Ari*_*man 17 r recode r-factor
我经常遇到这种情况,我认为必须有一个很好的成语.假设我有一个包含一系列属性的data.frame,包括"product".我还有一把钥匙,可以将产品转化为品牌+尺寸.产品代码1-3是Tylenol,4-6是Advil,7-9是拜耳,10-12是Generic.
什么是最快的(就人类时间而言)编码方式？
ifelse如果有3个或更少的类别,我倾向于使用嵌套的;如果有超过3个类型,则键入数据表并将其合并.任何更好的想法？Stata有一个非常漂亮的recode命令,虽然我相信它会促进数据代码混合有点过分.
dat <- structure(list(product = c(11L, 11L, 9L, 9L, 6L, 1L, 11L, 5L,
7L, 11L, 5L, 11L, 4L, 3L, 10L, 7L, 10L, 5L, 9L, 8L)), .Names = "product", row.names = c(NA,
-20L), class = "data.frame")
Mar*_*rek 19
您可以将变量转换为因子并按levels<-功能更改其级别.在一个命令中它可能像:
`levels<-`(
factor(dat$product),
list(Tylenol=1:3, Advil=4:6, Bayer=7:9, Generic=10:12)
)
步骤:
brands <- factor(dat$product)
levels(brands) <- list(Tylenol=1:3, Advil=4:6, Bayer=7:9, Generic=10:12)
huo*_*uon 14
可以使用列表作为关联数组来定义brand -> product code映射,即:
brands <- list(Tylenol=1:3, Advil=4:6, Bayer=7:9, Generic=10:12)
一旦你有了这个,你可以反过来创建一个product code -> brand列表(可能会占用大量内存),或者只是使用搜索功能:
find.key <- function(x, li, default=NA) {
ret <- rep.int(default, length(x))
for (key in names(li)) {
ret[x %in% li[[key]]] <- key
}
return(ret)
}
我确信有更好的方法来编写这个函数(for循环让我烦恼!),但至少它是矢量化的,所以它只需要一次通过列表.
使用它将是这样的:
> dat$brand <- find.key(dat$product, brands)
> dat
product brand
1 11 Generic
2 11 Generic
3 9 Bayer
4 9 Bayer
5 6 Advil
6 1 Tylenol
7 11 Generic
8 5 Advil
9 7 Bayer
10 11 Generic
11 5 Advil
12 11 Generic
13 4 Advil
14 3 Tylenol
15 10 Generic
16 7 Bayer
17 10 Generic
18 5 Advil
19 9 Bayer
20 8 Bayer
在recode和levels<-解决方案是非常好的,但他们也不止这一个显著慢(一旦你有find.key这种情况很容易换比人类recode和与票面levels<-):
> microbenchmark(
recode=recode(dat$product,recodes="1:3='Tylenol';4:6='Advil';7:9='Bayer';10:12='Generic'"),
find.key=find.key(dat$product, brands),
levels=`levels<-`(factor(dat$product),brands))
Unit: microseconds
expr min lq median uq max
1 find.key 64.325 69.9815 76.8950 83.8445 221.748
2 levels 240.535 248.1470 274.7565 306.8490 1477.707
3 recode 1636.039 1683.4275 1730.8170 1855.8320 3095.938
(我无法让switch版本正确地进行基准测试,但它似乎比上述所有版本更快,尽管对于人类而言比recode解决方案更糟糕.)
Ben*_*nes 12
我喜欢包中的recode功能car:
library(car)
dat$brand <- recode(dat$product,
recodes="1:3='Tylenol';4:6='Advil';7:9='Bayer';10:12='Generic'")
# > dat
# product brand
# 1 11 Generic
# 2 11 Generic
# 3 9 Bayer
# 4 9 Bayer
# 5 6 Advil
# 6 1 Tylenol
# 7 11 Generic
# 8 5 Advil
# 9 7 Bayer
# 10 11 Generic
# 11 5 Advil
# 12 11 Generic
# 13 4 Advil
# 14 3 Tylenol
# 15 10 Generic
# 16 7 Bayer
# 17 10 Generic
# 18 5 Advil
# 19 9 Bayer
# 20 8 Bayer
我经常使用以下技术:
key <- c()
key[1:3] <- "Tylenol"
key[4:6] <- "Advil"
key[7:9] <- "Bayer"
key[10:12] <- "Generic"
然后,
> key[dat$product]
[1] "Generic" "Generic" "Bayer" "Bayer" "Advil" "Tylenol" "Generic" "Advil" "Bayer" "Generic"
[11] "Advil" "Generic" "Advil" "Tylenol" "Generic" "Bayer" "Generic" "Advil" "Bayer" "Bayer"
"数据库方法"是为产品密钥定义保留单独的表(data.frame).它更有意义,因为你说你的产品键不仅转化为品牌,还转化为大小:
product.keys <- read.table(textConnection("
product brand size
1 Tylenol small
2 Tylenol medium
3 Tylenol large
4 Advil small
5 Advil medium
6 Advil large
7 Bayer small
8 Bayer medium
9 Bayer large
10 Generic small
11 Generic medium
12 Generic large
"), header = TRUE)
然后,您可以使用merge以下方式加入数据:
merge(dat, product.keys, by = "product")
# product brand size
# 1 1 Tylenol small
# 2 3 Tylenol large
# 3 4 Advil small
# 4 5 Advil medium
# 5 5 Advil medium
# 6 5 Advil medium
# 7 6 Advil large
# 8 7 Bayer small
# 9 7 Bayer small
# 10 8 Bayer medium
# 11 9 Bayer large
# 12 9 Bayer large
# 13 9 Bayer large
# 14 10 Generic small
# 15 10 Generic small
# 16 11 Generic medium
# 17 11 Generic medium
# 18 11 Generic medium
# 19 11 Generic medium
# 20 11 Generic medium
如您所知,行的顺序不会被保留merge.如果这是一个问题,该plyr包有一个join确保订单的功能:
library(plyr)
join(dat, product.keys, by = "product")
# product brand size
# 1 11 Generic medium
# 2 11 Generic medium
# 3 9 Bayer large
# 4 9 Bayer large
# 5 6 Advil large
# 6 1 Tylenol small
# 7 11 Generic medium
# 8 5 Advil medium
# 9 7 Bayer small
# 10 11 Generic medium
# 11 5 Advil medium
# 12 11 Generic medium
# 13 4 Advil small
# 14 3 Tylenol large
# 15 10 Generic small
# 16 7 Bayer small
# 17 10 Generic small
# 18 5 Advil medium
# 19 9 Bayer large
# 20 8 Bayer medium
最后,如果您的表很大并且速度是个问题,请考虑使用data.tables(来自data.table包)而不是data.frames.
这个需要一些打字,但如果你真的有一个庞大的数据集,这可能是要走的路.在talkstats.com上的Bryangoodrich和Dason教会了我这个.它使用哈希表或创建包含查找表的环境.我实际上将这个保留在我的.Rprofile(哈希函数)上,用于字典类型查找.
我将您的数据复制1000次以使其更大一些.
#################################################
# THE HASH FUNCTION (CREATES A ENW ENVIRONMENT) #
#################################################
hash <- function(x, type = "character") {
e <- new.env(hash = TRUE, size = nrow(x), parent = emptyenv())
char <- function(col) assign(col[1], as.character(col[2]), envir = e)
num <- function(col) assign(col[1], as.numeric(col[2]), envir = e)
FUN <- if(type=="character") char else num
apply(x, 1, FUN)
return(e)
}
###################################
# YOUR DATA REPLICATED 1000 TIMES #
###################################
dat <- dat <- structure(list(product = c(11L, 11L, 9L, 9L, 6L, 1L, 11L, 5L,
7L, 11L, 5L, 11L, 4L, 3L, 10L, 7L, 10L, 5L, 9L, 8L)), .Names = "product", row.names = c(NA,
-20L), class = "data.frame")
dat <- dat[rep(seq_len(nrow(dat)), 1000), , drop=FALSE]
rownames(dat) <-NULL
dat
#########################
# CREATE A LOOKUP TABLE #
#########################
med.lookup <- data.frame(val=as.character(1:12),
med=rep(c('Tylenol', 'Advil', 'Bayer', 'Generic'), each=3))
########################################
# USE hash TO CREATE A ENW ENVIRONMENT #
########################################
meds <- hash(med.lookup)
##############################
# CREATE A RECODING FUNCTION #
##############################
recoder <- function(x){
x <- as.character(x) #turn the numbers to character
rc <- function(x){
if(exists(x, env = meds))get(x, e = meds) else NA
}
sapply(x, rc, USE.NAMES = FALSE)
}
#############
# HASH AWAY #
#############
recoder(dat[, 1])
在这种情况下,散列很慢但是如果你有更多的级别来重新编码,那么它将比其他级别更快.