lm() called within mutate()

2020-04-07 05:51发布

I wonder if it is possible to use lm() within mutate() of dplyr package. Currently I have a dataframe of "date", "company", "return" and "market.ret" reproducible as 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$return <- rnorm(n.dates*n.stocks, 0, sd = 0.05)
names(x) <- c("date", "company", "return")
x <- group_by(x, date)    
x <- mutate(x, market.ret = mean(x$return, na.rm = TRUE))

Now for each company I would like to fit "return" by "market.ret", calculate the linear regression coefficient and store the slopes in a new column. I wish to do it with mutate(), but the below code does not work:

x <- group_by(x, company)
x <- mutate(x, beta = coef(lm(x$return~x$market.ret))[[2]])

The error reported by R is:

Error in terms.formula(formula, data = data) : 
invalid term in model formula

Thanks in advance for any suggestion!

标签: r lm dplyr
2条回答
【Aperson】
2楼-- · 2020-04-07 06:26

You seem to want to calculate a daily market return across all companies, and then regress return vs. market return for each company, across all days. If so, here's a solution using data.table; likely to be faster with very large datasets.

library(data.table) ## 1.9.2+
setDT(x)[ , market.ret := mean(return), by = date]
x[, beta := coef(lm(return ~ market.ret, data = .SD))[[2]], by = company]

where x is as shown below (using set.seed for reproducibility):

set.seed(1L)     # for reproducible example
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")
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姐就是有狂的资本
3楼-- · 2020-04-07 06:29

This seems to work for me:

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