R 计算滚动窗口协方差矩阵

我试图计算一系列资产的滚动窗口（移动1天）协方差矩阵

假设我的df看起来像这样：

df <- data.frame(x = c(1.5,2.3,4.7,3,8.4), y =c(5.3,2.4,8.4,1.3,2.5),z=c(2.5,1.3,6.5,4.3,2.8),u=c(1.1,2.5,4.3,2.5,6.3))

但一切都在循环中，因为我在数据集中有很多行

如果我想通过将滚动窗口移动1天，在滚动的基础上计算协方差矩阵，那么可能的

for

循环会是什么样子？或者我应该使用一些

apply

family函数吗

如果我想为上面的循环创建一个时间序列对象，那么哪个时间序列类更可取？现在我使用

fPortfolio

包中的

as.timeSeries

我就是搞不懂

---

致以最诚挚的问候

要创建滚动窗口，您可以使用

嵌入

## your data.frame
df <- data.frame(x=c(1.5,2.3,4.7,3,8.4), y=c(5.3,2.4,8.4,1.3,2.5), z=c(2.5,1.3,6.5,4.3,2.8), u=c(1.1,2.5,4.3,2.5,6.3))

## define windows
windowSize <- 3
windows <- embed(1:nrow(df), windowSize)

lapplyApproach <- function(df, windows) {
    ## convert window matrix to a list
    windowsList <- split(t(windows), rep(1:nrow(windows), each=ncol(windows)))
    ## edit: customize names: "from:to"
    names(windowsList) <- unlist(lapply(windowsList, function(x)paste(range(x), sep="", collapse=":")))
    return(lapply(windowsList, function(x)cov(df[x, ])))
}

forApproach <- function(df, windows) {
    l <- vector(mode="list", length=nrow(windows))
    for (i in 1:nrow(windows)) {
        l[[i]] <- cov(df[windows[i, ], ])
    }
    return(l)
}

## check results
all.equal(forApproach(df, windows), unname(lapplyApproach(df, windows)))
# TRUE

## test running time
library("rbenchmark")

## small example
benchmark(lapplyApproach(df, windows), forApproach(df, windows), order="relative")
#                         test replications elapsed relative user.self sys.self user.child sys.child
#2    forApproach(df, windows)          100   0.075     1.00     0.072        0          0         0                                                                          
#1 lapplyApproach(df, windows)          100   0.087     1.16     0.084        0          0         0

## a larger one
n <- 1e3
df <- data.frame(x=rnorm(1:n), y=rnorm(1:n), z=rnorm(1:n), u=rnorm(1:n))
windows <- embed(1:nrow(df), windowSize)
benchmark(lapplyApproach(df, windows), forApproach(df, windows), order="relative")
#                         test replications elapsed relative user.self sys.self user.child sys.child
#1 lapplyApproach(df, windows)          100  26.386    1.000    26.301    0.004          0         0                                                                          
#2    forApproach(df, windows)          100  26.932    1.021    26.838    0.000          0         0

##您的data.frame
df请展示您期望输出的样子。非常感谢！感谢：）关于如何“重命名”计算出的协方差矩阵，你有什么好的解决方案吗。例如，不是像“2011-01-01到2011 02-30”这样的$
3291
？最好的regards@user1665355如果问题解决了，请将问题标记为已回答。对于如何创建滚动窗口，而不是提前一天滚动，而是在间隔内滚动，您是否有任何清晰的建议。比如说，60天的数据用于创建协方差矩阵。然后，在60天后计算下一个cov.矩阵。再次感谢：）@user1665355增加
windowSize
？！
## your data.frame
df <- data.frame(x=c(1.5,2.3,4.7,3,8.4), y=c(5.3,2.4,8.4,1.3,2.5), z=c(2.5,1.3,6.5,4.3,2.8), u=c(1.1,2.5,4.3,2.5,6.3))

## define windows
windowSize <- 3
windows <- embed(1:nrow(df), windowSize)

lapplyApproach <- function(df, windows) {
    ## convert window matrix to a list
    windowsList <- split(t(windows), rep(1:nrow(windows), each=ncol(windows)))
    ## edit: customize names: "from:to"
    names(windowsList) <- unlist(lapply(windowsList, function(x)paste(range(x), sep="", collapse=":")))
    return(lapply(windowsList, function(x)cov(df[x, ])))
}

forApproach <- function(df, windows) {
    l <- vector(mode="list", length=nrow(windows))
    for (i in 1:nrow(windows)) {
        l[[i]] <- cov(df[windows[i, ], ])
    }
    return(l)
}

## check results
all.equal(forApproach(df, windows), unname(lapplyApproach(df, windows)))
# TRUE

## test running time
library("rbenchmark")

## small example
benchmark(lapplyApproach(df, windows), forApproach(df, windows), order="relative")
#                         test replications elapsed relative user.self sys.self user.child sys.child
#2    forApproach(df, windows)          100   0.075     1.00     0.072        0          0         0                                                                          
#1 lapplyApproach(df, windows)          100   0.087     1.16     0.084        0          0         0

## a larger one
n <- 1e3
df <- data.frame(x=rnorm(1:n), y=rnorm(1:n), z=rnorm(1:n), u=rnorm(1:n))
windows <- embed(1:nrow(df), windowSize)
benchmark(lapplyApproach(df, windows), forApproach(df, windows), order="relative")
#                         test replications elapsed relative user.self sys.self user.child sys.child
#1 lapplyApproach(df, windows)          100  26.386    1.000    26.301    0.004          0         0                                                                          
#2    forApproach(df, windows)          100  26.932    1.021    26.838    0.000          0         0