Sté*_*ent 4 r plyr dplyr data.table
比如说我有一个带有列的数据框,a我想a^i为几个值创建列i.
> dat <- data.frame(a=1:5)
> dat
a
1 1
2 2
3 3
4 4
5 5
举个例子,我想要的输出i=2:5:
a power_2 power_3 power_4 power_5
1 1 1 1 1 1
2 2 4 8 16 32
3 3 9 27 81 243
4 4 16 64 256 1024
5 5 25 125 625 3125
目前我得到的输出data.table如下:
DT <- data.table(dat)
exponents <- 2:5
DT[, paste0("power_",exponents):=lapply(exponents, function(p) a^p)]
怎么办plyr/ dplyr？当然,我可以通过键入power_i=a^i每个来做如下,i但这不是我想要的.
mutate(dat, power_2=a^2, power_3=a^3, ...)
已经提出了几个答案,并且已经通过@docendo discimus进行了比较.我只是加上比较data.table.
library(data.table)
library(dplyr)
set.seed(2015)
dat <- data.frame(a = sample(1000))
i <- 2:5
n <- c(names(dat), paste0("power_", i))
DT <- data.table(dat)
library(microbenchmark)
microbenchmark(
data.table = DT[, paste0("power_",i):=lapply(i, function(k) a^k)],
Henrik = dat %>% do(data.frame(., outer(.$a, i, `^`))) %>% setNames(n),
dd.do = dat %>% do(data.frame(., sapply(i, function(x) .$a^x))) %>% setNames(n),
dd.bc = dat %>% bind_cols(as.data.frame(lapply(i, function(x) .$a^x))) %>% setNames(n),
times = 30,
unit = "relative"
)
Unit: relative
expr min lq mean median uq max neval cld
data.table 1.022945 1.039674 1.108558 1.026319 1.083644 2.370180 30 a
Henrik 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 30 a
dd.do 1.149195 1.160735 1.167672 1.158141 1.150280 1.268279 30 a
dd.bc 14.350034 13.982658 13.737964 13.632361 13.606221 15.866711 30 b
通过两个base解决方案更新基准,Henrik2和josh(来自他的评论),这是最快的:
set.seed(2015)
dat <- data.frame(a = sample(1000))
microbenchmark(
data.table = DT[, paste0("power_",i):=lapply(i, function(k) a^k)],
Henrik = dat %>% do(data.frame(., outer(.$a, i, `^`))) %>% setNames(n),
Henrik2 = cbind(dat, outer(dat$a, setNames(i, paste0("power_", i)), `^`)),
dd.do = dat %>% do(data.frame(., sapply(i, function(x) .$a^x))) %>% setNames(n),
dd.bc = dat %>% bind_cols(as.data.frame(lapply(i, function(x) .$a^x))) %>% setNames(n),
josh = data.frame(dat, setNames(lapply(2:5, function(X) dat$a^X), paste0("power_", 2:5))),
times = 30,
unit = "relative"
)
# Unit: relative
# expr min lq mean median uq max neval cld
# data.table 1.991613 2.029778 1.982169 1.990417 1.946677 1.694030 30 bc
# Henrik 2.026345 2.017179 1.996419 2.003189 2.030176 1.733583 30 bc
# Henrik2 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 30 a
# dd.do 2.356886 2.375713 2.322452 2.348053 2.304826 2.101494 30 c
# dd.bc 37.445491 36.081298 34.791638 34.783854 34.787655 27.832116 30 d
# josh 1.725750 1.699887 1.641290 1.625331 1.637823 1.330598 30 b
一种可能性是使用outerin do,然后设置名称setNames
i <- 2:5
dat %>%
do(data.frame(., outer(.$a, i, `^`))) %>%
setNames(., c("a", paste0("power_", i)))
# a power_2 power_3 power_4 power_5
# 1 1 1 1 1 1
# 2 2 4 8 16 32
# 3 3 9 27 81 243
# 4 4 16 64 256 1024
# 5 5 25 125 625 3125
如果你首先命名'power vector'"i",你可以调用cbind而不是do和data.frame,并且dplyr在这种特殊情况下我看不到立即需要函数.
cbind(dat, outer(dat$a, setNames(i, paste0("power_", i)), `^`))
# a power_2 power_3 power_4 power_5
# 1 1 1 1 1 1
# 2 2 4 8 16 32
# 3 3 9 27 81 243
# 4 4 16 64 256 1024
# 5 5 25 125 625 3125
的base,非do代码是你的大样本数据的速度更快.我还添加了base@Josh O'Brien 的解决方案.
set.seed(2015)
dat <- data.frame(a = sample(1000))
microbenchmark(
data.table = DT[, paste0("power_",i):=lapply(i, function(k) a^k)],
Henrik = dat %>% do(data.frame(., outer(.$a, i, `^`))) %>% setNames(n),
Henrik2 = cbind(dat, outer(dat$a, setNames(i, paste0("power_", i)), `^`)),
dd.do = dat %>% do(data.frame(., sapply(i, function(x) .$a^x))) %>% setNames(n),
dd.bc = dat %>% bind_cols(as.data.frame(lapply(i, function(x) .$a^x))) %>% setNames(n),
josh = data.frame(dat, setNames(lapply(2:5, function(X) dat$a^X), paste0("power_", 2:5))),
times = 30,
unit = "relative"
)
# Unit: relative
# expr min lq mean median uq max neval cld
# data.table 1.991613 2.029778 1.982169 1.990417 1.946677 1.694030 30 bc
# Henrik 2.026345 2.017179 1.996419 2.003189 2.030176 1.733583 30 bc
# Henrik2 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 30 a
# dd.do 2.356886 2.375713 2.322452 2.348053 2.304826 2.101494 30 c
# dd.bc 37.445491 36.081298 34.791638 34.783854 34.787655 27.832116 30 d
# josh 1.725750 1.699887 1.641290 1.625331 1.637823 1.330598 30 b
这是一个使用选项do:
i <- 2:5
n <- c(names(dat), paste0("power_", i))
dat %>% do(data.frame(., sapply(i, function(x) .$a^x))) %>% setNames(n)
# a power_2 power_3 power_4 power_5
#1 1 1 1 1 1
#2 2 4 8 16 32
#3 3 9 27 81 243
#4 4 16 64 256 1024
#5 5 25 125 625 3125
另一种选择,使用bind_cols:
dat %>% bind_cols(as.data.frame(lapply(i, function(x) .$a^x))) %>% setNames(n)
# a power_2 power_3 power_4 power_5
#1 1 1 1 1 1
#2 2 4 8 16 32
#3 3 9 27 81 243
#4 4 16 64 256 1024
#5 5 25 125 625 3125
评论后编辑:
@Henrik的解决方案比我的快:
set.seed(2015)
dat <- data.frame(a = sample(1000))
i <- 2:5
n <- c(names(dat), paste0("power_", i))
library(microbenchmark)
microbenchmark(
Henrik = dat %>% do(data.frame(., outer(.$a, i, `^`))) %>% setNames(n),
dd.do = dat %>% do(data.frame(., sapply(i, function(x) .$a^x))) %>% setNames(n),
dd.bc = dat %>% bind_cols(as.data.frame(lapply(i, function(x) .$a^x))) %>% setNames(n),
times = 30,
unit = "relative"
)
Unit: relative
expr min lq median uq max neval
Henrik 1.000000 1.000000 1.000000 1.000000 1.000000 30
dd.do 1.138506 1.179104 1.173298 1.149581 2.660237 30
dd.bc 18.862923 18.702178 18.058984 17.537727 16.426538 30