Faster way to subset on rows of a data frame in R?

Question

I have been using these 2 methods interchangeably to subset data from a data frame in R.
Method 1
subset_df <- df[which(df$age>5) , ]
Method 2
subset_df <- subset(df, age>5)

I had 2 questions belonging to these.
1. Which one is faster considering I have very large data?
2. This post here Subsetting data frames in R suggests that there is in fact difference between above 2 methods. One of them handles NA accurately. Which one is safe to use then?

Answer 1

The question asks for a faster way to subset rows of a data frame. The fastest way is with data.table.

set.seed(1)  # for reproducible example
# 1 million rows - big enough?
df <- data.frame(age=sample(1:65,1e6,replace=TRUE),x=rnorm(1e6),y=rpois(1e6,25))

library(microbenchmark)
microbenchmark(result<-df[which(df$age>5),],
               result<-subset(df, age>5), 
               result<-df[df$age>5,],
               times=10)
# Unit: milliseconds
#                               expr       min        lq    median       uq      max neval
#  result <- df[which(df$age > 5), ]  77.01055  80.62678  81.43786 133.7753 145.4756    10
#      result <- subset(df, age > 5) 190.89829 193.04221 197.49973 203.7571 263.7738    10
#         result <- df[df$age > 5, ] 169.85649 171.02084 176.47480 185.9394 191.2803    10

library(data.table)
DT <- as.data.table(df)     # data.table
microbenchmark(DT[age > 5],times=10)
# Unit: milliseconds
#         expr      min       lq  median       uq      max neval
#  DT[age > 5] 29.49726 29.93907 30.1813 30.67168 32.81204    10

So in this simple case data.table is a little more than twice as fast as which(...), and more than 6 times faster than subset(...).

Answer 2

I re-write code by adding:

• subsetting operator [[;

• filter from "dplyr" package;

• function that uses standard evaluation.

# 1. Libraries
library(microbenchmark)
library(data.table)
library(dplyr)

# 2. Reproducibility
set.seed(1)

# 3. Create data structures (1e6 rows)

# 3.1. Data frame 
df <- data.frame(
  age = sample(1:65, 1e6, replace = TRUE),
  x = rnorm(1e6),
  y = rpois(1e6,25))

# 3.2. Data table
dt <- as.data.table(df)

# 4. Helper functions

# 4.1. Function that uses standard evaluation
# http://adv-r.had.co.nz/Computing-on-the-language.html
subset2_q <- function(x, condition) {
  r <- eval(condition, x, parent.frame())
  x[r, ]
}

subset2 <- function(x, condition) {
  subset2_q(x, substitute(condition))
}

# 5. Benchmarks
microbenchmark(

  # 5.1. Data frame (basic operations)
  df[which(df$age > 5), ],
  df[df$age > 5, ],
  subset(df, age > 5),
  df[df[['age']] > 5, ],

  # 5.2. Data frame (dplyr)
  df %>% filter(age > 5),

  # 5.3. Data table (basic)
  dt[age > 5],
  dt %>% filter(age > 5), 

  # 5.4. Data frame and table with 'subset2'
  dt %>% subset2(age > 5),
  df %>% subset2(age > 5),

  # 5.5. How many times
  times = 10)

# Results

expr      min       lq      mean   median       uq       max neval cld
 df[which(df$age > 5), ] 83.07726 88.77624 102.20981 90.08606 91.52631 212.10305    10   b
        df[df$age > 5, ] 72.17319 79.98209  80.68900 81.42234 82.33832  84.46876    10   b
     subset(df, age > 5) 84.95796 85.90815  88.79125 88.03345 89.49680  95.37453    10   b
   df[df[["age"]] > 5, ] 71.39021 80.22755  81.86848 81.33061 82.38236 104.97732    10   b
  df %>% filter(age > 5) 21.37622 21.97020  23.57504 22.27681 25.17569  29.64354    10  a 
             dt[age > 5] 20.26226 20.55946  36.58179 25.24155 29.68587 143.57794    10  a 
  dt %>% filter(age > 5) 21.35613 21.76579  25.57424 22.02750 30.99570  32.18407    10  a 
 dt %>% subset2(age > 5) 20.41449 20.57485  23.93314 20.70827 28.63391  31.15306    10  a 
 df %>% subset2(age > 5) 77.43044 79.63956  92.24558 80.80100 81.61990 197.36958    10   b

The best results were for data.table and df %>% filter(age > 5) operators. So, data.frame with dplyr can also be useful.