Hi There!
I have a time series data for 13 years with freqency of 23 per year. I have
plot the curve on R.
Is it possible to refer to the point inbetween the nodes...
eg. say the time series ts1 has ts1[1] 0.25 and ts1[2]=0.4. is it possible
to get the time when ts1 reach to 0.3??
--
View this
Petr -
your suggestion WORKS!
Thank you so much, really!
happy-Chaehan
On Wed, Mar 17, 2010 at 2:22 PM, Petr PIKAL wrote:
> Hi
>
>
> r-help-boun...@r-project.org napsal dne 17.03.2010 13:04:05:
>
> > Jim & Petr,
> > Thank you for your hint - I am really gratef
p=FALSE]
did solve the problem, so I got a data.frame with the indexes.
Yet, then I turned to the call
svp <- ksvm(x, y, type="nu-svc")
Error in .local(x, ...) : y must be a vector or a factor.
So then I followed your second advice, looking up the additional information
from help fil
Dear r-helpers,
I am getting a mismatch error between two variables:
svp <- ksvm(x, y, type="nu-svc")
Error in .local(x, ...) : x and y don't match.
and I suspect that it might be due to missing index in the y variable which
I defined as:
y <- (LVvar[,1])
I tried various methods to make
Don, that was great - thanks so much!
...and you are right rbind() being expensive, but my dataframe will always
stay that small (max. twice as large).
Chaehan
On Thu, Feb 11, 2010 at 4:50 PM, Don MacQueen wrote:
> You have a mistake in how you're setting up the object named "res
Ok, you're right - may I rephrase:
How should I modify the assignment of result <- latentVariableNames
so it produces the output without the first line?
I thought result <- NULL should do the job, but it didn't because of the
following names(result) assignment (which I understan
Dear r-helpers,
why do I get an output in the first iteration of the for-loop
which contains the string values of the input vector,
and how can I avoid that?
Here's the output (only line 1 is wrong)
latentVariable Indiv Group
1 rPlanning rIterat rTDD
2 rPlanning0.79 0.84
3
Dear Joe -
that was wonderful :-)
-Chaehan
On Thu, Feb 11, 2010 at 12:44 AM, Joe Cheng wrote:
> On Wed, Feb 10, 2010 at 3:37 PM, Chaehan So wrote:
>
>> how can I access currentName in my loop - evaluated as a variable although
>> it is a string?
>>
>
> I think yo
mes)
{
doSomethingWith(currentName)
}
}
On Wed, Feb 10, 2010 at 11:34 PM, David Winsemius wrote:
>
> On Feb 10, 2010, at 5:12 PM, Chaehan So wrote:
>
> Thank you, it works for the first problem!
>>
>> Yet for the second problem, how can I solve that in one dataframe (here:
>>
Dear R-helpers,
my little function below calculates the group score (tmpGroupMean) of an
item,
appends a "_mean" on its name and stores its value on this name.
However, it does not calculate the mean of these scores (LVMean) in the same
row correctly,
as you can see in the below output which stran
;testfile%s"
> lapply(sprintf(fileName, 1:2), read.csv, dec = ",")
>
> On Wed, Feb 10, 2010 at 5:28 PM, Chaehan So wrote:
> > Dear r-helpers,
> >
> > I am looking for an R-equivalent for the eval-function in javascript
> which
> > can
&
Dear r-helpers,
I am looking for an R-equivalent for the eval-function in javascript which
can
interpret a string as code on runtime, thereby allowing things like
for (i in c(1:2))
{
eval(items + "i") <- read.csv(eval(filename+ i), dec=",");
}
which would execute (with filename="testfile"):
i
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