Rollapply:by.column 有什么作用? [英] Rollapply: what does by.column do?

问题描述

我已经阅读了手册中对 rollapply 的 by.column 的描述，但我不明白如何使用它.见下文:

x=matrix(1:60,nrow=10)图书馆('动物园')rollapply(x,3,mean,fill=NA,align="right",by.column=FALSE)[1] 不适用 不适用 27 28 29 30 31 32 33 34

当我使用 by.column= FALSE 时:它适用于宽度 (3) 滚动行数 mean(x[1:3,])

现在，如果我使用 by.column=TRUE 然后我得到:

x=matrix(1:60,nrow=10)rollapply(x,3,mean,fill=NA,align="right",by.column=TRUE)[,1] [,2] [,3] [,4] [,5] [,6][1,] 呐呐呐呐呐呐呐呐[2,] 呐呐呐呐呐呐呐呐[3,] 2 12 22 32 42 52[4,] 3 13 23 33 43 53[5,] 4 14 24 34 44 54[6,] 5 15 25 35 45 55[7,] 6 16 26 36 46 56[8,] 7 17 27 37 47 57[9,] 8 18 28 38 48 58[10,] 9 19 29 39 49 59

我无法理解结果.谁能解释一下 by.column 的用途，并提供一个例子?

解决方案

by.column = TRUE (这是默认值) with FUN = mean 做滚动平均分别为每一列.结果的第 i 列将是:

rollapplyr(x[, i], 3, mean, fill = NA)

by.column = FALSE 一次将所有列输入到函数中，因此在这种情况下它与:

c(NA, NA, sapply(1:8, function(ix) mean(x[seq(ix, ix+2), ])))

I have read the description of by.column for rollapply in the manual but I couldn't understand how to use it. see below:

x=matrix(1:60,nrow=10)

library('zoo')
rollapply(x,3,mean,fill=NA,align="right",by.column=FALSE)

     [1] NA NA 27 28 29 30 31 32 33 34

when i use by.column= FALSE: it applies mean to width (3) rolling number of lines mean(x[1:3,])

now, if I use by.column=TRUE then I get:

x=matrix(1:60,nrow=10)

rollapply(x,3,mean,fill=NA,align="right",by.column=TRUE)

         [,1] [,2] [,3] [,4] [,5] [,6]

    [1,]  NA   NA   NA   NA   NA   NA

    [2,]  NA   NA   NA   NA   NA   NA

    [3,]  2   12   22   32   42   52

    [4,]  3   13   23   33   43   53

    [5,]  4   14   24   34   44   54

    [6,]  5   15   25   35   45   55

    [7,]  6   16   26   36   46   56

    [8,]  7   17   27   37   47   57

    [9,]  8   18   28   38   48   58

    [10,] 9   19   29   39   49   59

I can't make sense of the result. could anyone please explain what's the use of by.column and maybe provide an example?

解决方案

by.column = TRUE (which is the default) with FUN = mean does a rolling mean separately for each column. The ith column of the result would be:

rollapplyr(x[, i], 3, mean, fill = NA)

by.column = FALSE inputs all columns at once to the function so in this case it would be the same as:

c(NA, NA, sapply(1:8, function(ix) mean(x[seq(ix, ix+2), ])))

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