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

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本文介绍了在不使用list()的情况下,将dplyr中的NA替换为零的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

在dplyr中,我可以使用以下代码将NA替换为0.问题是这会将一个列表插入到我的数据框中,从而进一步破坏了行中的进一步分析.在这一点上,我什至都不了解列表或原子向量.我只想选择某些列,并将所有不适用的NA替换为零.并保持列的整数状态.

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"))

可以,但是可以将列转换为列表.如何在不将列转换为列表的情况下做到这一点?

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

这是在基数R中的操作方法.但是不确定如何将其处理为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

您正在使用哪个版本的dplyr?可能是旧的了. replace_na函数现在似乎在tidyr中.可行

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xy仍然是原子向量.经过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.

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