在dplyr中确定分组数据帧中最频繁因子的最快方法

Question

我在dplyr中总结一个数据框时,试图在一个组中找到几个因子变量中最常见的值.我需要一个执行以下操作的公式:

1. 在组中的一个变量的所有因子中找出最常用的因子水平(因此对于因子水平的计数基本上是"max()").
2. 如果几个最常用因子级别之间存在联系,请选择其中任何一个因子级别.
3. 返回因子级别名称(不是计数).

有几个公式可行.但是,我能想到的那些都很慢.快速的那些不方便一次应用于数据帧中的几个变量.我想知道是否有人知道一种与dplyr很好地集成的快速方法.

我尝试了以下方法:

生成样本数据(50000组,100个随机字母)

z <- data.frame(a = rep(1:50000,100), b = sample(LETTERS, 5000000, replace = TRUE))

str(z)
'data.frame':   5000000 obs. of  2 variables:
$ a: int  1 2 3 4 5 6 7 8 9 10 ...
$ b: Factor w/ 26 levels "A","B","C","D",..: 6 4 14 12 3 19 17 19 15 20 ...

"清洁" - 但缓慢的方法1

 y <- z %>% 
    group_by(a) %>% 
    summarise(c = names(table(b))[which.max(table(b))])

user    system  elapsed 
26.772  2.011   29.568 

"清洁" - 但缓慢的方法2

y <- z %>% 
    group_by(a) %>% 
    summarise(c = names(which(table(b) == max(table(b)))[1]))

user    system  elapsed 
29.329  2.029   32.361 

"清洁" - 但缓慢的方法3

y <- z %>% 
    group_by(a) %>% 
    summarise(c = names(sort(table(b),decreasing = TRUE)[1]))

user    system  elapsed 
35.086  6.905   42.485 

"凌乱"但快速的方法

y <- z %>% 
     group_by(a,b) %>% 
     summarise(counter = n()) %>% 
     group_by(a) %>% 
     filter(counter == max(counter))
y <- y[!duplicated(y$a),]
y <- y$counter <- NULL

user   system  elapsed 
7.061  0.330   7.664 

Answer 1

这是另一种选择dplyr:

set.seed(123)
z <- data.frame(a = rep(1:50000,100), 
                b = sample(LETTERS, 5000000, replace = TRUE), 
                stringsAsFactors = FALSE)

a <- z %>% group_by(a, b) %>% summarise(c=n()) %>% filter(row_number(desc(c))==1) %>% .$b 
b <- z %>% group_by(a) %>% summarise(c=names(which(table(b) == max(table(b)))[1])) %>% .$c 

我们确保这些是等效的方法:

> identical(a, b)
#[1] TRUE

更新

正如@docendodiscimus所提到的,你也可以这样做:

count(z, a, b) %>% slice(which.max(n))

以下是基准测试的结果:

library(microbenchmark)
mbm <- microbenchmark(
  steven = z %>% group_by(a, b) %>% summarise(c = n()) %>% filter(row_number(desc(c))==1),
  phil = z %>% group_by(a) %>% summarise(c = names(which(table(b) == max(table(b)))[1])),
  docendo = count(z, a, b) %>% slice(which.max(n)),
  times = 10
)

#Unit: seconds
#    expr       min        lq      mean    median        uq       max neval cld
#  steven  4.752168  4.789564  4.815986  4.813686  4.847964  4.875109    10  b 
#    phil 15.356051 15.378914 15.467534 15.458844 15.533385 15.606690    10   c
# docendo  4.586096  4.611401  4.669375  4.688420  4.702352  4.753583    10 a 

Answer 2

为什么选择dplyr？

#dummy data
set.seed(123)
z <- data.frame(a = rep(1:50000,100),
                b = sample(LETTERS, 5000000, replace = TRUE))

#result
names(sort(table(z$b),decreasing = TRUE)[1])
# [1] "S"

#time it
system.time(
  names(sort(table(z$b),decreasing = TRUE)[1])
)

# user  system elapsed 
# 0.36    0.00    0.36 

编辑:多列

#dummy data
set.seed(123)
z <- data.frame(a = rep(1:50000,100),
                b = sample(LETTERS, 5000000, replace = TRUE),
                c = sample(LETTERS, 5000000, replace = TRUE),
                d = sample(LETTERS, 5000000, replace = TRUE))

# check for multiple columns
sapply(c("b","c","d"), function(i)
  names(sort(table(z[,i]),decreasing = TRUE)[1])
  )
# b   c   d 
#"S" "N" "G" 

#time it
system.time(
  sapply(c("b","c","d"), function(i)
    names(sort(table(z[,i]),decreasing = TRUE)[1]))
  )
# user  system elapsed 
# 0.61    0.17    0.78 

Answer 3

data.table 仍然是最快的选择:

z <- data.frame(a = rep(1:50000,100), b = sample(LETTERS, 5000000, replace = TRUE))

标杆:

library(data.table)
library(dplyr)

#dplyr
system.time({
  y <- z %>% 
    group_by(a) %>% 
    summarise(c = names(which(table(b) == max(table(b)))[1]))  
})
 user  system elapsed 
14.52    0.01   14.70 

#data.table
system.time(
  setDT(z)[, .N, by=b][order(N),][.N,]
)
 user  system elapsed 
 0.05    0.02    0.06 

#@zx8754 's way - base R
system.time(
  names(sort(table(z$b),decreasing = TRUE)[1])
)
   user  system elapsed 
   0.73    0.06    0.81 

正如使用data.table可以看到的那样:

  setDT(z)[, .N, by=b][order(N),][.N,]

要么

  #just to get the name
  setDT(z)[, .N, by=b][order(N),][.N, b] 

似乎是最快的

所有列的更新:

使用@ zx8754的数据

set.seed(123)
z2 <- data.frame(a = rep(1:50000,100),
                b = sample(LETTERS, 5000000, replace = TRUE),
                c = sample(LETTERS, 5000000, replace = TRUE),
                d = sample(LETTERS, 5000000, replace = TRUE))

你可以这样做:

#with data.table
system.time(
 sapply(c('b','c','d'), function(x) {
  data.table(x = z2[[x]])[, .N, by=x][order(N),][.N, x] 
 }))
 user  system elapsed 
 0.34    0.00    0.34 

#with base-R
system.time(
  sapply(c("b","c","d"), function(i)
    names(sort(table(z2[,i]),decreasing = TRUE)[1]))
)
 user  system elapsed 
 4.14    0.11    4.26 

只是为了确认结果是一样的:

sapply(c('b','c','d'), function(x) {
  data.table(x = z2[[x]])[, .N, by=x][order(N),][.N, x] 
})
b c d 
S N G 

sapply(c("b","c","d"), function(i)
    names(sort(table(z2[,i]),decreasing = TRUE)[1]))
b   c   d 
"S" "N" "G"