按行查找最频繁的值

Question

我的问题如下：

我有一个包含多个因子变量的数据集，它们具有相同的类别。我需要找到每一行最常出现的类别。在平局的情况下，可以选择任意值，但如果我能对它有更多的控制权那就太好了。

我的数据集包含一百多个因素。但是，结构是这样的：

df = data.frame(id = 1:3
                var1 = c("red","yellow","green")
                var2 = c("red","yellow","green")
                var3 = c("yellow","orange","green")
                var4 = c("orange","green","yellow"))

df
#   id   var1   var2   var3   var4
# 1  1    red    red yellow orange
# 2  2 yellow yellow orange  green
# 3  3  green  green  green yellow

解决方案应该是数据框中的一个变量，例如 var5，它包含每行最常见的类别。可以是因子也可以是数值向量（以防数据需要先转换成数值向量）

在这种情况下，我想要这个解决方案：

df$var5
# [1] "red"    "yellow" "green" 

任何建议将不胜感激！提前致谢！

Answer 1

对于内部包，我创建了一个rowMode-函数，您可以在其中选择如何处理关联和缺失值：

rowMode <- function(x, ties = NULL, include.na = FALSE) {
  # input checks data
  if ( !(is.matrix(x) | is.data.frame(x)) ) {
    stop("Your data is not a matrix or a data.frame.")
  }
  # input checks ties method
  if ( !is.null(ties) && !(ties %in% c("random", "first", "last")) ) {
    stop("Your ties method is not one of 'random', 'first' or 'last'.")
  }
  # set ties method to 'random' if not specified
  if ( is.null(ties) ) ties <- "random"
  
  # create row frequency table
  rft <- table(c(row(x)), unlist(x), useNA = c("no","ifany")[1L + include.na])
  
  # get the mode for each row
  colnames(rft)[max.col(rft, ties.method = ties)]
}

几种可能的输出（基于不同的参数选项）：

> rowMode(DF[,-1])
 [1] "B" "E" "B" "E" "B" "C" "B" "E" "A" "E"
> rowMode(DF[,-1], ties = "first")
 [1] "B" "B" "B" "A" "B" "C" "B" "E" "A" "E"
> rowMode(DF[,-1], ties = "first", include.na = TRUE)
 [1] "B" NA  "B" NA  "B" "C" "B" "E" "A" "E"
> rowMode(DF[,-1], ties = "last", include.na = TRUE)
 [1] "B" NA  NA  NA  "B" "C" "B" "E" "D" "E"
> rowMode(DF[,-1], ties = "last")
 [1] "B" "C" "B" "E" "B" "C" "B" "E" "D" "E"

---

使用过的数据：

set.seed(2020)
DF <- data.frame(id = 1:10, matrix(sample(c(LETTERS[1:5], NA_character_), 60, TRUE), ncol = 6))

Answer 2

这是另一个基本 R 选项：

tab <- table(data.frame(as.vector(row(df[,-1L])), unlist(df[,-1L])))
colnames(tab)[max.col(tab, "first")]

或另一种data.table 方法：

melt(as.data.table(df), id.vars="id")[
    order(id, value), ri := rowid(rleid(value))][,
        value[which.max(ri)], id]$V1

计时码：

library(data.table)
set.seed(0L)
nr <- 1e5L
nc <- 4L
DF <- data.frame(id=1L:nr, as.data.frame(matrix(sample(letters, nr*nc, TRUE), ncol=nc)))
DT <- as.data.table(DF)

mtd0 <- function(df) apply(df,1,function(x) names(which.max(table(x))))

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

mtd_dt <- function(dt) melt(dt, id.vars="id")[, Mode(value), id]$V1

mtd_dt2 <- function(dt) melt(dt, id.vars="id")[
    order(id, value), ri := rowid(rleid(value))][,
        value[which.max(ri)], id]$V1

mtd2 <- function(df) {
    tab <- table(data.frame(as.vector(row(df[,-1L])), unlist(df[,-1L])))
    colnames(tab)[max.col(tab, "first")]
}

df = data.frame(id = 1:3,
    var1 = c("red","yellow","green"),
    var2 = c("red","yellow","green"),
    var3 = c("yellow","orange","green"),
    var4 = c("orange","green","yellow"))

a0 <- mtd0(df)
identical(a0, mtd_dt(as.data.table(df)))
#[1] TRUE

identical(a0, mtd2(df))
#[1] TRUE

identical(a0, mtd_dt2(as.data.table(df)))
#[1] TRUE

microbenchmark::microbenchmark(times=1L, mtd0(DF), mtd_dt(DT), mtd_dt2(DT), mtd2(DF))

时间安排：

Unit: milliseconds
        expr        min         lq       mean     median         uq        max neval
    mtd0(DF) 10083.9941 10083.9941 10083.9941 10083.9941 10083.9941 10083.9941     1
  mtd_dt(DT)  1056.2319  1056.2319  1056.2319  1056.2319  1056.2319  1056.2319     1
 mtd_dt2(DT)   168.6183   168.6183   168.6183   168.6183   168.6183   168.6183     1
    mtd2(DF)   519.2030   519.2030   519.2030   519.2030   519.2030   519.2030     1

Answer 3

如果您的数据非常大，您可能需要考虑使用data.table 包。

# Generate the data
nrow <- 10^5
id <- 1:nrow
colors <- c("red","yellow","green")
var1 <- sample(colors, nrow, replace = TRUE)
var2 <- sample(colors, nrow, replace = TRUE)
var3 <- sample(colors, nrow, replace = TRUE)
var4 <- sample(colors, nrow, replace = TRUE)

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

Chargaff 的解决方案很简单，并且在某些情况下效果很好。使用data.table 可以获得小幅性能提升（~20%）。

df <- data.frame(cbind(id, var1, var2, var3, var4))
system.time(apply(df, 1, Mode))
#   user  system elapsed
#  1.242   0.018   1.264

library(data.table)
dt <- data.table(cbind(id, var1, var2, var3, var4))
system.time(melt(dt, measure = patterns('var'))[, Mode(value1), by = id])
#   user  system elapsed
#  1.020   0.012   1.034

Answer 4

类似：

apply(df,1,function(x) names(which.max(table(x))))
[1] "red"    "yellow" "green" 

如果出现平局，which.max 取第一个最大值。来自 which.max 帮助页面：

确定位置，即（第一个）的索引 数值向量的最小值或最大值。

例如：

var4 <- c("yellow","green","yellow")
df <- data.frame(cbind(id, var1, var2, var3, var4))

> df
  id   var1   var2   var3   var4
1  1    red    red yellow yellow
2  2 yellow yellow orange  green
3  3  green  green  green yellow

apply(df,1,function(x) names(which.max(table(x))))
[1] "red"    "yellow" "green"