On 16/06/17 15:53, Sunny Singha wrote:
Hi,
I need to generate correlated (positive as well as negative) bivariate
exponential
distribution with rate of 1/5 or any rate.
I need some guidance here. Please help.
I believe this question is ill-posed.
(1) My understanding is that there is no "biva
Perfect. Thanks so much Adams. It is much appreciated.
Greg
On Fri, Jun 16, 2017 at 12:56 PM, Adams, Jean wrote:
> You could add size = log10_P to the aes() inside geom_point().
> Untested code below.
> See also http://ggplot2.tidyverse.org/reference/geom_point.html
>
> ggplot(mydata, aes(x =
R Users,
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You could add size = log10_P to the aes() inside geom_point().
Untested code below.
See also http://ggplot2.tidyverse.org/reference/geom_point.html
ggplot(mydata, aes(x = X, y = log10_P)) +
theme_bw() +
theme(panel.border=element_blank(), legend.position="top",
axis.text=element_text(size =
Hopefully someone has a cleaner approach, but this will work. You have to
remove the first column/row labels and print with text():
library(corrplot)
M <- cor(mtcars)
colnames(M)[1] <- ""
rownames(M)[1] <- ""
corrplot(M, method="circle", type="lower", diag=F)
text(1, dim(M)[1] - .1, "mpg", srt=90
Hi all;
I am running the following ggplot codes. Runs well. However, I need to
reflect the numeric values of the log10_P to the point size in the graph.
Your help highly appreciated,
Regards,
Greg
p <- ggplot(mydata, aes(x = X, y = log10_P)) +
theme_bw() +theme(panel.border=element_blank()
Dear All,
Please consider the following example
library(corrplot)
M <- cor(mtcars)
corrplot(M, method="circle", type="lower", diag=F)
Suppose that I want to have the label "mpg" at the top in black
and leave everything else in red.
How can I achieve that?
Cheers
Lorenzo
_
Peter,
thanks, very nice, this will work for me... could you also help with setting up
the code to run the on liner "approx(sort(x), seq(0,1,,length(x)), q)$y" on the
rows of a data frame using my example above? So if I cbind z and res,
df<-cbind(z,res)
the "x" in your one liner would be the fi
Never mind, I think i figured:
z<-df
apply(df,1,function(x) approx(sort(x[1:4]), seq(0,1,,length(x[1:4])), x[5])$y)
thanks again for the help
Andras Farkas,
On Friday, June 16, 2017 5:34 AM, Andras Farkas via R-help
wrote:
Peter,
thanks, very nice, this will work for me... could you
Peter,
thanks, very nice, this will work for me... could you also help with setting up
the code to run the on liner "approx(sort(x), seq(0,1,,length(x)), q)$y" on the
rows of a data frame using my example above? So if I cbind z and res,
df<-cbind(z,res)
the "x" in your one liner would be t
It would depend on which one of the 9 quantile definitions you are using. The
discontinuous ones aren't invertible, and the continuous ones won't be either,
if there are ties in the data.
This said, it should just be a matter of setting up the inverse of a piecewise
linear function. To set ide
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