On Feb 22, 2014, at 1:57 PM, Julie Royster wrote:
> Is there any way I can insert markers at the beginning and end of a large
> section of R statements and then copy all text in between to clipboard?
>
> I have trouble scrolling to select long sections, so if there is another way
> I would like
This is a function of the editor you use and the operating system you use,
neither of which is on topic here. That said, many popular GUI editors allow
you to select by holding the shift key down while moving the cursor with arrow
keys. Note that in the long run using the source function to have
Is there any way I can insert markers at the beginning and end of a large
section of R statements and then copy all text in between to clipboard?
I have trouble scrolling to select long sections, so if there is another way
I would like to know it!
Julie (for husband Larry)
[[alternative
Hi,
I am trying to put the expression which in LaTeX would be $\check Y$ as
a label on the axis of a plot.
How does one get the \check part of the above to produce a similar
symbol in plotmath in R? I looked around plotmath's help but could not
see this in the documentation.
Many thanks for any
Hi:
You were close...
library(ggplot2)
m <- mpg
# Set the factor labels with plotmath code (note the ==)
m$drv <- factor(m$drv, labels = c("sigma[0] == sqrt(2)",
"sigma[0] == 2 * sqrt(2)",
"sigma[0] == 3 * sqrt(2)"))
ggplot(m,
On Feb 22, 2014, at 1:06 PM, arun wrote:
> HI,
> Try ?curve
>
> fit <- lm(Mean_Percent_of_Range~log(No.ofPoints))
> coef(fit)
> #(Intercept) log(No.ofPoints)
> # -74.52645 46.14392
>
>
>
> plot(Mean_Percent_of_Range ~ No.ofPoints)
> curve(coef(fit)[[1]]+coef(fit)[[2]]*
On Feb 22, 2014, at 4:28 AM, PQuery wrote:
> Many thanks David,
>
> I will have a look on logistic regression for my case.
> Do you know about a good example regarding logistic regression ?
> I was thinking also of using Multiple Factor Analysis too (MFA - like in
> FactoMineR). However I am not
HI,
Try ?curve
fit <- lm(Mean_Percent_of_Range~log(No.ofPoints))
coef(fit)
# (Intercept) log(No.ofPoints)
# -74.52645 46.14392
plot(Mean_Percent_of_Range ~ No.ofPoints)
curve(coef(fit)[[1]]+coef(fit)[[2]]*log(x),add=TRUE,col=2)
A.K.
I realize this is a stupid questio
Hello:
Might someone have experience with video production? The
"animation" package in R can produce video clips. These could
presumably be merged and synced with audio using something like
Avidemux. I have not done this, but it looks like it might be a
relatively easy way for an R
Hi:
When converting some code to use plot = F in a call to hist(), I received
the following warning (with a much simplified example).
moo<-1:10
hist(moo, las=1, plot=F)
Warning message:
In hist.default(moo, las = 1, plot = F) : argument '...' is not made use of
>From the help page, I was able t
Hi everybody,
I have realized a multiple linear regression.
To know how well my model does in terms of prediction, I can compute prediction
intervals bands and decide if they are narrow enough to be of use. If they are
too wide, then they probably are not useful.
Using R, I have written these
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Hi, is it possible to add exposures to a glm with family=binomial()?
It's easy to do it for a Poisson/negative binomial: just multiply the
mean by the exposure, that is, offset(log(exposure)): but this obviously
wrong for a binomial/Bernoulli since the mean must be no bigger than 1.
My goal w
Many thanks David,
I will have a look on logistic regression for my case.
Do you know about a good example regarding logistic regression ?
I was thinking also of using Multiple Factor Analysis too (MFA - like in
FactoMineR). However I am not sure how successful this is going to be.
Best,
P.
-
Hello,
I would like to show in my facet labels the equivalent in LaTex of
$\sigma_{0}= \sqrt{2}$. I think I'm close below, but not yet as it shows
$(\sigma_{0}, \sqrt{2})$
m <- mpg
levels(m$drv) <- c("sigma[0]=sqrt(2)", "sigma[0]=2 * sqrt(2)", "sigma[0]= 3
* sqrt(2)")
ggplot(m, aes(x = displ, y
Second.
Rui Barradas
Em 21-02-2014 23:44, Rolf Turner escreveu:
On 22/02/14 11:53, Greg Snow wrote:
Why are you testing your data for normality? For large sample sizes
the normality tests often give a meaningful answer to a meaningless
question (for small samples they give a meaningless a
Hello,
Inline
Em 21-02-2014 23:13, Rolf Turner escreveu:
On 22/02/14 11:04, Rui Barradas wrote:
Hello,
Not answering directly to your question, if the sample size is a
documented problem with shapiro.test and you want a normality test, why
don't you use ?ks.test?
m <- mean(HP_TrinityK25$V2)
Dear list members.
We are trying see effect of N on yield of three wheat genotypes under late
and early planting conditions in saline and non-saline environments. Our
experimental structure as follows:
Districts (2: DIST01 and DIST02) - Not randomly selected
ENV (Saline and Non-Saline) - One sal
Hi,
May be this helps:
res.i <- NULL
for(i in seq_along(x)){ res.i <- c(res.i,x[i]-y[i])}
#or using your nested loop:
res.ij <- NULL
for(i in seq_along(x)){
for(j in seq_along(y)){
if(i==j){
res.ij <- c(res.ij,x[i]-y[j])
}
}}
identical(x-y,res.i)
#[1] TRUE
identical(res.i,res.ij)
#[1] TR
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