how to get the output of proc tabulate (SAS) in R - r

Question

Is there any R function which could give me directly the same output of proc tabulate ??
var1<-c(rep("A",4),rep("B",4))
var2<-c(rep("C",4),rep("D",4))
var3<-c(rep("E",2),rep("F",4),rep("G",2))
dataset<-data.frame(var1,var2,var3)
proc tabulate data=dataset;
class var1 var2 var3;
table var1*var2 ,var3 all (n rowpctn);
run;
The output that I want is like this:

Answer 1

Here is a way with R -
Create a column of 1s - n
Expand the data to fill the missing combinations - complete
Reshape to 'wide' format - pivot_wider
Create the 'Total' column by getting the row wise sum - rowSums
Add the percentage by looping across the 'var3' columns
library(dplyr)
library(tidyr)
library(stringr)
dataset %>%
mutate(n = 1, var3 = str_c('var3_', var3)) %>%
complete(var1, var2, var3, fill = list(n = 0)) %>%
pivot_wider(names_from = var3, values_from = n, values_fn = sum) %>%
mutate(Total = rowSums(across(where(is.numeric)))) %>%
group_by(var1) %>%
mutate(across(starts_with('var3'),
~ case_when(. == 0 ~ '0(0%)',
TRUE ~ sprintf('%d(%d%%)', ., 100 * mean(. != 0))))) %>%
ungroup
-output
# A tibble: 4 × 6
var1 var2 var3_E var3_F var3_G Total
<chr> <chr> <chr> <chr> <chr> <dbl>
1 A C 2(50%) 2(50%) 0(0%) 4
2 A D 0(0%) 0(0%) 0(0%) 0
3 B C 0(0%) 0(0%) 0(0%) 0
4 B D 0(0%) 2(50%) 2(50%) 4
Update
Based on the comments by #IceCreamToucan, there was a bug, which is corrected in the below code
dataset %>%
mutate(n = 1, var3 = str_c('var3_', var3)) %>%
complete(var1, var2, var3, fill = list(n = 0)) %>%
pivot_wider(names_from = var3, values_from = n, values_fn = sum) %>%
mutate(Total = rowSums(across(where(is.numeric))),
100 * across(starts_with('var3'), ~ . != 0,
.names = "{.col}_perc")/rowSums(across(starts_with('var3'), ~ .!= 0)),
across(matches('var3_[A-Z]$'), ~ case_when(. == 0 ~ '0(0%)',
TRUE ~ sprintf('%d(%.f%%)', ., get(str_c(cur_column(), '_perc')))))) %>%
select(-ends_with('perc'))

Answer 2

Here's a more generic version, where I define a function.
var1<-c(rep("A",4),rep("B",4))
var2<-c(rep("C",4),rep("D",4))
var3<-c(rep("E",2),rep("F",4),rep("G",2))
df<-data.frame(var1,var2,var3)
df_tabulate(df, id_cols = c(var1, var2), names_from = var3)
#> # A tibble: 4 × 6
#> var1 var2 var3_E var3_F var3_G Total
#> <chr> <chr> <chr> <chr> <chr> <dbl>
#> 1 A C 2(50.0%) 2(50.0%) 0(0.0%) 4
#> 2 A D 0(0%) 0(0%) 0(0%) 0
#> 3 B C 0(0%) 0(0%) 0(0%) 0
#> 4 B D 0(0.0%) 2(50.0%) 2(50.0%) 4
You can define the function using janitor
library(janitor, warn.conflicts = FALSE)
library(dplyr, warn.conflicts = FALSE)
library(rlang)
library(tidyr)
df_tabulate <- function(df, id_cols, names_from){
id_cols <- enquo(id_cols)
if (quo_is_call(id_cols, 'c'))
id_cols <- call_args(id_cols)
else
id_cols <- ensym(id_cols)
names_from_chr <- as_label(enquo(names_from))
counts <- df %>%
mutate(g = eval(call2(paste, !!!id_cols, sep = ',')),
col = paste0(names_from_chr, '_', {{ names_from }})) %>%
tabyl(g, col) %>%
adorn_totals('col')
percs <- adorn_percentages(counts) %>%
adorn_pct_formatting()
rbind(counts, percs) %>%
group_by(g) %>%
summarise(across(-Total, ~ paste0(first(.), '(', last(.), ')')),
Total = as.numeric(first(Total))) %>%
separate(g, into = as.character(id_cols)) %>%
complete(!!!id_cols) %>%
mutate(across(starts_with(names_from_chr), ~ coalesce(., '0(0%)')),
across(Total, ~ coalesce(., 0)))
}

Answer 3

Here it is as a single pipeline with discrete simple steps. Long, to be sure, but if you wanted many tables like this you could store it as a function.
library(tidyverse)
library(janitor)
dataset %>%
mutate(across(var1:var2, as.factor)) %>%
count(var1, var2, var3, .drop = FALSE) %>%
unite(vars, var1, var2) %>%
pivot_wider(names_from = var3, values_from = n) %>%
select(-`NA`) %>%
replace(is.na(.), 0) %>%
adorn_totals("col") %>%
adorn_percentages(,,,,-c(vars, Total)) %>%
adorn_pct_formatting(digits = 0,,,,-c(vars, Total)) %>%
adorn_ns(position = "front",,,-c(vars, Total)) %>%
separate(vars, into = c("var1", "var2"))
#> # A tibble: 4 x 6
#> var1 var2 E F G Total
#> <chr> <chr> <chr> <chr> <chr> <dbl>
#> 1 A C 2 (50%) 2 (50%) 0 (0%) 4
#> 2 A D 0 (-) 0 (-) 0 (-) 0
#> 3 B C 0 (-) 0 (-) 0 (-) 0
#> 4 B D 0 (0%) 2 (50%) 2 (50%) 4
This replaces the questionable 0/0 = 0% with simply - for a cleaner(IMO) result.