Conduct rolling regression in windows of 5 years for each group in R

Question

I have data with variables industry_sales , year , and industry .

I want to conduct time series regressions for each industry in rolling windows of 5 years, with the dependent variable being industry_sales and the independent variable being year . For example, if data ranges from years 2000-2010, then the rolling windows for each regression and industry group would be 2000-2004, 2001-2005, and so on.

I had an attempt but the code fails:

library(tidyverse)
library(zoo)

# Group the data by industry
data_grouped <- data %>% 
  group_by(industry) 
glimpse(data_grouped)

# Define a function to run a regression
regression_func <- function(x) {
  lm(industry_sales ~ year, data = x)
}

# Perform rolling linear regression on each group, with a window of 5 years
results_list <- data_grouped %>% 
  do({
    rollapply(data_grouped$year, width = 5, FUN = regression_func, by = 1)
})

dput() output of the data is as below, any help would be appreciated:

structure(list(year = c(2010, 2011, 2012, 2013, 2014, 2015, 2016, 
2017, 2018, 2019, 2020, 2021, 2022, 2010, 2011, 2012, 2013, 2014, 
2015, 2016, 2017, 2018, 2019, 2020, 2021, 2022, 2010, 2011, 2012, 
2013, 2014, 2015, 2016, 2017, 2018, 2019, 2020, 2021, 2022, 2010, 
2011, 2012, 2013, 2014, 2015, 2016, 2017, 2018, 2019, 2020, 2021, 
2022, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017, 2018, 2019, 
2020, 2021, 2022, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017, 
2018, 2019, 2020, 2021, 2022, 2010, 2011, 2012, 2013, 2014, 2015, 
2016, 2017, 2018, 2019, 2020, 2021, 2022, 2010, 2011, 2012, 2013, 
2014, 2015, 2016, 2017, 2018, 2019, 2020, 2021, 2022), industry_sales = c(853.218, 
1001.856, 987.741, 990.679, 1061.804, 985.904, 1073.43, 1305.779, 
1314.722, 1188.122, 1327.245, 1711.143, 577.358, 25731.969, 26484.263, 
23085.259, 24842.509, 30563.672, 34947.361, 40066.425, 38293.972, 
40484.096, 37160.342, 29109.755, 35227.521, 16522.915, 1072310.255, 
1314543.982, 1288932.701, 1245467.184, 1169743.543, 882325.562, 
804108.99, 966832.252, 1099655.803, 999685.99, 725094.688, 797072.967, 
59906.738, 131274.107, 167635.266, 170003.001, 161892.01, 149722.289, 
108169.702, 92950.89, 120837.727, 132130.143, 121670.84, 118206.596, 
147732.623, 542.17, 201842.321, 209721.793, 215722.823, 223158.977, 
221191.213, 223076.061, 229199.539, 242995.929, 256145.995, 257282.085, 
234455.443, 214560.191, 13964.787, 88190.421, 90191.422, 80848.451, 
93999.434, 92325.688, 99138.466, 98682.173, 95106.591, 97622.097, 
94254.621, 81145.338, 87421.329, 12116.901, 247825.263, 262355, 
235423.37, 236697.987, 245287.766, 222221.931, 219657.82, 222990.939, 
224672.689, 162633.722, 139853.429, 148534.588, 4463.853, 28177.129, 
29298.618, 29251.077, 31696.968, 31636.033, 33064.57, 35745.408, 
32034.079, 31978.646, 31661.88, 28572.857, 34190.976, 3400.497
), industry = c("Agriculture, Forestry and Fishing", "Agriculture, Forestry and Fishing", 
"Agriculture, Forestry and Fishing", "Agriculture, Forestry and Fishing", 
"Agriculture, Forestry and Fishing", "Agriculture, Forestry and Fishing", 
"Agriculture, Forestry and Fishing", "Agriculture, Forestry and Fishing", 
"Agriculture, Forestry and Fishing", "Agriculture, Forestry and Fishing", 
"Agriculture, Forestry and Fishing", "Agriculture, Forestry and Fishing", 
"Agriculture, Forestry and Fishing", "Construction", "Construction", 
"Construction", "Construction", "Construction", "Construction", 
"Construction", "Construction", "Construction", "Construction", 
"Construction", "Construction", "Construction", "Manufacturing", 
"Manufacturing", "Manufacturing", "Manufacturing", "Manufacturing", 
"Manufacturing", "Manufacturing", "Manufacturing", "Manufacturing", 
"Manufacturing", "Manufacturing", "Manufacturing", "Manufacturing", 
"Mining", "Mining", "Mining", "Mining", "Mining", "Mining", "Mining", 
"Mining", "Mining", "Mining", "Mining", "Mining", "Mining", "Retail Trade", 
"Retail Trade", "Retail Trade", "Retail Trade", "Retail Trade", 
"Retail Trade", "Retail Trade", "Retail Trade", "Retail Trade", 
"Retail Trade", "Retail Trade", "Retail Trade", "Retail Trade", 
"Services", "Services", "Services", "Services", "Services", "Services", 
"Services", "Services", "Services", "Services", "Services", "Services", 
"Services", "Transportation, Communications, Electric, Gas and Sanitary service", 
"Transportation, Communications, Electric, Gas and Sanitary service", 
"Transportation, Communications, Electric, Gas and Sanitary service", 
"Transportation, Communications, Electric, Gas and Sanitary service", 
"Transportation, Communications, Electric, Gas and Sanitary service", 
"Transportation, Communications, Electric, Gas and Sanitary service", 
"Transportation, Communications, Electric, Gas and Sanitary service", 
"Transportation, Communications, Electric, Gas and Sanitary service", 
"Transportation, Communications, Electric, Gas and Sanitary service", 
"Transportation, Communications, Electric, Gas and Sanitary service", 
"Transportation, Communications, Electric, Gas and Sanitary service", 
"Transportation, Communications, Electric, Gas and Sanitary service", 
"Transportation, Communications, Electric, Gas and Sanitary service", 
"Wholesale Trade", "Wholesale Trade", "Wholesale Trade", "Wholesale Trade", 
"Wholesale Trade", "Wholesale Trade", "Wholesale Trade", "Wholesale Trade", 
"Wholesale Trade", "Wholesale Trade", "Wholesale Trade", "Wholesale Trade", 
"Wholesale Trade")), row.names = c(NA, -104L), class = c("tbl_df", 
"tbl", "data.frame"))

Answer 1

Here's a solution using slider::slide2() instead of zoo::rollapply() . This produces a named, nested list with results for 5-yr periods nested within industries.

library(tidyverse)
library(slider)

rolling_lm <- function(x, y) {
  res <- slide2(x, y, \(x, y) lm(y ~ x), .after = 5, .complete = TRUE)
  names(res) <- slide_chr(x, \(x) paste(min(x), "-", max(x)), .after = 5, .complete = TRUE)
  res[!is.null(res)]
}

results_list <- data %>%
  split(.$industry) %>%
  map(\(grp) rolling_lm(grp$year, grp$industry_sales))

# preview first 3 results
results_list[1:3]

$`Agriculture, Forestry and Fishing`
$`Agriculture, Forestry and Fishing`$`2010 - 2015`

Call:
lm(formula = y ~ x)

Coefficients:
(Intercept)            x  
  -47676.99        24.18  


$`Agriculture, Forestry and Fishing`$`2011 - 2016`

Call:
lm(formula = y ~ x)

Coefficients:
(Intercept)            x  
   -23345.5         12.1  


$`Agriculture, Forestry and Fishing`$`2012 - 2017`

Call:
lm(formula = y ~ x)

Coefficients:
(Intercept)            x  
 -100379.38        50.36  

Note, this assumes there's one row per year with no missing or duplicated years. If that's not the case, slide2_index() or slide2_period() might be more useful; see the introductory vi.nette .

Answer 2

There are several problems:

• The code in the question is passing a sub-vector of data$year, the window, to the regression function but the regression function is defined there to accept a data frame argument, not a vector of years.
• it is not clear what the ultimate output desired is. Presumably it is a data frame with industry, year and various statistics as columns or if there is only one statistic maybe a 2d result with years as rows and industry as columns. See below for some ways to get these.

1) We will assume that for each regression we want one row with the industry, last year of the window, intercept, slope and R^2. Modify c(...) to change which statistics you want. For example, c(year = tail(yr, 1), unlist(broom::glance(fm))) if you wanted a large number of statistics.

This groups by industry and then uses group_modify to create the rows for each industry.

library(dplyr, exclude = c("filter", "lag"))
library(zoo)

out <- data %>%
  group_by(industry) %>%
  group_modify(~ {
    .x$year %>%
       rollapplyr(5, function(yr) {
         fm <- lm(industry_sales ~ year, .x, subset = year %in% yr)
         c(year = tail(yr, 1), intercept = coef(fm)[[1]], slope = coef(fm)[[2]], 
           r.squared = summary(fm)$r.squared)
       }) %>%
       as.data.frame
  }) %>% 
  ungroup

giving:

> out
# A tibble: 72 × 5
   industry                           year intercept  slope r.squared
   <chr>                             <dbl>     <dbl>  <dbl>     <dbl>
 1 Agriculture, Forestry and Fishing  2014   -80707.  40.6     0.704 
 2 Agriculture, Forestry and Fishing  2015    -7481.   4.22    0.0433
 3 Agriculture, Forestry and Fishing  2016   -32534.  16.7     0.362 
 4 Agriculture, Forestry and Fishing  2017  -128244.  64.2     0.605 
 5 Agriculture, Forestry and Fishing  2018  -165315.  82.6     0.741 
 6 Agriculture, Forestry and Fishing  2019  -129070.  64.6     0.503 
 7 Agriculture, Forestry and Fishing  2020   -77455.  39.0     0.317 
 8 Agriculture, Forestry and Fishing  2021  -164845.  82.3     0.428 
 9 Agriculture, Forestry and Fishing  2022   193469. -95.2     0.134 
10 Construction                       2014 -1587815. 802.      0.208 
# … with 62 more rows
# ℹ Use `print(n = ...)` to see more rows

2) If you just wanted a single statistic -- the slope, say -- then it might be more convenient to have a 2d zoo series with one column per industry and one row per year. Note that the slope is shift invariant so we can just use 1:5 as the predictor in each case and we will get the same slope as if we had used year.

library(zoo)
z <- read.zoo(data, split = "industry")
out2 <- rollapplyr(z, 5, function(x) coef(lm(x ~ seq(5)))[[2]])

giving:

> out2[1:3, 1:3]
     Agriculture, Forestry and Fishing Construction Manufacturing
2014                           40.5995     802.1652      12578.98
2015                            4.2159    2440.4609     -98362.60
2016                           16.6603    4406.7184    -133278.90

If you wanted this as a data frame use fortify.zoo(out2) or in long form fortify.zoo(out2, melt = TRUE) .