Aggregate by multiple groups and keep the result for each level in R

Question

Is there any possibility to get expected result in simplest way (without intermediate steps, as I did in my code)? I was looking for it and couldn't find any solution... So, for example: df <- data.frame( A = c("01","01","01","01","01","01","02","02","02","02","02","02"), B = c("12","12","12","12","14","14","18","18","18","22","22","22"), C = c("01","01","02","02","01","02","01","02","05","01","02","02"), D = c("1","2","1","2","2","1","2","1","3","2","1","3"), value_1 = c(25,14,1,15,0,15,0,16,18,74,112,36), value_2 = c(2,0,48,12,0,47,95,14,188,65,14,47), value_3 = c(78,5,14,26,9,0,4,0,0,8,0,4)) I would like to summarise values for each level (A,B,C,D) and combination of them to show the sum of each level in the final table. A <- df %>% group_by(A) %>% summarise(across(value_1:value_3, sum)) B <- df %>% group_by(A,B) %>% summarise(across(value_1:value_3, sum)) C <- df %>% group_by(A,B,C) %>% summarise(across(value_1:value_3, sum)) AD <- df %>% group_by(A,D) %>% summarise(across(value_1:value_3, sum)) BD <- df %>% group_by(A,B,D) %>% summarise(across(value_1:value_3, sum)) result <-bind_rows(A, B, C, AD, BD) result[is.na(result)] <- "00" result <- result[,c("A","B","C","D", "value_1", "value_2", "value_3")] result <- result[order(result$A, result$B, result$C, result$D),] My data has much more levels, so how could I do it without making intermediate steps like: B, C.. to get the result like this: > head(result) # A tibble: 6 x 7 A B C D value_1 value_2 value_3 1 01 00 00 00 70 109 132 2 01 00 00 1 41 97 92 3 01 00 00 2 29 12 40 4 01 12 00 00 55 62 123 5 01 12 00 1 26 50 92 6 01 12 00 2 29 12 31 Is there any possibility to do it with .groups argument or using aggregate function instead of combination group_by and summarise? JavaScript questions and answers, JavaScript questions pdf, JavaScript question bank, JavaScript questions and answers pdf, mcq on JavaScript pdf, JavaScript questions and solutions, JavaScript mcq Test , Interview JavaScript questions, JavaScript Questions for Interview, JavaScript MCQ (Multiple Choice Questions)

Answer 1

df <- data.frame( A = c("01","01","01","01","01","01","02","02","02","02","02","02"), B = c("12","12","12","12","14","14","18","18","18","22","22","22"), C = c("01","01","02","02","01","02","01","02","05","01","02","02"), D = c("1","2","1","2","2","1","2","1","3","2","1","3"), value_1 = c(25,14,1,15,0,15,0,16,18,74,112,36), value_2 = c(2,0,48,12,0,47,95,14,188,65,14,47), value_3 = c(78,5,14,26,9,0,4,0,0,8,0,4)) grp_vars <- list("A", c("A", "B"), c("A", "B", "C")) library(tidyverse) map_df(grp_vars, ~ df %>% group_by(across(.x)) %>% summarise(across(value_1:value_3, sum), .groups = "drop")) %>% mutate(across(where(is.character), ~ replace_na(.x, replace = "00"))) %>% relocate(where(is.numeric), .after = where(is.character)) #> # A tibble: 15 x 6 #> A B C value_1 value_2 value_3 #> #> 1 01 00 00 70 109 132 #> 2 02 00 00 256 423 16 #> 3 01 12 00 55 62 123 #> 4 01 14 00 15 47 9 #> 5 02 18 00 34 297 4 #> 6 02 22 00 222 126 12 #> 7 01 12 01 39 2 83 #> 8 01 12 02 16 60 40 #> 9 01 14 01 0 0 9 #> 10 01 14 02 15 47 0 #> 11 02 18 01 0 95 4 #> 12 02 18 02 16 14 0 #> 13 02 18 05 18 188 0 #> 14 02 22 01 74 65 8 #> 15 02 22 02 148 61 4 Created on 2022-03-31 by the reprex package (v2.0.1)