用dplyr将某些值设置为NA

问题描述

我试图找出一个简单的方法来使用dplyr（data set = dat，variable = x）来做这样的事情：

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

 c $ c 

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，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.
这里有一些比较不同的答案，这仅仅是当然的：
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
（我正在使用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)
由于我们对基准测试非常感兴趣，特别是在data.table-vs- dplyr讨论我提供了另一个基准的3个答案使用微基准和akrun的数据。请注意，我修改了 dplyr1 作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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