try this version instead:
gini - function(x, unbiased = TRUE, na.rm = FALSE){
if (!is.numeric(x)){
warning('x' is not numeric; returning NA)
return(NA)
}
if (!na.rm any(na.ind - is.na(x)))
stop('x' contain NAs)
if (na.rm)
x - x[!na.ind]
n - length(x)
mu - mean(x)
N - if (unbiased) n * (n - 1) else n * n
ox - x[order(x)]
dsum - drop(crossprod(2 * 1:n - n - 1, ox))
dsum / (mu * N)
}
gini(c(100,0,0,0))
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: MICHELE DE MEO [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Monday, June 11, 2007 4:13 PM
Subject: [R] Gini coefficient in R
If I use the Ineq library and the Gini function in this way:
Gini(c(100,0,0,0))
I obtain the result 0.75 instead of 1 (that is the perfect
inequality).
I think Gini's formula in Ineq is based on a formula as reported
here:
http://mathworld.wolfram.com/GiniCoefficient.html
but in the case of perfect inequality:
x_1=...=x_n-1 =0
x_n0
these formula are equal to 1 - 1/n, not to 1.
I don't know where I'm wrong
--
Michele De Meo
http://micheledemeo.blogspot.com/
[[alternative HTML version deleted]]
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