将两个 xts 对象（矩阵）合并到 R 中的单个数组中

发布于 2024-11-02 06:35:27 字数 921 浏览 0 评论 0原文

我有两个 xts 对象。

> require(quantmod)

> getSymbols("GLD;SLV")
[1] "GLD" "SLV"

> head(SLV, n=2)
           SLV.Open SLV.High SLV.Low SLV.Close SLV.Volume SLV.Adjusted
2007-01-03   129.96   131.16  124.70    125.58    7480000        12.56
2007-01-04   126.04   127.97  125.45    125.80    3220000        12.58

> head(GLD, n=2)
           GLD.Open GLD.High GLD.Low GLD.Close GLD.Volume GLD.Adjusted
2007-01-03    63.58    64.02   62.00     62.28    8061900        62.28
2007-01-04    62.07    62.50   61.52     61.65    4858600        61.65

> NROW(SLV) == NROW(GLD)
[1] TRUE

如何将这些单独的二维矩阵“堆叠”到单个三维数组中？

大致如下：

metal <- xts_stacking_function (GLD, SLV)

class(metal)
[1] "array"

metal[1,,1]
              GLD.Open GLD.High GLD.Low GLD.Close GLD.Volume GLD.Adjusted
2007-01-03    63.58    64.02      62     62.28    8061900        62.28

原文

I have two xts objects.

> require(quantmod)

> getSymbols("GLD;SLV")
[1] "GLD" "SLV"

> head(SLV, n=2)
           SLV.Open SLV.High SLV.Low SLV.Close SLV.Volume SLV.Adjusted
2007-01-03   129.96   131.16  124.70    125.58    7480000        12.56
2007-01-04   126.04   127.97  125.45    125.80    3220000        12.58

> head(GLD, n=2)
           GLD.Open GLD.High GLD.Low GLD.Close GLD.Volume GLD.Adjusted
2007-01-03    63.58    64.02   62.00     62.28    8061900        62.28
2007-01-04    62.07    62.50   61.52     61.65    4858600        61.65

> NROW(SLV) == NROW(GLD)
[1] TRUE

How can you 'stack' these separate two-dimensional matrices into a single three-dimensional array?

Something along the lines of the following:

metal <- xts_stacking_function (GLD, SLV)

class(metal)
[1] "array"

metal[1,,1]
              GLD.Open GLD.High GLD.Low GLD.Close GLD.Volume GLD.Adjusted
2007-01-03    63.58    64.02      62     62.28    8061900        62.28

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

枫林﹌晚霞¤ 2024-11-09 06:35:27

我不认为 xts 对象支持 3d 数组。您可能需要一个像下面这样的丑陋的解决方案。将所有内容放入数组中会将内容强制为数值。但至少这保留了日期索引，尽管格式不同，因为数组只能有一种数据类型。

require(quantmod)
getSymbols("GLD;SLV")
GLD <- cbind(index(GLD), as.matrix(GLD))
SLV <- cbind(index(SLV), as.matrix(SLV))
C <- array(,c(dim(GLD),2))
C[,,1] <- GLD
C[,,2] <- SLV

I don't think xts objects support 3d arrays. You'll probably need an ugly solution like the one below. Putting everything into an array coerces things to numeric values. But at least this preserves the date index, albeit in a different format because arrays can only have one data type.

require(quantmod)
getSymbols("GLD;SLV")
GLD <- cbind(index(GLD), as.matrix(GLD))
SLV <- cbind(index(SLV), as.matrix(SLV))
C <- array(,c(dim(GLD),2))
C[,,1] <- GLD
C[,,2] <- SLV

回复收藏 0 原文

◇流星雨 2024-11-09 06:35:27

鉴于您的最新评论，我认为您可以通过以下两种方式之一执行您想要的操作：

将 lapply 与 xts 对象列表一起使用
，将 eapply 与存储在自己的 xts 对象中使用 #2 使用 getSymbols 的示例

（请注意，eapply 返回一个列表）：

library(quantmod)
myEnv <- new.env()
getSymbols("GLD;SLV", env=myEnv)
eapply(myEnv, function(x) head(Cl(x),3))
# $SLV
#            SLV.Close
# 2007-01-03    125.58
# 2007-01-04    125.80
# 2007-01-05    121.80
# 
# $GLD
#            GLD.Close
# 2007-01-03     62.28
# 2007-01-04     61.65
# 2007-01-05     60.17

Given your latest comment, I think you could do what you want in one of two ways:

use lapply with a list of xts objects
use eapply with xts objects stored in their own environment

An example of #2 using getSymbols (note that eapply returns a list):

library(quantmod)
myEnv <- new.env()
getSymbols("GLD;SLV", env=myEnv)
eapply(myEnv, function(x) head(Cl(x),3))
# $SLV
#            SLV.Close
# 2007-01-03    125.58
# 2007-01-04    125.80
# 2007-01-05    121.80
# 
# $GLD
#            GLD.Close
# 2007-01-03     62.28
# 2007-01-04     61.65
# 2007-01-05     60.17

回复收藏 0 原文

~没有更多了~