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

This hack should work fine. Gives you the value as well as the count of mode:

Mode <- function(x){
a = table(x) # x is a vector
return(a[which.max(a)])
}

Answer 2

There is package modeest which provide estimators of the mode of univariate unimodal (and sometimes multimodal) data and values of the modes of usual probability distributions.

mySamples <- c(19, 4, 5, 7, 29, 19, 29, 13, 25, 19)

library(modeest)
mlv(mySamples, method = "mfv")

Mode (most likely value): 19 
Bickel's modal skewness: -0.1 
Call: mlv.default(x = mySamples, method = "mfv")

For more information see this page

Answer 3

R has so many add-on packages that some of them may well provide the [statistical] mode of a numeric list/series/vector.

However the standard library of R itself doesn't seem to have such a built-in method! One way to work around this is to use some construct like the following (and to turn this to a function if you use often...):

mySamples <- c(19, 4, 5, 7, 29, 19, 29, 13, 25, 19)
tabSmpl<-tabulate(mySamples)
SmplMode<-which(tabSmpl== max(tabSmpl))
if(sum(tabSmpl == max(tabSmpl))>1) SmplMode<-NA
> SmplMode
[1] 19

For bigger sample list, one should consider using a temporary variable for the max(tabSmpl) value (I don't know that R would automatically optimize this)

Reference: see "How about median and mode?" in this KickStarting R lesson
This seems to confirm that (at least as of the writing of this lesson) there isn't a mode function in R (well... mode() as you found out is used for asserting the type of variables).

Answer 4

You could also calculate the number of times an instance has happened in your set and find the max number. e.g.

> temp <- table(as.vector(x))
> names (temp)[temp==max(temp)]
[1] "1"
> as.data.frame(table(x))
r5050 Freq
1     0   13
2     1   15
3     2    6
> 

Answer 5

This works pretty fine

> a<-c(1,1,2,2,3,3,4,4,5)
> names(table(a))[table(a)==max(table(a))]

Answer 6

Based on @Chris's function to calculate the mode or related metrics, however using Ken Williams's method to calculate frequencies. This one provides a fix for the case of no modes at all (all elements equally frequent), and some more readable method names.

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

  # Get unique values
  ux <- unique(x)
  n <- length(ux)

  # Get frequencies of all unique values
  frequencies <- tabulate(match(x, ux))
  modes <- frequencies == max(frequencies)

  # Determine number of modes
  nmodes <- sum(modes)
  nmodes <- ifelse(nmodes==n, 0L, nmodes)

  if (method %in% c("one", "mode", "") | is.na(method)) {
    # Return NA if not exactly one mode, else return the mode
    if (nmodes != 1) {
      return(NA)
    } else {
      return(ux[which(modes)])
    }
  } else if (method %in% c("n", "nmodes")) {
    # Return the number of modes
    return(nmodes)
  } else if (method %in% c("all", "modes")) {
    # Return NA if no modes exist, else return all modes
    if (nmodes > 0) {
      return(ux[which(modes)])
    } else {
      return(NA)
    }
  }
  warning("Warning: method not recognised.  Valid methods are 'one'/'mode' [default], 'n'/'nmodes' and 'all'/'modes'")
}

Since it uses Ken's method to calculate frequencies the performance is also optimised, using AkselA's post I benchmarked some of the previous answers as to show how my function is close to Ken's in performance, with the conditionals for the various ouput options causing only minor overhead: