在 mts 对象上使用 Apply 系列函数 [英] Using Apply family of functions on mts objects

问题描述

在 mts 对象上使用 apply(或 sapply)会在发送到函数时删除其时间序列属性.我应该如何在 mts 对象中的每个时间序列上应用相同的函数(带有 ts 输入和 ts 输出)并返回它(最好是 mts)[我的意思是除了使用 for 循环之外]?

Using apply (or sapply) on an mts object removes its time series properties when sending to function. How should I apply same function (with ts input and ts output) on each of times series in an mts object and return it (preferably as mts) [I mean besides using for loops]?

例如假设我编写了一个返回时间序列趋势的函数(使用 stl)

For example suppose I write a function that returns the trend of a time series (using stl)

myfunc <- function(x) {
      return(stl(x,"per")$time.series[,2])
}

现在是一个示例 mts

Now for a sample mts

z <- ts(matrix(rnorm(90), 30, 3), start=c(1961, 1), frequency=4)
class(z)

只发送一个时间序列是正确的:

Sending only one of the time series works correct:

myfunc(z[,1]) # works correctly, returns the trend of first series

我的函数不是为多个时间序列设计的，所以:

My function is not designed for multiple time series so:

myfunc(z) # will not work returning the error below

Error in stl(x, "per") : only univariate series are allowed

在 mts 对象上使用 apply 将每个时间序列作为向量发送，而不保留其时间序列属性 (tsp):

Using apply on the mts object send each of the time series as a vector, not preserving its time series properties (tsp):

apply(z,2,myfunc) # will not work returning the error below

Error in stl(x, "per") : 
series is not periodic or has less than two periods

推荐答案

解决这个问题的一个简单方法是使用索引而不是干净的 apply :

A simple way around this, is to work with the indices instead of a clean apply :

sapply(seq_len(ncol(z)),function(i) myfunc(z[,i]))

apply 将干净的向量放入函数中，因为它首先将对象转换为矩阵.通过使用为时间序列对象定义的 [ 函数，您可以确保每次都提取有效的时间序列.

apply puts clean vectors inside the function, because it first converts an object to a matrix. By using the [ function defined for time series objects, you are sure that you extract a valid time series each time.

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