Question

我正在尝试编写一个行为如下的函数，但它证明非常困难：

DF <- data.frame(x = seq(1,10), y = rep(c('a','b','c','d','e'),2))
> DF
    x y
1   1 a
2   2 b
3   3 c
4   4 d
5   5 e
6   6 a
7   7 b
8   8 c
9   9 d
10 10 e

>OverLapSplit(DF,nsplits=2,overlap=2)
[[1]]
  x y
1 1 a
2 2 b
3 3 c
4 4 d
5 5 e
6 6 a

[[2]]
   x y
1  5 a
2  6 b
3  7 c
4  8 d
5  9 e
6 10 a

>OverLapSplit(DF,nsplits=1)
[[1]]
    x y
1   1 a
2   2 b
3   3 c
4   4 d
5   5 e
6   6 a
7   7 b
8   8 c
9   9 d
10 10 e

>OverLapSplit(DF,nsplits=2,overlap=4)
[[1]]
  x y
1 1 a
2 2 b
3 3 c
4 4 d
5 5 e
6 6 a
7 7 b

[[2]]
   x y
1  4 e
2  5 a
3  6 b
4  7 c
5  8 d
6  9 e
7 10 a

>OverLapSplit(DF,nsplits=5,overlap=1)
[[1]]
  x y
1 1 a
2 2 b
3 3 c

[[2]]
  x y
1 3 c
2 4 d
3 5 e

[[3]]
  x y
1 5 e
2 6 a
3 7 b

[[4]]
  x y
1 7 b
2 8 c
3 9 d

[[5]]
   x y
1  8 d
2  9 e
3 10 f

如果您尝试OverLapSplit(DF,nsplits=2,overlap=1)

之类的内容，我还没有想过会发生什么

可能如下：

[[1]]
  x y
1 1 a
2 2 b
3 3 c
4 4 d
5 5 e

[[2]]
   x y
1  5 a
2  6 b
3  7 c
4  8 d
5  9 e
6 10 a

谢谢！

Answer 1

尝试类似：

OverlapSplit <- function(x,nsplit=1,overlap=2){
    nrows <- NROW(x)
    nperdf <- ceiling( (nrows + overlap*nsplit) / (nsplit+1) )
    start <- seq(1, nsplit*(nperdf-overlap)+1, by= nperdf-overlap )

    if( start[nsplit+1] + nperdf != nrows )
        warning("Returning an incomplete dataframe.")

    lapply(start, function(i) x[c(i:(i+nperdf-1)),])
}

用nsplit分割的数量！（nsplit = 1返回2个数据帧）。如果重叠拆分不真正适合数据帧，这将呈现不完整的最后一个数据帧，并发出警告。

> OverlapSplit(DF,nsplit=3,overlap=2)
[[1]]
  x y
1 1 a
2 2 b
3 3 c
4 4 d

[[2]]
  x y
3 3 c
4 4 d
5 5 e
6 6 a

[[3]]
  x y
5 5 e
6 6 a
7 7 b
8 8 c

[[4]]
    x y
7   7 b
8   8 c
9   9 d
10 10 e

一个有警告的人

> OverlapSplit(DF,nsplit=1,overlap=1)
[[1]]
  x y
1 1 a
2 2 b
3 3 c
4 4 d
5 5 e
6 6 a

[[2]]
    x    y
6   6    a
7   7    b
8   8    c
9   9    d
10 10    e
NA NA <NA>

Warning message:
In OverlapSplit(DF, nsplit = 1, overlap = 1) :
  Returning an incomplete dataframe.

Answer 2

这使用莱迪思图形中的木瓦理念，因此利用包lattice中的代码生成间隔，然后使用循环将原始DF分解为正确的子集。

我不确定overlap = 1是什么意思 - 我认为你的意思是1个样本/观察重叠。如果是这样，下面的代码就是这样做的。

OverlapSplit <- function(x, nsplits = 1, overlap = 0) {
    stopifnot(require(lattice))
    N <- seq_len(nr <- nrow(x))
    interv <- co.intervals(N, nsplits, overlap / nr)
    out <- vector(mode = "list", length = nrow(interv))
    for(i in seq_along(out)) {
        out[[i]] <- x[interv[i,1] < N & N < interv[i,2], , drop = FALSE]
    }
    out
}

给出了：

> OverlapSplit(DF, 2, 2)
[[1]]
  x y
1 1 a
2 2 b
3 3 c
4 4 d
5 5 e
6 6 a

[[2]]
    x y
5   5 e
6   6 a
7   7 b
8   8 c
9   9 d
10 10 e

> OverlapSplit(DF)
[[1]]
    x y
1   1 a
2   2 b
3   3 c
4   4 d
5   5 e
6   6 a
7   7 b
8   8 c
9   9 d
10 10 e

> OverlapSplit(DF, 4, 1)
[[1]]
  x y
1 1 a
2 2 b
3 3 c

[[2]]
  x y
3 3 c
4 4 d
5 5 e

[[3]]
  x y
6 6 a
7 7 b
8 8 c

[[4]]
    x y
8   8 c
9   9 d
10 10 e

Answer 3

只是为了清楚我在这里做了什么：

#Load Libraries
library(PerformanceAnalytics)
library(quantmod)

#Function to Split Data Frame
OverlapSplit <- function(x,nsplit=1,overlap=0){
    nrows <- NROW(x)
    nperdf <- ceiling( (nrows + overlap*nsplit) / (nsplit+1) )
    start <- seq(1, nsplit*(nperdf-overlap)+1, by= nperdf-overlap )

    if( start[nsplit+1] + nperdf != nrows )
        warning("Returning an incomplete dataframe.")

    lapply(start, function(i) x[c(i:(i+nperdf-1)),])
}

#Function to run regression on 30 days to predict the next day
FL <- as.formula(Next(HAM1)~HAM1+HAM2+HAM3+HAM4)
MyRegression <- function(df,FL) {
  df <- as.data.frame(df)
  model <- lm(FL,data=df[1:30,])
  predict(model,newdata=df[31,])
}

#Function to roll the regression
RollMyRegression <- function(data,ModelFUN,FL) {
  rollapply(data, width=31,FUN=ModelFUN,FL,
    by.column = FALSE, align = "right", na.pad = FALSE)
}

#Load Data
data(managers)

#Split Dataset
split.data <- OverlapSplit(managers,2,30)
sapply(split.data,dim)

#Run rolling regression on each split
output <- lapply(split.data,RollMyRegression,MyRegression,FL)
output
unlist(output)

通过这种方式，您可以使用并行版本的lapply替换lapply，并稍微提高您的速度。

当然，考虑到处理器的数量和数据集的大小，现在存在优化分割/重叠的问题。

将数据帧拆分为重叠的数据帧

3 个答案: