我在使用data.table时搜索了一种计算向量的所有向量级别的累积和（列表）的有效方法。
dataframe / data.table DT最初由四个变量组成，一个名为 experience 。目标是一个向量，其中包含体验中的因子级别的累积计数条件性的另外两个变量 id 和 cl 。值得注意的是，因子经验具有比数据集中更多的因子水平（这是必要的属性）。
数据看起来像
id trial experience cl
1: 1 1 000A A
2: 1 2 000A A
3: 1 3 000B B
4: 1 4 111A A
5: 1 5 001B B
6: 2 1 100B B
7: 2 2 111A A
8: 2 3 100B B
9: 2 4 010A A
10: 2 5 011B B
经验的因子水平为16
levels(DT$experience)
# [1] "000A" "001A" "010A" "011A" "100A" "101A" "110A" "111A"
# [9] "000B" "001B" "010B" "011B" "100B" "101B" "110B" "111B"
我们想要计算的是以 id 和 cl 为条件的体验的累积计数。考虑前三行：对于 id = 1，第一个经验值是000A，所以计数器变量 c000A = 1.第二个经验值也是000A，所以计数器 c000A = 2.但是现在第三个经验值是000B，因此前一个计数器 c000A 保持2，但另一个计数器 c000B = 1，在那之前是0。
遵循这个逻辑，我们想要的结果如下：
id trial experience cl c000A c001A c010A c011A c100A c101A c110A c111A c000B c001B c010B c011B c100B c101B c110B c111B
1: 1 1 000A A 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2: 1 2 000A A 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
3: 1 3 000B B 2 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
4: 1 4 111A A 2 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0
5: 1 5 001B B 2 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0
6: 2 1 100B B 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
7: 2 2 111A A 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0
8: 2 3 100B B 0 0 0 0 0 0 0 1 0 0 0 0 2 0 0 0
9: 2 4 010A A 0 0 1 0 0 0 0 1 0 0 0 0 2 0 0 0
10: 2 5 011B B 0 0 1 0 0 0 0 1 0 0 0 1 2 0 0 0
注意：将16个条目 c000A，...，c111B 分配给不同的列并不重要。如果结果是一个有16个条目的向量作为c000A，c001A，...，c110B，c111B来保存累积计数就足够了。
我使用的当前代码是以下两步法。它既不漂亮也不优雅。
foo <- function(DT){
# tabulate experience for each trial
# store in an auxiliary variables <s000A, s001A, ..., s110B, s111B>
DT[, paste(sep="","s",levels(DT$experience)) := as.list(table(experience)), by = c("id","cl","trial")]
# sum each of the s____ variables by id
DT[, "c000A" := cumsum(s000A), by = id] # this is clumsy
DT[, "c001A" := cumsum(s001A), by = id]
DT[, "c010A" := cumsum(s010A), by = id]
DT[, "c011A" := cumsum(s011A), by = id]
DT[, "c100A" := cumsum(s100A), by = id]
DT[, "c101A" := cumsum(s101A), by = id]
DT[, "c110A" := cumsum(s110A), by = id]
DT[, "c111A" := cumsum(s111A), by = id]
DT[, "c000B" := cumsum(s000B), by = id]
DT[, "c001B" := cumsum(s001B), by = id]
DT[, "c010B" := cumsum(s010B), by = id]
DT[, "c011B" := cumsum(s011B), by = id]
DT[, "c100B" := cumsum(s100B), by = id]
DT[, "c101B" := cumsum(s101B), by = id]
DT[, "c110B" := cumsum(s110B), by = id]
DT[, "c111B" := cumsum(s111B), by = id]
}
对于具有n = 1e + 4次试验和2次ID的数据集，此代码采用：
system.time(foo(DT))
# User System verstrichen
# 9.78 0.00 10.05
library("data.table")
library("R.utils")
# Sample dataframe DF with n=1e+4
n <- 1e+4 #to test change this to n=5
DT <- data.table(id = rep(1:2,each=n), trial = rep(1:n,2), experience = c("000A","000A","000B","111A","001B","100B","111A","100B","010A","011B"), cl = c("A","A","B","A","B","B","A","B","A","B")) # experience needs to be a factor w more levels
DT$experience <- factor(DT$experience, levels = paste(sep="", intToBin(0:7), rep(c("A","B"),each=8)))
setkey(DT,id,trial,cl) #set the data.table keys
谁拥有更快更优雅的解决方案？
谢谢！ 亚娜
library("microbenchmark")
benchmk <- microbenchmark(
DT2 <- foo2(DT),
DT3a <- foo3a(DT),
DT3b <- foo3b(DT),
times=100L
)
print(benchmk)
# with n=1e+4
#
# unit milliseconds
# expr min lq median uq max neval
# DT2 <- foo2(DT) 46.96745 52.17469 74.72479 120.93339 212.7912 100
# DT3a <- foo3a(DT) 25.21907 26.57921 28.84702 34.89401 121.3164 100
# DT3b <- foo3b(DT) 19.82076 20.80570 22.87369 30.83561 148.0520 100
# with n=1e+5
#
# unit milliseconds
# expr min lq median uq max neval
# DT2 <- foo2(DT) 386.93890 445.0184 481.4660 534.9619 1160.6151 100
# DT3a <- foo3a(DT) 144.45937 154.5672 170.6048 233.6362 494.8972 100
# DT3b <- foo3b(DT) 95.91988 100.5313 110.4060 125.1678 364.5651 100
foo2对应Eddi的代码
foo2 <- function(DT){
DT[, counter := 1:.N]
DT[, dummy := 1]
RE <- dcast.data.table(DT, counter+id ~ experience, value.var = 'dummy', fill = 0)[,lapply(.SD, cumsum), by = id, .SDcols = c(-1,-2)]
RE[, setdiff(levels(DT$experience), unique(DT$experience)) := 0]
setcolorder(RE, c("id",levels(DT$experience)))
}
foo3a使用级别
与Arun的第一个代码对应foo3a <- function(DT){
ex = levels(DT$experience)
DT[, c(ex) := 0L]
tmp = DT[, list(list(.I)), by=experience]
tmp[, experience := as.character(experience)] ## convert to char
for(i in seq(nrow(tmp))) {
set(DT, i=tmp$V1[[i]], j=tmp$experience[i], val=1L)
}
DT[, c(ex) := lapply(.SD, cumsum), by=id, .SDcols=ex]
}
foo3b对应使用字符
的Arun代码foo3b <- function(DT){
ex = levels(DT$experience)
DT[, c(ex) := 0L]
tmp = DT[, list(list(.I)), by=experience]
tmp[, experience := as.character(experience)] ## convert to char
for(i in seq(nrow(tmp))) {
set(DT, i=tmp$V1[[i]], j=tmp$experience[i], val=1L)
}
ex = as.character(unique(DT$experience)) ## rewrite 'ex'
DT[, c(ex) := lapply(.SD, cumsum), by=id, .SDcols=ex]
}
答案 0 :(得分：4)
这个怎么样？
首先创建所有列并将它们初始化为0L。
ex = levels(DT$experience)
DT[, c(ex) := 0L]
现在，按experience分组并获取列表中每个experience对应的行号，如下所示：
tmp = DT[, list(list(.I)), by=experience]
tmp[, experience := as.character(experience)] ## convert to char
然后，您可以循环搜索每一列并使用set与V1的相应行（来自列experience）和列（来自tmp列） } 1将DT分配给for(i in seq(nrow(tmp))) {
set(DT, i=tmp$V1[[i]], j=tmp$experience[i], val=1L)
}
中的相应列，如下所示：
cumsum最后id在每列上DT[, c(ex) := lapply(.SD, cumsum), by=id, .SDcols=ex]
：
dcast.data.table总共需要0.013秒（as.character(unique(DT$experience))解决方案，这也很好，耗时0.027秒）。
如果您在最后一行使用ex而不是cumsum，则可能会节省更多时间..因为有些列全部为0且您没有{{1他们。那就是：
ex = as.character(unique(DT$experience)) ## rewrite 'ex'
DT[, c(ex) := lapply(.SD, cumsum), by=id, .SDcols=ex]
答案 1 :(得分：2)
或许这样的事情：
# add some extra variables
DT[, counter := 1:.N]
DT[, dummy := 1]
dcast.data.table(DT, counter+id ~ experience, value.var = 'dummy', fill = 0)[,
lapply(.SD, cumsum), by = id, .SDcols = c(-1,-2)]
# id 000A 010A 111A 000B 001B 011B 100B
# 1: 1 1 0 0 0 0 0 0
# 2: 1 2 0 0 0 0 0 0
# 3: 1 2 0 0 1 0 0 0
# 4: 1 2 0 1 1 0 0 0
# 5: 1 2 0 1 1 1 0 0
# ---
#19996: 2 2000 999 1999 1000 1000 999 1999
#19997: 2 2000 999 2000 1000 1000 999 1999
#19998: 2 2000 999 2000 1000 1000 999 2000
#19999: 2 2000 1000 2000 1000 1000 999 2000
#20000: 2 2000 1000 2000 1000 1000 1000 2000
如果你愿意，你可以cbind回来。