try
is.na()
regards,
Albert
Am Freitag, den 04.07.2008, 08:29 +0200 schrieb Keld Jørn Simonsen:
> Hi
>
> I would like to sum a number of time series, some of them having NA's
>
> Standard action is here that if I sum a value with a NA, then the result
> is NA. I would like it to just keep the
does e.g.
x <- c(8,5,4,2)
cumsum(x) / 1:length(x)
do what you need?
Albert
Am Mittwoch, den 21.05.2008, 20:36 -0500 schrieb Jacques Wagnor:
> Dear List,
>
> Does there exist a function that calculates a cumulative average?
> Neither running() from library(gregmisc) nor running.mean() from
> libra
Hi useRs,
in the following graphic...
http://www.survey4all.org/tmp/with_margin.png
I drawed a function and added two axes afterwards with "axis". I could
not find a way to erase the margin between the axes and the plotting
region, so I solved the problem with "text" and "segments", which looks
li
Am Montag, den 17.12.2007, 21:22 +0100 schrieb Scionforbai:
> > My question: is there a way to add axes the usual way (as tried for
the
> > first graphic), but to erase the margin, so that the axes start at
point
> > (0/0) in my case.
>
> Not really sure if this is what you ask, but maybe you shou
yes, that's it, thank's a lot to all respondents!
Am Montag, den 17.12.2007, 18:12 -0500 schrieb Duncan Murdoch:
> On 17/12/2007 4:19 PM, Albert Greinoecker wrote:
> > Am Montag, den 17.12.2007, 21:22 +0100 schrieb Scionforbai:
> >>> My question: is there a way to a
Hi Michael,
a solution would be not to draw axes directly, but afterwards with the
axis-command:
plot(length, col=color, axes=F)
axis(2,at=, labels=, cex.axis=0.6, las=1);
# left axis
axis(1,at=, labels=, cex.axis=0.6, las=1);
# bottom axis
cex.axis...relativefont size
las...text orientation
se
z[z > 1]
regards,
Albert
Am Donnerstag, den 24.01.2008, 10:54 -0300 schrieb Juan Pablo Fededa:
> Dear Contributors:
>
> I have the next vector:
>
> "Z"
>
> z = (526, 0.1, 110, 0.2, 34, 0.4, 614, 0.5, 1, 14, 15)
>
>
> I want to generate a vector containing the values higher than 1
> correspo
try:
aggregate(d[,2:3], by=list(d$foo), FUN=sum)
cheers,
Albert
Am Dienstag, den 18.03.2008, 11:41 +0100 schrieb Hans Ekbrand:
> Consider the following data.frame, d.
>
> d <- data.frame(c("a","b","c","d","b","a"), c(1,4,85,3,4,1), c(7,6,2,4,15,33))
> names(d) <- c("foo", "bar", "baz")
>
> To
does
> x[!x == remove.value]
do what you need?
[1] 2 6 9 10
regards,
Albert
Am Dienstag, den 18.03.2008, 05:21 -0700 schrieb Josh Roofchop:
> This should be an easy one, but I haven't been able to figure it out,
>
> x<-c(2, 4, 6, 9, 10)
> #how do I remove a single value
> x<-x[-2] # will r
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