Fastest way of determining most frequent factor in a grouped data frame in dplyr

Question

I am trying to find the most frequent value within a group for several factor variables while summarizing a data frame in dplyr. I need a formula that does the following:

1. Find the most frequently used factor level among all factors for one variable in a group (so basically "max()" for counts of factor levels).
2. If there is a tie between several most-used-factor levels, pick any one of those factors-levels.
3. Return the factor-level name (not number of counts).

There are several formulas that work. However, those that I could think of are all slow. Those that are fast are not convenient to apply to several variables in a data frame at once. I was wondering if somebody knows a fast method that integrates nicely with dplyr.

I tried the following:

generating sample data (50000 groups with 100 random letters)

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 ...

"Clean"-but-slow approach 1

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

user    system  elapsed 
26.772  2.011   29.568 

"Clean"-but-slow approach 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 

"Clean"-but-slow approach 3

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

user    system  elapsed 
35.086  6.905   42.485 

"Messy"-but-fast approach

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

Here's another option with 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 

We make sure these are equivalent approaches:

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

Update

As per mentioned by @docendodiscimus, you could also do:

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

---

Here are the results on the benchmark:

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

Why 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 

EDIT: multiple columns

#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 is still the fastest choice for this:

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

Benchmarking:

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 

As it can be seen using data.table with this:

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

or

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

seems to be the fastest

Update for all columns:

Using @zx8754 's data

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))

You could do:

#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 

And just to confirm results are the same:

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" 

Answer 4

Following LyzandeR's suggestion, I'll add another answer:

require(data.table)
setDT(z)[, .N, by=.(a,b)][order(-N), .(b=b[1L]), keyby=a]