February 07, 2013 2:09 PM
To: 'James Jong'
Subject: RE: [R] Sourcing my file does not print command outputs
James,
Your code seems to have ââ¦â sitting on a line all by itself (maybe should
be at the end of the precedi
Ummm...
Have you actually read ?load to see how do this using the
"environment" argument?
-- Bert
On Thu, Feb 7, 2013 at 1:58 PM, Shi, Tao wrote:
> In fact, they have objects with exact same names. They were created by the
> same code at different time points. The code is not version-control
What are the *RDS counterparts? What is the difference?
James
On Thu, Feb 7, 2013 at 5:00 PM, Michael Weylandt wrote:
>
>
> On Feb 7, 2013, at 9:51 PM, James Jong wrote:
>
> > Thanks Stephan. I can't believe I didn't try that first. I greatly
> > appreciate it.
> >
> > James
> >
> >
> > On Th
On Feb 7, 2013, at 9:51 PM, James Jong wrote:
> Thanks Stephan. I can't believe I didn't try that first. I greatly
> appreciate it.
>
> James
>
>
> On Thu, Feb 7, 2013 at 4:45 PM, Stephan Kolassa wrote:
>
>> Have you looked at ?save and ?load?
I generally prefer their *RDS counterparts whi
>> Notice that the '[[' function is superior in every way to the '$' function.
> I'm curious to know whether you can point to some consolidated comparison?
Two problems are that List$name allows abbreviations of name and does not allow
variable names.
With List$name you get:
> List <- list(Ele
Thanks Stephan. I can't believe I didn't try that first. I greatly
appreciate it.
James
On Thu, Feb 7, 2013 at 4:45 PM, Stephan Kolassa wrote:
> Have you looked at ?save and ?load?
>
> As I already wrote here: http://stackoverflow.com/**
> questions/14761496/saving-and-**loading-a-model-in-r
Thanks Dan,
Here is a minimal full example, where dotplot fails to print anything if I
source the file. I have tried restarting R and directly sourcing this file
without doing anything else first (my .Rprofile is empty). This is running
on Linux with the latest stable version of R.
==
Have you looked at ?save and ?load?
As I already wrote here:
http://stackoverflow.com/questions/14761496/saving-and-loading-a-model-in-r
Best,
Stephan
On 07.02.2013 22:33, James Jong wrote:
Say I train a model in caret, e.g.:
RFmodel <- train(X,Y,method='rf',trControl=myCtrl,tuneLength=1)
I donât know where resamps comes from and I donât know what parameters you
provided to source(), so you havenât provided a reproducible example. But,
did you call source with print.eval=TRUE ?
Dan
Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Per
Say I train a model in caret, e.g.:
RFmodel <- train(X,Y,method='rf',trControl=myCtrl,tuneLength=1)
How can I save this to disk and load it later in R?
How about an object of the class "resamples"?
resamps <- resamples(
list( RF = RFmodel,
SVM = SVMmodel,
Thanks. Interestingly I am having the same problem with
dotplot(resamps, metric = "Accuracy")
Nothing shows up if I source the file. But I can visualize it I call this
command from the command line... Any thoughts?
Thanks again,
James
On Thu, Feb 7, 2013 at 3:09 PM, Nordlund, Dan (DSHS/RDA) <
... but you need to load them into different environments in case they
have objects with the same name, right?
-- Bert
On Thu, Feb 7, 2013 at 1:25 PM, R. Michael Weylandt
wrote:
> On Thu, Feb 7, 2013 at 5:53 PM, Shi, Tao wrote:
>> Hi list,
>>
>> Is there a easy way to compare objects in two dif
On Thu, Feb 7, 2013 at 5:53 PM, Shi, Tao wrote:
> Hi list,
>
> Is there a easy way to compare objects in two different workspace files (i.e.
> .RData files) in R? I can use some generic file compare softwares (e.g.
> BeyondCompare) to binary comparison, but when it says they're different you
>
On Thu, Feb 7, 2013 at 2:07 PM, Ken Weiss wrote:
> Greetings,
>
> I am having an interesting problem and I wonder if anyone else has
> seen this behavior.
>
> I am running R 2.11.1 with SNOW 0.3-3 on a Dell cluster running CentOS 5.5.
Simplest suggestion -- can you update to current R and use th
You can solve this problem by actually *reading* the error
message. Evidently the object "point" is a marked point pattern
whose marks are in the form of a data frame.
So either:
(1) Do not create "point" as a marked point pattern --- i.e.
don't add a "marks" component when you create
?unique
Please learn to search for yourself, e.g. via:
??unique
or the tons of other ways.
-- Bert
On Thu, Feb 7, 2013 at 12:50 PM, Dimitri Liakhovitski
wrote:
> Hello!
>
> I have a data frame with several rows, for example:
>
> x=as.data.frame(matrix(c(1,2,3,
>1,2,3,
>1,2,2,
>
Hi David,
The following is an interesting observation ...
On Friday, February 8, 2013, David Winsemius wrote:
>
> On Feb 7, 2013, at 10:20 AM, Winfried Moser wrote:
>
>
Notice that the '[[' function is superior in every way to the '$' function.
>
I'm curious to know whether you can point t
Everyone:
In the last few months I've noticed an increasing number of questions to
this list from students who are new to the use of R and who have limited (in
any) local support on R and specifically support for R syntax.
My text
(http://www.springer.com/statistics/social+sciences+%26+law/
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of John Kane
> Sent: Thursday, February 07, 2013 10:57 AM
> To: James Jong; r-help@r-project.org
> Subject: Re: [R] Sourcing my file does not print command outputs
>
> From ?source
You can leave your script as it was and use source()'s
print.eval=TRUE or echo=TRUE arguments to get the
autoprinting done as it is at the command line
(echo=TRUE also prints the commands).
E.g., for the file "/tmp/x.R" containing the 2 lines
d <- mtcars[,c("am","gear","hp","mpg")]
summary(lm(
thanks! I didn't know, that x$y isn't a variable, but an expression!
with the suggested notation d[[x]] i get a matrix and have to extract the
diagonale.
d <- data.frame(d1=c(1,0,0), d2=c(0,1,0), d3=c(0,0,1))
d$nr <- NA
sapply(1:3, function(x) ifelse(d[[x]]==**1,x,d$nr))
[,1] [,2] [,3]
[1,]1
Hello,
Sorry, there's a bug in my first reply. Corrected:
fun <- function(x, y){
df3 <- cbind(x, y[setdiff(names(y), names(x))])
df3[order(names(df3))]
}
fun(df1, df2)
Rui Barradas
Em 07-02-2013 19:16, Rui Barradas escreveu:
Hello,
The following function will give what you s
On Feb 7, 2013, at 10:20 AM, Winfried Moser wrote:
> Dear Listers,
>
> I try to change the structure of my data. i have an indicator-matrix and
> want to end up with a factor.
>
> i have
>
> v1 v2 v3
> 1 0 0
> 0 1 0
> 0 0 1
>
> and want
>
> v1 v2 v3 v4
> 1 0 0 1
> 0 1
thanks, that worked!
2013/2/7 arun
> Hi,
> You could get the result:
> dat1 <- read.table(text = "
> v1 v2 v3
> 1 0 0
> 0 1 0
> 0 0 1
> ",sep="",header=TRUE)
>
> dat1$v4<-apply(mapply(`==`,dat1,1),2,which)
> dat1
> # v1 v2 v3 v4
> #1 1 0 0 1
> #2 0 1 0 2
> #3 0 0 1 3
thank's, that worked!
2013/2/7 jim holtman
> try this:
>
> > x <- read.table(text = "v1 v2 v3
> + 1 0 0
> + 0 1 0
> + 0 0 1", header = TRUE)
> >
> > x$v4 <- apply(x, 1, function(a) which(a == 1))
> > x
> v1 v2 v3 v4
> 1 1 0 0 1
> 2 0 1 0 2
> 3 0 0 1 3
> >
> >
> >
>
>
>
On Feb 7, 2013, at 10:58 AM, Aminreza Aamini wrote:
> Hi,
> I am comparing SAS commands with R.
> consider i want to assess the effect of one exposure(like CAT where 1
> equals high and 0 equals low) and one cofounding variable (like
> cholesterol leve,CHL) on a diseasee variable (like CHD).
> In
Hello,
The following function will give what you seem to want.
fun <- function(x, y){
df3 <- x
df3 <- cbind(df3, df2[setdiff(names(y), names(x))])
df3[order(names(df3))]
}
fun(df1, df3)
Hope this helps,
Rui Barradas
Em 07-02-2013 18:36, Anika Masters escreveu:
#I
Hi,
You could use ?merge() or ?join() from library(plyr)
merge(df1,df2,all=TRUE)
# col2 col3 col6 col8 col9 col1
#1 2 3 6 NA NA 1
#2 NA NA NA NA NA NA
A.K.
- Original Message -
From: Anika Masters
To: r-help@r-project.org
Cc:
Sent: Thursday, February 7, 2
Hi,
I am comparing SAS commands with R.
consider i want to assess the effect of one exposure(like CAT where 1
equals high and 0 equals low) and one cofounding variable (like
cholesterol leve,CHL) on a diseasee variable (like CHD).
In SAS to get a numerical value for the odds ratio, consider, for
e
>From ?source
"Note that running code via source differs in a few respects from entering it
at the R command line. Since expressions are not executed at the top level,
auto-printing is not done. So you will need to include explicit print calls for
things you want to be printed"
John Kane
Kingst
One rather extremem limitation in Stata is that there is
always only just one "active" or loaded dataset.
So all the commands that involve more than one dataset
operate on the loaded "master" dataset and one or more
other datasets that are stored somewhere.
Merge in Stata joins the loaded master d
I have an environment variable `$R_HISTFILE` pointing to
`/home/my.username/.RHistory` and the following code in my `.Rprofile` in
my home directory:
.Last <- function() {
if (!any(commandArgs()=='--no-readline') && interactive()){
require(utils)
I looked at the documentation of source() and summary(), and I could not
find the reason why calling something like:
> summary(resamps)
from the command line, works (it prints the summary)
whereas calling
summary(resampls)
from a file that I source with source("my_file.r") does not print anyth
Hi list,
Is there a easy way to compare objects in two different workspace files (i.e.
.RData files) in R? I can use some generic file compare softwares (e.g.
BeyondCompare) to binary comparison, but when it says they're different you
can't tell where the difference are from.
Thanks!
Tao
_
Hi,
You could get the result:
dat1 <- read.table(text = "
v1 v2 v3
1 0 0
0 1 0
0 0 1
",sep="",header=TRUE)
dat1$v4<-apply(mapply(`==`,dat1,1),2,which)
dat1
# v1 v2 v3 v4
#1 1 0 0 1
#2 0 1 0 2
#3 0 0 1 3
A.K.
- Original Message -
From: Winfried Moser
To: r-
I know that within sum of squares, DB, sillhouette and cophenetic are
indicators of clustering quality, but what indicators I need to observe when
I choose attributes for kmeans?
--
View this message in context:
http://r.789695.n4.nabble.com/Feature-selection-for-kmeans-tp4657830.html
Sent fro
On Feb 7, 2013, at 10:47 AM, "Lopez, Dan" wrote:
> Hi,
>
> Is anyone aware if there is a package that allows one to connect to Oracle
> Business Intelligence (aka OBIEE)?
>
> I want to be able to pull in a summary table contained in an "answer" (aka
> OBIEE "query").
>
> Dan
> Lawrence Live
#I have 2 dataframes df1 & df2. Each has a subset of all possible column names.
#How do I combine the 2 tables so that they contain all column names?
data1 <- c('2','3', '6', '8' , '9')
data2 <- c('1', '2', '3', '6')
df1 <- data.frame(matrix(data=NA, nrow=1, ncol=length(data1),
dimnames=list(NULL
On 07/02/2013 1:20 PM, Winfried Moser wrote:
Dear Listers,
I try to change the structure of my data. i have an indicator-matrix and
want to end up with a factor.
i have
v1 v2 v3
1 0 0
0 1 0
0 0 1
and want
v1 v2 v3 v4
1 0 0 1
0 1 0 2
0 0 1 3
amongst other thing
try this:
> x <- read.table(text = "v1 v2 v3
+ 1 0 0
+ 0 1 0
+ 0 0 1", header = TRUE)
>
> x$v4 <- apply(x, 1, function(a) which(a == 1))
> x
v1 v2 v3 v4
1 1 0 0 1
2 0 1 0 2
3 0 0 1 3
>
>
>
On Thu, Feb 7, 2013 at 1:20 PM, Winfried Moser wrote:
> Dear Listers,
>
> I try
On 06/02/2013 8:33 PM, ivo welch wrote:
is it possible to throw a stop() that is so hard that it will escape
even tryCatch?
You can signal a condition that is not an error, and if tryCatch hasn't
been written to catch it, it will stop. For example,
g <- function() {
tryCatch(f(), error=f
#I have 2 dataframes df1 & df2. Each has a subset of all possible column
names. #How do I combine the 2 tables so that they contain all column
names? data1 <- c('2','3', '6', '8' , '9') data2 <- c('1', '2',
'3', '6') df1
<- data.frame(matrix(data=NA, nrow=1, ncol=length(data1),
dimnames=lis
> For merge, when merging dataset A with dataset B on some
> set of key variables, it is possible to specify if the
> join should be done 1:1, 1:n, n:1 or n:m. Stata creates
> a data structure that contains information about which
> observations could or could not be matched and which
> matches did
Dear Franklin Bretschneider,
Thank you so much for your reply and explanation about the filter using the
stats and signal package.
I decided to opt the filter method in signal package. I have a simple
question about the cut off frequency here.
I have 30 minute collected tidal data and I want to
Dear Listers,
I try to change the structure of my data. i have an indicator-matrix and
want to end up with a factor.
i have
v1 v2 v3
1 0 0
0 1 0
0 0 1
and want
v1 v2 v3 v4
1 0 0 1
0 1 0 2
0 0 1 3
amongst other things i tried the following
d <- data.frame(d1=c(1
Liang:
In nonlinear models especially (and more generally, also), "small" p
values are not reliable indicators of whether a fit is or is
not"good." I would strongly suggest that you consult with your local
statistician -- this is a (complicated, as it depends on the meaning
of "good") statistical
I gave a small number of examples in my first email.
For append, in Stata any variable that is missing in the
other dataset is automatically inserted with all values set
to the missing value. In R, one would first have to compare
the columns in both data frames and generate columns
apropriately.
Leigh Michelle Williams gsu.edu> writes:
>
> I'm new to R and have not been able to find what I am looking for online.
> I found the system command but that
> seems to be used for getting lists of directories, etc.
>
> I am trying to replicate something that I used to do in SAS. I am simulating
I think this is because some decimal numbers with a fixed
number of digits after the decimal point cannot be represented
as base-2 floating point numbers with a limited number of
digits. So when one adds them up, the result is slightly
different from the sum:
huh1 = seq(.1,.7,.15)
huh1 - 0.55
[1
Hi Leigh,
On Thu, Feb 7, 2013 at 10:35 AM, Leigh Michelle Williams
wrote:
> I'm new to R and have not been able to find what I am looking for online. I
> found the system command but that seems to be used for getting lists of
> directories, etc.
The "etc." includes many things in this case!
>
On Thu, Feb 7, 2013 at 4:16 PM, christel lacaze
wrote:
>
> Hi there,
>
> I've got a set of 10 numeric variables called Mood1 to Mood10 in a dataset
> called mood.
That's where you went wrong in the first place. Don't use variable
names for indexing purposes.
You should have created one variab
On Feb 7, 2013, at 9:12 AM, John Smith wrote:
Hi Gerrit,
as I said in my original email, I already know both the
merge and the rbind commands, but I think that many
standard situations (I have given just a few) require
rather clumsy ad-hoc programming. So I was wondering
if there are any packa
R-FAQ 7.31
Berend
On 07-02-2013, at 17:41, "J. Braeken" wrote:
> Dear R-list,
>
>
> We stumbled upon some weird problem when performing a simple indexing
> operation.
> Below some example code to illustrate the issue
>
> #FAILS TO FIND .55 Oo
> huh1 = seq(.1,.7,.15);huh1
> # [1] 0.10 0.25
Hi José
Yes indeed that does help.
Thanks,
Derek
On 2/7/2013 5:01 AM, Jose Iparraguirre wrote:
Hi, Derek,
Try and modify the value for the overlap argument.
Hope this helps,
José
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf O
Dear R-list,
We stumbled upon some weird problem when performing a simple indexing operation.
Below some example code to illustrate the issue
#FAILS TO FIND .55 Oo
huh1 = seq(.1,.7,.15);huh1
# [1] 0.10 0.25 0.40 0.55 0.70
huh1 == .25
# [1] FALSE TRUE FALSE FALSE FALSE
huh1 == .55
# [1] FALSE FA
Hi there,
I've got a set of 10 numeric variables called Mood1 to Mood10 in a dataset
called mood. I'm trying to create a set of 10 new variables called m1 to m10
so that m1=Mood1*1, m2=Mood2*2, etc to m10=Mood10*10
Trawling through the internet, I eventually tried the following code:
for (i
I'm new to R and have not been able to find what I am looking for online. I
found the system command but that seems to be used for getting lists of
directories, etc.
I am trying to replicate something that I used to do in SAS. I am simulating
some data (which I can do in R with a particular pac
Hi,
I didn't fully understand the logic.
You could get the result by:
list1<-lapply(mapply(cbind,lapply(1:2,function(i)
E[,,i]),lapply(c(1,3),function(i) C[,i,]),SIMPLIFY=FALSE),function(x)
x[,c(TRUE,apply(matrix(!x[,-1]%in% x[,1],nrow=5),2,all))])
Fnew<-array(unlist(list1),dim=c(dim(list1[[1]]
Hi Gerrit,
as I said in my original email, I already know both the
merge and the rbind commands, but I think that many
standard situations (I have given just a few) require
rather clumsy ad-hoc programming. So I was wondering
if there are any packages or existing code that would
make it easier to
Greetings,
I am having an interesting problem and I wonder if anyone else has
seen this behavior.
I am running R 2.11.1 with SNOW 0.3-3 on a Dell cluster running CentOS 5.5.
I create my cluster using:
cluster<- makeCluster(nodes,type="SOCK",port=10191) # nodes is a
vector of compute nodes
Hi Rolf,
sorry for this late answer and thanks for your kind explanation and
relative R code. I really appreciate.
In reality the concept that I'm trying to address is a bit more complex.
I'm fitting a model y vs 6 predictors with MARS / RandomForest /
Multiple Linear Regression Models having 140 o
Thanks, Ellison. Another question is if this p-value is a good parameter to
test if the fitting is good, as you this test is only for the null that the
coefficient is 0 (a is 0 in y=exp(a*x), right?)?
On Thu, Feb 7, 2013 at 10:48 AM, S Ellison wrote:
>
>
> > After applying the NLS for a model l
Hi,
Is anyone aware if there is a package that allows one to connect to Oracle
Business Intelligence (aka OBIEE)?
I want to be able to pull in a summary table contained in an "answer" (aka
OBIEE "query").
Dan
Lawrence Livermore National Lab
WF Analyst
[[alternative HTML version delet
q()?
E.g.,
> f <- function(x) { switch(sign(x)+2, q(), stop("Oops"), NULL) ; sqrt(x) }
> tryCatch(f(2), error=function(e)"error")
[1] 1.414214
> tryCatch(f(0), error=function(e)"error")
[1] "error"
> tryCatch(f(-2), error=function(e)"error")
Save workspace image? [y/n/c]: c
> for(i
Thanks.
-Roy M.
On Feb 6, 2013, at 11:55 PM, Prof Brian Ripley wrote:
> On 06/02/2013 22:47, Roy Mendelssohn - NOAA Federal wrote:
>> Hi All:
>>
>> We are developing (actually updating) a piece of code that runs in
>> R.
> Due to some limitations of some the libraries we use, at the present
> t
Hello, John,
as a start take a look at
?merge
And to (maybe) get a bit overwhelmed at first sight use
RSiteSearch( "merge")
Hth -- Gerrit
On Thu, 7 Feb 2013, John Smith wrote:
I know that the basic approach to append or merge data frames is using the
rbind and merge commands.
However,
> After applying the NLS for a model like y=exp(a*x), and I get
> a result showing the summary as:
> Estimate Std. Error t value Pr(>|t|)
> 2.6720 1.4758 1.811 0.3212
>
> My question is what this t-statistics tests? And what's the
> meaning of Pr?
t is (estimate/std.err) and can be us
> If
> another user does not want to get the full taxonomic
> description of that particular fly, he/she is welcome to
> ignore it along with all the other taxa he/she is not referencing.
All true, but the OP was asking about modifying a file that is part of an
installed. package.
That is sure
Hello,
I am writing cause I have a problem when try to create confidence envelopes
in spatstat.
I have a point data representing firms in my study area.
When I try to create confidence envelopes the error message turns up:
[> p <- envelope(point, fun=Kest, nsim=99)
Error in marks.ppp(Y, dfok = FA
Janesh,
This might help get you started:
http://biostatmatt.com/archives/78
(apologies for linking to my own blog)
Regards,
Matt
--
Message: 51
Date: Wed, 6 Feb 2013 18:50:43 -0600
From: Janesh Devkota
To: r-help@r-project.org
Subject: [R] low pass filter analysi
Hi,
I need assitance with performing a Bayesian Ordered Logistic Regression in R.
Would you be able to assist?
Aruna
Sent from my BlackBerry® wireless device available from bmobile.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/list
Hello,
Try the following.
> maintainer("spatstat")
[1] "Adrian Baddeley "
Hope this helps,
Rui Barradas
Em 07-02-2013 02:05, Hiroshi Saito escreveu:
Dear sir,
Which mailing list is appropriate to ask for kstest in spatstat?
Regards,
Hiroshi Saito
*
Hiroshi Saito:
E
I know that the basic approach to append or merge data frames is using the
rbind and merge commands.
However, if I understand things correctly, for both commands one needs to do
quite some additional programming to get e.g. behavior as with the Stata append
and morge commands or to achieve some
Hello there!
Remmy is my name and I came across an interesting code in R on this web
https://stat.ethz.ch/pipermail/r-help/2004-May/050424.html , there are some
parts that I have not followed and I would like to get some comments if you may!
Please comment on the highlighted two parts of the c
car::spreadLevelPlot is designed for just this purpose.
Something similar is one of the plots produced by plot(lm(...))
HTH
On 2/6/2013 9:42 AM, N. Janz wrote:
To detect heteroscedasticity for a multiple linear OLS regression (no
time dependencies):
What if the residuals vs. fitted values plo
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