[R] Density

[David L Carlson]

The numbers are there, they just aren't listed by the default print method
for density.  When you type the object name, den0, R runs
print.density(den0) which provides summary statistics. You need to look at
the manual page (?density) and pay close attention to the section labeled
"Value" which provides information about what values are returned by the
function.

str(den0)
den0$x
den0$y
plot(den0$x, den0$y, typ="l")

----------------------------------------------
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77843-4352


> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
> project.org] On Behalf Of li li
> Sent: Wednesday, August 08, 2012 9:03 PM
> To: r-help
> Subject: [R] Density
> 
> Dear all,
>    Given a set of observations X, I want to evaluate the kernel density
> estimator
> at these observed values. If I do the following, I do not get the those
> estimated values directly.
> Can anyone familiar with this give an idea on how to find out the
> estimated
> density values for X?
> 
> > X <- rnorm(100)
> 
> > density(X)-> den0
> 
> > den0
> 
> 
> Call:
> 
> density.default(x = X)
> 
> 
> Data: X (100 obs.); Bandwidth 'bw' = 0.354
> 
> 
>        x                 y
> 
>  Min.   :-3.2254   Min.   :0.0002658
> 
>  1st Qu.:-1.6988   1st Qu.:0.0359114
> 
>  Median :-0.1721   Median :0.1438772
> 
>  Mean   :-0.1721   Mean   :0.1635887
> 
>  3rd Qu.: 1.3545   3rd Qu.:0.2866889
> 
>  Max.   : 2.8812   Max.   :0.3776935
> 
> 
> I did write the code for the kernel density
> 
> estimator myself. So once I find a proper bandwidth,
> 
> I can use the following function. However, it would be nicer
> 
> if there is a more direct way.
> 
> 
> > fhat <- function(x, X){
> 
> +          h <- density(X, bw="SJ")$bw
> 
> +          n <- length(X)
> 
> +          1/(n*h)*sum(dnorm((x-X)/h))}
> 
> >
> 
> > est <- numeric(length(X))
> 
> > for (i in 1:length(X)){est[i] <- fhat(x=X[i], X=X)}
> 
> >
> 
> > est
> 
> 
> 
> 
> Thanks in advance.
> 
>       Hannah
> 
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> 
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