Question

我有一个“日期”，“公司”和“返回”的数据框，可通过以下代码重现：

library(dplyr)
n.dates <- 60
n.stocks <- 2
date <- seq(as.Date("2011-07-01"), by=1, len=n.dates)
symbol <- replicate(n.stocks, paste0(sample(LETTERS, 5), collapse = ""))
x <- expand.grid(date, symbol)
x$return <- rnorm(n.dates*n.stocks, 0, sd = 0.05)
names(x) <- c("date", "company", "return")

通过这个数据框，我可以计算每日市场平均回报，并将该结果添加到新栏目“market.ret”。

x <- group_by(x, date)    
x <- mutate(x, market.ret = mean(x$return, na.rm = TRUE))

现在我想将不同公司的所有数据分组（在这种情况下为2）。

x <- group_by(x, company)

执行此操作后，我想通过“market.ret”拟合“返回”并计算线性回归系数并将斜率存储在新列中。如果我想对给定公司内的整个数据集进行拟合，那么我可以简单地调用lm（）：

group_by(x, company) %>%
do(data.frame(beta = coef(lm(return ~ market.ret,data = .))[2])) %>%
left_join(x,.)

然而，我实际上想要在“滚动”的基础上进行线性回归，即在20天的尾随期间分别进行每一天。我想使用rollapply（）但不知道如何将两列传递给函数。非常感谢任何帮助或建议。

注意：下面是我用来计算20天回滚标准差的代码，可能会有所帮助：

sdnoNA <- function(x){return(sd(x, na.rm = TRUE))}
x <- mutate(x, sd.20.0.d = rollapply(return, FUN = sdnoNA, width = 20, fill = NA))

Answer 1

## lms is a function which calculate the linear regression coefficient
lms <- function(y, x){
s = which(is.finite(x * y))
y = y[s]
x = x[s]
return(cov(x, y)/var(x))
}

## z is a dataframe which stores our final result
z <- data.frame()

## x has to be ungrouped
x <- ungroup(x)

## subset with "filter" and roll with "rollapply"
symbols <- unique(x$company)
for(i in 1:length(symbols)){
temp <- filter(x, company == symbols[i])
z <- rbind(z, mutate(temp, beta = rollapply(temp[, c(3, 4)], 
                                          FUN = function(x) lms(x[, 1], x[, 2]),
                                          width = 20, fill = NA,
                                          by.column = FALSE, align = "right")))
}

## final result
print(z)

Answer 2

这是一个dplyr解决方案

#####
# setup data as OP (notice the fix when computing the market return)
library(dplyr)
set.seed(41797642)
n.dates <- 60
n.stocks <- 2
date <- seq(as.Date("2011-07-01"), by=1, len=n.dates)
symbol <- replicate(n.stocks, paste0(sample(LETTERS, 5), collapse = ""))
x <- expand.grid(date, symbol)
x$return <- rnorm(n.dates*n.stocks, 0, sd = 0.05)
names(x) <- c("date", "company", "return")

x <- x %>%
  group_by(date) %>%
  mutate(market.ret = mean(return))

#####
# compute coefs using rollRegres
library(rollRegres)
func <- . %>% {
    roll_regres.fit(x = cbind(1, .$market.ret),
                    y = .$return, width = 20L)$coefs }
out <- x %>%
  group_by(company) %>%
  # make it explicit that data needs to be sorted
  arrange(date, .by_group = TRUE) %>%
  do(cbind(reg_col = select(., market.ret, return) %>% func,
           date_col = select(., date))) %>%
  ungroup

head(out[!is.na(out$reg_col.1), ], 5)
#R # A tibble: 5 x 4
#R company reg_col.1 reg_col.2 date
#R   <fct>       <dbl>     <dbl> <date>
#R 1 SNXAD    -0.0104      0.746 2011-07-20
#R 2 SNXAD    -0.00953     0.755 2011-07-21
#R 3 SNXAD    -0.0124      0.784 2011-07-22
#R 4 SNXAD    -0.0167      0.709 2011-07-23
#R 5 SNXAD    -0.0148      0.691 2011-07-24
tail(out[!is.na(out$reg_col.1), ], 5)
#R # A tibble: 5 x 4
#R company  reg_col.1 reg_col.2 date
#R   <fct>        <dbl>     <dbl> <date>
#R 1 UYLTS   -0.00276       0.837 2011-08-25
#R 2 UYLTS    0.0000438     0.928 2011-08-26
#R 3 UYLTS    0.000250      0.936 2011-08-27
#R 4 UYLTS   -0.000772      0.886 2011-08-28
#R 5 UYLTS    0.00173       0.902 2011-08-29

它非常接近the answer here，尽管使用rollRegres软件包也非常接近this answer。

用dplyr滚动回归

2 个答案: