numtonv(wavethresh)
numtonv()所属R语言包：wavethresh
Convert an index number into a node vector object.
转换的索引号到节点矢量对象。
译者：生物统计家园网 机器人LoveR
描述----------Description----------
Convert an index number into a node vector object.
转换成node vector对象的索引号。
用法----------Usage----------
numtonv(number, nlevels)
参数----------Arguments----------
参数：number
The index number of a particular basis within a wavelet object.
一个特定的基础内的小波对象的索引号。
参数：nlevels
The number of levels that the wavelet object has (can often be discovered using the nlevels function).
的数目，小波对象具有的水平（通常可以使用nlevels函数）被发现。
Details
详细信息----------Details----------
A basis within a (e.g. non-decimated) wavelet object (such as a wst.object) is represented in WaveThresh by a nv or node vector.
内（如不消灭的）小波对象（如wst.object）在WaveThresh表示一个nv或节点矢量的基础。
A packet-ordered non-decimated wavelet transform object wst for short) which is the transform of a vector of length n contains n bases. Each basis can be indexed from 0 to (n-1) .
数据包排序，非抽取小波变换对象wst的简称）是变换的矢量的长度nn碱基。每个基础上可索引是从0(n-1)。
A wst.object is simply a fully populated binary tree. There are nlevels levels in the tree with a split at each level. The root of the tree is at level 0, there are two branches at level 1, four at level 2, eight at level 3 and so on. A path through the tree can be constructed by starting at the root and choosing "left" or "right" at each possible branch. For certain data situations this path is constructed using minimum entropy algorithms (for examples MaNoVe). This function (numtonv takes the numerical representation of a path and converts it into a node.vector form suitable for passing to InvBasis to invert the representation according to a basis specicified by number.
Awst.object简直是一个完全填充的二进制树。有在树NLEVELS水平与在每个级别分割。树的根的级别是0，有两个分支1级，2级，3级8等。可以构造一个路径树的根开始，然后选择“左”或“右”在每一个可能的分支。对于某些数据的情况下，这条道路是用最小熵算法（为例子MaNoVe“）。此功能（numtonv需要的数值表示的路径，并将其转换成node.vector形式适于传递给InvBasis反转表示根据到数字specicified基准。
The least significant digit in number corresponds to deciding on the left/right decision at the fine leaves of the tree (high-frequency structure) and the most significant digit in number corresponds to deciding on the left/right decision at the root. Therefore gradually incrementing number from 0 to 2^{nlevels}-1 steps through all possible bases in the wst object ranging from all decisions being made "left" to all decisions being made "right". The "number" dividied by 2^{nlevels} corresponds exactly to the binary number epsilon in Nason and Silverman (1995).
重要数字的数量至少对应的左/右的决定，以决定在美好的树的叶子（高频结构）以及最重要的数字的数量对应于左/右的决定，决定根。因此逐渐递增的数字从0到2^{nlevels}-1步骤，通过所有可能的碱基wst对象范围从“左”的所有决定作出“正确”的所有决定。 2^{nlevels}的“数字”dividied的完全对应利晨和Silverman（1995）的二进制数小量。
值----------Value----------
An object of class nv (node vector). This contains information about a path through a wavelet object (a basis in a wavelet object).
一个对象的类nv（节点矢量）。这包含信息通过小波对象的路径（一个基础在小波的对象）。
RELEASE----------RELEASE----------
Version 3.6.0 Copyright Guy Nason 1995
版本3.6.0版权盖利晨1995年
（作者）----------Author(s)----------
G P Nason
参见----------See Also----------
wst, wst.object, MaNoVe, nv.object, InvBasis, nlevels.
wst，wst.object，MaNoVe，nv.object，InvBasis，nlevels。
实例----------Examples----------
#[]
# Generate some test data[生成一些测试数据。]
#[]
test.data <- example.1()$y
#[]
# Make it noisy[它吵]
#[]
ynoise <- test.data + rnorm(512, sd=0.1)
#[]
# Do packet ordered non-decimated wavelet transform[做包下令非抽取小波变换]
#[]
ynwst <- wst(ynoise)
#[]
# Now threshold the coefficients[现在阈值的系数]
#[]
ynwstT <- threshold(ynwst)
#[]
# Select basis number 9 (why not?)[选择基础数字9（为什么不呢？）]
#[]
NodeVector9 <- numtonv(9, nlevels(ynwstT))
#[]
# Let's print it out to see what it looks like[让我们把它打印出来看是什么样子]
# (nb, if you're repeating this examples, the basis might be different[（注意，如果你重复这个例子的基础上可能会有所不同]
# as you may have generated different pseudo random noise to me)[正如你可能已经产生了不同的伪随机噪声给我）]
#[]
NodeVector9
# Level : 8 Action is R (getpacket Index: 1 )[等级：8行动为R（getpacket指数：1）]
# Level : 7 Action is L (getpacket Index: 2 )[等级：7动作的L（getpacket指数：2）]
# Level : 6 Action is L (getpacket Index: 4 )[等级：6行动是L（getpacket指数：4）]
# Level : 5 Action is R (getpacket Index: 9 )[等级：5行动为R（getpacket指数：9）]
# Level : 4 Action is L (getpacket Index: 18 )[等级：4行动是L（getpacket指数：18）]
# Level : 3 Action is L (getpacket Index: 36 )[等级：3行动是L（getpacket指数：36）]
# Level : 2 Action is L (getpacket Index: 72 )[等级：2动作是L（getpacket指数：72）]
# Level : 1 Action is L (getpacket Index: 144 )[等级：1动作是L（getpacket指数：144）]
# Level : 0 Action is L (getpacket Index: 288 )[等级：0行动是L（getpacket指数：288）]
# There are 9 reconstruction steps[有9个重建步骤]
#[]
# The print-out describes the tree through ynwstT that corresponds to[的打印输出描述树通过ynwstT对应于]
# basis 9.[依据9。]
#[]
# The NodeVector9 and ynwstT objects could now be supplied to[现在可以提供给，NodeVector9和ynwstT对象]
# InvBasis.wst for inverting ynwstT according[InvBasis.wst，为反相ynwstT根据]
# to the NodeVector9 or basis number 9.[NodeVector9或根据第9号。]
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。
注2：由于是机器人自动翻译，难免有不准确之处，使用时仔细对照中、英文内容进行反复理解，可以帮助R语言的学习。
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发表于 2012-10-1 17:28:31
