Is there a built-in function for finding the mode?

Question

In R, mean() and median() are standard functions which do what you'd expect. mode() tells you the internal storage mode of the object, not the value that occurs the most in its argument. But is there is a standard library function that implements the statistical mode for a vector (or list)?

Answer 1

I can't vote yet but Rasmus Bååth's answer is what I was looking for. However, I would modify it a bit allowing to contrain the distribution for example fro values only between 0 and 1.

estimate_mode <- function(x,from=min(x), to=max(x)) {
  d <- density(x, from=from, to=to)
  d$x[which.max(d$y)]
}

We aware that you may not want to constrain at all your distribution, then set from=-"BIG NUMBER", to="BIG NUMBER"

Answer 2

Below is the code which can be use to find the mode of a vector variable in R.

a <- table([vector])

names(a[a==max(a)])

Answer 3

I found Ken Williams post above to be great, I added a few lines to account for NA values and made it a function for ease.

Mode <- function(x, na.rm = FALSE) {
  if(na.rm){
    x = x[!is.na(x)]
  }

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

Answer 4

A small modification to Ken Williams' answer, adding optional params na.rm and return_multiple.

Unlike the answers relying on names(), this answer maintains the data type of x in the returned value(s).

stat_mode <- function(x, return_multiple = TRUE, na.rm = FALSE) {
  if(na.rm){
    x <- na.omit(x)
  }
  ux <- unique(x)
  freq <- tabulate(match(x, ux))
  mode_loc <- if(return_multiple) which(freq==max(freq)) else which.max(freq)
  return(ux[mode_loc])
}

To show it works with the optional params and maintains data type:

foo <- c(2L, 2L, 3L, 4L, 4L, 5L, NA, NA)
bar <- c('mouse','mouse','dog','cat','cat','bird',NA,NA)

str(stat_mode(foo)) # int [1:3] 2 4 NA
str(stat_mode(bar)) # chr [1:3] "mouse" "cat" NA
str(stat_mode(bar, na.rm=T)) # chr [1:2] "mouse" "cat"
str(stat_mode(bar, return_mult=F, na.rm=T)) # chr "mouse"

Thanks to @Frank for simplification.

Answer 5

I would use the density() function to identify a smoothed maximum of a (possibly continuous) distribution :

function(x) density(x, 2)$x[density(x, 2)$y == max(density(x, 2)$y)]

where x is the data collection. Pay attention to the adjust paremeter of the density function which regulate the smoothing.

Answer 6

A quick and dirty way of estimating the mode of a vector of numbers you believe come from a continous univariate distribution (e.g. a normal distribution) is defining and using the following function:

estimate_mode <- function(x) {
  d <- density(x)
  d$x[which.max(d$y)]
}

Then to get the mode estimate:

x <- c(5.8, 5.6, 6.2, 4.1, 4.9, 2.4, 3.9, 1.8, 5.7, 3.2)
estimate_mode(x)
## 5.439788