Create column with rolling linear model coefficients-CodePudding

I need to perform a rolling linear regression of 24 months between a variable and time and save the B1 coefficient on a column. I can do the rolling regression but I don't know how to save the B1 results in my original dataframe. Here is a reproducible example:

date <- seq(as.Date("2011-01-01"), as.Date("2012-01-01"), by="days")

set.seed(8)
df <- as.data.frame(date) %>%
  mutate(value = runif(366, min = 0, max = 100))

#Estimate rolling linear regression coef when t = 24
nr <- nrow(df)
reg <- function(ix) coef(lm(value ~ date, df, subset = ix)); rollapplyr(1:nr, 24, reg, fill = NA)

#How to save B1 results in a column of the data frame?

Does anybody knows how to do this? Thanks in advance!

CodePudding user response：

It's perhaps better practice to lm(.) on values instead of reaching out to df for it. Also, rollapplyr can work on a frame, row-wise, using by.column=FALSE, though one catch is that it converts everything to the highest class; to work with that, we need to preemptively convert date to numeric.

reg2 <- function(X) coef(lm(value ~ date, data = as.data.frame(X)))

reg2 <- function(X) setNames(coef(lm(value ~ date, data = data.frame(X))), c("intercept", "coef_date"))
df %>%
  bind_cols(rollapplyr(transmute(., value, date=as.numeric(date)),
                       24, FUN = reg2, by.column = FALSE, fill = NA)) %>%
  tail()
#           date    value intercept  coef_date
# 361 2011-12-27 10.48895 -7016.084  0.4613276
# 362 2011-12-28 45.51299 -1602.472  0.1080812
# 363 2011-12-29 35.17561 -2635.868  0.1753899
# 364 2011-12-30 44.89097 -8457.421  0.5550818
# 365 2011-12-31 58.28106 -3354.462  0.2222435
# 366 2012-01-01 22.91829  6228.027 -0.4029083

CodePudding user response：

Assuming that B1 means the slope of the regression line:

df %>% 
  mutate(slope = rollapplyr(1:nr, 24, function(x) reg(ix)[[2]], fill = NA))