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我有一个非常基本的问题,但我不知道如何实现它:我有一个返回的 data.frame,其中每个工具的返回是逐行的:

 tmp<-as.data.frame(t(data.frame(a=rnorm(250,0,1),b=rnorm(250,0,1))))

现在我想计算每一行的每周收益,即每个后续五个元素的乘积(=1+return)。这样我就得到了 50 个不重叠的每周回报。

我怎样才能做到这一点?我希望我能把我的问题告诉你。谢谢

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2 回答 2

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嗯,这并不完全清楚,但包zoo和功能rollapply应该让你得到你想要的(我想!):

tmp2 <- rollapply(t(tmp) , width = 5 , FUN = function(x) prod( 1 + x ) , by = 5 , by.column = TRUE )
head( tmp2 )
                  a             b
 [1,] -6.2279765402  0.2035895290
 [2,]  0.4088722374  3.3903379722
 [3,]  0.2112797361 -0.3812740172
 [4,]  0.7713020593 -0.0559832485
 [5,] -0.4930095796 -2.1750123558
 [6,]  6.2076460590  0.0188344154

我曾经t()转置数据框,参数设置width窗口的大小,by参数设置窗口移动时移动多少元素,所以我们的宽度为5,每次移动5个位置,结果为50返回。同样,当我这样做时,R 是矢量化function(x) prod( 1 + x )的,如果 x 的每个元素超过一个,它会为 x 的每个元素添加一个,并且由于我们在函数中将宽度设置为 5,因此有 5 个元素x

并且测试我们发现行'a'的前五列的乘积是:

 1+ tmp[1,1:5]
          V1      V2       V3       V4       V5
a -0.4655032 2.02795 2.133612 1.029273 3.004145

prod( 1+ tmp[1,1:5] )
[1] -6.227977
于 2013-04-23T09:14:39.390 回答
0

因此,据我所知,您想计算每行每五个后续列的乘积。这是代码:

# Sample data
set.seed(3)
tmp <- as.data.frame(t(data.frame(a=rnorm(250,0,1),b=rnorm(250,0,1))))

# Compute products per row in steps of five and rbind results afterwards
tmp.agg <- do.call("rbind", lapply(seq(nrow(tmp)), function(i) {
  sapply(seq(1, length(tmp), 5), function(j) {
    Reduce(prod, tmp[i,j:(j+4)] + 1)
  })
}))

tmp.agg
            [,1]         [,2]       [,3]         [,4]        [,5]        [,6]         [,7]        [,8]      [,9]      [,10]      [,11]       [,12]
[1,] -0.01642594 -0.004438307 -0.0231848 -0.046496247 -0.08965697 -0.07132495 -0.144778621 -0.02727861 0.1148542 0.17021314 -0.3852979 -0.46081791
[2,]  0.15293410  0.031532305  0.1023034  0.003880801 -0.01795657 -0.03010335 -0.000979364  0.48269767 0.1357064 0.04855989  0.4002406  0.06878954
          [,13]         [,14]         [,15]        [,16]      [,17]         [,18]       [,19]        [,20]        [,21]       [,22]      [,23]        [,24]
[1,] 0.05578327 -0.0461665415 -0.0009288497  0.054057705 -0.4094909 -0.0004848274  0.22665096 -0.003439506 -0.161383874  0.07839329 0.06254542 1.320146e-05
[2,] 0.26472709 -0.0001023569  0.0565177862 -0.009908618 -0.3056234 -0.0798260012 -0.03187966  0.001753645 -0.003536846 -0.05118659 0.72050663 1.726878e-01
          [,25]      [,26]       [,27]      [,28]       [,29]      [,30]       [,31]      [,32]       [,33]      [,34]         [,35]      [,36]       [,37]
[1,] 0.09297968 -0.3897201 -0.03092538  3.4957663 0.316273690 0.25769516 -1.54040152 0.04903115 -1.72941882 -0.7814886 -6.448821e-03 -7.1289278  0.01151266
[2,] 0.03609944  0.5364990  0.01610708 -0.0652631 0.007055414 0.02600772 -0.03116456 2.30021712  0.01136907  0.1532950 -2.141868e-05 -0.7549226 -0.13295228
             [,38]       [,39]       [,40]       [,41]        [,42]      [,43]      [,44]       [,45]      [,46]       [,47]      [,48]       [,49]
[1,] -0.4453617085  1.22240858  0.01303989 -0.30123118 -6.736829848 -0.0988588 -0.2314011 0.120263690 0.05006402  0.17631687 0.03386454 -0.14113772
[2,]  0.0005811551 -0.08559412 -6.35558415  0.01240032  0.001522634  0.1929070 -0.3015167 0.000790891 2.07795076 -0.05711388 0.17521929  0.03909323
           [,50]
[1,] -0.11744356
[2,]  0.09676705
于 2013-04-23T09:20:24.367 回答