在R中,mean()和median()是标准函数,它们执行您所期望的功能。Mode()告诉您对象的内部存储模式,而不是参数中出现次数最多的值。但是是否存在一个标准库函数来实现向量(或列表)的统计模式?


当前回答

我将使用density()函数来确定一个(可能是连续的)分布的平滑最大值:

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

其中x是数据集合。注意调节平滑的密度函数的调节参数。

其他回答

我发现Ken Williams上面的帖子很棒,我添加了几行来解释NA值,并使其成为一个函数。

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)))])
}

我浏览了所有这些选项,开始想知道它们的相对特性和性能,所以我做了一些测试。如果其他人也好奇,我在这里分享我的结果。

我不想为这里发布的所有函数而烦恼,我选择了一个基于一些标准的示例:函数应该对字符、因子、逻辑和数字向量都有效,它应该适当地处理na和其他有问题的值,输出应该是“合理的”,即没有数字作为字符或其他类似的愚蠢行为。

我还添加了一个我自己的函数,它是基于与chrispy相同的想法,除了适应更一般的用途:

library(magrittr)

Aksel <- function(x, freq=FALSE) {
    z <- 2
    if (freq) z <- 1:2
    run <- x %>% as.vector %>% sort %>% rle %>% unclass %>% data.frame
    colnames(run) <- c("freq", "value")
    run[which(run$freq==max(run$freq)), z] %>% as.vector   
}

set.seed(2)

F <- sample(c("yes", "no", "maybe", NA), 10, replace=TRUE) %>% factor
Aksel(F)

# [1] maybe yes  

C <- sample(c("Steve", "Jane", "Jonas", "Petra"), 20, replace=TRUE)
Aksel(C, freq=TRUE)

# freq value
#    7 Steve

最后,我通过微基准测试在两组测试数据上运行了五个函数。函数名指的是它们各自的作者:

Chris的函数被设置为method="modes"和na。rm=TRUE默认值,以使其更具可比性,但除此之外,这里使用的函数是由它们的作者提供的。

In matter of speed alone Kens version wins handily, but it is also the only one of these that will only report one mode, no matter how many there really are. As is often the case, there's a trade-off between speed and versatility. In method="mode", Chris' version will return a value iff there is one mode, else NA. I think that's a nice touch. I also think it's interesting how some of the functions are affected by an increased number of unique values, while others aren't nearly as much. I haven't studied the code in detail to figure out why that is, apart from eliminating logical/numeric as a the cause.

计算包含离散值的向量“v”的MODE的一个简单方法是:

names(sort(table(v)))[length(sort(table(v)))]

抱歉,我可能把它理解得太简单了,但这不是可以工作的吗?(我的机器上的1E6值在1.3秒内):

t0 <- Sys.time()
summary(as.factor(round(rnorm(1e6), 2)))[1]
Sys.time()-t0

你只需要用你的向量替换“round(rnorm(1e6),2)”。

下面是一个查找模式的函数:

mode <- function(x) {
  unique_val <- unique(x)
  counts <- vector()
  for (i in 1:length(unique_val)) {
    counts[i] <- length(which(x==unique_val[i]))
  }
  position <- c(which(counts==max(counts)))
  if (mean(counts)==max(counts)) 
    mode_x <- 'Mode does not exist'
  else 
    mode_x <- unique_val[position]
  return(mode_x)
}