Jim,
Wow, that does it! I think I can use strsplit and unlist
to convert the string of row names into a R list.
thank you!
-david
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PLEASE do read the posting guide http:
vibhava wrote
>
> Dear R Users,
> I am a beginner in R programming and need some help
> with a simple plotting problem that i am having. My dataset consist of
> three columns: first one has data_id, second is the date and third is the
> actual data itself corresponding to eac
Hello again.
I don't understand what didn't work.
First, it seems better to use 'nrow', the result is the same
stopifnot(length(x[,1]) == nrow(x))
Then your multiple OR condition.
#if((x[i,1] || x[i,2] || x[i,3] || x[i,4]) < 5)
x <- matrix(1:24, ncol=4)
for(i in 1:nrow(x))
if(any(x
Try this:
> x <- read.table(text = " P1 P2 P3 P4
+ 1 0011
+ 2 0111
+ 3 1000
+ 4 0000
+ 5 1111 ", header = TRUE)
> labs <- apply(x, 1, function(.row){
+ indx <- which(.row == 1)
+ if (length(indx) > 0) return(pas
Agreed with David - if you don't even know how many points there are between
your observations, you can't do any meaningful interpolation.
Perhaps something like
out <- seq(min(x), max(x), length.out = 20)
But that has it's own presumptions.
Perhaps talk a little about your data and how the
Hi everyone,
I need to recode multiple columns in a dataframe into a single column in a
variety of different ways. Often the values will be TRUE/FALSE and I want a
list of the columns that are true as in the Result column below:
P1 P2 P3 P4 Result
1 0011P3,P4
2
Here is the code:
Y=mydata
X=cbind(1:length(dY))
results=auto.arima(dY, xreg=dX)
Of course I could first do a linear regression external to auto.arima,
and then combine the forecasts from both of the linear regression part and
the residual arima part...
But it would be great if the "au
Happy holidays!
Could you please help with the error message "No regressors provided" when
using "auto.arima"?
What could have happened?
Thanks a lot!
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On Dec 25, 2011, at 4:12 PM, Alemtsehai Abate wrote:
Dear Michael,
Thanks for your prompt response.
na.approx in zoo does fine if we have some time metric. But in my
case, we
don't know the time points which the data values correspond to.
The given values in x and y are irregularly observed
MerryXmas() to R-help!
No one answered this query, but I found an easy solution that others may
find useful, using a wrapper for contour() that calls contourLines()
and then replots the contours using polygon(). It could use a little
tuning (easier way to specify fill.alpha) to make it a gener
Dear Michael,
Thanks for your prompt response.
na.approx in zoo does fine if we have some time metric. But in my case, we
don't know the time points which the data values correspond to.
The given values in x and y are irregularly observed values at unequal
distances in time.
My objective is just to
Hi Rolf,
Thank you for clarifying. After reading the help page over more carefully, I
realize that I misunderstood what chisq.test. was doing. I thought it was using
simulation whenever the expected frequencies were given using the "p" argument,
whereas it used the asymptotic chi-squared distrib
In addition to the general tools of the XML package,
I also had code that read documents with a similar structure
to the ones Andy illustrated. I put them and simple examples
of using them at the bottom of
http://www.omegahat.org/RSXML/
page.
D.
On 12/23/11 5:50 PM, Ben Tupper wrote:
> Hi
On 24.12.2011 12:03, reena wrote:
It didn't work. :(
What did not work???
Please do not misuse the R-help mailing list! Its posting guide clearly
asks you to cite the thread and specify reproducible examples that make
other able to help.
Best,
Uwe Ligges
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