Question

我有一个包含大量列的数据框。对于数据帧的每一行，我想得到NA的列数。问题是我只对一些列感兴趣，并希望（有效地）将这些列调出来。

使用mutate，我在下面的假样本中的方式给了我正确的答案。

library(stringr)

df  <- data_frame(
         id = 1:10
       , name = fruit[1:10]
       , word1 = c(words[1:5],NA,words[7:10])
       , word2 = words[11:20]
       , word3 = c(NA,NA,NA,words[25],NA,NA,words[32],NA,NA,words[65])
    ) %>%
    mutate(
        n_words = 
            as.numeric(!is.na(word1)) + 
            as.numeric(!is.na(word2)) + 
            as.numeric(!is.na(word3)) 
    )

然而，即使像这样的玩具例子，打字和阅读也很痛苦 - 当我有超过3列的数量时，它基本上没用。是否有更多的R / dplyr-y方式来编写它，可能使用select()样式语法（例如。n_words = !count_blank(word1:word3)）？

我考虑使用summarize() sans分组，但是，我需要我正在计算的列中的数据，如果我将它们添加到group_by，我就在同一条船上呼唤所有专栏。

Answer 1

您可以对所选列使用is.na()，然后rowSums()结果：

library(stringr)
df <- data_frame(
  id = 1:10
  , name = fruit[1:10]
  , word1 = c(words[1:5],NA,words[7:10])
  , word2 = words[11:20]
  , word3 = c(NA,NA,NA,words[25],NA,NA,words[32],NA,NA,words[65]))

df$word_count <- rowSums( !is.na( df [,3:5]))

df
      id         name    word1     word2   word3 n_words
   <int>        <chr>    <chr>     <chr>   <chr>   <dbl>
1      1        apple        a    actual    <NA>       2
2      2      apricot     able       add    <NA>       2
3      3      avocado    about   address    <NA>       2
4      4       banana absolute     admit   agree       3
5      5  bell pepper   accept advertise    <NA>       2
6      6     bilberry     <NA>    affect    <NA>       1
7      7   blackberry  achieve    afford alright       3
8      8 blackcurrant   across     after    <NA>       2
9      9 blood orange      act afternoon    <NA>       2
10    10    blueberry   active     again   awful       3

修改

使用dplyr你可以这样做：

df %>% 
    select(3:5) %>% 
    is.na %>% 
    `!` %>% 
    rowSums

Answer 2

另一个dplyr解决方案：

library(stringr)

## define count function

count_na <- function(x) sum(!is.na(x))

df$count_na <- df %>%

  select(starts_with("word")) %>%

    apply(., 1, count_na)

## A tibble: 10 × 6
      id         name    word1     word2   word3 n_words
   <int>        <chr>    <chr>     <chr>   <chr>   <int>
1      1        apple        a    actual    <NA>       2
2      2      apricot     able       add    <NA>       2
3      3      avocado    about   address    <NA>       2
4      4       banana absolute     admit   agree       3
5      5  bell pepper   accept advertise    <NA>       2
6      6     bilberry     <NA>    affect    <NA>       1
7      7   blackberry  achieve    afford alright       3
8      8 blackcurrant   across     after    <NA>       2
9      9 blood orange      act afternoon    <NA>       2
10    10    blueberry   active     again   awful       3

Answer 3

library(dplyr)
library(stringr)

df  <- data_frame(
  id = 1:10
  , name = fruit[1:10]
  , word1 = c(words[1:5],NA,words[7:10])
  , word2 = words[11:20]
  , word3 = c(NA,NA,NA,words[25],NA,NA,words[32],NA,NA,words[65])
) 

library(purrr)
# Rowwise sum of NAs
df %>% by_row(~ sum(is.na(.)), .collate = 'cols')

# Rowwise sum of non-NAs for word columns
df %>% 
  select(starts_with('word')) %>% 
  by_row(~ sum(!is.na(.)), .collate = 'cols')

计算一行中某些单元格中的值不是NA（在R中）

3 个答案:

修改