On Mon, 14 Nov 2016, Marc Schwartz wrote:
On Nov 14, 2016, at 11:26 AM, Charles C. Berry <ccbe...@ucsd.edu> wrote:
On Mon, 14 Nov 2016, Bert Gunter wrote:
[stuff deleted]
Hi,
Both gsub() and strsplit() are using regex based pattern matching
internally. That being said, they are ultimately calling .Internal code,
so both are pretty fast.
For comparison:
## Create a 1,000,000 character vector
set.seed(1)
Vec <- paste(sample(letters, 1000000, replace = TRUE), collapse = "")
nchar(Vec)
[1] 1000000
## Split the vector into single characters and tabulate
table(strsplit(Vec, split = "")[[1]])
a b c d e f g h i j k l
38664 38442 38282 38496 38540 38623 38548 38288 38143 38493 38184 38621
m n o p q r s t u v w x
38306 38725 38705 38144 38529 38809 38575 38355 38386 38364 38904 38310
y z
38265 38299
## Get just the count of "a"
table(strsplit(Vec, split = "")[[1]])["a"]
a
38664
nchar(gsub("[^a]", "", Vec))
[1] 38664
## Check performance
system.time(table(strsplit(Vec, split = "")[[1]])["a"])
user system elapsed
0.100 0.007 0.107
system.time(nchar(gsub("[^a]", "", Vec)))
user system elapsed
0.270 0.001 0.272
So, the above would suggest that using strsplit() is somewhat faster
than using gsub(). However, as Chuck notes, in the absence of more
exhaustive benchmarking, the difference may or may not be more
generalizable.
Whether splitting on fixed strings rather than treating them as
regex'es (i.e.`fixed=TRUE') makes a big difference seems to depend on
what you split:
First repeating what Marc did...
system.time(table(strsplit(Vec, split = "",fixed=TRUE)[[1]])["a"])
user system elapsed
0.132 0.010 0.139
system.time(table(strsplit(Vec, split = "",fixed=FALSE)[[1]])["a"])
user system elapsed
0.130 0.010 0.138
... fixed=TRUE hardly matters. But the idiom I proposed...
system.time(sum(lengths(strsplit(paste0("X", Vec, "X"),"a",fixed=TRUE)) - 1))
user system elapsed
0.017 0.000 0.018
system.time(sum(lengths(strsplit(paste0("X", Vec, "X"),"a",fixed=FALSE)) - 1))
user system elapsed
0.104 0.000 0.104
... is 5 times faster with fixed=TRUE for this case.
This result matchea Marc's count:
sum(lengths(strsplit(paste0("X", Vec, "X"),"a",fixed=FALSE)) - 1)
[1] 38664
Chuck
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