Qn(robustbase)
Qn()所属R语言包：robustbase
Robust Location-Free Scale Estimate More Efficient than MAD
鲁棒位置的规模测算的效率比MAD
译者：生物统计家园网 机器人LoveR
描述----------Description----------
Compute the robust scale estimator Qn, an efficient alternative to the MAD.
强大的规模计算，估计Qn，一个有效的替代的MAD。
See the references for more.
更多的参考。
用法----------Usage----------
Qn(x, constant = 2.21914, finite.corr = missing(constant))
s_Qn(x, mu.too = FALSE, ...)
参数----------Arguments----------
参数：x
numeric vector of observations.
数字观测向量。
参数：constant
number by which the result is multiplied; the default achieves consistency for normally distributed data. Note that until Nov. 2010, “thanks” to a typo in the very first papers, a slightly wrong default constant, 2.2219, was used instead of the correct one which is equal to 1 / (sqrt(2) * qnorm(5/8)). If you need the old slightly off version for historical reproducibility, you can use Qn.old(). Note that the relative difference is only about 1 in 1000, and that the correction should not affect the finite sample corrections for n <= 9.
号码的结果乘以默认的实现为正态分布的数据的一致性。请注意，直到2010年11月，“感谢”一个错字在第一篇论文中，稍有错误的默认不变，2.2219，是用来代替等于1 / (sqrt(2) * qnorm(5/8))这是正确的。如果你需要老版本稍微偏离历史的重现性，你可以使用Qn.old()。需要注意的是相对不同的是1000年大约只有1，修正不应该影响有限样本的修正n <= 9。
参数：finite.corr
logical indicating if the finite sample bias correction factor should be applied. Defaults to TRUE unless constant is specified.
逻辑表明，如果在有限样本偏差校正系数应适用。默认为TRUE除非constant指定。
参数：mu.too
logical indicating if the median(x) should also be returned for s_Qn().
逻辑说明，如果median(x)也应该返回s_Qn()。
参数：...
potentially further arguments for s_Qn() passed to Qn().
有可能进一步参数为s_Qn()传递给Qn()。
Details
详细信息----------Details----------
As the (default, consistency) constant needed to be corrected, the finite sample correction has been based on a much more extensive simulation, and on a 3rd or 4th degree polynomial model in 1/n for odd or even n, respectively.
（默认情况下，一致性）需要纠正的常数，是根据有限样本校正在更广泛的模拟，在1/n为奇数或偶数n，分别在第三或第四次多项式模型。
值----------Value----------
Qn() returns a number, the Qn robust scale estimator, scaled to be consistent for σ^2 and i.i.d. Gaussian observatsions, optionally bias corrected for finite samples.
Qn()返回一个数字，Qn强大的规模估计，按比例是一致的σ^2和IID高斯observatsions，选择偏压校正有限样本。
s_Qn(x, mu.too=TRUE) returns a length-2 vector with location (μ) and scale; this is typically only useful for covOGK(*, sigmamu = s_Qn).
s_Qn(x, mu.too=TRUE)返回一个长度为2个向量的位置（μ）和规模，这是典型的只适用于covOGK(*, sigmamu = s_Qn)。
（作者）----------Author(s)----------
Original Fortran code:
Christophe Croux and Peter Rousseeuw <a href="mailto:rousse@wins.uia.ac.be">rousse@wins.uia.ac.be</a>.
<br>
Port to C and R: Martin Maechler, <a href="mailto:maechler@R-project.org">maechler@R-project.org</a>
参考文献----------References----------
Alternatives to the Median Absolute Deviation, Journal of the American Statistical Association 88, 1273–1283.
Time-Efficient Algorithms for Two Highly Robust Estimators of Scale, Computational Statistics, Vol. 1, ed. Dodge and Whittaker, Physica-Verlag Heidelberg, 411–428;<br> also available from http://win-www.uia.ac.be/u/statis/abstract/Timeff92.htm.
参见----------See Also----------
mad for the "most robust" but much less efficient scale estimator; Sn for a similar faster but less efficient alternative; scaleTau2.
mad“最强大”，但估计要少得多有效规模; Sn一个类似的快，但效率较低的替代品，“scaleTau2。
实例----------Examples----------
set.seed(153)
x <- sort(c(rnorm(80), rt(20, df = 1)))
s_Qn(x, mu.too = TRUE)
Qn(x, finite.corr = FALSE)
转载请注明:出自 生物统计家园网(http://www.biostatistic.net)。
注：
注1：为了方便大家学习，本文档为生物统计家园网机器人LoveR翻译而成，仅供个人R语言学习参考使用，生物统计家园保留版权。
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发表于 2012-9-27 22:14:02
