在不使用 list() 的情况下，在 dplyr 中将 NA 替换为零 [英] Replace NA with Zero in dplyr without using list()

问题描述

在 dplyr 中，我可以使用以下代码将 NA 替换为 0.问题是这会在我的数据框中插入一个列表，这会破坏进一步的分析.在这一点上，我什至不理解列表或原子向量或任何一个.我只想选择某些列，并将所有出现的 NA 替换为零.并保持列整数状态.

In dplyr I can replace NA with 0 using the following code. The issue is this inserts a list into my data frame which screws up further analysis down the line. I don't even understand lists or atomic vectors or any of that at this point. I just want to pick certain columns, and replace all occurrences of NA with zero. And maintain the columns integer status.

library(dplyr)
df <- tibble(x = c(1, 2, NA), y = c("a", NA, "b"), z = list(1:5, NULL, 10:20))
df
df %>% replace_na(list(x = 0, y = "unknown"))

这可行，但会将列转换为列表.如何在不将列转换为列表的情况下执行此操作?

That works but transforms the column into a list. How do I do it without transforming the column into a list?

这里是如何在基础 R 中执行此操作.但不确定如何将其转换为 mutate 语句:

And here's how to do it in base R. But not sure how to work this into a mutate statement:

df$x[is.na(df$x)] <- 0

推荐答案

您使用的是什么版本的 dplyr?它可能是一个旧的.replace_na 函数现在似乎在 tidyr 中.这有效

What version of dplyr are you using? It might be an old one. The replace_na function now seems to be in tidyr. This works

library(tidyr)
df <- tibble::tibble(x = c(1, 2, NA), y = c("a", NA, "b"), z = list(1:5, NULL, 10:20))
df %>% replace_na(list(x = 0, y = "unknown")) %>% str()
# Classes ‘tbl_df’, ‘tbl’ and 'data.frame': 3 obs. of  3 variables:
#  $ x: num  1 2 0
#  $ y: chr  "a" "unknown" "b"
#  $ z:List of 3
#   ..$ : int  1 2 3 4 5
#   ..$ : NULL
#   ..$ : int  10 11 12 13 14 15 16 17 18 19 ...

我们可以看到 NA 值已被替换，列 x 和 y 仍然是原子向量.使用 tidyr_0.7.2 测试.

We can see the NA values have been replaced and the columns x and y are still atomic vectors. Tested with tidyr_0.7.2.

这篇关于在不使用 list() 的情况下，在 dplyr 中将 NA 替换为零的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！