19

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，尽管如果我可以更好地控制它会很棒。

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。它可以是一个因子或一个数值向量（如果需要先将数据转换为数值向量）

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！

4

4 回答 4

24

就像是 ：

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

如果有平局，which.max 取第一个最大值。从 which.max 帮助页面：

确定位置，即数字向量的（第一个）最小值或最大值的索引。

前任 ：

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"
于 2013-11-14T16:34:24.903 回答
3

如果您的数据很大，您可能需要考虑使用该data.table软件包。

# 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，在某些情况下效果很好。您可以使用data.table.

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
于 2016-02-18T14:11:19.927 回答
1

对于内部包，我制作了一个rowMode-function，您可以在其中选择如何处理关系和缺失值：

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"

使用数据：

set.seed(2020)
DF <- data.frame(id = 1:10, matrix(sample(c(LETTERS[1:5], NA_character_), 60, TRUE), ncol = 6))
于 2020-07-26T12:26:29.550 回答
0

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：

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
于 2020-07-21T23:56:08.857 回答