Thanks all for the enlightenment.
So, it does make sense that mean() produces NaN and median()/sd() NA,
from a calculation point of view at least.
But I still think it also makes sense that the mean of NA is NA as well,
be it only for consistency with other functions. That's just my opinion
of
These are great, thanks.
I always forget about paste().
===
Richard Sherman
rss@gmail.com
> On Aug 22, 2018, at 17:56, Fox, John wrote:
>
> fm <- vector("character",6)
> fm[1]<- "mpg ~ hp"
> for(i in 2:6)fm[i]<- paste0(fm[i-1]," + I(hp^", i,")")
_
Dear Bert,
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Bert Gunter
> Sent: Wednesday, August 22, 2018 8:38 PM
> To: Jim Lemon
> Cc: rss@gmail.com; R-help
> Subject: Re: [R] graphing repeated curves
>
> I do not think this does what the OP w
I do not think this does what the OP wants -- it does not produce
polynomials of the form desired.
John Fox's solution using poly() seems to me to be the right approach, but
I will show what I think is a considerably simpler way to build up the
polynomial expressions just as an example of one way
Dear Richard,
How about this:
ord <- order(mtcars$hp)
mtcars$hp <- mtcars$hp[ord]
mtcars$mpg <- mtcars$mpg[ord]
plot(mpg ~ hp, data=mtcars)
for (p in 1:6){
m <- lm(mpg ~ poly(hp, p), data=mtcars)
lines(mtcars$hp, fitted(m), lty=p, col=p)
}
legend("topright", legend=1:6, lty=1:6, col=1:6,
Hi Richard,
This may be what you want:
data(mtcars)
m<-list()
for(i in 1:6) {
rhterms<-paste(paste0("I(hp^",1:i,")"),sep="+")
lmexp<-paste0("lm(mpg~",rhterms,",mtcars)")
cat(lmexp,"\n")
m[[i]]<-eval(parse(text=lmexp))
}
plot(mpg~hp,mtcars,type="n")
for(i in 1:6) abline(m[[i]],col=i)
Jim
On
Hi all,
I have a simple graphing question that is not really a graphing question, but a
question about repeating a task.
I’m fiddling with some of McElreath’s Statistical Rethinking, and there’s a
graph illustrating extreme overfitting (a number of polynomial terms in x equal
to the number of
The Bioconductor project does things their own way. Please use their support
channels for help with their packages. https://www.bioconductor.org/help/
On August 22, 2018 2:27:47 PM PDT, Spencer Brackett
wrote:
>Hello all,
>
>Once the R package TCGAbiolinks or biocLite(“TCGAbiolinks”) is don
On Wed, 22 Aug 2018, Bert Gunter wrote:
groups = Summary.Type, ...
in your call will then do the job.
As an aside, this is a good example of why you should adhere to this format
for data analysis in R.
Bert,
Progress and retreat. I'm putting this aside for a day or so because I
need t
Hello all,
Once the R package TCGAbiolinks or biocLite(“TCGAbiolinks”) is done
unpacking, how do I go about apply any particular analysis with it?
Currently, I am unable to access the consule and edit any further. Would I
therefore have to open up a new script?
Spencer Brackett
[[a
On Wed, 22 Aug 2018, Bert Gunter wrote:
(I know that you said your post may already be "out of date", but ...)
Bert,
Still reading ?xyplot/?barchart.
But ?barchart says:
"Formally, if groups is specified, then groups along with subscripts is
passed to the panel function, ..."
which, as I
(I know that you said your post may already be "out of date", but ...)
" Despite additional reading of barchart() examples and help pages I'm
still
missing how to get grouping working and use the years in the dataframe as
labels on the x-axis."
But ?barchart says:
"Formally, if groups is specif
On Wed, 22 Aug 2018, Rich Shepard wrote:
Correcting the barchard() command fixed the main issue; getting the second
set of bars is still eluding me, but I'll continue working on fixing this.
I'll get the years as the x-axis labels rather than year number in
sequence from 1 to 29.
Despite add
On Wed, 22 Aug 2018, Bert Gunter wrote:
See inline.
Bert,
Will do. Sent a reply before seeing this. More to follow.
Thanks,
Rich
__
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https://stat.ethz.ch/mailman/listinfo/r-help
P
Dear Erick,
This is great!!
Many thanks for resolving the problem.
Ogbos
On Wed, Aug 22, 2018 at 5:44 PM Eric Berger wrote:
> Hi Ogbos,
> I took a closer look at your code.
> Here's a modified version (using dummy data) that seems to do what you
> want.
> Hopefully this will make it clear what
See inline.
-- Bert
On Wed, Aug 22, 2018 at 9:17 AM Rich Shepard
wrote:
> On Wed, 22 Aug 2018, Bert Gunter wrote:
>
> > No reproducible example (see posting guide below) so minimal help.
>
> Hi Bert,
>
>I thought the header and six data rows of the dataframe plus the syntax
> of
> the com
Hi Ogbos,
I took a closer look at your code.
Here's a modified version (using dummy data) that seems to do what you want.
Hopefully this will make it clear what you need to to.
nn <- 100
lDf <- data.frame(Li=rnorm(nn),CR=rnorm(nn))
fit<-lm(Li~CR, data=lDf)
a<-summary(fit)
N <- nrow(lDf)
C <- 50
I think that one can usefully look at this question from the
point of view of what "NaN" and "NA" are abbreviations for
(at any rate, according to the understanding I have adopted
since many years -- maybe over-simplified).
NaN: Mot a Number
NA: Not Available
So NA is typically used for missing v
You have an extra comma ... it should be
Li[sample(1:N, size = S, replace = TRUE)]
i.e. no comma after the closing parenthesis
On Wed, Aug 22, 2018 at 7:20 PM, Ogbos Okike
wrote:
> Hello Eric,
> Thanks for this.
>
> I tried it. It went but another problem prevents the code from running.
> s
Hello Erick,
Thanks again.
Another line indicated error:
source("script.R")
Error in eval(predvars, data, env) :
numeric 'envir' arg not of length one
Thank you for additional assitance.
Ogbos
On Wed, Aug 22, 2018 at 5:23 PM Eric Berger wrote:
> You have an extra comma ... it should be
>
>
Hello Eric,
Thanks for this.
I tried it. It went but another problem prevents the code from running.
source("script.R")
Error in Li[sample(1:N, size = S, replace = TRUE), ] :
incorrect number of dimensions
The error is coming from the line:
subsample <- Li[sample(1:N, size=S, replace=TRUE), ]
On Wed, 22 Aug 2018, Bert Gunter wrote:
No reproducible example (see posting guide below) so minimal help.
Hi Bert,
I thought the header and six data rows of the dataframe plus the syntax of
the command I used were sufficient. Regardless, here's the dput() output:
structure(list(Year = c(1
Li is defined as d1$a which is a vector. You should use
N <- length(Li)
HTH,
Eric
On Wed, Aug 22, 2018 at 6:02 PM, Ogbos Okike
wrote:
> Kind R-users,
> I run a simple regression. I am interested in using the Monte Carlo to test
> the slope parameter.
> Here is what I have done:
> d1<-read.tab
No reproducible example (see posting guide below) so minimal help.
Remove the quotes from your formula. Why did you think they should be
there? -- see ?formula.
Read the relevant portions of ?xyplot carefully (again?). You seemed to
have missed:
"*Primary variables:* The x and y variables should
Hi,
It might even be worthwhile to review this recent thread on R-Devel:
https://stat.ethz.ch/pipermail/r-devel/2018-July/076377.html
which touches upon a subtly related topic vis-a-vis NaN handling.
Regards,
Marc Schwartz
> On Aug 22, 2018, at 10:55 AM, Bert Gunter wrote:
>
> ... And FW
Kind R-users,
I run a simple regression. I am interested in using the Monte Carlo to test
the slope parameter.
Here is what I have done:
d1<-read.table("Lightcor",col.names=c("a"))
d2<-read.table("CRcor",col.names=c("a"))
Li<-d1$a
CR<-d2$a
fit<-lm(Li~CR)
a<-summary(fit)
a gives the slope as 88.
I've not before created bar charts, only scatter plots and box plots.
Checking in Deepayan's book, searching the web, and looking at ?barchart has
not shown me the how to get the results I need.
The dataframe looks like this:
head(stage_heights)
Year Med Max
1 1989 91.17 93.32
2 1990
... And FWIW (not much, I agree), note that if z = numeric(0) and sum(z) =
0, then mean(z) = NaN makes sense, as length(z) = 0, so dividing by 0 gives
NaN. So you can see the sorts of issues you may need to consider.
Bert Gunter
"The trouble with having an open mind is that people keep coming alo
Actually, the dissonance is a bit more basic.
After xxx(, na.rm=TRUE) with all NA's in ... you have numeric(0). So
what you see is actually:
> z <- numeric(0)
> mean(z)
[1] NaN
> median(z)
[1] NA
> sd(z)
[1] NA
> sum(z)
[1] 0
etc.
I imagine that there may be more of these little inconsistenc
On 22/08/2018 10:33 AM, Ivan Calandra wrote:
Dear useRs,
I have just noticed that when input is only NA with na.rm=TRUE, mean()
results in NaN, whereas median() and sd() produce NA. Shouldn't it all
be the same? I think NA makes more sense than NaN in that case.
The mean can be defined as sum(
Dear useRs,
I have just noticed that when input is only NA with na.rm=TRUE, mean()
results in NaN, whereas median() and sd() produce NA. Shouldn't it all
be the same? I think NA makes more sense than NaN in that case.
x <- c(NA, NA, NA) mean(x, na.rm=TRUE) [1] NaN median(x, na.rm=TRUE) [1]
N
Some string manipulation can convert the formula to a named vector such as
the one shown at the end of your post.
library(gsubfn)
# input
fo <- y ~ 2 - 1.1 * x1 + x3 - x1:x3 + 0.2 * x2:x2
pat <- "([+-])? *(\\d\\S*)? *\\*? *([[:alpha:]]\\S*)?"
ch <- format(fo[[3]])
m <- matrix(strapplyc(ch, pat)[
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