How to get all possible total combinations in r without repetition?

Question

enter image description here

I have a dataset with 1000 samples where I try to understand the diet pattern of students. I want to understand, how many have taken only: 1. only apple 2. only Banana 3. Only Orange 4. All three fruits 5. Apple + Banana 6. apple + orange 7. banana+ orange

It doesn’t seem clear what your question is. Are you using a specific language or database for this? Generically, you can produce all combinations of three parameters with three nested `for` loops. — crosen9999, Jul 04 '22 at 06:32
@crosen9999. Thank you for the suggestion. I am sorry for the ambiguity of my question. I am quite new to r language and have not been able to make a loop. — user142632, Jul 04 '22 at 09:28
Welcome to SO. Please do not post data as image. We need the data to show you how you can do it and you will hardly find anyone who's going to copy it. You can use `dput` and add the output to your question. Often it also helps when people see that you tried solving the issue yourself. Maybe this post helps: [How to make a great R reproducible example](https://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example). — Jan, Jul 04 '22 at 21:08

score 2 · Answer 1 · answered Jul 08 '22 at 10:37

2

df %>%
   pivot_longer(-student_id) %>%
   group_by(student_id)%>%
   summarise(name = toString(name[value>0]))%>%
   count(name)

# A tibble: 5 x 2
  name                      n
  <chr>                 <int>
1 Apple                     5
2 Apple, Banana, orange     1
3 Apple, orange             2
4 Banana                    3
5 orange                    2

answered Jul 08 '22 at 10:37

Onyambu

67,392
3
24
53

@onyambu.Thank you very much. Makes it much easier – user142632 Jul 10 '22 at 02:07

score 0 · Answer 2 · answered Jul 08 '22 at 10:35

You could do:

library(tidyverse)

data <-
  tibble(student = c(1,2,3,4,5),
        apple = c(1,0,0,1,1),
        banana = c(0,0,1,0,1),
        orange = c(0,1,0,1,1))

data |>
  pivot_longer(-student, names_to = "fruit") |>
  filter(value == 1) |>
  group_by(student) |>
  summarise(fruit = paste(fruit, collapse = "+")) |>
  count(fruit)

Output:

# A tibble: 5 × 2
  fruit                   n
  <chr>               <int>
1 apple                   1
2 apple+banana+orange     1
3 apple+orange            1
4 banana                  1
5 orange                  1

All combinations will show up using the full data.

score 0 · Answer 3 · answered Jul 08 '22 at 22:08

Here is a base R option using table + paste

as.data.frame(
  table(
    trimws(
      do.call(
        paste,
        as.data.frame(
          ifelse(df[-1] > 0,
            names(df[-1])[col(df[-1])], 
            ""
          )
        )
      )
    )
  )
)

which gives

                 Var1 Freq
1               apple    5
2       apple  orange    2
3 apple banana orange    1
4              banana    3
5              orange    2

Or

as.data.frame(
  table(
    apply(
      as.data.frame(ifelse(df[-1] > 0, names(df[-1])[col(df[-1])], NA)), 
      1, 
      function(x) toString(na.omit(x))
    )
  )
)

which gives

                   Var1 Freq
1                 apple    5
2 apple, banana, orange    1
3         apple, orange    2
4                banana    3
5                orange    2

Data

df <- data.frame(
  student = 1:13,
  apple = c(0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0),
  banana = c(1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1),
  orange = c(0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0)
)

How to get all possible total combinations in r without repetition?

3 Answers3