如果我们想要获得两个向量的所有组合,我们可以使用rep
/回收规则:
x <- 1:4
y <- 1:2
cbind(rep(x, each = length(y)), rep(y, length(x)))
# [,1] [,2]
# [1,] 1 1
# [2,] 1 2
# [3,] 2 1
# [4,] 2 2
# [5,] 3 1
# [6,] 3 2
# [7,] 4 1
# [8,] 4 2
但expand.grid
更好 - 它为我们处理所有重复。
expand.grid(x, y)
# Var1 Var2
# 1 1 1
# 2 2 1
# 3 3 1
# 4 4 1
# 5 1 2
# 6 2 2
# 7 3 2
# 8 4 2
是否有一个简单的版本用于连接字符串?像paste.grid
一样?我有一个命名对象,其中许多对象的名称如x_y_z
,其中x
,y
和z
的变化类似x
和{{1}上面。
例如,假设y
可以是x
或"avg"
,"median"
可以是y
或"male"
,"female"
}可以是z
或"height"
。我们怎样才能简明地得到这三种组合的所有8种组合?
使用"weight"
是一件痛苦的事:
rep
重新调整x <- c("avg", "median")
y <- c("male", "female")
z <- c("height", "weight")
paste(rep(x, each = length(y) * length(z)),
rep(rep(y, each = length(z)), length(x)),
rep(z, length(x) * length(y)), sep = "_")
有点笨拙(可能效率低下):
expand.grid
我错过了什么吗?有更好的方法吗?
答案 0 :(得分:8)
是的,这是interaction
做的
levels(interaction(x,y,z,sep='_'))
实施与您的rep
代码几乎相同。
输出:
[1] "avg_female_height" "median_female_height" "avg_male_height" "median_male_height" "avg_female_weight" [6] "median_female_weight" "avg_male_weight" "median_male_weight"
答案 1 :(得分:6)
使用data.table的CJ
交叉加入函数:
library(data.table)
CJ(x,y,z)[, paste(V1,V2,V3, sep = "_")]
#[1] "avg_female_height" "avg_female_weight" "avg_male_height" "avg_male_weight"
#[5] "median_female_height" "median_female_weight" "median_male_height" "median_male_weight"
或者apply
方法的变体是:
do.call(paste, c(expand.grid(x, y, z), sep = "_"))
#[1] "avg_male_height" "median_male_height" "avg_female_height" "median_female_height"
#[5] "avg_male_weight" "median_male_weight" "avg_female_weight" "median_female_weight"
答案 2 :(得分:4)
基本的(microbenchmark::microbenchmark
)基准测试通过使用显示了相当显着的加速:
library(tidyr)
library(magrittr)
df <- data.frame(x, y, z)
df %>%
complete(x, y, z) %>%
unite("combo", x, y, z, sep = "_")
有点慢,但可能是apply
技术的更直接和矢量化变体:
df <- expand.grid(x, y, z)
df$combo <- paste(df$Var1, df$Var1, df$Var3, sep = "_")
有人应该采用data.table
方式...
set.seed(21034)
x <- sample(letters, 4, TRUE)
y <- sample(letters, 4, TRUE)
z <- sample(letters, 4, TRUE)
a <- sample(letters, 4, TRUE)
library(data.table)
library(microbenchmark)
library(magrittr)
library(tidyr)
microbenchmark(times = 25L,
DT1 = CJ(x, y, z, a)[ , paste(V1, V2, V3, V4, sep = "_")],
DT2 = CJ(x, y, z, a)[ , do.call(paste, c(.SD, sep = "_"))],
app1 = do.call(paste, c(expand.grid(x, y, z, a), sep = "_")),
app2 = paste((df <- expand.grid(x, y, z, a))$Var1,
df$Var2, df$Var3, sep = "_"),
magg_outer = outer(x, y, paste, sep = "_") %>%
outer(z, paste, sep = "_") %>%
outer(a, paste, sep = "_") %>% as.vector,
magg_tidy = data.frame(x, y, z, a) %>%
complete(x, y, z, a) %>%
unite("combo", x, y, z, a, sep = "_"),
interaction = levels(interaction(x, y, z, a, sep = "_")),
original = apply(expand.grid(x, y, z, a), 1, paste, collapse = "_"),
rep = paste(rep(x, each = (ny <- length(y)) * (nz <- length(z)) *
(na <- length(a))),
rep(rep(y, each = nz * na), (nx <- length(x))),
rep(rep(z, each = na), nx * ny), sep = "_"),
Reduce = Reduce(function(x, y) paste(rep(x, each = length(y)),
rep(y, length(x)), sep = "_"),
list(x, y, z, a)))
# Unit: microseconds
# expr min lq mean median uq max neval cld
# DT1 529.578 576.6400 624.00002 589.8270 604.9845 5449.287 1000 d
# DT2 561.028 606.4220 639.94659 620.4335 636.2735 5484.514 1000 d
# app1 201.043 225.4475 240.36960 233.4795 243.7090 4244.687 1000 b
# app2 196.692 225.6130 244.33543 234.0455 243.7925 4110.605 1000 b
# magg_outer 164.352 194.1395 205.30300 204.4220 211.1990 456.122 1000 b
# magg_tidy 1872.228 2038.1560 2150.98234 2067.8770 2126.1025 21891.884 1000 f
# interaction 254.885 295.1935 313.54392 306.6680 316.8095 4196.465 1000 c
# original 852.018 935.4960 976.24388 954.5115 972.5550 4973.724 1000 e
# rep 50.737 54.1515 60.22671 55.3660 56.9220 3823.655 1000 a
# Reduce 58.395 65.3860 68.46049 66.8920 68.5640 158.184 1000 a
set.seed(21034)
x <- sprintf("%03d", sample(100))
y <- sprintf("%03d", sample(100))
z <- sprintf("%02d", sample(10))
a <- sprintf("%02d", sample(10))
library(data.table)
library(microbenchmark)
library(magrittr)
library(tidyr)
microbenchmark(times = 25L,
DT1 = CJ(x, y, z, a)[ , paste(V1, V2, V3, V4, sep = "_")],
DT2 = CJ(x, y, z, a)[ , do.call(paste, c(.SD, sep = "_"))],
app1 = do.call(paste, c(expand.grid(x, y, z, a), sep = "_")),
app2 = paste((df <- expand.grid(x, y, z, a))$Var1,
df$Var2, df$Var3, sep = "_"),
magg_outer = outer(x, y, paste, sep = "_") %>%
outer(z, paste, sep = "_") %>%
outer(a, paste, sep = "_") %>% as.vector,
magg_tidy = data.frame(x, y, z, a) %>%
complete(x, y, z, a) %>%
unite("combo", x, y, z, a, sep = "_"),
interaction = levels(interaction(x, y, z, a, sep = "_")),
original = apply(expand.grid(x, y, z, a), 1, paste, collapse = "_"),
rep = paste(rep(x, each = (ny <- length(y)) * (nz <- length(z)) *
(na <- length(a))),
rep(rep(y, each = nz * na), (nx <- length(x))),
rep(rep(z, each = na), nx * ny), sep = "_"),
Reduce = Reduce(function(x, y) paste(rep(x, each = length(y)),
rep(y, length(x)), sep = "_"),
list(x, y, z, a)))
# Unit: milliseconds
# expr min lq mean median uq max neval cld
# DT1 360.6528 467.8408 517.4579 520.1484 549.1756 861.1567 25 ab
# DT2 355.0438 504.9642 572.0732 551.9106 615.6621 927.3210 25 b
# app1 727.4513 766.3053 926.1888 910.3998 957.7610 1690.1540 25 c
# app2 472.5724 567.1121 633.5304 600.3779 634.3158 1135.7535 25 b
# magg_outer 384.0112 475.5070 600.6317 525.8936 676.7134 927.6736 25 b
# magg_tidy 520.6428 602.5028 695.5500 680.8821 748.8746 1180.1107 25 bc
# interaction 353.7317 481.4732 531.0035 518.7084 585.0872 693.5171 25 ab
# original 4965.1156 5358.8704 5914.3560 5780.6609 6074.7470 9024.6476 25 d
# rep 206.0964 236.5811 273.1093 252.8179 285.0910 455.1776 25 a
# Reduce 322.0695 390.2595 446.3948 424.9185 508.5235 621.1878 25 ab
答案 3 :(得分:2)
使用outer()
怎么样?你的两个例子变成了
x <- 1:4
y <- 1:2
as.vector(outer(x, y, paste, sep = "_"))
## [1] "1_1" "2_1" "3_1" "4_1" "1_2" "2_2" "3_2" "4_2"
library(magrittr)
x <- c("avg", "median")
y <- c("male", "female")
z <- c("height", "weight")
outer(x, y, paste, sep = "_") %>% outer(z, paste, sep = "_") %>% as.vector
## [1] "avg_male_height" "median_male_height" "avg_female_height" "median_female_height" "avg_male_weight"
## [6] "median_male_weight" "avg_female_weight" "median_female_weight"
使用Reduce()
:
Reduce(function(a, b) outer(a, b, paste, sep = "_"), list(x, y, z)) %>% as.vector
然而,这并不高效。使用microbenchmark
,我发现使用rep()
的解决方案速度提高了大约10倍。