使用 dplyr 将某些值设置为 NA [英] Set certain values to NA with dplyr

问题描述

我正在尝试找出一种简单的方法来使用 dplyr(数据集 = dat，变量 = x)执行此类操作:

I'm trying to figure out a simple way to do something like this with dplyr (data set = dat, variable = x):

day$x[dat$x<0]=NA

应该很简单，但这是我目前能做的最好的.有没有更简单的方法?

Should be simple but this is the best I can do at the moment. Is there an easier way?

dat =  dat %>% mutate(x=ifelse(x<0,NA,x))

推荐答案

你可以使用 replace 比 ifelse 快一点:

You can use replace which is a bit faster than ifelse:

dat <-  dat %>% mutate(x = replace(x, x<0, NA))

您可以通过使用 which 为 replace 提供索引来加快速度:

You can speed it up a bit more by supplying an index to replace using which:

dat <- dat %>% mutate(x = replace(x, which(x<0L), NA))

在我的机器上，这将时间缩短到三分之一，见下文.

On my machine, this cut the time to a third, see below.

这里是对不同答案的一些比较，当然只是指示性的:

Here's a little comparison of the different answers, which is only indicative of course:

set.seed(24)
dat <- data.frame(x=rnorm(1e6))
system.time(dat %>% mutate(x = replace(x, x<0, NA)))
       User      System     elapsed
       0.03        0.00        0.03 
system.time(dat %>% mutate(x=ifelse(x<0,NA,x)))
       User      System     elapsed
       0.30        0.00        0.29 
system.time(setDT(dat)[x<0,x:=NA])
       User      System     elapsed
       0.01        0.00        0.02 
system.time(dat$x[dat$x<0] <- NA)
       User      System     elapsed
       0.03        0.00        0.03 
system.time(dat %>% mutate(x = "is.na<-"(x, x < 0)))
       User      System     elapsed
       0.05        0.00        0.05 
system.time(dat %>% mutate(x = NA ^ (x < 0) * x))
       User      System     elapsed
       0.01        0.00        0.02 
system.time(dat %>% mutate(x = replace(x, which(x<0), NA)))
       User      System     elapsed
       0.01        0.00        0.01 

(我使用的是 dplyr_0.3.0.2 和 data.table_1.9.4)

(I'm using dplyr_0.3.0.2 and data.table_1.9.4)

因为我们总是对基准测试非常感兴趣，特别是在 data.table-vs-dplyr 讨论过程中，我使用 microbenchmark 和 akrun 的数据提供了 3 个答案的另一个基准.请注意，我将 dplyr1 修改为我的答案的更新版本:

Since we're always very interested in benchmarking, especially in the course of data.table-vs-dplyr discussions I provide another benchmark of 3 of the answers using microbenchmark and the data by akrun. Note that I modified dplyr1 to be the updated version of my answer:

set.seed(285)
dat1 <- dat <- data.frame(x=sample(-5:5, 1e8, replace=TRUE), y=rnorm(1e8))
dtbl1 <- function() {setDT(dat)[x<0,x:=NA]}
dplr1 <- function() {dat1 %>% mutate(x = replace(x, which(x<0L), NA))}
dplr2 <- function() {dat1 %>% mutate(x = NA ^ (x < 0) * x)}
microbenchmark(dtbl1(), dplr1(), dplr2(), unit='relative', times=20L)
#Unit: relative
#    expr      min       lq   median       uq      max neval
# dtbl1() 1.091208 4.319863 4.194086 4.162326 4.252482    20
# dplr1() 1.000000 1.000000 1.000000 1.000000 1.000000    20
# dplr2() 6.251354 5.529948 5.344294 5.311595 5.190192    20

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