使用
zoo::na.fill(0)
可以将所有 NA 填充为 0 ,但这会改变变量属性,
例如,变量amount
的属性在dataframe md中是num
,在zoo::na.fill(0)
之后属性更改为chr
。有什么方法可以像na.fill
一样简单地填充na并保持可变属性吗?
md <- data.frame(cat=c('a','b','d',NA,'E',NA),
subcat=c('A','C',NA,NA,NA,'D'),
amount=c(1,2,NA,5,NA,8))
md %>% zoo::na.fill(0)
这是一个不使用zoo的解决方案。
library(dplyr)
md2 <- md %>%
mutate(across(where(is.factor), as.character)) %>%
mutate(across(where(is.character), function(x) { replace(x, is.na(x), "0") } )) %>%
mutate(across(where(is.numeric), function(x) { replace(x, is.na(x), 0) } )) %>%
mutate(across(where(is.character), as.factor))
如果你愿意,你可以将其包装成一个自定义函数,以像zoo的fill-na方法一样方便地使用它,例如
FillNA <- function(df){
df2 <- df %>%
mutate(across(where(is.factor), as.character)) %>%
mutate(across(where(is.character), function(x) { replace(x, is.na(x), "0") } )) %>%
mutate(across(where(is.numeric), function(x) { replace(x, is.na(x), 0) } )) %>%
mutate(across(where(is.character), as.factor))
return(df2)
}
这里验证类型:
> str(md)
'data.frame': 6 obs. of 3 variables:
$ cat : Factor w/ 4 levels "a","b","d","E": 1 2 3 NA 4 NA
$ subcat: Factor w/ 3 levels "A","C","D": 1 2 NA NA NA 3
$ amount: num 1 2 NA 5 NA 8
str(FillNA(md))
'data.frame': 6 obs. of 3 variables:
$ cat : Factor w/ 5 levels "0","a","b","d",..: 2 3 4 1 5 1
$ subcat: Factor w/ 4 levels "0","A","C","D": 2 3 1 1 1 4
$ amount: num 1 2 0 5 0 8
更新代码,标记一下即可
md %>% mutate(across(cat:amount,~ replace(.x,is.na(.x),0)))