Two things I'd try:
run your code in a clean R session, started at the command line with R
--vanilla and just loading that exact dataset and running that line of
code, in case there are issues with other things in your session.
Upgrade to the current version of R.
Dear Sarah,
I upgraded
This is probably due to a numerical value that is coded as a factor. Have a
look at the example below.
In ggplot2 one line per group is plotted. The default grouping is the
combination of all factors. In the first example only B, in the second and
third example both A and B. This leaves just
Dear R helpers,
Please have a look at the following : -
Note : My goal is to find and replace all Inf's in a data array with 0.
t-data.frame(A=c(Inf,0,0),B=c(1,2,3))
t
A B
1 Inf 1
2 0 2
3 0 3
str(t)
'data.frame':3 obs. of 2 variables:
$ A: num Inf 0 0
$ B: num 1 2 3
Dear David,
Many thanks for your reply.
On Fri, Jul 15, 2011 at 5:24 PM, David Winsemius dwinsem...@comcast.netwrote:
On Jul 15, 2011, at 5:20 AM, Ashim Kapoor wrote:
Dear R helpers,
Please have a look at the following : -
Note : My goal is to find and replace all Inf's in a data
ttt - data.frame(A = c(Inf, 0, 0), B = c(1, 2, 3))
apply(ttt, 2, function(x) {x[is.infinite(x)] - 0; x})
Ok thank you. That does work. What does
apply(ttt, 1, function(x) x[is.infinite(x)] - 0 )
this return. I get all 0's,but can you explai why ?
Thank you.
Ashim
[[alternative
Ok Josh. Many thanks for your effort.
Ashim : )
On Mon, Jul 18, 2011 at 1:34 PM, Joshua Wiley jwiley.ps...@gmail.comwrote:
On Mon, Jul 18, 2011 at 12:22 AM, Ashim Kapoor ashimkap...@gmail.com
wrote:
ttt - data.frame(A = c(Inf, 0, 0), B = c(1, 2, 3))
apply(ttt, 2, function(x) {x
Dear R-helpers,
In the following example I compute ret and returns the SAME way. In ret I
use compute returns for EACH column and in returns I do it for the whole
data frame. Could someone please tell me why I see a lagged result,by which
I mean ret and returns are different by one lag.
NA
Dennis
On Thu, Jul 28, 2011 at 11:09 PM, Ashim Kapoor ashimkap...@gmail.com
wrote:
Dear R-helpers,
In the following example I compute ret and returns the SAME way. In ret I
use compute returns for EACH column and in returns I do it for the whole
data frame. Could someone please tell
Dear All,
How would we do this problem looping over seq(1:2) ?
Thank you,
Ashim
On Thu, Aug 4, 2011 at 10:59 AM, Richard Ma xuanlong...@uts.edu.au wrote:
Thank you so much GlenB!
I got it done using your method.
I'm just curious how did you get this idea? Cause for me, this looks so
How would we do this problem looping over seq(1:2) ?
To extend the example in the corresponding nabble post : -
sub1-list(x=a,y=ab)
sub2-list(x=c,y=ad)
lst-list(sub1=sub1,sub2=sub2)
for ( t in seq(1:2) ) print(lst[[t]]$y)
So I can print out the sub1$y/sub2$y but it's not clear how to
On Thu, Aug 4, 2011 at 1:02 PM, Joshua Wiley jwiley.ps...@gmail.com wrote:
On Thu, Aug 4, 2011 at 12:12 AM, Ashim Kapoor ashimkap...@gmail.com
wrote:
How would we do this problem looping over seq(1:2) ?
Because this goes to an email list serv, it is good practice to quote
the original
On Thu, Aug 4, 2011 at 2:12 PM, Joshua Wiley jwiley.ps...@gmail.com wrote:
On Thu, Aug 4, 2011 at 1:40 AM, Ashim Kapoor ashimkap...@gmail.com
wrote:
On Thu, Aug 4, 2011 at 1:02 PM, Joshua Wiley jwiley.ps...@gmail.com
wrote:
On Thu, Aug 4, 2011 at 12:12 AM, Ashim Kapoor ashimkap
Thank you Joshua.
Ashim.
On Thu, Aug 4, 2011 at 2:53 PM, Joshua Wiley jwiley.ps...@gmail.com wrote:
On Thu, Aug 4, 2011 at 2:08 AM, Ashim Kapoor ashimkap...@gmail.com
wrote:
On Thu, Aug 4, 2011 at 2:12 PM, Joshua Wiley jwiley.ps...@gmail.com
wrote:
On Thu, Aug 4, 2011 at 1:40 AM
Dear R-helpers,
Please look at the following minimal code.
\documentclass[a4paper]{article}
\begin{document}
==
library(zoo)
a-zoo(1:4,order.by=Sys.time()+1:4)
str(a)
@
\end{document}
When I do R CMD Sweave,followed by pdflatex ,and view the final pdf, the
letter surrounding the phrase zoo,
a
On Tue, Sep 20, 2011 at 4:31 PM, Duncan Murdoch murdoch.dun...@gmail.comwrote:
On 11-09-20 3:08 AM, Ashim Kapoor wrote:
Dear R-helpers,
Please look at the following minimal code.
\documentclass[a4paper]{**article}
\begin{document}
==
library(zoo)
a-zoo(1:4,order.by=Sys.time()**+1:4
Dear R-helpers,
www.statistik.lmu.de/~*leisch*/*Sweave*/*Sweave*-Rnews-2003-2.pdf
says that R CMD build creates the pdf from a .Rnw file. But when I do R CMD
build followed by R CMD check it gives me a warning that there is a vignette
without a PDF. My query is : when
R CMD build works it
/**pipermail/r-devel/2011-May/**060775.htmlhttps://stat.ethz.ch/pipermail/r-devel/2011-May/060775.html
Regards,
Enrico
Am 21.09.2011 11:01, schrieb Ashim Kapoor:
Dear R-helpers,
www.statistik.lmu.de/~*leisch***/*Sweave*/*Sweave*-Rnews-2003-**2.pdfhttp://www.statistik.lmu.de/%7E*leisch*/*Sweave
the procedure.
Thank you Enrico that was very interesting to hear.
Regards,
Ashim
Am 21.09.2011 12:38, schrieb Ashim Kapoor:
Dear Enrico,
Many thanks for your reply.I was doing R CMD check pkg ( where pkg is the
directory name and NOT the tar ball ). May I ask another query or should
Dear R-helpers,
Please have a look at the following. f1 is the same as f2 except that it has
some values replaced by NA. But it's corresponding file is slightly bigger
than the file containing f2. Could someone please tell me if this is an
anomaly ?
load(file1)
ls()
[1] f1
load(file2)
ls()
Dear R-helpers,
It seems to me that a character zoo cannot be coerced to a numeric zoo.
Below is a minimal example. Can someone tell me what I have done wrong?
z-zoo(1:4,order.by=1:4)
coredata(z)-as.character(coredata(z))
str(z)
zoo series from 1 to 4
Data: chr [1:4] 1 2 3 4
Index: int
)
However, I prefer to use this sort of line of code:
z.Num - zoo(as.double(z.Str), index(z.Str))
finding it a little more transparent.
Hope this helps,
Michael Weylandt
On Tue, Sep 27, 2011 at 5:56 AM, Ashim Kapoor ashimkap...@gmail.comwrote:
Dear R-helpers,
It seems to me that a character
) != z
Consider this:
x = letters[1:5]
x2 - x1 - x
mode(x)
x1[1:5] - 1:5 # Modify the values of x1 without changing the mode
print(x1)
x2 - 1:5 # Replace x2
print(x2)
Michael Weylandt
On Tue, Sep 27, 2011 at 6:41 AM, Ashim Kapoor ashimkap...@gmail.comwrote:
Dear Michael,
I don't
Apologies,it should be ans-cumsum(vec1)
On Thu, Oct 13, 2011 at 4:58 PM, Ashim Kapoor ashimkap...@gmail.com wrote:
I don't have access to R so I can't test my example but I think this will
work.
vec ( as defined by you)
# flip the false and the trues
vec1-ifelse(vec==FALSE,TRUE,FALSE
I don't have access to R so I can't test my example but I think this will work.
vec ( as defined by you)
# flip the false and the trues
vec1-ifelse(vec==FALSE,TRUE,FALSE)
ans-cumsum(vec)
Regards,
Ashim
On Thu, Oct 13, 2011 at 4:45 PM, syrvn ment...@gmx.net wrote:
Hello!
If I have a vector
Dear R-helpers,
I want to split the following vector into 2 vectors by the last occurance
of a .
dput(rownames(sensext))
c(pat, cash_bank_bal, invest_abroad, pat.1, cash_bank_bal.1,
invest_abroad.1, pat.2, cash_bank_bal.2, invest_abroad.2,
pat.3, cash_bank_bal.3, invest_abroad.3, pat.4,
Assuming we want to split off the number at the end try this which
splits on those dots which are followed by a digit:
strsplit(r, \\.(?=\\d), perl = TRUE)
Dear Gabor,
Thank you very much. That works very well. I don't completely understand
it though. A few words on what the (?=\\d) is
See the info on zero width lookahead assertions on the ?regex page.
Thank you again.
Best Regards,
Ashim
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
Dear R-helpers,
I want to append a Ctrl-M character to a string and then save it to a text
file.
mystring-This is a test.
# How do I add a Ctrl-M to it in the end ??
cat(mystring,file=testfile)
Many thanks,
Ashim
[[alternative HTML version deleted]]
---
Sent from my phone. Please excuse my brevity.
David Winsemius dwinsem...@comcast.net wrote:
On Nov 10, 2011, at 9:35 PM, Ashim Kapoor wrote:
Dear R-helpers,
I want to append a Ctrl-M character to a string and then save it to
a text
file.
mystring
Dear all,
I have the following data, which has \\n in place of \n. I introduced \n's
in the csv file so that I could use it in barchart in lattice. When I did
that and read it into R using read.csv, it read it as \\n. My question is
how do I introduce \n in the middle of a long string of quoted
Dear all,
I want to draw ticks on the 3rd and 4th row of a lattice. How do I do this
? In my search of the help, I discovered a parameter alternating,which kind
of says where the ticks will be but does not suffice for me.
I am running this command : -
barchart(X03/1000~time|Company,
at 11:25 PM, Ashim Kapoor ashimkap...@gmail.com
wrote:
Dear all,
I have the following data, which has \\n in place of \n. I introduced
\n's
in the csv file so that I could use it in barchart in lattice. When I did
that and read it into R using read.csv, it read it as \\n. My question
Dear David,
Thank you. That was very clear. I will try to make minimal examples in the
future.
Best Regards,
Ashim
On Fri, Nov 18, 2011 at 6:50 PM, David Winsemius dwinsem...@comcast.netwrote:
On Nov 18, 2011, at 1:51 AM, Ashim Kapoor wrote:
Dear Dennis,
Many thanks.I was wondering
Dear Mr Sarkar,
My apologies for not being articulate enough. Your solution worked
perfectly.
Many thanks,
Ashim
On Tue, Nov 22, 2011 at 9:12 PM, Deepayan Sarkar
deepayan.sar...@gmail.comwrote:
On Fri, Nov 18, 2011 at 11:22 AM, Ashim Kapoor ashimkap...@gmail.com
wrote:
Dear all,
I want
On Wed, May 18, 2011 at 1:44 PM, Karl Ove Hufthammer k...@huftis.orgwrote:
Ajay Ohri wrote:
What is the appropriate software package for dumping say 20 PDFS in a
folder, then creating data visualization with frequency counts of
certain words as well as measure correlation within each file
Dear R helpers,
I have 2 questions : -
1. My excel sheet has a column with dates like 01/03/1980 which is formatted
as 03/80 when I read this into R it reads as Mar-80. How can I read it in
the source format ?
2.
v-c(Mar-80)
as.Date(v,format=%b-%y)
[1] NA
Could someone please tell me where
You get the NA since it is indeterminate as to the date; paste on a 1
for the day
Alright Jim,
Many thanks.
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do
Dear All,
I wanted to install the reshape package which in turn requires the plyr.
When I tried to install plyr it says it needs
ERROR: this R is version 2.10.1, package 'plyr' requires R = 2.11.0
My question is how do I upgrade my R ? I have Ubuntu 10.04.2 LTS.
Many thanks for your help.
Dear R helpers,
I have molten data which is : -
t3
Year variablevalue
1 2005 ICICI.Bank 27488370
2 2006 ICICI.Bank 43166850
3 2007 ICICI.Bank 59515300
4 2008 ICICI.Bank 63085760
5 2009 ICICI.Bank 54526330
6 2005 HDFC.Bank
reshape_0.8.3 plyr_1.4
loaded via a namespace (and not attached):
[1] digest_0.4.2 tools_2.13.0
Sarah
On Thu, Jun 30, 2011 at 8:31 AM, Ashim Kapoor ashimkap...@gmail.com
wrote:
Dear R helpers,
I have molten data which is : -
t3
Year variablevalue
1 2005
8S 4M4
> Web: socserv.mcmaster.ca/jfox
>
>
>
>
> > -Original Message-
> > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Ashim
> > Kapoor
> > Sent: June 4, 2016 5:33 AM
> > To: peter dalgaard <pda...@gmail.com>
> > Cc:
ander Version 2.2-0: Sat Jun 4 14:58:57 2016
[2] NOTE:
Rcmdr Version 2.2-0
[3] NOTE: The dataset Anscombe has 51 rows and 4 columns.
[4] ERROR:
could not find function "grid"
Best Regards,
Ashim
On Sat, Jun 4, 2016 at 2:50 PM, peter dalgaard <pda...@gmail.com> wrote:
>
>
Dear all,
I am new to using Rcmdr. From the menu I chose the following : -
data(Anscombe, package="car")
densityPlot( ~ income, data=Anscombe, bw="SJ", adjust=1, kernel="gaussian")
This gave me the following error:-
[1] NOTE: R Commander Version 2.2-0: Sat Jun 4 12:32:09 2016
[2] NOTE:
Rcmdr
he link [1].
> >
> > 1. http://svitsrv25.epfl.ch/R-doc/library/car/html/Ask.html
> >
> > On Wed, Jun 22, 2016 at 6:57 AM, Rolf Turner <r.tur...@auckland.ac.nz>
> > wrote:
> >
> >> On 22/06/16 13:06, Ashim Kapoor wrote:
> >>
> >&g
Dear Sir,
Thank you.
Best Regards,
Ashim
On Wed, Jun 22, 2016 at 10:47 AM, Fox, John <j...@mcmaster.ca> wrote:
> Dear Ashim,
>
> > -Original Message-----
> > From: Ashim Kapoor [mailto:ashimkap...@gmail.com]
> > Sent: June 21, 2016 10:10 PM
> > To: Fox,
Dear All,
my details:-
> sessionInfo()
R version 3.3.0 (2016-05-03)
Platform: x86_64-pc-linux-gnu (64-bit)
Running under: Ubuntu 15.10
locale:
[1] LC_CTYPE=en_IN LC_NUMERIC=C LC_TIME=en_IN
[4] LC_COLLATE=en_IN LC_MONETARY=en_INLC_MESSAGES=en_IN
[7] LC_PAPER=en_IN
/16 13:06, Ashim Kapoor wrote:
>
>> Dear All,
>>
>> my details:-
>>
>>> sessionInfo()
>>>
>> R version 3.3.0 (2016-05-03)
>> Platform: x86_64-pc-linux-gnu (64-bit)
>> Running under: Ubuntu 15.10
>>
>> locale:
>
-
> John Fox, Professor
> McMaster University
> Hamilton, Ontario
> Canada L8S 4M4
> Web: socserv.mcmaster.ca/jfox
>
>
> > -Original Message-----
> > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Ashim
> > Kapoor
> > Sen
> -
> David L Carlson
> Department of Anthropology
> Texas A University
> College Station, TX 77840-4352
>
>
>
>
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Ashim
> Kapoor
>
Dear all,
For treatment contrasts I would :
library(car)
data(Prestige)
attach(Prestige)
levels(type)
contrasts(type) <-contr.treatment(levels(type),base =2 )
Alternatively to change my level I would do :-
type = relevel(type,ref="prof")
Now I want the sigma constraint.
For this the LAST level
e the
names of a few (active) statistics help list ?
On Sat, Dec 3, 2016 at 1:33 AM, David Winsemius <dwinsem...@comcast.net>
wrote:
>
> > On Dec 2, 2016, at 9:09 AM, David Winsemius <dwinsem...@comcast.net>
> wrote:
> >
> >>
> >> On Dec 2, 2016,
erts but it has something to
> do with contrasts.
>
> ?contrasts
>
> Number 44.556 is mean for wool A and tension L and this is the level to
> which every other level is compared.
>
> Cheers
> Petr
>
> > -Original Message-
> > From: R-help [mailto:r-
Dear all,
Here is a small example : -
> model <- aov(breaks ~ wool * tension, data = warpbreaks)
> summary.lm(model)
Call:
aov(formula = breaks ~ wool * tension, data = warpbreaks)
Residuals:
Min 1Q Median 3Q Max
-19.5556 -6.8889 -0.6667 7.1944 25.
> John Fox, Professor
> McMaster University
> Hamilton, Ontario
> Canada L8S 4M4
> Web: socserv.mcmaster.ca/jfox
>
>
>
> > -Original Message-
> > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Ashim
> Kapoor
> &g
On Sun, Dec 4, 2016 at 10:03 AM, Ashim Kapoor <ashimkap...@gmail.com> wrote:
> Dear Sir,
>
> Many thanks for the explanation. Prior to your email (with some help from
> a friend of mine) I was able to figure this one out. If we look at the
> model : -
>
> y = intercep
>
> For full understanding you will need to read in a text more about
> sets of linear contrasts and their algebra.
> I recommend Section 10.3 in mine, of course.
>
> Statistical Analysis and Data Display:
> An Intermediate Course with Examples in R
> Heiberger, Richard M., Ho
Dear All,
Suppose I do :-
head(warpbreaks)
model1<- aov(breaks ~ wool*tension,data = warpbreaks)
summary(model1)
There is significant interaction. So I need to test for simple effects of
wool at each level of tension and vice versa. I can do a subset and then
do a one way anova for each level
Dear All,
I am currently trying to use the Markov Switching Models(MSwM) package in
R.I am not sure as to how to use the switch parameter. From what I
understand, switch is a boolean vector which says what parameters are
regime dependent.
My confusion is that what is the order of the parameters
>
> On 2017-04-17 7:58 AM, Ashim Kapoor wrote:
>
>> Dear All,
>>
>> set.seed(123)
>> qqplot(rbinom(n=100,size=100,p=.05), rbinom(n=100,size=100,p=.05) )
>>
>> I expect to see 1 clear line,but I don't. What am I misunderstanding?
>>
>
>
&
uot;1 clear line" quite well - for
> most of the values.
>
>
> B.
>
>
>
>
> > On Apr 17, 2017, at 9:01 AM, Spencer Graves <spencer.graves@
> effectivedefense.org> wrote:
> >
> >
> >
> > On 2017-04-17 7:58 AM, Ashim Kapoor wrote:
Dear All,
set.seed(123)
qqplot(rbinom(n=100,size=100,p=.05), rbinom(n=100,size=100,p=.05) )
I expect to see 1 clear line,but I don't. What am I misunderstanding?
Best Regards,
Ashim
[[alternative HTML version deleted]]
__
ed",
>lwd = 0.5)
> ... and use your original qqplot()
>
>
> B.
>
>
> > On Apr 18, 2017, at 12:47 AM, Ashim Kapoor <ashimkap...@gmail.com>
> wrote:
> >
> > Dear Boris,
> >
> > Thank you for your reply.
> >
> > > d
=100, size=100, p=0.05),
>rbinom(n=100, size=100, p=0.05) )
>
> qqline(rbinom(n=100,size=100,p=.05),
>distribution = function(probs) { qbinom(probs, size=100, prob=0.05)
> },
>col = "red",
>lwd = 0.5)
>
>
>
>
> B.
&
Dear all,
I am not able to understand the interplay of absolute vs relative and
tolerance in the use of all.equal
If I want to find out if absolute differences between 2 numbers/vectors are
bigger than a given tolerance I would do:
all.equal(1,1.1,scale=1,tol= .1)
If I want to find out if
7, 9, 8, 7, 2, 5, 4, 3, 7,
7, 2, 5, 4, 3, 3, 6, 10, 4, 9, 6, 5, 3, 4, 5, 5, 6, 6, 7, 3,
4, 8, 6, 4, 5, 1, 5, 9, 3, 6, 2, 4, 5, 5, 3, 3, 3, 3, 5, 4, 4,
5, 5, 1, 4, 5, 8, 7, 4, 3, 3, 5, 5, 4, 6, 5, 4, 7, 4, 4, 3, 3,
8, 4, 6, 7, 3, 4, 3, 5, 5, 7, 3, 6, 9, 7, 4, 3, 2, 6)
On Wed, Apr 19, 2017 at 1
Dear All,
My query is:
Do we always need to use perl = TRUE option when doing ignore.case=TRUE?
A small example :
my_text =
"RECOVERY OFFICER-II\nDEBTS RECOVERY TRIBUNAL-III\n RC No. 162/2015\nSBI
VS RAMESH GUPTA.\nDated: 01.03.2016 Item no.01\n
Present: Ms. Sonakshi,
Dear All,
Here is a small example:
library(skmeans)
library(tm)
data("crude")
#Examine the first document
inspect(crude[[1]])
dtm <- DocumentTermMatrix(crude, control =
list(removePunctuation = TRUE,
removeNumbers = TRUE,
Dear Enrico,
Many thanks and Best Regards,
Ashim.
On Thu, Jun 8, 2017 at 5:11 PM, Enrico Schumann <e...@enricoschumann.net>
wrote:
>
> Zitat von Ashim Kapoor <ashimkap...@gmail.com>:
>
>
> Dear All,
>>
>> My query is:
>>
>> Do we always ne
Here is a small reproducible example:
data <-
structure(list(V1 = structure(1:3, .Label = c("Name1", "Name2",
"Name3"), class = "factor"), V2 = structure(c(1L, 3L, 2L), .Label =
c("nam1",
"name-1", "name_12"), class = "factor"), V3 = structure(1:3, .Label =
c("nam2",
"nam_34", "name-2"), class =
quot;Name2",..: 1 2 3
> $ V2: Factor w/ 3 levels "nam1","name-1",..: 1 3 2
> $ V3: Factor w/ 3 levels "nam2","nam_34",..: 1 2 3
> $ V4: Factor w/ 3 levels "","nam3","nam_56": 2 3 1
> $ V5: Factor w/ 2 level
Dear All,
I have a file data.txt as follows:
Name_1,A,B,C
Name_2,E,F
Name_3,I,J,I,K,L,M
I will read this with:
my_data<- read.csv("data.txt",header=FALSE,col.names=paste0("V",
seq(1:10)),fill=TRUE)
Then the file will have 10 columns. I am assuming that each row in data.txt
will have at the max
Dear all,
I have made a dlm model,where I am getting a perfect prediction.
Here is a link to the output:
http://pasteboard.co/9IxVQwjm6.png
The query and code is on:
https://stats.stackexchange.com/questions/276449/perfect-prediction-in-case-of-a-univariate-ar1-model-using-dlm
Can someone
On Sun, Apr 30, 2017 at 10:05 PM, Duncan Murdoch <murdoch.dun...@gmail.com>
wrote:
> On 30/04/2017 12:26 PM, Ashim Kapoor wrote:
>
>> Dear All,
>>
>> This answer is very clear. Many thanks.
>>
>> I am now confused about how str*ucture works. Where can
and chr make sense, what does language mean? I think I
need to read some more.
Many thanks,
Ashim
On Tue, Apr 25, 2017 at 3:14 PM, Martin Maechler <maech...@stat.math.ethz.ch
> wrote:
> >>>>> Ashim Kapoor <ashimkap...@gmail.com>
> >>>>> on Tue, 25 Ap
asurement error on the state values, hence the
> conditional distribution of theta_t given y_t is just the point value of
> y_t...
>
> -pd
>
> > On 4 May 2017, at 12:05 , Ashim Kapoor <ashimkap...@gmail.com> wrote:
> >
> > Dear all,
> >
> &g
Dear Peter,
Many thanks,
Ashim
On Mon, Oct 2, 2017 at 3:30 PM, peter dalgaard <pda...@gmail.com> wrote:
> The first one, i.e. "optimal"; check help for match.arg() for the idiom.
>
> -pd
>
>
> > On 2 Oct 2017, at 11:48 , Ashim Kapoor <ashimkap...@gmail.co
Dear All,
>From :
?maps::map
we have :
map(database = "world", regions = ".", exact = FALSE, boundary = TRUE,
interior = TRUE, projection = "", parameters = NULL, orientation =
NULL,
fill = FALSE, col = 1, plot = TRUE, add = FALSE, namesonly = FALSE,
xlim = NULL, ylim
Dear all,
I want to :
1. Estimate a weighted 2D kernel.
2. Paint a heatmap on a ggmap.
Here is some reproducible data / code (I got it from the internet) :
s_rit <- structure(list(score = c(45, 60, 38, 98, 98, 53, 90, 42, 96,
45, 89, 18, 66, 2, 45, 98, 6, 83, 63, 86, 63, 81, 70, 8, 78,
15, 7,
Dear All,
I am trying to use the function ses from the forecast package.
>From its help I have :
Usage:
ses(y, h = 10, level = c(80, 95), fan = FALSE, initial = c("optimal",
"simple"), alpha = NULL, lambda = NULL, biasadj = FALSE, x = y, ...)
My query is that if I do not mention
Dear All,
I want to save the XML representation of a model using PMML. Then I want to
read the model and predict using the model and a new dataset.
This is described in this blog post :
https://www.r-bloggers.com/predictive-modeling-using-r-and-the-openscoring-engine-a-pmml-approach/
I am able
Dear All,
I posted this query on stack overflow but I received no replies so I am
posting it here so that someone here can take a look at this.
My query is that suppose the variable I am faceting by has 40 categories
and I want 3 categories per page( 1 row x 3 columns),then the last category
Dear all,
I am doing a one way between subjects anova in an unbalanced data set.
Suppose we have "a" levels of the one factor. I want to merge the not so
significantly different levels into the same cluster.
Can I do a Tukey Kramer HSD and then use the following algorithm:
For i in 2 : "a"
1) / p ?
Which one of the following is the final transformation ? Please clarify.
> boxplot(breaks ^ -.2 ~v)
> boxplot((breaks^-.2 -1)/(-.2)~v)
>
Best Regards,
Ashim
On Sun, Jan 7, 2018 at 10:59 AM, Ashim Kapoor <ashimkap...@gmail.com> wrote:
> Dear All,
>
> w
Dear All,
we need to do :
library(car) for the spreadLevelPlot function
I forgot to say that.
Apologies,
Ashim
On Sun, Jan 7, 2018 at 10:37 AM, Ashim Kapoor <ashimkap...@gmail.com> wrote:
> Dear All,
>
> I want a transformation which will make the spread of the r
Dear All,
I want a transformation which will make the spread of the response at all
combinations
of 2 factors the same.
See for example :
boxplot(breaks ~ tension * wool, warpbreaks)
The closest I can do is :
spreadLevelPlot(breaks ~tension , warpbreaks)
spreadLevelPlot(breaks ~ wool ,
ty
> Hamilton, Ontario, Canada
> Web: socialsciences.mcmaster.ca/jfox/
>
>
>
> > -Original Message-
> > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Ashim
> > Kapoor
> > Sent: Sunday, January 7, 2018 12:08 AM
> > To: r-help@r-proje
ly because the data are balanced.
>
> Best,
> John
>
> -
> John Fox, Professor Emeritus
> McMaster University
> Hamilton, Ontario, Canada
> Web: http://socserv.mcmaster.ca/jfox/
>
>
>
>
> On 2018-01-09, 10:18 AM, "Ashim Kapoor"
t; sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Thu, Aug 23, 2018 at 2:12 AM Ashim Kapoor
> wrote:
>
>> Dear All,
>>
>> I am not sure about the summary of the function ca.jo. I have posted m
Dear All,
I am not sure about the summary of the function ca.jo. I have posted my
query here :-
https://stats.stackexchange.com/questions/363188/interpreting-the-names-used-in-the-output-of-johansen-test-in-package-urca-in-r
I did not receive any reply so I am posting my query here.
Many
Dear All,
Here is a minimal working example :
library(AMModels)
mymodels <- amModelLib(description = "Simple AM Model library")
dataset1 <- 1:100
dataset2 <- seq(2,200,2)
model1 <- lm(dataset2~1)
model2 <- lm(dataset2~ dataset1)
m1 <- amModel(model = model1,comment = "Initial model.")
m2 <-
le index values.
>
> Do you want
>
> x[ !( x > 100 ) ]
>
> ?
>
> On April 18, 2018 6:13:30 AM CDT, Ashim Kapoor <ashimkap...@gmail.com>
> wrote:
> >Dear All,
> >
> >Here is a reprex:
> >
> >> x<- 1:100
> >> x[-wh
Dear All,
Here is a reprex:
> x<- 1:100
> x[-which(x>100)]
integer(0)
In words, I am finding out which indices correspond to values in x which
are greater than 100 ( there are no such items ) . Then I remove those
indices. I should get back the x that I started with since there are no
items
Dear All,
Here is a reprex:
set.seed(123)
b <- arima.sim(list(order = c(1,0,0),ar= .9),n=1000,sd=1)
arima(b)
Call:
arima(x = b)
Coefficients:
intercept
0.2250
s.e. 0.0688
sigma^2 estimated as 4.735: log likelihood = -2196.4, aic = 4396.81
>
Should sigma^2 not be equal to
s.e. 0.0145 0.2783
>
> sigma^2 estimated as 1.002: log likelihood = -1420.82, aic = 2847.63
>
>
> Bill Dunlap
> TIBCO Software
> wdunlap tibco.com
>
> On Tue, Nov 13, 2018 at 4:02 AM, Ashim Kapoor
> wrote:
>
>> Dear All,
>>
>> Here
coefficients, as requested/expected.
>
> HTH,
> Eric
>
>
> On Wed, Nov 14, 2018 at 12:08 PM Ashim Kapoor
> wrote:
>
>> Dear Eric and William,
>>
>> Why do the 1st and 2nd incantation of arima return sigma^2 as 5.233 vs
>> .?
>> The help for arima
Dear All,
I am trying to use this package --->
https://cran.r-project.org/web/packages/MARSS/index.html
I am reading this book which shows some examples based on the above package
---> https://nwfsc-timeseries.github.io/atsa-labs/
In a few words, the incantation MARSS(...) estimates the
Dear all,
I have created a time varying parameters regression. When I do that I have
a parameter which is AR1. I am not able to recover this parameter.
My query is posted here :
https://stats.stackexchange.com/questions/377295/unable-to-recover-time-varying-ar1-parameter-from-state-space-model
Dear all,
I have a query with regards to the package dlm. My query is : will dlm
return the same results if I give it the same data set ?
Here is a MWE ( created from :
https://sites.ualberta.ca/~sfossati/e509/files/other/dlm_ex.R )
library(dlm)
# simulate AR(1) process
# phi = .8, sig2 = .25
saying it
> can be *run* on Ubuntu 18.04.1.)
>
> Hope this helps,
>
> Rui Barradas
>
>
> Às 06:51 de 18/09/2018, Ashim Kapoor escreveu:
> > Dear All,
> >
> > I was reading this page --->
> > https://cran.r-project.org/bin/linux/ubuntu/README.html
Dear All,
I was reading this page --->
https://cran.r-project.org/bin/linux/ubuntu/README.html
It says: R 3.4 packages for Ubuntu on i386 and amd64 are available for all
stable Desktop releases of Ubuntu prior to Bionic Beaver (18.04) until
their official end of life date.
The page also shows
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