如何重定义R语言base::mean()函数以集成is.finite()功能，实现NA、NaN和Inf值的自动剔除

Got it, let's tweak your mean function to handle all three problematic values (NA, NaN, Inf) in one go. Here's the adjusted code:

Mean.new <- function(x) {
  # Filter to keep only finite values (automatically excludes NA, NaN, Inf)
  valid_values <- x[is.finite(x)]
  # Calculate mean of the cleaned subset
  base::mean(valid_values)
}

How this works: #

• The is.finite() function checks each element in your vector: it returns TRUE only for real numbers (so it rejects NA, NaN, and Inf all at once).
• We use this check to subset your input x down to just valid numerical values, then compute the mean on that filtered set.
• Unlike relying solely on na.rm=TRUE, this approach covers all three edge cases without needing extra conditional checks.

Test it to confirm: #

Let's run a quick example to make sure it behaves as expected:

test_data <- c(5, 10, NA, NaN, Inf, 15)
Mean.new(test_data) # Should return (5+10+15)/3 = 10

This will give you the mean of only the valid observations, exactly what you need for your tabular() table.

内容的提问来源于stack exchange，提问作者corkinabottle