这是一种方法，但它只工作，因为在sex和vaccine变量中有相同数量的因子。如果在数据集中有一个具有不同数量因子的变量，则由于行数不同，对list_cbind的调用将失败。

library(dplyr)
library(purrr)

set.seed(10)
cluster <- sample(1:4, 1, size = 20)
sex <- sample(0:1, 1, size = 20)
vaccine <- sample(0:1, 1, size = 20)
df <- as.data.frame(cbind(cluster,sex,vaccine))
df <- as.data.frame(lapply(df, as.factor))

# function to calculate the counts by variable, this assumes that `cluster` will be a grouping variable in all cases.
# `arrange(cluster)` ensures that resultant data frames are in the same row order provided that the number of factor levels for each variable is the same.

fun <- function(var){

  # create a summary variable name based on the input variable name
  sum_var <- paste0("count_", as.name(var))
  
  df1 <- 
    df |>
    summarise(!!sum_var := n(), .by = c(cluster, {{var}})) |> 
    arrange(cluster)
  
  return(df1)
  
  }

# vector for all other names in the data set
arg1 = names(df)[-1]

# use purrr::map to loop through the variables and bind the resulting  dataframes. Finally a bit of tidying up to remove the duplicated cluster columns. 
 
map(arg1, fun) |> 
  list_cbind() |> 
  rename(cluster = cluster...1) |> 
  select(-starts_with("cluster..."))

#> New names:
#> • `cluster` -> `cluster...1`
#> • `cluster` -> `cluster...4`
#>   cluster sex count_sex vaccine count_vaccine
#> 1       1   1         1       1             1
#> 2       1   0         1       0             1
#> 3       2   1         4       1             3
#> 4       2   0         1       0             2
#> 5       3   0         6       0             3
#> 6       3   1         2       1             5
#> 7       4   0         3       1             2
#> 8       4   1         2       0             3

创建于2023-06-14带有reprex v2.0.2

尝试merge处理生成的tibles列表。使用非对称数据集进行说明。
colnames(df)[-1]排除 cluster，留下 sex 和 vaccine 用于计数。

library(dplyr)

setNames(purrr::reduce(lapply(colnames(df)[-1], \(x) 
           df %>% 
             group_by(cluster, !!rlang::sym(x)) %>% 
             count()), merge, by=1:2, all=T), 
  c(colnames(df)[1], "values", paste0(colnames(df)[-1], "_count")))
  cluster values sex_count vaccine_count
1       1      0         1             4
2       1      1         4             1
3       2      0         4             2
4       2      1         1             3
5       3      0         5             5
6       3      1         3             3
7       4      0        NA             1
8       4      1         2             1

数据

df <- structure(list(cluster = structure(c(2L, 3L, 1L, 3L, 3L, 1L, 
1L, 1L, 2L, 3L, 3L, 3L, 3L, 2L, 2L, 3L, 1L, 4L, 4L, 2L), levels = c("1", 
"2", "3", "4"), class = "factor"), sex = structure(c(1L, 1L, 
2L, 2L, 2L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 2L, 2L, 
2L, 1L), levels = c("0", "1"), class = "factor"), vaccine = structure(c(2L, 
1L, 1L, 2L, 2L, 1L, 2L, 1L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 2L, 2L), levels = c("0", "1"), class = "factor")), class = "data.frame", 
row.names = c(NA, -20L))