我有如下表格
product=c("a","b","c")
min=c(1,5,3)
max=c(1,7,7)
dd=data.frame(product,min,max)
> dd
product min max
1 a 1 1
2 b 5 7
3 c 3 7
我想创建一个如下所示的表格。我想为产品的最小值和最大值之间(包括最小值和最大值)之间的每个值创建一行
product mm
a 1
b 5
b 6
b 7
c 3
c 4
c 5
c 6
c 7
我如何使用 R 来做到这一点？有什么套餐可以快速见效吗？
最佳答案
尝试
library(data.table)
setDT(dd)[, list(mm=min:max), by = product]
# product mm
#1: a 1
#2: b 5
#3: b 6
#4: b 7
#5: c 3
#6: c 4
#7: c 5
#8: c 6
#9: c 7
或者更快的选择是 seq.int(min, max, 1L) 正如@David Arenburg 所建议的
setDT(dd)[, list(mm = seq.int(min, max, 1L)), by = product]
基准
library(stringi)
set.seed(24)
product <- unique(stri_rand_strings(1e5,4))
min1 <- sample(1:10, length(product), replace=TRUE)
max1 <- sample(11:15, length(product), replace=TRUE)
dd <- data.frame(product, min1, max1)
dd2 <- copy(dd)
josilber <- function(){res1 <- data.frame(product=rep(dd$product,
dd$max1-dd$min1+1),
mm=unlist(mapply(seq, dd$min1, dd$max1)))
}
akrun <- function(){as.data.table(dd2)[, list(mm = seq.int(min1, max1,
1L)), by = product]}
Ananda <- function() {stack(lapply(split(dd[-1], dd[1]),
function(x) seq(x[[1]], x[[2]])))}
jiber <- function(){res <- by(dd[,-1], dd[,1], function(x)
seq(x$min1, x$max1) )
res <- as.data.frame(unlist(res))
data.frame(product=gsub("[0-9]", "", rownames(res)), mm=res[,1])}
system.time(akrun())
# user system elapsed
# 0.129 0.001 0.129
system.time(josilber())
# user system elapsed
# 0.762 0.002 0.764
system.time(Ananda())
# user system elapsed
#45.449 0.191 45.636
system.time(jiber())
# user system elapsed
# 48.013 8.218 56.291
library(microbenchmark)
microbenchmark(josilber(), akrun(), times=20L, unit='relative')
#Unit: relative
# expr min lq mean median uq max neval cld
#josilber() 6.39757 6.713236 5.570836 5.901037 5.603639 3.970663 20 b
# akrun() 1.00000 1.000000 1.000000 1.000000 1.000000 1.000000 20 a
关于R从两列创建一个序列表，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/30602821/