On Wed, 25 Jun 2014 14:16:08 -0700 (PDT)
Jeff Newmiller wrote:
> The brokenness of your perm.broken function arises from the attempted
> use of sapply to bind matrices together, which is not something
> sapply does.
>
> perm.fixed <- function( x ) {
>if ( length( x ) == 1 ) return( matrix( x
So my company has hired a few young McKinsey guys from overseas for a
couple of weeks to help us with a production line optimization. They
probably charge what I make in a year, but that's OK because I just
never have the time to really dive into one particular time, and I have
to hand it to the co
Hello Lib,
I think what you're trying to do is very easy using ggplot2 -- easy, that
is, once you got your hear around ggplot2 in the first place. The layering
you mention is the core feature of ggplot2. Fortunately it is
well-documented including a thin, overpriced book from Springer (which I
hav
On Sun, 09 Feb 2014 12:28:11 +
Rui Barradas wrote:
> Hello,
>
> Inside the function try
>
> dots <- list(...)
Hi guys,
thanks a lot. I knew it HAD to be something ultra-simple, like most
things in R.
Regards,
robert
__
R-help@r-project.org mai
Hello all,
To set options in a package I'm putting together I'd like to write a
function like options, that is:
my.options <- function(...) {
# ...
}
Now I'd like to access the named arguments that were passed to my
funtion within that function. How does that work? formals() doesn't do
it, neith
Hi guys,
like so often, the answert came to me minutes after posting. pmatch()
does exactly what I need. match() gives the values of the elements,
but not their positions.
Thanks,
robert
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https://stat.ethz.ch/mailman
Hello all,
given two vectors X and Y I'd like to receive a vector Z which
contains, for each element of X, the index of the corresponding
element in Y (or NA if that element isn't in Y).
Example:
x <- c(4,5,6)
y <- c(10,1,5,12,4,13,14)
z <- findIndexIn(x, y)
z
[1] 5 3 NA
1st element of z is 5,
Hello all,
I have a dataframe that looks like this:
head(df)
datey
1 2010-09-27 1356
2 2010-10-04 1968
3 2010-10-11 2602
4 2010-10-17 3116
5 2010-10-24 3496
6 2010-10-31 3958
I need a function that, given any date, returns the y value
corresponding to the given date or the last day b
Hello Andrea,
I don't know if I can help you (probably not, I'm a beginner myself),
but you that you should make it a lot easier for those that can if you
post a self-contained script in this forum that shows what you're
trying to do. Use dput() to dump your dataset in text form.
Good luck,
rober
Hello all,
I have a bunch of aggregated measurement data. The data describe two
different physical properties that correlate, and I want to estimate
the coefficients (slope and intercept) from the dataset.
This is of course easy, I've done it, and I got the expected result.
But here's the thing:
On Tue, May 15, 2012 at 1:59 PM, Bryan Hanson wrote:
> I see that no one has replied on this, so I'll take a stab.
Hi, Ryan!
> This is probably a matter of personal taste, but I would suggest a somewhat
> different and simpler approach. What you have done is not strictly an ANOVA,
> it's a li
Hello all,
here's a real-world example: I'm measuring a quantity (d) at five
sites (site1 thru site5) on a silicon wafer. There is a clear
site-dependence of the measured value. To find out if this is a
measurement artifact I measured the wafer four times: twice in the
normal position (posN), and
Hello Thierry,
thanks for your answer! There is one thing, however, that I don't
understand. The values labeled "B" in my data are generated with
1/20th the variance of the others, yet the standard error and
confidence intervals are the same for all levels of the factor. How
come?
> summary(mod_
Hello all,
I'm very satisfied to say that my grip on both R and statistics is
showing the first hints of firmness, on a very greenhorn level.
I'm faced with a problem that I intend to analyze using ANOVA, and to
test my understanding of a primitive, one-way ANOVA I've written the
self-contained p
On Wed, May 9, 2012 at 4:19 PM, R. Michael Weylandt
wrote:
> Good reproducible example ;-)
>
> Easiest is probably just:
>
> cbind(tencor, ThisRowMean = rowMeans(tencor[, 4:8]))
Actually, after frying my brain on tapply() and sapply() I found that
just plain apply() does what I need:
tencor$mea
Hello,
me again.
I have a data frame that looks like this (actual dput output at bottom):
> head(tencor)
datelot wf.id s1 s2 s3 s4 s5
1 08.05.2012 W0X3H0 9 1238 1263 1244 1200 1183
2 08.05.2012 W0X3H010 1367 1396 1371 1325 1311
3 08.05.2012 W0X3H011 1383 1417 13
On Tue, May 8, 2012 at 3:38 PM, R. Michael Weylandt
wrote:
> So this actually looks like something of a tricky one: if you wouldn't
> mind sending the result of dput(head(agg)) I can confirm, but here's
> my hunch:
Hi Michael,
while I'm trying to get my head around the rest of your post, here's
Hello all,
I am doing an aggregation where the aggregating function returns not a
single numeric value but a vector of two elements using return(c(val1,
val2)). I don't know how to access the individual columns of that
vector in the resulting dataframe though. How is this done correctly?
Thanks, r
sed when you signed up?
> That's a frequent cause of moderation.
>
> Sarah
>
> On Fri, May 4, 2012 at 1:09 PM, Robert Latest wrote:
>> Hello,
>>
>> is it possible to replace the text of tick marks in a plot?
>> Specifically, I'd like to have a ppnor
Hello,
is it possible to replace the text of tick marks in a plot?
Specifically, I'd like to have a ppnorm plot in which the theoretical
quantiles are not expressed in terms of standard deviations, but in
actual percentages. Anybody who's seen a probability plot in MINITAB
knows what I'm talking a
Hello all,
followup to yesterday's question: Part of my confusion was caused by
my embarrassing mistake of overwriting the "ppk" function with another
object, which of course broke the next iteration of the loop.
Secondly, I got exactly what I wanted like this:
aggregate.zoo <- function(series)
Hello all,
please look at my code below. The problems start where it says #
PROBLEMS START HERE. Some sample data is at the very bottom.
This is the disgnostic output from the script:
> source('load.R')
ts.null
1 NA
2 NA
3 NA
4 NA
5 NA
6 NA
[1] "Adding data" "VS1A
h
>
> ?aggregate
>
> you will need to specify "by" to aggregate by month
>
> Steve
>
> On Thu, Apr 12, 2012 at 7:10 AM, Robert Latest wrote:
>>
>> Hi all,
>>
>> I'm just getting started in R. My problem is the following:
>>
Hi all,
I'm just getting started in R. My problem is the following:
I have a data frame (v1) with lots of production data measurements.
Each row contains a single measurement ('ARI_MIT') with a timestamp. I
want to "lump" the data by months with their mean and standard
deviation.
I have already
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