模式并不是在所有情况下都有用。所以函数应该处理这种情况。试试下面的函数。

Mode <- function(v) {
  # checking unique numbers in the input
  uniqv <- unique(v)
  # frquency of most occured value in the input data
  m1 <- max(tabulate(match(v, uniqv)))
  n <- length(tabulate(match(v, uniqv)))
  # if all elements are same
  same_val_check <- all(diff(v) == 0)
  if(same_val_check == F){
    # frquency of second most occured value in the input data
    m2 <- sort(tabulate(match(v, uniqv)),partial=n-1)[n-1]
    if (m1 != m2) {
      # Returning the most repeated value
      mode <- uniqv[which.max(tabulate(match(v, uniqv)))]
    } else{
      mode <- "Two or more values have same frequency. So mode can't be calculated."
    }
  } else {
    # if all elements are same
    mode <- unique(v)
  }
  return(mode)
}

输出,

x1 <- c(1,2,3,3,3,4,5)
Mode(x1)
# [1] 3

x2 <- c(1,2,3,4,5)
Mode(x2)
# [1] "Two or more varibles have same frequency. So mode can't be calculated."

x3 <- c(1,1,2,3,3,4,5)
Mode(x3)
# [1] "Two or more values have same frequency. So mode can't be calculated."

为了生成模式，我写了下面的代码。

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)))
}

让我们试试吧:

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

在r邮件列表中发现了这个，希望对你有帮助。我也是这么想的。您将希望table()数据，排序，然后选择第一个名称。这有点粗俗，但应该有用。

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

假设你的观测值是来自实数的类，当你的观测值是2,2,3,3时，你期望模态为2.5，然后你可以用mode = l1 + I * (f1-f0) / (2f1 -f0 - f2)来估计模态，其中l1..最频繁类的下限，f1..最频繁类的频率，f0..在最频繁类之前的类的频率，f2..在最频繁类之后的类的频率，i..分类间隔，如在1,2,3中给出:

#Small Example
x <- c(2,2,3,3) #Observations
i <- 1          #Class interval

z <- hist(x, breaks = seq(min(x)-1.5*i, max(x)+1.5*i, i), plot=F) #Calculate frequency of classes
mf <- which.max(z$counts)   #index of most frequent class
zc <- z$counts
z$breaks[mf] + i * (zc[mf] - zc[mf-1]) / (2*zc[mf] - zc[mf-1] - zc[mf+1])  #gives you the mode of 2.5


#Larger Example
set.seed(0)
i <- 5          #Class interval
x <- round(rnorm(100,mean=100,sd=10)/i)*i #Observations

z <- hist(x, breaks = seq(min(x)-1.5*i, max(x)+1.5*i, i), plot=F)
mf <- which.max(z$counts)
zc <- z$counts
z$breaks[mf] + i * (zc[mf] - zc[mf-1]) / (2*zc[mf] - zc[mf-1] - zc[mf+1])  #gives you the mode of 99.5

如果你想要最频繁的级别，并且你有多个最频繁的级别，你可以得到所有的级别，例如:

x <- c(2,2,3,5,5)
names(which(max(table(x))==table(x)))
#"2" "5"

下面是可以用来找到R中矢量变量的模式的代码。

a <- table([vector])

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

我还不能投票，但Rasmus Bååth的答案是我一直在寻找的。 但是，我将稍微修改一下，允许将分布限制在0到1之间。

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

我们知道你可能根本不想约束你的分布，那么设置from=-"BIG NUMBER"， to="BIG NUMBER"