1

I want to get coefficient of regression for each data frame in a list of dataframes with a rolling period but somehow I am getting very different result from what I am looking for. I have tried the following code: my data looks like this 

library("zoo") ## for rollapply()
data <- list(mtcars,mtcars,mtcars)

fapplyFunction <- function(x){
coef(lm(mpg ~ drat, data=as.data.frame(x)))}

coef_list <- lapply(data, rollapply, 20, fapplyFunction, partial = FALSE, by.column = FALSE)

I wish to get regression result for each element for rolling windows as a list ,which I can later bind

I am new to R. Any help would be much appreciated.

philomath
  • 23
  • 6

1 Answers1

0

Providing a data.frame as the first rollapply argument will apply FUN to every column of the data.frame separately. Operating on data from two columns simultaneously can be achieved by moving a rolling window across the the sequence of row numbers in the data.frame.

lapply(data, function(x)
    rollapply(1:nrow(x), 20, function(i) coef(lm(mpg ~ drat, data = x[i, ]))))
#[[1]]
#      (Intercept)     drat
# [1,]   -11.70889 8.981350
# [2,]   -12.09923 9.124252
# [3,]   -11.47530 9.015324
# [4,]   -11.91551 9.124458
# [5,]   -12.51405 9.094820
# [6,]   -12.10843 8.994363
# [7,]   -15.57941 9.937651
# [8,]   -14.06719 9.511583
# [9,]   -14.42693 9.684131
#[10,]   -11.68393 8.789089
#[11,]   -12.12158 8.954089
#[12,]   -13.12850 9.243443
#[13,]   -12.81957 9.095040
# 
#[[2]]
#      (Intercept)     drat
# [1,]   -11.70889 8.981350
# [2,]   -12.09923 9.124252
# [3,]   -11.47530 9.015324
# [4,]   -11.91551 9.124458
# [5,]   -12.51405 9.094820
# [6,]   -12.10843 8.994363
# [7,]   -15.57941 9.937651
# [8,]   -14.06719 9.511583
# [9,]   -14.42693 9.684131
#[10,]   -11.68393 8.789089
#[11,]   -12.12158 8.954089
#[12,]   -13.12850 9.243443
#[13,]   -12.81957 9.095040
# 
#[[3]]
#      (Intercept)     drat
# [1,]   -11.70889 8.981350
# [2,]   -12.09923 9.124252
# [3,]   -11.47530 9.015324
# [4,]   -11.91551 9.124458
# [5,]   -12.51405 9.094820
# [6,]   -12.10843 8.994363
# [7,]   -15.57941 9.937651
# [8,]   -14.06719 9.511583
# [9,]   -14.42693 9.684131
#[10,]   -11.68393 8.789089
#[11,]   -12.12158 8.954089
#[12,]   -13.12850 9.243443
#[13,]   -12.81957 9.095040
Maurits Evers
  • 49,617
  • 4
  • 47
  • 68
  • @G.Grothendieck Yes, you're right, I was mistaken; I made an edit to clarify. It doesn't change the solution nor approach though. – Maurits Evers Aug 17 '18 at 13:29
  • rollapply only applies to each column separately if by.column = TRUE but the code in the question uses by.column = FALSE. – G. Grothendieck Aug 17 '18 at 13:51
  • @G.Grothendieck By default (and in my example) we have `by.column = TRUE`. The goal in the original question was to extract coefficients based on a linear model that considers rolling data from two columns. My solution does exactly that. Perhaps you want to post a different solution to illustrate an alternative approach. – Maurits Evers Aug 17 '18 at 14:02
  • @philomath Please consider closing the question by setting the green check mark next to the answer. That way you will help future SO readers identify relevant Q&As and keep SO tidy. – Maurits Evers Aug 20 '18 at 02:34
  • 1
    @MauritsEvers done. I am new to stack overflow.Thanks. – philomath Aug 20 '18 at 06:05