rolling regression with dplyr

问题

I have a dataframe of "date", "company" and "return", reproducible by the code below:

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")

With this dataframe, I can calculate the daily market average return and add that result into a new column "market.ret".

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

Now I want to group all my data by different companies (2 in this case).

x <- group_by(x, company)

After doing this, I would like to fit "return" by "market.ret" and calculate the linear regression coefficient and store the slopes in a new column. If I want to do the fitting for the whole data set within a given company, then I can simply call lm():

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

However, I actually want to do the linear regression on a "rolling" basis, i.e. for each day separately over a 20-day trailing period. I want to use rollapply() but do not know how to pass two columns into the function. Any help or suggestion is greatly appreciated.

Note: Below is the code that I used for calculating 20-day rolling standard deviation of returns which might be helpful:

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

回答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)

回答2:

Here is a dplyr solution

#####
# 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

It is very close to the answer here which is fairly close to this answer though using the rollRegres package.

来源：https://stackoverflow.com/questions/24770478/rolling-regression-with-dplyr

标签

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dplyr