[R] problem for wtd.quantile()

roger koenker rkoenker at uiuc.edu
Thu Mar 16 14:58:16 CET 2006

```Certainly an improvement, but probably not what is really
wanted... I get:

> rq(x ~ 1, weights=w,tau = c(.01,.25,.5,.75,.99))
Call:
rq(formula = x ~ 1, tau = c(0.01, 0.25, 0.5, 0.75, 0.99), weights = w)

Coefficients:
tau= 0.01 tau= 0.25 tau= 0.50 tau= 0.75 tau= 0.99
(Intercept)         1         1         2         3         5

Degrees of freedom: 5 total; 4 residual

The first observation x=1 has weight .33  so it should be the
.25 quantile, unless there is some interpolation going on....

url:    www.econ.uiuc.edu/~roger            Roger Koenker
email    rkoenker at uiuc.edu            Department of Economics
vox:     217-333-4558                University of Illinois
fax:       217-244-6678                Champaign, IL 61820

On Mar 16, 2006, at 7:34 AM, Liaw, Andy wrote:

> Perhaps you're looking for this?
>
>> ?wtd.quantile
>> wtd.quantile(x,weights=w, normwt=TRUE)
>   0%  25%  50%  75% 100%
>    1    2    2    3    5
>
> Andy
>
> From: Jing Yang
>>
>> Dear R-users,
>>
>> I don't know if there is a problem in wtd.quantile (from
>> library "Hmisc"):
>> --------------------------------
>> x <- c(1,2,3,4,5)
>> w <- c(0.5,0.4,0.3,0.2,0.1)
>> wtd.quantile(x,weights=w)
>> -------------------------------
>> The output is:
>>   0%  25%  50%  75% 100%
>> 3.00 3.25 3.50 3.75 4.00
>>
>> The version of R I am using is: 2.1.0
>>
>> Best,Jing
>>
>>
>
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