一起使用 `mutate_at` 和 `na_if` 仅对某些列用 NA 替换零 [英] Using `mutate_at` and `na_if` together to replace zeros with NA for only some columns

问题描述

我的数据采用以下格式:

My data takes this format:

library(tidyverse)
df <- mtcars
df <- df %>% mutate(vs_doubled = vs * 2) %>% select(mpg, cyl, vs, am, vs_doubled)

head(df)


#>    mpg cyl vs am vs_doubled
#> 1 21.0   6  0  1          0
#> 2 21.0   6  0  1          0
#> 3 22.8   4  1  1          2
#> 4 21.4   6  1  0          2
#> 5 18.7   8  0  0          0
#> 6 18.1   6  1  0          2

我正在尝试使用 mutate_at 和 na_if 将 0 值设置为 NA——但仅限于特定列(vs"和am").我想将vs_doubled"列保留为零.

I'm trying to use mutate_at and na_if to set 0 values as NA--but only for specific columns ("vs" and "am"). I would like to leave the column "vs_doubled" with zeros in it.

我不太正确，因为以下行不起作用:

I haven't quite got it right, because the following line doesn't work:

df <- df %>% mutate_at(.vars = c("vs", "am"), .funs = na_if(y = 0))

推荐答案

更新

从 dplyr 1.0.0 我们可以使用 across :

From dplyr 1.0.0 we can use across :

library(dplyr)
df %>% mutate(across(c(vs,am), na_if, 0)) %>% head

#   mpg cyl vs am vs_doubled
#1 21.0   6 NA  1          0
#2 21.0   6 NA  1          0
#3 22.8   4  1  1          2
#4 21.4   6  1 NA          2
#5 18.7   8 NA NA          0
#6 18.1   6  1 NA          2

<小时>

原答案

在之前版本的 dplyr 中，我们可以使用 mutate_at :

In the previous versions of dplyr we can use mutate_at :

df %>%  mutate_at(vars(vs,am), ~na_if(.,0)) %>% head

或者另一种方式是

df %>% mutate_at(vars(vs,am), na_if, 0)

~ 是 purrr 风格的公式语法，而 . 表示列的值.它是匿名函数调用的替代方法，您可以使用匿名函数调用将上述函数编写为

~ is purrr-styled formula syntax whereas . represents value of the column. It's an alternative to anonymous function calls with which you would have written the above function as

df %>%  mutate_at(vars(vs,am), function(x) na_if(x, 0)) 

另外所示的替代方式不需要 ~，我们可以直接传递带有附加参数的函数(对于 y 此处为 0).

Also the alternative way shown does not require ~ and we can directly pass the function with additional arguments (which is 0 here for y).

当然还有其他方法可以不使用 na_if

And of course there are other ways to do this without using na_if

df %>% mutate_at(vars(vs, am), ~replace(., . == 0, NA)) 

或与基础 R 相同

cols <- c("vs", "am")
df[cols] <- lapply(df[cols], function(x) replace(x, x == 0, NA))

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