xbs*_*bsd 19 r reshape data.table
我在SO上已经阅读了一些关于类似问题的参考,但是还没有找到解决方案,并且想知道是否有任何方法只使用data.table来执行以下操作.
我将使用一个简化的示例,但在实践中,我的数据表有> 1000列,类似于var1,var2,... var1000等.
dt <- data.table(uid=c("a","b"), var1=c(1,2), var2=c(100,200))
我正在寻找一种解决方案,可以让我得到一个类似于reshape融化功能的输出 -
> melt(dt, id=c("uid"))
uid variable value
1 a var1 1
2 b var1 2
3 a var2 100
4 b var2 200
也就是说,除了uid之外的所有列都列在单个列下,并且在相邻列中具有相应的值.我已经尝试过使用list等的组合,但可能会遗漏一些明显的东西.
dt中的所有uid都是唯一的.
提前致谢.
And*_*eas 18
对于data.table重塑,请尝试以下操作:
dt[, list(variable = names(.SD), value = unlist(.SD, use.names = F)), by = uid]
语法的代价是值得的; 该功能运行得非常快!
stack通常表现优异melt.
解决这个问题的直接方法stack是:
dt[, stack(.SD), by = "uid"]
当然,您可以.SDcols根据需要指定.然后,使用setnames()将名称更改为您想要的任何名称.
(自我提升警报)
我写了一些函数并将它们放在一个名为"splitstackshape"的包中.其中一个函数被调用Stacked(),并且在 "splitstackshape"软件包的1.2.0版本中,应该可以非常快速地工作.
它与仅仅堆叠a中的所有剩余列有点不同data.table.它更类似于基础R而reshape()不是melt()"reshape2".这是一个实际的例子Stacked().
我已经创建了一个相当大的data.table测试.我们要堆叠50个数字列,我们要堆叠50个因子列.我还进一步优化了@ Andreas的答案.
set.seed(1)
m1 <- matrix(rnorm(10000*50), ncol = 50)
m2 <- matrix(sample(LETTERS, 10000*50, replace = TRUE), ncol = 50)
colnames(m1) <- paste("varA", sprintf("%02d", 1:50), sep = "_")
colnames(m2) <- paste("varB", sprintf("%02d", 1:50), sep = "_")
dt <- data.table(uid = 1:10000, m1, m2)
test1 <- function() Stacked(dt, "uid", c("varA", "varB"), "_")
## merged.stack
test2 <- function() merged.stack(dt, "uid", c("varA", "varB"), "_")
## unlist(..., use.names = TRUE) -- OPTIMIZED
test3 <- function() {
list(cbind(dt[, "uid", with = FALSE],
dt[, list(variable = rep(names(.SD), each = nrow(dt)),
value = unlist(.SD)),
.SDcols = 2:51]),
cbind(dt[, "uid", with = FALSE],
dt[, list(variable = rep(names(.SD), each = nrow(dt)),
value = unlist(.SD)),
.SDcols = 52:101]))
}
## unlist(..., use.names = FALSE) -- OPTIMIZED
test4 <- function() {
list(cbind(dt[, "uid", with = FALSE],
dt[, list(variable = rep(names(.SD), each = nrow(dt)),
value = unlist(.SD, use.names = FALSE)),
.SDcols = 2:51]),
cbind(dt[, "uid", with = FALSE],
dt[, list(variable = rep(names(.SD), each = nrow(dt)),
value = unlist(.SD, use.names = FALSE)),
.SDcols = 52:101]))
}
## Andreas's current answer
test5 <- function() {
list(dt[, list(variable = names(.SD),
value = unlist(.SD, use.names = FALSE)),
by = uid, .SDcols = 2:51],
dt[, list(variable = names(.SD),
value = unlist(.SD, use.names = FALSE)),
by = uid, .SDcols = 52:101])
}
library(microbenchmark)
microbenchmark(Stacked = test1(), merged.stack = test2(),
unlist.namesT = test3(), unlist.namesF = test4(),
AndreasAns = test5(), times = 3)
# Unit: milliseconds
# expr min lq median uq max neval
# Stacked 391.3251 393.0976 394.8702 421.4185 447.9668 3
# merged.stack 764.3071 769.6935 775.0799 867.2638 959.4477 3
# unlist.namesT 1680.0610 1761.9701 1843.8791 1881.9722 1920.0653 3
# unlist.namesF 215.0827 242.7748 270.4669 270.6944 270.9218 3
# AndreasAns 16193.5084 16249.5797 16305.6510 16793.3832 17281.1154 3
^^我不确定为什么Andreas目前的答案在这里很慢.我所做的"优化"基本上unlist没有使用by,这对"varB"(因子)列产生了巨大的影响.
手动方法仍然比"splitstackshape"中的函数更快,但这些是我们正在谈论的毫秒,以及一些相当紧凑的单行代码!
作为参考,这是输出的Stacked()外观.它list是"堆叠" data.table的,每个堆叠变量的一个列表项.
test1()
# $varA
# uid .time_1 varA
# 1: 1 01 -0.6264538
# 2: 1 02 -0.8043316
# 3: 1 03 0.2353485
# 4: 1 04 0.6179223
# 5: 1 05 -0.2212571
# ---
# 499996: 10000 46 -0.6859073
# 499997: 10000 47 -0.9763478
# 499998: 10000 48 0.6579464
# 499999: 10000 49 0.7741840
# 500000: 10000 50 0.5195232
#
# $varB
# uid .time_1 varB
# 1: 1 01 D
# 2: 1 02 A
# 3: 1 03 S
# 4: 1 04 L
# 5: 1 05 T
# ---
# 499996: 10000 46 A
# 499997: 10000 47 W
# 499998: 10000 48 H
# 499999: 10000 49 U
# 500000: 10000 50 W
而且,这是merged.stack输出的样子.它类似于reshape(..., direction = "long")从基础R 使用时的效果.
test2()
# uid .time_1 varA varB
# 1: 1 01 -0.6264538 D
# 2: 1 02 -0.8043316 A
# 3: 1 03 0.2353485 S
# 4: 1 04 0.6179223 L
# 5: 1 05 -0.2212571 T
# ---
# 499996: 10000 46 -0.6859073 A
# 499997: 10000 47 -0.9763478 W
# 499998: 10000 48 0.6579464 H
# 499999: 10000 49 0.7741840 U
# 500000: 10000 50 0.5195232 W
您可能想尝试melt_我的包Kmisc.melt_本质上是reshape2:::melt.data.frame用C语言完成的大部分繁琐工作的重写,并避免尽可能多的复制和类型强制,以便快速实现.
一个例子:
## devtools::install_github("Kmisc", "kevinushey")
library(Kmisc)
library(reshape2)
library(microbenchmark)
n <- 1E6
big_df <- data.frame( stringsAsFactors=FALSE,
x=sample(letters, n, TRUE),
y=sample(LETTERS, n, TRUE),
za=rnorm(n),
zb=rnorm(n),
zc=rnorm(n)
)
all.equal(
melt <- melt(big_df, id.vars=c('x', 'y')),
melt_ <- melt_(big_df, id.vars=c('x', 'y'))
)
## we don't convert the 'variable' column to factor by default
## if we do, we see they're identical
melt_$variable <- factor(melt_$variable)
stopifnot( identical(melt, melt_) )
microbenchmark( times=5,
melt=melt(big_df, id.vars=c('x', 'y')),
melt_=melt_(big_df, id.vars=c('x', 'y'))
)
给我
Unit: milliseconds
expr min lq median uq max neval
melt 916.40436 931.60031 999.03877 1102.31090 1160.3598 5
melt_ 61.59921 78.08768 90.90615 94.52041 182.0879 5
运气好的话,这对你的数据来说足够快了.