table。tbl <- table(rep(1:nrow(dat),2),unlist(dat))
head(tbl, 2)
# a b x y
# 1 1 0 1 0
# 2 0 1 0 1
或者可能更有效的选项是
library(Matrix)
sM <- sparse.model.matrix(~ -1 + x1 +x2, dat,
contrasts.arg = lapply(dat, contrasts, contrasts = FALSE))
colnames(sM) <- sub(".*\\d", "", colnames(sM))
head(sM, 2)
# 2 x 4 sparse Matrix of class "dgCMatrix"
# a b x y
#1 1 . 1 .
#2 . 1 . 1
将其转换为矩阵，即可将其转换为二进制。
head(as.matrix(sM),2)
# a b x y
#1 1 0 1 0
#2 0 1 0 1
已经有一些好的解决方案发布了，但是对于性能来说都不是最优的。我们可以通过循环每个输入列，并且循环每个输入列中的每个因子级别索引，然后进行直接的整数比较来优化性能。这不是最简洁或最优雅的代码，但是它相当直接和快速。
do.call(cbind,lapply(dat,function(col)
`colnames<-`(do.call(cbind,lapply(seq_along(levels(col)),function(i)
as.integer(as.integer(col)==i)
)),levels(col))
));
性能：
library(Matrix);
library(data.table);
library(microbenchmark);
bgoldst <- function(dat) do.call(cbind,lapply(dat,function(col) `colnames<-`(do.call(cbind,lapply(seq_along(levels(col)),function(i) as.integer(as.integer(col)==i))),levels(col))));
akrun1 <- function(dat) table(rep(1:nrow(dat),2),unlist(dat));
akrun2 <- function(dat) sparse.model.matrix(~-1+x1+x2,dat,contrasts.arg=lapply(dat,contrasts,contrasts=FALSE));
davidar <- function(dat) { dat[,rowid:=.I]; dcast(melt(dat,id='rowid'),rowid~value,length); }; ## requires a data.table
dataminer <- function(dat) t(apply(dat,1,function(x) as.numeric(unique(unlist(dat))%in%x)));
N <- 100L; dat <- data.frame(x1=rep(c('a','b'),N),x2=rep(c('x','y'),N)); datDT <- setDT(copy(dat));
identical(unname(bgoldst(dat)),matrix(as.vector(akrun1(dat)),ncol=4L));
## [1] TRUE
identical(unname(bgoldst(dat)),unname(matrix(as.integer(as.matrix(akrun2(dat))),ncol=4L)));
## [1] TRUE
identical(bgoldst(dat),as.matrix(davidar(datDT)[,rowid:=NULL]));
## [1] TRUE
identical(unname(bgoldst(dat)),matrix(as.integer(dataminer(dat)),ncol=4L));
## [1] TRUE
N <- 100L;
dat <- data.frame(x1=rep(c('a','b'),N),x2=rep(c('x','y'),N)); datDT <- setDT(copy(dat));
microbenchmark(bgoldst(dat),akrun1(dat),akrun2(dat),davidar(datDT),dataminer(dat));
## Unit: microseconds
## expr min lq mean median uq max neval
## bgoldst(dat) 67.570 92.374 106.2853 99.6440 121.2405 188.596 100
## akrun1(dat) 581.182 652.386 773.6300 690.6605 916.4625 1192.299 100
## akrun2(dat) 4429.208 4836.119 5554.5902 5145.3135 5977.0990 11263.537 100
## davidar(datDT) 5064.273 5498.555 6104.7621 5664.9115 6203.9695 11713.856 100
## dataminer(dat) 47577.729 49529.753 55217.3726 53190.8940 60041.9020 74346.268 100
N <- 1e4L;
dat <- data.frame(x1=rep(c('a','b'),N),x2=rep(c('x','y'),N)); datDT <- setDT(copy(dat));
microbenchmark(bgoldst(dat),akrun1(dat),akrun2(dat),davidar(datDT));
## Unit: milliseconds
## expr min lq mean median uq max neval
## bgoldst(dat) 1.775617 1.820949 2.299493 1.84725 1.972124 8.362336 100
## akrun1(dat) 38.954524 41.109257 48.409613 45.60304 52.147633 162.365472 100
## akrun2(dat) 16.915832 17.762799 21.288200 19.20164 23.775180 46.494055 100
## davidar(datDT) 36.151684 38.366715 42.875940 42.38794 45.916937 58.695008 100
N <- 1e5L;
dat <- data.frame(x1=rep(c('a','b'),N),x2=rep(c('x','y'),N)); datDT <- setDT(copy(dat));
microbenchmark(bgoldst(dat),akrun1(dat),akrun2(dat),davidar(datDT));
## Unit: milliseconds
## expr min lq mean median uq max neval
## bgoldst(dat) 17.16473 22.97654 35.01815 26.76662 31.75562 152.6188 100
## akrun1(dat) 501.72644 626.14494 671.98315 680.91152 727.88262 828.8313 100
## akrun2(dat) 212.12381 242.65505 298.90254 272.28203 357.65106 429.6023 100
## davidar(datDT) 368.04924 461.60078 500.99431 511.54921 540.39358 638.3840 100
data.frame（而不是矩阵）一样的数据，您也可以重新整理数据。library(data.table)
setDT(dat)[, rowid := .I] # Creates a row index
res <- dcast(melt(dat, id = "rowid"), rowid ~ value, length) # long/wide format
head(res)
# rowid a b x y
# 1 1 1 0 1 0
# 2 2 0 1 0 1
# 3 3 1 0 1 0
# 4 4 0 1 0 1
# 5 5 1 0 1 0
# 6 6 0 1 0 1
一些基准测试
dat = data.frame(
x1 = rep(c('a', 'b'), 1e3),
x2 = rep(c('x', 'y'), 1e3)
)
library(data.table)
library(Matrix)
library(microbenchmark)
dat2 <- copy(dat)
microbenchmark("akrun1 : " = table(rep(1:nrow(dat),2),unlist(dat)),
"akrun2 : " = sparse.model.matrix(~ -1 + x1 +x2, dat, contrasts.arg = lapply(dat, contrasts, contrasts = FALSE)),
"DatamineR : " = t(apply(dat,1, function(x) as.numeric(unique(unlist(dat)) %in% x))),
"David Ar : " = {setDT(dat2)[, rowid := .I] ; dcast(melt(dat2, id = "rowid"), rowid ~ value, length)},
times = 10L)
# Unit: milliseconds
# expr min lq mean median uq max neval cld
# akrun1 : 3.826075 4.061904 6.654399 5.165376 11.26959 11.82029 10 a
# akrun2 : 5.269531 5.713672 8.794434 5.943422 13.34118 20.01961 10 a
# DatamineR : 3199.336286 3343.774160 3410.618547 3385.756972 3517.22133 3625.70909 10 b
# David Ar : 8.092769 8.254682 11.030785 8.465232 15.44893 19.83914 10 a
apply解决方案效率极低，在更大的数据集上会花费很长时间。在排除apply解决方案的情况下比较更大的数据集。
dat = data.frame(
x1 = rep(c('a', 'b'), 1e4),
x2 = rep(c('x', 'y'), 1e4)
)
dat2 <- copy(dat)
microbenchmark("akrun1 : " = table(rep(1:nrow(dat),2),unlist(dat)),
"akrun2 : " = sparse.model.matrix(~ -1 + x1 +x2, dat, contrasts.arg = lapply(dat, contrasts, contrasts = FALSE)),
#"DatamineR : " = t(apply(dat,1, function(x) as.numeric(unique(unlist(dat)) %in% x))),
"David Ar : " = {setDT(dat2)[, rowid := .I] ; dcast(melt(dat2, id = "rowid"), rowid ~ value, length)},
times = 100L)
# Unit: milliseconds
# expr min lq mean median uq max neval cld
# akrun1 : 38.66744 41.27116 52.97982 42.72534 47.17203 161.0420 100 b
# akrun2 : 17.02006 18.93534 27.27582 19.35580 20.72022 153.2397 100 a
# David Ar : 34.15915 37.91659 46.11050 38.58536 41.40412 149.0038 100 b
看起来 Matrix 包在处理更大的数据集时表现出色。
当存在更多列/唯一值的情况下，比较不同的场景可能是值得的。
apply的替代方法。head(t(apply(dat,1, function(x) as.numeric(unique(unlist(dat)) %in% x))))
[,1] [,2] [,3] [,4]
[1,] 1 0 1 0
[2,] 0 1 0 1
[3,] 1 0 1 0
[4,] 0 1 0 1
[5,] 1 0 1 0
[6,] 0 1 0 1