根据 R 中的条件替换 NA

Question

我有一个数据集，我想在那些缺失值数量大于或等于 n 的列中用空字符串替换 NA。例如，n = 500.

set.seed(2022)

synthetic <- tibble(
  col1 = runif(1000),
  col2 = runif(1000),
  col3 = runif(1000)
)

na_insert <- c(sample(nrow(synthetic), 500, replace = FALSE))

synthetic[na_insert, 1] <- NA

我正在尝试做但最终失败的事情：

synthetic %>% 
  mutate(across(everything(), ~ replace_na(sum(is.na(.x)) >= 500, "")))

我在这个原始练习中做错了什么？

Answer 1

您可以使用 where 和类似 purrr 的函数：

library(dplyr)

synthetic %>% 
    mutate(across(where(~sum(is.na(.x)) >= 500), ~coalesce(as.character(.x), "")))

这个returns

# A tibble: 1,000 x 3
   col1                  col2   col3
   <chr>                <dbl>  <dbl>
 1 ""                   0.479 0.139 
 2 "0.647259329678491"  0.410 0.770 
 3 ""                   0.696 0.805 
 4 ""                   0.863 0.803 
 5 "0.184729989385232"  0.146 0.652 
 6 "0.635790845612064"  0.634 0.0830
 7 ""                   0.305 0.527 
 8 "0.0419759317301214" 0.297 0.275 
 9 ""                   0.883 0.698 
10 "0.757252902723849"  0.115 0.933 
# ... with 990 more rows

Answer 2

使用ifelse函数：

library(dplyr)

synthetic |> 
  mutate_all(~ifelse(
    sum(is.na(.)) >= 500 & is.na(.),
    "",
    .
  ))

输出：

# A tibble: 1,000 x 3
   col1                  col2   col3
   <chr>                <dbl>  <dbl>
 1 ""                   0.479 0.139 
 2 "0.647259329678491"  0.410 0.770 
 3 ""                   0.696 0.805 
 4 ""                   0.863 0.803 
 5 "0.184729989385232"  0.146 0.652 
 6 "0.635790845612064"  0.634 0.0830
 7 ""                   0.305 0.527 
 8 "0.0419759317301214" 0.297 0.275 
 9 ""                   0.883 0.698 
10 "0.757252902723849"  0.115 0.933

编辑：

使用 across 而不是 mutate_all:

synthetic |>
  mutate(across(everything(),
                ~ ifelse(sum(is.na(
                  .
                )) >= 500 & is.na(.),
                "",
                .)))

Answer 3

library(data.table)
n <- 500
# convert all to character
setDT(synthetic)[, names(synthetic) := lapply(.SD, as.character)]
# find columns with >= 500 NA's
cols <- which(colSums(is.na(synthetic)) >= n)
# fast!! replace all NA in the found columns to ""
for(col in cols) set(synthetic, 
                     i = which(is.na(synthetic[[col]])), 
                     j = col, 
                     value = "")

根据 R 中的条件替换 NA

Replace NAs based on conditions in R

r

tidyselect