0

What I have:

library(magrittr)
set.seed(1234)
what_i_have <- tibble::tibble(
    A = c(0, 1) |> sample(5, replace = TRUE),
    B = c(0, 1) |> sample(5, replace = TRUE),
    C = c(0, 1) |> sample(5, replace = TRUE)
)

It looks like this:

> what_i_have
# A tibble: 5 x 3
      A     B     C
  <dbl> <dbl> <dbl>
1     1     1     1
2     1     0     1
3     1     0     1
4     1     0     0
5     0     1     1

What I want:

what_i_want <- what_i_have %>% .[apply(., 1, function(row) row |> unique() |> length() == 2),]

It looks like this:

# A tibble: 4 x 3
      A     B     C
  <dbl> <dbl> <dbl>
1     1     0     1
2     1     0     1
3     1     0     0
4     0     1     1

My question is: is there a tidyverse way to do the things above?

I tried this:

what_i_have |> 
    dplyr::rowwise() |> 
    dplyr::filter_all(function(row) row |> unique() |> length() == 2)

but it returns the following empty tibble and I do not know why

# A tibble: 0 x 3
# Rowwise: 
# … with 3 variables: A <dbl>, B <dbl>, C <dbl>

Thank you.

AndrewGB
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Chris LAM
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1 Answers1

2

Here is one option with tidyverse. Here, I treat each row as a vector (via c_across), then get the number of distinct values using n_distinct and return TRUE for the rows that have 2 unique values.

library(tidyverse)

what_i_have %>%
  rowwise %>%
  filter(n_distinct(c_across(everything())) == 2)

Output

      A     B     C
  <dbl> <dbl> <dbl>
1     0     1     1
2     1     0     1
3     1     0     0
4     1     1     0

A mixed method approach with apply could be:

what_i_have %>% 
  filter(apply(., 1, \(x)length(unique(x)))==2)

Data

what_i_have  <-
  structure(
    list(
      A = c(0, 1, 1, 1, 1),
      B = c(1, 0, 0, 1, 1),
      C = c(1, 1, 0, 1, 0)
    ),
    class = c("tbl_df", "tbl", "data.frame"),
    row.names = c(NA,-5L)
  )
AndrewGB
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