collapse::na_omit：

collapse::na_omit(myinput, cols = paste0("value", 2:4), prop = 1)

#>    sample value1 value2 value3 value4
#> 1 sample1      A     NA     13     11
#> 2 sample2      B      5     NA      4
#> 3 sample3      A      7      7     NA
#> 4 sample4   <NA>     NA     NA      9

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1，000，000 x 21数据集的基准：

collapse速度快5 - 6倍：

#Using data from https://stackoverflow.com/a/48830183/13460602
library(dplyr)
library(collapse)

mb <- microbenchmark::microbenchmark(
  rowSums = df[rowSums(is.na(df[, 2:21])) < 3,],
  "dplyr+rowSums" = df %>%
    filter(rowSums(is.na(pick(2:21))) < 3),
  if_all = df |>
    filter(!if_all(2:21, is.na)),
  collapse = na_omit(df, cols = 2:21, prop = 1)
)

型
x1c 0d1x的数据

我们可以使用is.na和rowSums来计算每行中NA s的数量。

base R

myinput[rowSums(is.na(myinput[, c("value2", "value3", "value4")])) < 3,]
#    sample value1 value2 value3 value4
# 1 sample1      A     NA     13     11
# 2 sample2      B      5     NA      4
# 3 sample3      A      7      7     NA
# 4 sample4   <NA>     NA     NA      9

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下载

library(dplyr)
myinput %>%
  filter(rowSums(is.na(pick(value1, value2, value3))) < 3)
#    sample value1 value2 value3 value4
# 1 sample1      A     NA     13     11
# 2 sample2      B      5     NA      4
# 3 sample3      A      7      7     NA
# 4 sample5      C     NA     NA     NA

型

使用dplyr::if_all，您可以：

library(dplyr, warn = FALSE)

myinput |>
  filter(!if_all(value2:value4, is.na))
#>    sample value1 value2 value3 value4
#> 1 sample1      A     NA     13     11
#> 2 sample2      B      5     NA      4
#> 3 sample3      A      7      7     NA
#> 4 sample4   <NA>     NA     NA      9

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创建于2023-12-18带有reprex v2.0.2

在filter中使用across

library(dplyr)

myinput %>% 
  filter(rowSums(across(value2:value4, ~ is.na(.x))) < 3)
   sample value1 value2 value3 value4
1 sample1      A     NA     13     11
2 sample2      B      5     NA      4
3 sample3      A      7      7     NA
4 sample4   <NA>     NA     NA      9

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