R因子的算术运算 [英] Arithmetic operations on R factors

问题描述

我有一个R数据框，我试图从另一列中减去一列.我使用$运算符提取了列，但是列的类是'factor'，R不会对因子执行算术运算.有特殊功能可以做到这一点吗?

I have an R dataframe and I'm trying to subtract one column from another. I extract the columns using the $ operator but the class of the columns is 'factor' and R won't perform arithmetic operations on factors. Are there special functions to do this?

推荐答案

如果您确实希望使用该因子的级别，那么您所做的事情非常错误或太聪明了.

If you really want the levels of the factor to be used, you're either doing something very wrong or too clever for its own good.

如果您拥有的因子是包含存储在因子级别中的数字，那么您首先要使用as.numeric(as.character(...))将其强制为数字:

If what you have is a factor containing numbers stored in the levels of the factor, then you want to coerce it to numeric first using as.numeric(as.character(...)):

dat <- data.frame(f=as.character(runif(10)))

您可以在此处看到访问因子索引和分配因子内容之间的区别:

You can see the difference between accessing the factor indices and assigning the factor contents here:

> as.numeric(dat$f)
 [1]  9  7  2  1  4  6  5  3 10  8
> as.numeric(as.character(dat$f))
 [1] 0.6369432 0.4455214 0.1204000 0.0336245 0.2731787 0.4219241 0.2910194
 [8] 0.1868443 0.9443593 0.5784658

定时与另一种方法(仅在级别上进行转换)表明，如果级别不是每个元素唯一的，则转换会更快:

Timings vs. an alternative approach which only does the conversion on the levels shows it's faster if levels are not unique to each element:

dat <- data.frame( f = sample(as.character(runif(10)),10^4,replace=TRUE) )
library(microbenchmark)
microbenchmark(
  as.numeric(as.character(dat$f)),
  as.numeric( levels(dat$f) )[dat$f] ,
  as.numeric( levels(dat$f)[dat$f] ),
  times=50
  )

                              expr     min      lq  median      uq     max
1  as.numeric(as.character(dat$f)) 7835865 7869228 7919699 7998399 9576694
2 as.numeric(levels(dat$f))[dat$f]  237814  242947  255778  270321  371263
3 as.numeric(levels(dat$f)[dat$f]) 7817045 7905156 7964610 8121583 9297819

因此，如果length(levels(dat$f)) < length(dat$f)，请使用as.numeric(levels(dat$f))[dat$f]以获得可观的速度增益.

Therefore, if length(levels(dat$f)) < length(dat$f), use as.numeric(levels(dat$f))[dat$f] for a substantial speed gain.

如果length(levels(dat$f))大约等于length(dat$f)，则没有速度增益:

If length(levels(dat$f)) is approximately equal to length(dat$f), there is no speed gain:

dat <- data.frame( f = as.character(runif(10^4) ) )
library(microbenchmark)
microbenchmark(
  as.numeric(as.character(dat$f)),
  as.numeric( levels(dat$f) )[dat$f] ,
  as.numeric( levels(dat$f)[dat$f] ),
  times=50
  )

                              expr     min      lq  median      uq      max
1  as.numeric(as.character(dat$f)) 7986423 8036895 8101480 8202850 12522842
2 as.numeric(levels(dat$f))[dat$f] 7815335 7866661 7949640 8102764 15809456
3 as.numeric(levels(dat$f)[dat$f]) 7989845 8040316 8122012 8330312 10420161

这篇关于R因子的算术运算的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！