将列中以逗号分隔的字符串拆分为单独的行
我有一个数据框，如下所示：
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'))
如您所见，director列中的某些条目是以逗号分隔的多个名称。我想将这些条目拆分为单独的行，同时保持另一列的值。例如，上面数据框中的第一行应分为两行，director列中的每一行都有一个名称，AB列中的每个名称都为“A”。
6 个答案:
答案 0 :(得分：75)
有几种选择：
1）data.table的两种方式：
library(data.table)
# method 1 (preferred)
setDT(v)[, lapply(.SD, function(x) unlist(tstrsplit(x, ",", fixed=TRUE))), by = AB
][!is.na(director)]
# method 2
setDT(v)[, strsplit(as.character(director), ",", fixed=TRUE), by = .(AB, director)
][,.(director = V1, AB)]
2）dplyr / tidyr组合：或者，您也可以使用dplyr / tidyr组合：
library(dplyr)
library(tidyr)
v %>%
mutate(director = strsplit(as.character(director), ",")) %>%
unnest(director)
3）仅限tidyr ：使用tidyr 0.5.0（及更高版本），您也可以使用separate_rows：
separate_rows(v, director, sep = ",")
您可以使用convert = TRUE参数自动将数字转换为数字列。
4）基础R：
# if 'director' is a character-column:
stack(setNames(strsplit(df$director,','), df$AB))
# if 'director' is a factor-column:
stack(setNames(strsplit(as.character(df$director),','), df$AB))
答案 1 :(得分：62)
这个老问题经常被用作欺骗目标（用r-faq标记）。截至今天，已经回答了三次，提供了6种不同的方法，但缺乏基准作为指导哪种方法是最快的 1 。
基准测试解决方案包括
- Matthew Lundberg's base R approach但根据Rich Scriven's comment修改，
- Jaap's两种
data.table方法和两种dplyr/tidyr方法， - Ananda's
splitstackshapesolution， - 以及Jaap的
data.table方法的两个其他变体。
总共有8种不同的方法使用microbenchmark包对6种不同大小的数据框进行了基准测试（参见下面的代码）。
OP给出的样本数据仅包含20行。为了创建更大的数据帧，这20行简单地重复1次，10次，100次，1000次，10000次和100000次，这给出了最多200万行的问题大小。
基准测试结果
基准测试结果表明，对于足够大的数据帧，所有data.table方法都比任何其他方法更快。对于行数超过5000行的数据帧，Jaap的data.table方法2和变体DT3速度最快，速度比最慢的方法快。
值得注意的是，两个tidyverse方法和splistackshape解决方案的时间非常相似，以至于难以对图表中的曲线进行分类。它们是所有数据帧大小中最慢的基准测试方法。
对于较小的数据帧，Matt的基本R解决方案和data.table方法4似乎比其他方法具有更少的开销。
代码
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")
library(data.table)
library(magrittr)
为问题大小n
的基准运行定义函数
run_mb <- function(n) {
# compute number of benchmark runs depending on problem size `n`
mb_times <- scales::squish(10000L / n , c(3L, 100L))
cat(n, " ", mb_times, "\n")
# create data
DF <- data.frame(director = rep(director, n), AB = rep(AB, n))
DT <- as.data.table(DF)
# start benchmarks
microbenchmark::microbenchmark(
matt_mod = {
s <- strsplit(as.character(DF$director), ',')
data.frame(director=unlist(s), AB=rep(DF$AB, lengths(s)))},
jaap_DT1 = {
DT[, lapply(.SD, function(x) unlist(tstrsplit(x, ",", fixed=TRUE))), by = AB
][!is.na(director)]},
jaap_DT2 = {
DT[, strsplit(as.character(director), ",", fixed=TRUE),
by = .(AB, director)][,.(director = V1, AB)]},
jaap_dplyr = {
DF %>%
dplyr::mutate(director = strsplit(as.character(director), ",")) %>%
tidyr::unnest(director)},
jaap_tidyr = {
tidyr::separate_rows(DF, director, sep = ",")},
cSplit = {
splitstackshape::cSplit(DF, "director", ",", direction = "long")},
DT3 = {
DT[, strsplit(as.character(director), ",", fixed=TRUE),
by = .(AB, director)][, director := NULL][
, setnames(.SD, "V1", "director")]},
DT4 = {
DT[, .(director = unlist(strsplit(as.character(director), ",", fixed = TRUE))),
by = .(AB)]},
times = mb_times
)
}
针对不同的问题规模运行基准
# define vector of problem sizes
n_rep <- 10L^(0:5)
# run benchmark for different problem sizes
mb <- lapply(n_rep, run_mb)
准备绘图数据
mbl <- rbindlist(mb, idcol = "N")
mbl[, n_row := NROW(director) * n_rep[N]]
mba <- mbl[, .(median_time = median(time), N = .N), by = .(n_row, expr)]
mba[, expr := forcats::fct_reorder(expr, -median_time)]
创建图表
library(ggplot2)
ggplot(mba, aes(n_row, median_time*1e-6, group = expr, colour = expr)) +
geom_point() + geom_smooth(se = FALSE) +
scale_x_log10(breaks = NROW(director) * n_rep) + scale_y_log10() +
xlab("number of rows") + ylab("median of execution time [ms]") +
ggtitle("microbenchmark results") + theme_bw()
会话信息＆amp;包版本（摘录）
devtools::session_info()
#Session info
# version R version 3.3.2 (2016-10-31)
# system x86_64, mingw32
#Packages
# data.table * 1.10.4 2017-02-01 CRAN (R 3.3.2)
# dplyr 0.5.0 2016-06-24 CRAN (R 3.3.1)
# forcats 0.2.0 2017-01-23 CRAN (R 3.3.2)
# ggplot2 * 2.2.1 2016-12-30 CRAN (R 3.3.2)
# magrittr * 1.5 2014-11-22 CRAN (R 3.3.0)
# microbenchmark 1.4-2.1 2015-11-25 CRAN (R 3.3.3)
# scales 0.4.1 2016-11-09 CRAN (R 3.3.2)
# splitstackshape 1.4.2 2014-10-23 CRAN (R 3.3.3)
# tidyr 0.6.1 2017-01-10 CRAN (R 3.3.2)
1 我的好奇心被this exuberant comment Brilliant激怒了！更快的订单！到a question的tidyverse答案，该答案已作为此问题的副本而关闭。
答案 2 :(得分：47)
为原始data.frame v命名，我们有：
> s <- strsplit(as.character(v$director), ',')
> data.frame(director=unlist(s), AB=rep(v$AB, sapply(s, FUN=length)))
director AB
1 Aaron Blaise A
2 Bob Walker A
3 Akira Kurosawa B
4 Alan J. Pakula A
5 Alan Parker A
6 Alejandro Amenabar B
7 Alejandro Gonzalez Inarritu B
8 Alejandro Gonzalez Inarritu B
9 Benicio Del Toro B
10 Alejandro González Iñárritu A
11 Alex Proyas B
12 Alexander Hall A
13 Alfonso Cuaron B
14 Alfred Hitchcock A
15 Anatole Litvak A
16 Andrew Adamson B
17 Marilyn Fox B
18 Andrew Dominik B
19 Andrew Stanton B
20 Andrew Stanton B
21 Lee Unkrich B
22 Angelina Jolie B
23 John Stevenson B
24 Anne Fontaine B
25 Anthony Harvey A
请注意使用rep构建新的AB列。这里，sapply返回每个原始行中的名称数。
答案 3 :(得分：29)
晚会，但另一个通用的替代方法是使用{“1}}来自我的”splitstackshape“包，其中包含cSplit个参数。将其设置为direction以获得您指定的结果：
"long"答案 4 :(得分：0)
devtools::install_github("yikeshu0611/onetree")
library(onetree)
dd=spread_byonecolumn(data=mydata,bycolumn="director",joint=",")
head(dd)
director AB
1 Aaron Blaise A
2 Bob Walker A
3 Akira Kurosawa B
4 Alan J. Pakula A
5 Alan Parker A
6 Alejandro Amenabar B
答案 5 :(得分：0)
当前建议使用 base 中的strsplit生成的另一个基准测试将一列中用逗号分隔的字符串拆分为单独的行，因为在各种尺寸中最快：
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会对计时产生重大影响。
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，类似
的方法(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
为strsplit 导演返回一个unique，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，这不是要求的。
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