使用tidyverse-通过pick对'var 3'到'var 5'列执行操作来创建COUNT，并获得行求和（rowSums-因为它是二进制的，0值不会相加），然后循环across列，用列名替换1（cur_column()），默认情况下TRUE将是NA，通过粘贴将tibble列转换为具有unnest_wider和unite的适当列，同时删除具有na.rm = TRUE的NA，将NAMES_转换为单列

library(dplyr) # version >= 1.1.0
library(tidyr)
library(stringr) 
df %>% 
  mutate(COUNT = rowSums(pick(var3:var5)),
   NAMES = across(var3:var5, ~ case_when(as.logical(.x) ~ cur_column()))) %>% 
  unnest_wider(NAMES, names_sep = "_") %>%
  unite(NAMES, starts_with("NAMES_"), sep = ", ",  na.rm = TRUE)

• 输出

# A tibble: 10 × 8
    var1  var2  var3  var4  var5  var6 COUNT NAMES             
   <int> <int> <int> <int> <int> <int> <dbl> <chr>             
 1     0     0     1     1     0     0     2 "var3, var4"      
 2     0     0     1     1     1     0     3 "var3, var4, var5"
 3     1     1     1     0     1     1     2 "var3, var5"      
 4     0     1     0     0     0     1     0 ""                
 5     0     0     0     1     0     1     1 "var4"            
 6     0     1     1     0     0     0     1 "var3"            
 7     0     1     0     0     1     1     1 "var5"            
 8     1     1     1     0     1     0     2 "var3, var5"      
 9     0     1     1     1     1     1     3 "var3, var4, var5"
10     0     0     0     1     1     0     2 "var4, var5"

下面是一个基本的R方法：

cols  <- names(df)[3:5] # "var3" "var4" "var5"

df$COUNT  <- rowSums(df[cols])

df$NAMES  <- apply(df[cols], 1, \(x) {
    col_names  <- paste(
       cols[as.logical(x)], collapse = ";"
    )
    if(col_names == "") col_names  <- NA_character_
    col_names
}
)

#    var1 var2 var3 var4 var5 var6 COUNT          NAMES
# 1     1    1    0    1    0    1     1           var4
# 2     0    1    1    0    0    0     1           var3
# 3     1    1    1    1    1    0     3 var3;var4;var5
# 4     1    1    0    1    0    0     1           var4
# 5     0    0    1    1    1    0     3 var3;var4;var5
# 6     0    1    0    1    1    0     2      var4;var5
# 7     1    1    0    1    1    0     2      var4;var5
# 8     1    0    1    1    0    1     2      var3;var4
# 9     1    0    1    0    0    0     1           var3
# 10    0    0    0    0    1    0     1           var5

或者，您可以对data.table使用相同的方法：

setDT(df)

df[, COUNT := rowSums(.SD), .SDcols = cols]

df[, 
    NAMES := apply(.SD, 1, \(x) {
        col_names  <- paste(
        cols[as.logical(x)], collapse = ";"
    )
    if(col_names == "") col_names  <- NA_character_
    col_names
    }
    ), .SDcols = cols
]

# same output

编辑：我注意到输出与@akrun的响应不同，所以我用set.seed(100)重新运行了它，以获得重现性。