目前可以推荐使用strsplitfrom base产生的另一个基准将列中的逗号分隔字符串拆分为单独的行，因为它在各种大小范围内都是最快的：
s <- strsplit(v$director, ",", fixed=TRUE)
s <- data.frame(director=unlist(s), AB=rep(v$AB, lengths(s)))
请注意，使用fixed=TRUE对时间有重大影响。
比较方法：
met <- alist(base = {s <- strsplit(v$director, ",") #Matthew Lundberg
s <- data.frame(director=unlist(s), AB=rep(v$AB, sapply(s, FUN=length)))}
, baseLength = {s <- strsplit(v$director, ",") #Rich Scriven
s <- data.frame(director=unlist(s), AB=rep(v$AB, lengths(s)))}
, baseLeFix = {s <- strsplit(v$director, ",", fixed=TRUE)
s <- data.frame(director=unlist(s), AB=rep(v$AB, lengths(s)))}
, cSplit = s <- cSplit(v, "director", ",", direction = "long") #A5C1D2H2I1M1N2O1R2T1
, dt = s <- setDT(v)[, lapply(.SD, function(x) unlist(tstrsplit(x, "," #Jaap
, fixed=TRUE))), by = AB][!is.na(director)]
#, dt2 = s <- setDT(v)[, strsplit(director, "," #Jaap #Only Unique
# , fixed=TRUE), by = .(AB, director)][,.(director = V1, AB)]
, dplyr = {s <- v %>% #Jaap
mutate(director = strsplit(director, ",", fixed=TRUE)) %>%
unnest(director)}
, tidyr = s <- separate_rows(v, director, sep = ",") #Jaap
, stack = s <- stack(setNames(strsplit(v$director, ",", fixed=TRUE), v$AB)) #Jaap
#, dt3 = {s <- setDT(v)[, strsplit(director, ",", fixed=TRUE), #Uwe #Only Unique
# by = .(AB, director)][, director := NULL][, setnames(.SD, "V1", "director")]}
, dt4 = {s <- setDT(v)[, .(director = unlist(strsplit(director, "," #Uwe
, fixed = TRUE))), by = .(AB)]}
, dt5 = {s <- vT[, .(director = unlist(strsplit(director, "," #Uwe
, fixed = TRUE))), by = .(AB)]}
)
图书馆：
library(microbenchmark)
library(splitstackshape) #cSplit
library(data.table) #dt, dt2, dt3, dt4
#setDTthreads(1) #Looks like it has here minor effect
library(dplyr) #dplyr
library(tidyr) #dplyr, tidyr
数据：
v0 <- data.frame(director = c("Aaron Blaise,Bob Walker", "Akira Kurosawa",
"Alan J. Pakula", "Alan Parker", "Alejandro Amenabar", "Alejandro Gonzalez Inarritu",
"Alejandro Gonzalez Inarritu,Benicio Del Toro", "Alejandro González Iñárritu",
"Alex Proyas", "Alexander Hall", "Alfonso Cuaron", "Alfred Hitchcock",
"Anatole Litvak", "Andrew Adamson,Marilyn Fox", "Andrew Dominik",
"Andrew Stanton", "Andrew Stanton,Lee Unkrich", "Angelina Jolie,John Stevenson",
"Anne Fontaine", "Anthony Harvey"), AB = c('A', 'B', 'A', 'A', 'B', 'B', 'B', 'A', 'B', 'A', 'B', 'A', 'A', 'B', 'B', 'B', 'B', 'B', 'B', 'A'))
计算和计时结果：
n <- 10^(0:5)
x <- lapply(n, function(n) {v <- v0[rep(seq_len(nrow(v0)), n),]
vT <- setDT(v)
ti <- min(100, max(3, 1e4/n))
microbenchmark(list = met, times = ti, control=list(order="block"))})
y <- do.call(cbind, lapply(x, function(y) aggregate(time ~ expr, y, median)))
y <- cbind(y[1], y[-1][c(TRUE, FALSE)])
y[-1] <- y[-1] / 1e6 #ms
names(y)[-1] <- paste("n:", n * nrow(v0))
y #Time in ms
# expr n: 20 n: 200 n: 2000 n: 20000 n: 2e+05 n: 2e+06
#1 base 0.2989945 0.6002820 4.8751170 46.270246 455.89578 4508.1646
#2 baseLength 0.2754675 0.5278900 3.8066300 37.131410 442.96475 3066.8275
#3 baseLeFix 0.2160340 0.2424550 0.6674545 4.745179 52.11997 555.8610
#4 cSplit 1.7350820 2.5329525 11.6978975 99.060448 1053.53698 11338.9942
#5 dt 0.7777790 0.8420540 1.6112620 8.724586 114.22840 1037.9405
#6 dplyr 6.2425970 7.9942780 35.1920280 334.924354 4589.99796 38187.5967
#7 tidyr 4.0323765 4.5933730 14.7568235 119.790239 1294.26959 11764.1592
#8 stack 0.2931135 0.4672095 2.2264155 22.426373 289.44488 2145.8174
#9 dt4 0.5822910 0.6414900 1.2214470 6.816942 70.20041 787.9639
#10 dt5 0.5015235 0.5621240 1.1329110 6.625901 82.80803 636.1899
注意，方法如
(v <- rbind(v0[1:2,], v0[1,]))
# director AB
#1 Aaron Blaise,Bob Walker A
#2 Akira Kurosawa B
#3 Aaron Blaise,Bob Walker A
setDT(v)[, strsplit(director, "," #Jaap #Only Unique
, fixed=TRUE), by = .(AB, director)][,.(director = V1, AB)]
# director AB
#1: Aaron Blaise A
#2: Bob Walker A
#3: Akira Kurosawa B
strsplit为unique director返回 a并且可能与
tmp <- unique(v)
s <- strsplit(tmp$director, ",", fixed=TRUE)
s <- data.frame(director=unlist(s), AB=rep(tmp$AB, lengths(s)))
但据我了解，这没有被问到。
