在给定时间段内计算 returns - R
Calculating returns over given periods - R
我目前正在尝试计算不同时间范围（1、5、20、50、200、250 天）内的股票 returns，但我还找不到方便的解决方案。据我所知，Quantmod 仅提供预设 returns。
因此，我使用了一个解决方案 ，我修改了它以获得 returns，但与以下函数没有区别：
sret = function(x,n){(apply(lag(zoo(x), c(-n,0), na.pad = TRUE), 1L, diff)/lag(zoo(x), c(-n,0)))}
我现在的问题是我不能在 xts 系列中使用它，因为结果显然计算了两个值：一个具有预期的 n-lag，另一个具有分母中的当前值。有趣的是，只有正确的值显示在数据框中。所以我计算如下：
#Calculate returns
cCDAX$R1 = sret(cCDAX$Close, 1)
cCDAX$R5 = sret(cCDAX$Close, 5)
这给了我以下值：
Date Close Volume R1 R5
1 2010-01-04 523.96 137055000 NA NA
2 2010-01-05 523.64 168916800 -0.0006107336 NA
3 2010-01-06 524.33 145659600 0.0013176992 NA
4 2010-01-07 523.83 182195400 -0.0009535979 NA
5 2010-01-08 525.55 214804700 0.0032835080 NA
6 2010-01-11 525.93 189962700 0.0007230520 3.759829e-03
7 2010-01-12 517.59 191580300 -0.0158576236 -1.155374e-02
8 2010-01-13 519.71 185076700 0.0040959060 -8.811245e-03
9 2010-01-14 522.48 167065200 0.0053298955 -2.577172e-03
10 2010-01-15 513.14 208268000 -0.0178762823 -2.361336e-02
11 2010-01-18 516.37 112098400 0.0062945785 -1.817732e-02
12 2010-01-19 520.56 159323200 0.0081143366 5.738132e-03
13 2010-01-20 510.21 167641400 -0.0198824343 -1.827943e-02
14 2010-01-21 501.77 190062800 -0.0165422081 -3.963788e-02
15 2010-01-22 496.67 240544400 -0.0101640194 -3.209650e-02
16 2010-01-25 491.91 199198900 -0.0095838283 -4.736913e-02
17 2010-01-26 494.76 188213100 0.0057937428 -4.956201e-02
18 2010-01-27 492.25 193048200 -0.0050731668 -3.520119e-02
19 2010-01-28 484.26 229885500 -0.0162315896 -3.489647e-02
20 2010-01-29 489.82 252945300 0.0114814356 -1.379185e-02
当我直接在控制台中输入公式时，每天的returns如下所示：
lag-1 lag0
1 NA NA
2 -0.0006107336 -0.0006111069
3 0.0013176992 0.0013159651
4 -0.0009535979 -0.0009545081
5 0.0032835080 0.0032727619
显然，由于这些值（即使不是这样出现）实际上有两个值，所以我无法在之后将它们转换为 xts 对象。没有 xts 对象，我无法 运行 我的时间序列分析。问题肯定是分母，但我需要 c(-n, 0) 才能得到正确的计算。我尝试了多种方式，比如
sret = function(x,n){(apply(lag(zoo(x), c(-n,0), na.pad = TRUE), 1L, diff)/lag(zoo(x), c(-n)))}
sret = function(x,n){(apply(lag(zoo(x), c(-n,0), na.pad = TRUE), 1L, diff)/lag(zoo(x), n))}
sret = function(x,n){(apply(lag(zoo(x), c(-n,0), na.pad = TRUE), 1L, diff)/lag(x, c(-n,0)))}
sret = function(x,n){(apply(lag(zoo(x), c(-n,0), na.pad = TRUE), 1L, diff)/lag(x, n))}
并没有真正奏效，所以上面的版本（也在下面）是唯一提供正确值的版本，但是之后无法处理......有没有人有解决方案来抑制或延迟0 滞后？
sret = function(x,n){(apply(lag(zoo(x), c(-n,0), na.pad = TRUE), 1L, diff)/lag(zoo(x), c(-n,0)))}
在任何操作之前，带有 dput(head(cCDAX, 20)) 的 cCDAX 的输出如下所示：
> dput(head(cCDAX, 20))
structure(list(Date = structure(c(1262559600, 1262646000, 1262732400,
1262818800, 1262905200, 1263164400, 1263250800, 1263337200, 1263423600,
1263510000, 1263769200, 1263855600, 1263942000, 1264028400, 1264114800,
1264374000, 1264460400, 1264546800, 1264633200, 1264719600), class = c("POSIXct",
"POSIXt"), tzone = ""), Close = c(523.96, 523.64, 524.33, 523.83,
525.55, 525.93, 517.59, 519.71, 522.48, 513.14, 516.37, 520.56,
510.21, 501.77, 496.67, 491.91, 494.76, 492.25, 484.26, 489.82
), Volume = c(137055000L, 168916800L, 145659600L, 182195400L,
214804700L, 189962700L, 191580300L, 185076700L, 167065200L, 208268000L,
112098400L, 159323200L, 167641400L, 190062800L, 240544400L, 199198900L,
188213100L, 193048200L, 229885500L, 252945300L)), row.names = c(NA,
20L), class = "data.frame")
我目前运行宁的过程如下（我省略了不同的滞后）：
library(vars)
library(fpp2)
library(nortest)
library(ggpubr)
library(xts)
library(highfrequency)
library(quantmod)
library(pracma)
library(zoo)
# clear all
rm(list=ls())
#Moving Average Function
mav = function(x,n){filter(x, rep(1/n,n), sides = 1)}
#Standard Deviation Function
vari = function(x,n){rollapply(x, width = n, FUN = sd, fill = NA, align = c("right"))}
#Return function
sret = function(x,n){(apply(lag(zoo(x), c(-n,0), na.pad = TRUE), 1L, diff)/lag(zoo(x), c(-n,0)))}
#Loading data, transfering the Date column in an actual date
cCDAX = read.csv("./CDAX_Clean.csv", header=TRUE, sep=",", dec=".")
cCDAX$Date = as.POSIXct(cCDAX$Date, format = "%d.%m.%Y")
#Adding moving averages for the closing prices
cCDAX$MA5C = mav(cCDAX[,"Close"], 5)
#Calculate standard deviations for the closing prices
cCDAX$SD5C = vari(cCDAX $Close, 5)
#Calculate returns
cCDAX$R1 = sret(cCDAX$Close, 1)
cCDAX$R5 = sret(cCDAX$Close, 5)
#Calculate standard deviation of returns
cCDAX$SD5R = vari(cCDAX $R1, 5)
#Adding moving averages for the daily volumes
cCDAX$MA5V = mav(cCDAX[,"Volume"], 5)
#Calculate standard deviations for the closing prices
cCDAX$SD5V = vari(cCDAX$Volume, 5)
#Calculate change in daily volume
cCDAX$VC1 = sret(cCDAX$Volume, 1)
cCDAX$VC5 = sret(cCDAX$Volume, 5)
#Calculate standard deviation of volume change
cCDAX$SD5VC = vari(cCDAX $VC1, 5)
#Creating a time series; with omitted variables should be [,2:13] instead of [,2:45]
CDAX_ts = as.xts(cCDAX[,2:45], order.by = cCDAX[,1])
我可能遗漏了一些东西，但下面的代码是问题要求的吗？它为 class "xts" 的对象定义了一个泛型 sret 和一个方法。 class "zoo"（但 xts）对象的方法可以用相同的方式定义。那么调用函数就可以了
library(xts)
library(quantmod)
sret <- function(x, ...) UseMethod("sret", x)
sret.default <- function(x, n = 1){
m <- length(x)
y <- rep(NA_real_, m)
y[(n + 1):m] <- x[seq_len(m - n)]
(x - y)/y
}
sret.data.frame <- function(x, n){
i <- sapply(x, is.numeric)
x[i] <- lapply(x[i], sret.default, n = n)
x
}
sret.xts <- function(x, n = 1){
y <- lag(x, n, na.pad = TRUE)
(x - y)/y
}
getSymbols("AAPL", from = Sys.Date() - 20)
sret(AAPL, 5)
sret(AAPL$AAPL.Open, 2)
# AAPL.Open
#2021-03-15 NA
#2021-03-16 NA
#2021-03-17 0.021281733
#2021-03-18 -0.022949219
#2021-03-19 -0.034612185
#2021-03-22 -0.021191681
#2021-03-23 0.027811562
#2021-03-24 0.020273555
#2021-03-25 -0.031704877
#2021-03-26 -0.020523490
#2021-03-29 0.017344850
#2021-03-30 -0.001998143
#2021-03-31 0.000000000
#2021-04-01 0.028707770
1) diff.zoo 和 diff.xts 有一个默认为 TRUE 的 arithmetic= 参数，但如果为 FALSE，则采用比率而不是差值.它也可以同时用于所有列。
library(quantmod)
getSymbols("FB")
k <- 2
diff(Ad(FB), k, arith = FALSE) - 1 # returns over k days of Adjusted Close
diff(FB, k, arith = FALSE) - 1 # returns over k days for each column
2) 这也有效并且仅使用基数 R。如果不需要填充，请在示例中省略 NA 部分。 k 是需要 return 的天数。
xx <- c(5, 3, 6, 2, 1)
k <- 2
c(rep(NA, k), exp(diff(log(xx), 2)) - 1)
或一次在多个列上。 BOD是R自带的数据框
rbind(matrix(NA, k, ncol(BOD)), exp(diff(log(as.matrix(BOD)), k)) - 1)
我目前正在尝试计算不同时间范围（1、5、20、50、200、250 天）内的股票 returns，但我还找不到方便的解决方案。据我所知，Quantmod 仅提供预设 returns。
因此，我使用了一个解决方案
sret = function(x,n){(apply(lag(zoo(x), c(-n,0), na.pad = TRUE), 1L, diff)/lag(zoo(x), c(-n,0)))}
我现在的问题是我不能在 xts 系列中使用它，因为结果显然计算了两个值：一个具有预期的 n-lag，另一个具有分母中的当前值。有趣的是，只有正确的值显示在数据框中。所以我计算如下：
#Calculate returns
cCDAX$R1 = sret(cCDAX$Close, 1)
cCDAX$R5 = sret(cCDAX$Close, 5)
这给了我以下值：
Date Close Volume R1 R5
1 2010-01-04 523.96 137055000 NA NA
2 2010-01-05 523.64 168916800 -0.0006107336 NA
3 2010-01-06 524.33 145659600 0.0013176992 NA
4 2010-01-07 523.83 182195400 -0.0009535979 NA
5 2010-01-08 525.55 214804700 0.0032835080 NA
6 2010-01-11 525.93 189962700 0.0007230520 3.759829e-03
7 2010-01-12 517.59 191580300 -0.0158576236 -1.155374e-02
8 2010-01-13 519.71 185076700 0.0040959060 -8.811245e-03
9 2010-01-14 522.48 167065200 0.0053298955 -2.577172e-03
10 2010-01-15 513.14 208268000 -0.0178762823 -2.361336e-02
11 2010-01-18 516.37 112098400 0.0062945785 -1.817732e-02
12 2010-01-19 520.56 159323200 0.0081143366 5.738132e-03
13 2010-01-20 510.21 167641400 -0.0198824343 -1.827943e-02
14 2010-01-21 501.77 190062800 -0.0165422081 -3.963788e-02
15 2010-01-22 496.67 240544400 -0.0101640194 -3.209650e-02
16 2010-01-25 491.91 199198900 -0.0095838283 -4.736913e-02
17 2010-01-26 494.76 188213100 0.0057937428 -4.956201e-02
18 2010-01-27 492.25 193048200 -0.0050731668 -3.520119e-02
19 2010-01-28 484.26 229885500 -0.0162315896 -3.489647e-02
20 2010-01-29 489.82 252945300 0.0114814356 -1.379185e-02
当我直接在控制台中输入公式时，每天的returns如下所示：
lag-1 lag0
1 NA NA
2 -0.0006107336 -0.0006111069
3 0.0013176992 0.0013159651
4 -0.0009535979 -0.0009545081
5 0.0032835080 0.0032727619
显然，由于这些值（即使不是这样出现）实际上有两个值，所以我无法在之后将它们转换为 xts 对象。没有 xts 对象，我无法 运行 我的时间序列分析。问题肯定是分母，但我需要 c(-n, 0) 才能得到正确的计算。我尝试了多种方式，比如
sret = function(x,n){(apply(lag(zoo(x), c(-n,0), na.pad = TRUE), 1L, diff)/lag(zoo(x), c(-n)))}
sret = function(x,n){(apply(lag(zoo(x), c(-n,0), na.pad = TRUE), 1L, diff)/lag(zoo(x), n))}
sret = function(x,n){(apply(lag(zoo(x), c(-n,0), na.pad = TRUE), 1L, diff)/lag(x, c(-n,0)))}
sret = function(x,n){(apply(lag(zoo(x), c(-n,0), na.pad = TRUE), 1L, diff)/lag(x, n))}
并没有真正奏效，所以上面的版本（也在下面）是唯一提供正确值的版本，但是之后无法处理......有没有人有解决方案来抑制或延迟0 滞后？
sret = function(x,n){(apply(lag(zoo(x), c(-n,0), na.pad = TRUE), 1L, diff)/lag(zoo(x), c(-n,0)))}
在任何操作之前，带有 dput(head(cCDAX, 20)) 的 cCDAX 的输出如下所示：
> dput(head(cCDAX, 20))
structure(list(Date = structure(c(1262559600, 1262646000, 1262732400,
1262818800, 1262905200, 1263164400, 1263250800, 1263337200, 1263423600,
1263510000, 1263769200, 1263855600, 1263942000, 1264028400, 1264114800,
1264374000, 1264460400, 1264546800, 1264633200, 1264719600), class = c("POSIXct",
"POSIXt"), tzone = ""), Close = c(523.96, 523.64, 524.33, 523.83,
525.55, 525.93, 517.59, 519.71, 522.48, 513.14, 516.37, 520.56,
510.21, 501.77, 496.67, 491.91, 494.76, 492.25, 484.26, 489.82
), Volume = c(137055000L, 168916800L, 145659600L, 182195400L,
214804700L, 189962700L, 191580300L, 185076700L, 167065200L, 208268000L,
112098400L, 159323200L, 167641400L, 190062800L, 240544400L, 199198900L,
188213100L, 193048200L, 229885500L, 252945300L)), row.names = c(NA,
20L), class = "data.frame")
我目前运行宁的过程如下（我省略了不同的滞后）：
library(vars)
library(fpp2)
library(nortest)
library(ggpubr)
library(xts)
library(highfrequency)
library(quantmod)
library(pracma)
library(zoo)
# clear all
rm(list=ls())
#Moving Average Function
mav = function(x,n){filter(x, rep(1/n,n), sides = 1)}
#Standard Deviation Function
vari = function(x,n){rollapply(x, width = n, FUN = sd, fill = NA, align = c("right"))}
#Return function
sret = function(x,n){(apply(lag(zoo(x), c(-n,0), na.pad = TRUE), 1L, diff)/lag(zoo(x), c(-n,0)))}
#Loading data, transfering the Date column in an actual date
cCDAX = read.csv("./CDAX_Clean.csv", header=TRUE, sep=",", dec=".")
cCDAX$Date = as.POSIXct(cCDAX$Date, format = "%d.%m.%Y")
#Adding moving averages for the closing prices
cCDAX$MA5C = mav(cCDAX[,"Close"], 5)
#Calculate standard deviations for the closing prices
cCDAX$SD5C = vari(cCDAX $Close, 5)
#Calculate returns
cCDAX$R1 = sret(cCDAX$Close, 1)
cCDAX$R5 = sret(cCDAX$Close, 5)
#Calculate standard deviation of returns
cCDAX$SD5R = vari(cCDAX $R1, 5)
#Adding moving averages for the daily volumes
cCDAX$MA5V = mav(cCDAX[,"Volume"], 5)
#Calculate standard deviations for the closing prices
cCDAX$SD5V = vari(cCDAX$Volume, 5)
#Calculate change in daily volume
cCDAX$VC1 = sret(cCDAX$Volume, 1)
cCDAX$VC5 = sret(cCDAX$Volume, 5)
#Calculate standard deviation of volume change
cCDAX$SD5VC = vari(cCDAX $VC1, 5)
#Creating a time series; with omitted variables should be [,2:13] instead of [,2:45]
CDAX_ts = as.xts(cCDAX[,2:45], order.by = cCDAX[,1])
我可能遗漏了一些东西，但下面的代码是问题要求的吗？它为 class "xts" 的对象定义了一个泛型 sret 和一个方法。 class "zoo"（但 xts）对象的方法可以用相同的方式定义。那么调用函数就可以了
library(xts)
library(quantmod)
sret <- function(x, ...) UseMethod("sret", x)
sret.default <- function(x, n = 1){
m <- length(x)
y <- rep(NA_real_, m)
y[(n + 1):m] <- x[seq_len(m - n)]
(x - y)/y
}
sret.data.frame <- function(x, n){
i <- sapply(x, is.numeric)
x[i] <- lapply(x[i], sret.default, n = n)
x
}
sret.xts <- function(x, n = 1){
y <- lag(x, n, na.pad = TRUE)
(x - y)/y
}
getSymbols("AAPL", from = Sys.Date() - 20)
sret(AAPL, 5)
sret(AAPL$AAPL.Open, 2)
# AAPL.Open
#2021-03-15 NA
#2021-03-16 NA
#2021-03-17 0.021281733
#2021-03-18 -0.022949219
#2021-03-19 -0.034612185
#2021-03-22 -0.021191681
#2021-03-23 0.027811562
#2021-03-24 0.020273555
#2021-03-25 -0.031704877
#2021-03-26 -0.020523490
#2021-03-29 0.017344850
#2021-03-30 -0.001998143
#2021-03-31 0.000000000
#2021-04-01 0.028707770
1) diff.zoo 和 diff.xts 有一个默认为 TRUE 的 arithmetic= 参数，但如果为 FALSE，则采用比率而不是差值.它也可以同时用于所有列。
library(quantmod)
getSymbols("FB")
k <- 2
diff(Ad(FB), k, arith = FALSE) - 1 # returns over k days of Adjusted Close
diff(FB, k, arith = FALSE) - 1 # returns over k days for each column
2) 这也有效并且仅使用基数 R。如果不需要填充，请在示例中省略 NA 部分。 k 是需要 return 的天数。
xx <- c(5, 3, 6, 2, 1)
k <- 2
c(rep(NA, k), exp(diff(log(xx), 2)) - 1)
或一次在多个列上。 BOD是R自带的数据框
rbind(matrix(NA, k, ncol(BOD)), exp(diff(log(as.matrix(BOD)), k)) - 1)