是否有一个内置的功能找到模式？

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)?

One more solution, which works for both numeric & character/factor data:

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

On my dinky little machine, that can generate & find the mode of a 10M-integer vector in about half a second.

If your data set might have multiple modes, the above solution takes the same approach as which.max, and returns the first-appearing value of the set of modes. To return all modes, use this variant (from @digEmAll in the comments):

Modes <- function(x) {
  ux <- unique(x)
  tab <- tabulate(match(x, ux))
  ux[tab == max(tab)]
}

found this on the r mailing list, hope it's helpful. It is also what I was thinking anyways. You'll want to table() the data, sort and then pick the first name. It's hackish but should work.

names(sort(-table(x)))[1]

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

Here, another solution:

freq <- tapply(mySamples,mySamples,length)
#or freq <- table(mySamples)
as.numeric(names(freq)[which.max(freq)])

I've written the following code in order to generate the mode.

MODE <- function(dataframe){
    DF <- as.data.frame(dataframe)

    MODE2 <- function(x){      
        if (is.numeric(x) == FALSE){
            df <- as.data.frame(table(x))  
            df <- df[order(df$Freq), ]         
            m <- max(df$Freq)        
            MODE1 <- as.vector(as.character(subset(df, Freq == m)[, 1]))

            if (sum(df$Freq)/length(df$Freq)==1){
                warning("No Mode: Frequency of all values is 1", call. = FALSE)
            }else{
                return(MODE1)
            }

        }else{ 
            df <- as.data.frame(table(x))  
            df <- df[order(df$Freq), ]         
            m <- max(df$Freq)        
            MODE1 <- as.vector(as.numeric(as.character(subset(df, Freq == m)[, 1])))

            if (sum(df$Freq)/length(df$Freq)==1){
                warning("No Mode: Frequency of all values is 1", call. = FALSE)
            }else{
                return(MODE1)
            }
        }
    }

    return(as.vector(lapply(DF, MODE2)))
}

Let's try it:

MODE(mtcars)
MODE(CO2)
MODE(ToothGrowth)
MODE(InsectSprays)

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

The following function comes in three forms:

method = "mode" [default]: calculates the mode for a unimodal vector, else returns an NA
method = "nmodes": calculates the number of modes in the vector
method = "modes": lists all the modes for a unimodal or polymodal vector

modeav <- function (x, method = "mode", na.rm = FALSE)
{
  x <- unlist(x)
  if (na.rm)
    x <- x[!is.na(x)]
  u <- unique(x)
  n <- length(u)
  #get frequencies of each of the unique values in the vector
  frequencies <- rep(0, n)
  for (i in seq_len(n)) {
    if (is.na(u[i])) {
      frequencies[i] <- sum(is.na(x))
    }
    else {
      frequencies[i] <- sum(x == u[i], na.rm = TRUE)
    }
  }
  #mode if a unimodal vector, else NA
  if (method == "mode" | is.na(method) | method == "")
  {return(ifelse(length(frequencies[frequencies==max(frequencies)])>1,NA,u[which.max(frequencies)]))}
  #number of modes
  if(method == "nmode" | method == "nmodes")
  {return(length(frequencies[frequencies==max(frequencies)]))}
  #list of all modes
  if (method == "modes" | method == "modevalues")
  {return(u[which(frequencies==max(frequencies), arr.ind = FALSE, useNames = FALSE)])}  
  #error trap the method
  warning("Warning: method not recognised.  Valid methods are 'mode' [default], 'nmodes' and 'modes'")
  return()
}

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"

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)])
}

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.