Efficient Means of Identifying Number of Distinct

问题:

library(dplyr)

I have the following data set

set.seed(123)
n <- 1e6
d <- data.frame(a = letters[sample(5, n, replace = TRUE)], b = letters[sample(5, n, replace = TRUE)], c = letters[sample(5, n, replace = TRUE)],  d = letters[sample(5, n, replace = TRUE)]) 

And I would like to count the number of distinct letters in each row. To do this I use

sapply(as.data.frame(t(d)), function(x) n_distinct(x))

However because this approach is implementing a loop, it is slow. Do you have an suggestions on how to speed this up?

My laptop is a piece of junk so...

system.time(sapply(as.data.frame(t(d)), function(x) n_distinct(x)))
  user  system elapsed 
185.78    0.86  208.08 

回答1:

If the different values are not so many, you can try:

d<-as.matrix(d)
uniqueValues<-unique(as.vector(d))
Reduce("+",lapply(uniqueValues,function(x) rowSums(d==x)>0))

For the example you provided, this is much faster than other solutions and yields the same result.

回答2:

You can try,

system.time(colSums(apply(d, 1, function(i) !duplicated(i))))
#user  system elapsed 
#6.50    0.02    6.53 

回答3:

Here are some options that are faster (on my machine) than the OP's method (included methods in the other posts)

system.time({ #@nicola's function
 d<-as.matrix(d)
 uniqueValues<-unique(as.vector(d))
 Reduce("+",lapply(uniqueValues,function(x) rowSums(d==x)>0))
})
#   user  system elapsed 
#  0.61    0.00    0.61 

system.time(colSums(apply(d, 1, function(i) !duplicated(i)))) #@Sotos function
#   user  system elapsed 
#  8.16    0.00    8.18 


system.time(apply(d, 1, function(x) sum(!duplicated(x))))
#  user  system elapsed 
#  8.19    0.01    8.25 



system.time(apply(d, 1, uniqueN)) #uniqueN from `data.table`
#   user  system elapsed 
#  15.59    0.03   15.74 


system.time(apply(d, 1, n_distinct)) #n_distinct from `dplyr`
#  user  system elapsed 
# 31.50    0.04   53.82 

system.time(sapply(as.data.frame(t(d)), function(x) n_distinct(x)))
#   user  system elapsed 
# 70.12    0.36   72.03