Median for all numeric and mode for all characters in R

user_13484989 0
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With a dataset like original:

id <- c("JF", "GH", "GH", "ANN", "GH", "ROG", "JF")
group <- c("most", "least", "most", "least", "least", "most", "least")
NP <- c(4,6,18,1,3,12,8)
iso_USA <- c(1, 0, 0, 0, 0, 1, 1)
iso_CHN <- c(0, 1, 1, 0, 0, 0, 0)
color <- c("blue", "orange", "blue", "blue", "red", "orange", "black")

original <- data.frame(id, group, NP, iso_USA, iso_CHN, color)


numeric <- unlist(lapply(original, is.numeric))  
numeric <- names(original[ , numeric])

char <- unlist(lapply(original, is.character))  
char <- names(original[ , char])
char <- char[-1]   #remove id from variables of interest

I want to group by "group" and calculate the median for the numeric variables and the mode for the character variables. Therefore, the data looks like original2. Note that my actual dataset has way more columns than the mock version presented here:

group <- c("least", "most")
NP <- c(6,12)
iso_USA <- c(0,1)
iso_CHN <- c(0, 0)
color <- c("orange", "blue")

original2 <- data.frame(group, NP, iso_USA, iso_CHN, color)

Any clue?

dplyr r
2 Answers

Using dplyr's across functionality and the the accepted answer at the FAQ about implementing a mode function:

Mode <- function(x) {
  ux <- unique(x)
  ux[which.max(tabulate(match(x, ux)))]
}

library(dplyr)
original %>%
  select(-id) %>%
  group_by(group) %>%
  summarize(
    across(where(is.numeric), median),
    across(where(is.character), Mode)
  )
# # A tibble: 2 × 6
#   group    NP iso_USA iso_CHN color 
#   <chr> <dbl>   <dbl>   <dbl> <chr> 
# 1 least   4.5       0       0 orange
# 2 most   12         1       0 blue
user_9030610

Here's a base R solution.

Little EDIT (thx @Gergor Thomas for your valuable input)

For this I also need

* a function that can tell me if elements are castable to numeric and - obviously -

  • a Mode function (which I will copy blatantly from @Gregor Thomas's solution).
Mode <- function(x){ ux <- unique(x)
  ux[which.max(tabulate(match(x, ux)))] }

Usage

aggregate( original[,3:6], by=list(original$group), function(x){ 
  'if'( is.numeric(x),median(x),Mode(x) ) } )

  Group.1   NP iso_USA iso_CHN  color
1   least  4.5       0       0 orange
2    most 12.0       1       0   blue
user_94620950
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