You haven't indicated what information you want to convey with this gradient.
You also are using arrays where you should be using vectors, usually stored in
a data frame.
Here is one way using the contributed package ggplot2:
library(ggplot2)
DF <- data.frame( V1=1:10, V2=11:20, C=21:30 )
p <-
> On Feb 4, 2017, at 9:14 AM, Riyas MJ wrote:
>
> Hi all,
>
> I am a new user of R. I just did my first real program.
> I would like to know how to put a gradient (like rainbow() or topo.colors,
> etc) to a* line* graph.
>
> Example:
> ar1=array(data=1:10,dim=9)
> ar2=array(data=11:20,dim=9)
>
> On 03 Feb 2017, at 21:07 , Art U wrote:
>
> Hello,
>
> I have a vector of coefficients from backward selection model that looks
> like:
>
> [,1]
> (Intercept) -0.15
> s1[, 1] 2.65
> s1[, 2] 1.81
> s1[, 5] 2.35
>
>
> I'd like to get a new vector that contains zero
Hi all,
I am a new user of R. I just did my first real program.
I would like to know how to put a gradient (like rainbow() or topo.colors,
etc) to a* line* graph.
Example:
ar1=array(data=1:10,dim=9)
ar2=array(data=11:20,dim=9)
plot(ar1,ar2,type="l",col="red",lwd=3)
Instead of a red color, I woul
Many thanks for the prompt response.
Best Regards,
Ankush Sharma,PhD
Visiting CASyM Postdoctoral Research fellow (CASyM Consortium, EU-FP7)
LISM, Institute of Clinical Physiology, Siena (Italy)
Experimental Oncology Unit (UOS),
I
nstitute of Clinical Physiology
- National Research Council,
Hello,
I am trying to answer the question:
"What is the probability of a certain day's low forecast (of 21) being the
month's overall low?"
I decided to attempt it in R
Searching pointed me to extreme value analysis.
I created a csv file of the last 8 years lows for the month:
year month
Dear useRs,
We are pleased to announce you a new version of "fitdistrplus" on CRAN :
https://cran.r-project.org/web/packages/fitdistrplus/index.html
"fitdistrplus" is a package dedicated to help the fit of a parametric
distribution to non-censored or censored data.
The main new features in this
Hello,
I have a vector of coefficients from backward selection model that looks
like:
[,1]
(Intercept) -0.15
s1[, 1] 2.65
s1[, 2] 1.81
s1[, 5] 2.35
I'd like to get a new vector that contains zeroes for variables that were
not included in the final model. For example
In the help page of ?tapply it says that the first argument (X) is "an
atomic object, typically a vector."
However, tapply seems to be able to handle list objects. For example:
###
l <- as.list(1:10)
is.atomic(l) # FALSE
index <- c(rep(1,5),rep(2,5))
tapply(l,index,unlist)
> tap
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