kindly help me fine what is the mistake with following:
my aim is to compute those steps and obtain a vector with values (3,5)
but i am geting (NAN,NAN)
Q<-matrix(c(5,-3,-3,2),2,2)
b<-rbind(0,1)
H0<-diag(2)
x0<-rbind(0,0)
d0<-b
g0<--b
a0<--(t(g0)%*%d0)/(t(d0)%*%Q%*%d0)
x1<-x0+a0[,1]*d0
dx0<-x1-x
Hi
Do not extract columns from your data frame. Just put correct names to columns.
names(BHD)[29] <- "O2"
Then you can construct time series objects on fly
with(BHD, ts(O2, frequency=12))
if it is monthly time series and use it in stl directly
stl(with(BHD, ts(O2, frequency=12)))
Or you can
On Jan 12, 2014, at 10:03 PM, vasanthi P wrote:
> I have very basic knowledge in R. Kindly help me to convert dot file to gml
> file. Can u suggest me some datasets of social networks consists of edge
> weight and edge timestamp.
You are requested not to crosspost to StackOverflow and Rhelp.
>
I have very basic knowledge in R. Kindly help me to convert dot file to gml
file. Can u suggest me some datasets of social networks consists of edge
weight and edge timestamp.
Thanks and regards,
Vasanthi P.
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___
Library "tables" and tabular function is neato.
I'm trying to figure out how to get percents other than just row and
columns. I'd like a percent of a factor.
library(tables)
c=data.frame(
gender=c(1,1,1,1,2,2,2,2),
race=c(3,3,4,4,4,4,4,4)
)
tabular(
Factor(g
Hi all,
Do you know how to change the color of the regression line, confidence interval
etc. when plotting the "effect" function:
library(effects)
plot(effect("explanatory variable", model.name),xlab,ylab)
I tried to include the "col" argument of the plot function but it didn't work.
Than
On 01/13/2014 10:19 AM, Martha Zorn wrote:
I am having problems with adding legend to choropleth map using GISTools. The
map that I get does not include a legend (see attached). Thanks for any help
you can provide.
Below is my code:
#read in sample
at001<-read.csv("C:/Users/martha/Dropbox/s
sounds like you are referring to library(effects)...in which case try:
plot(effect("x1", model), color="red")
From: David Winsemius
To: karine heerah
Cc: "r-help@r-project.org"
Sent: Sunday, January 12, 2014 8:35 PM
Subject: Re: [R] how to change the col
Hi Karine,
Yes, please do follow Davids advice in the future.
The answer to your question about colors is in the effects package
documentation. Please see ?plot.eff, particularly the "color"
argument.
Best,
Ista
On Sun, Jan 12, 2014 at 11:35 PM, David Winsemius
wrote:
>
> On Jan 12, 2014, at 7
On Jan 12, 2014, at 7:26 PM, karine heerah wrote:
> Hi all,
> Do you know how to change the color of the regression line, confidence
> interval etc. when plotting the "effect" function:
> plot(effect("explanatory varible", model.name),xlab,ylab)
> I tried to include the "col" argument of the
Hi all,
Do you know how to change the color of the regression line, confidence interval
etc. when plotting the "effect" function:
plot(effect("explanatory varible", model.name),xlab,ylab)
I tried to include the "col" argument of the plot function but it didn't work.
Thanks,Karine
On 01/13/2014 10:19 AM, Martha Zorn wrote:
I am having problems with adding legend to choropleth map using GISTools. The
map that I get does not include a legend (see attached). Thanks for any help
you can provide.
Below is my code:
#read in sample
at001<-read.csv("C:/Users/martha/Dropbox/s
Hi,
May be this helps:
mapply(grep,'^Ab$',test.list)
#or
lapply(test.list,function(x) grep("^Ab$",x))
A.K.
On Sunday, January 12, 2014 9:31 PM, Hermann Norpois wrote:
Hello,
I want to detect Ab not Abc. For a normal vector
test
[1] "A" "Ab" "GG" "GA" "H" "Abc" "Gz" "HU"
> grep ("
Hello,
I want to detect Ab not Abc. For a normal vector
test
[1] "A" "Ab" "GG" "GA" "H" "Abc" "Gz" "HU"
> grep ("^Ab$", test)
[1] 2
works well.
For
test.list
[[1]]
[1] "A" "Ab" "GG" "GA"
[[2]]
[1] "H" "Abc" "Gz" "HU"
grep ("^Ab$", test.list)
integer(0)
doest not work.
Why?
How
I am having problems with adding legend to choropleth map using GISTools. The
map that I get does not include a legend (see attached). Thanks for any help
you can provide.
Below is my code:
#read in sample
at001<-read.csv("C:/Users/martha/Dropbox/shiny/sample.csv")
a <- matrix(at001$npovr, n
Hello,
You should address your questions to R-Help, the odds of getting more
and better answers are greater.
As for your question, the trick is to use function ?lower.tri. (There's
also an upper.tri) Something like the following.
x <- 1:6
m <- matrix(0, nrow = 4, ncol = 4)
m[lower.tri(m)]
On Jan 6, 2014, at 11:16 AM, Walter Anderson wrote:
> On 01/06/2014 11:14 AM, Sarah Goslee wrote:
>> Hi Walter,
>>
>> I can't reproduce your results. Please provide some data that
>> demonstrates the problem.
>>
>> http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-ex
Have a look at
http://davegiles.blogspot.ie/2013/10/more-on-distribution-of-r-squared.html
John
On 10 January 2014 20:32, Troels Ring wrote:
> In R package "psychometrics" an estimate of SE of R squared of /sersq <-
> sqrt((4*rsq*(1-rsq)^2*(n-k-1)^2)/((n^2-1)*(n+3))) with n sample size,
> and
Hi Mathew,
You must have noticed that I used:
output <-vector() # instead of output=array(NA,c(length(pick.a),
length(pick.d)))
The below methods should give the results:
Method1:
res <- t(sapply(pick.a,function(x) sapply(pick.d,function(y) {ts1 <-
sum(t.sham(m.sham,x,y)); tc1 <- sum(t.co
Works like a charm! Thank you so much!
One more comment: I either had to add a second input parameter "..." to
gcomma or remove the "..." from the code, otherwise I got an error.
--
View this message in context:
http://r.789695.n4.nabble.com/ggplot2-axis-label-German-formatting-tp4683477p46834
Hi,
You can write a function to format the labels, like this:
gcomma <- function(x) format(x, ..., big.mark = ".", decimal.mark =
",", scientific = FALSE)
p.new + scale_x_continuous(labels = gcomma)
Best,
Ista
On Sun, Jan 12, 2014 at 6:02 AM, Stageexp wrote:
> Hi all I have a problem with for
> Wouldn't with(dd, EVYEAR==2012 & EVMONTH=='02')
> be sufficient when using with()?
It probably would be sufficient to get the right answer, but I
thought the OP was wondering why there was a difference.
Comparing the results of his original code with new code
would help uncover the reason.
Bill
[Also inline]
On 12-Jan-2014 17:45:03 Rui Barradas wrote:
> Hello,
>
> Inline.
>
> Em 12-01-2014 10:48, gj1989lh escreveu:
>> Hi,
>>
>>
>> How can I subscribe this mailing list?
>
> Apparently, you already have.
> Welcome.
Well, apparently not. "gj1989lh" is not listed in the R-help
subscri
Hello,
Inline.
Em 12-01-2014 10:48, gj1989lh escreveu:
Hi,
How can I subscribe this mailing list?
Apparently, you already have.
Welcome.
thx
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Don't post in html, please.
Rui Barradas
__
R-hel
Hi all I have a problem with formatting my ggplot2 graph.
Let's look at this example:
library(ggplot2)
library(scales)
x <- rnorm(100, mean=100, sd = 1) * 100
y <- rnorm(100, mean=100, sd = 1) * 100
df <- data.frame(x,y)
p.new <- ggplot(df,aes(x,y)) +
geom_point()
print(p.new)
This is
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained,
Hi,
How can I subscribe this mailing list?
thx
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.
On Mon, 06-Jan-2014 at 07:38PM +, William Dunlap wrote:
|> You could compare the outputs of
|> z1 <- with(dd, dd$EVYEAR==2012 & dd$EVMONTH=='02')
Wouldn't with(dd, EVYEAR==2012 & EVMONTH=='02')
be sufficient when using with()?
|> (which is like subset()) and that of
|> z2 <- dd$EVY
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