R语言中按因子水平出现次数筛选data.frame列的方法

Hey folks, let's solve this specific R problem where you need to clean up a data frame with two-level factor columns—keeping only those columns where both factor levels show up more than once, and dropping any columns where one level appears ≤1 times (we'll also retain non-factor columns like row IDs by default).

Example Data #

First, let's set up the sample data frame we're working with:

df <- data.frame(
  row.name = 1:4,
  Factor1 = c("dog", "dog", "dog", "dog"),
  Factor2 = c("dog", "dog", "cat", "cat"),
  Factor3 = c("cat", "cat", "dog", "dog"),
  Factor4 = c("cat", "dog", "dog", "dog"),
  stringsAsFactors = TRUE
)

In this example, we want to retain Factor2 and Factor3 (both levels appear twice), while dropping Factor1 (only "dog" exists) and Factor4 ("cat" only shows up once).

Working Solutions #

After tweaking a method from Ryan, here are two solid approaches that get the job done perfectly:

1. Using dplyr (Tidyverse Style)

If you prefer the tidyverse syntax, use select_if to filter columns conditionally:

library(dplyr)

filtered_df <- df %>% 
  select_if(~ !is.factor(.x) | sum(tabulate(.x) > 1) >= 2)

How it works:

• For non-factor columns (like row.name), we keep them automatically with !is.factor(.x).
• For factor columns, tabulate(.x) counts how many times each level appears. We then sum how many levels have a count greater than 1—if that sum is ≥2, both levels meet our requirement, so we keep the column.

2. Using Base R

If you want to stick to base R without extra packages, use sapply to create a boolean subset vector:

filtered_df <- df[, sapply(df, function(x) !is.factor(x) | sum(tabulate(.x) > 1) >= 2)]

How it works:

• This uses the exact same logic as the dplyr version. sapply iterates over each column, runs the check, and returns TRUE/FALSE for each column. We use that vector to keep only the columns where the check passes.

Result #

Either method will give you the filtered data frame you need:

row.name Factor2 Factor3
1        1     dog     cat
2        2     dog     cat
3        3     cat     dog
4        4     cat     dog

内容的提问来源于stack exchange，提问作者plik