r: replace_na() 不替换 NAs

Question

I have a variable with NAs called df$salesContribution .

Using dplyr I've created a statement below, but can't figure out why my df$salesContribution is returning NAs still:

df<- df %>% 
  mutate(salesContribution = as.numeric(salesContribution)) %>% 
  replace_na(0)

Is the 0 not registering?

Answer 1

replace_na() will not work if the variable is a factor, and the replacement is not already a level for your factor. If this is the issue, you can add another level to your factor variable for 0 before running replace_na(), or you can convert the variable to numeric or character first.

Answer 2

看起来你想要

df$salesContribution <- df$salesContribution %>% as.numeric() %>% replace_na(0)

Answer 3

I found a blog response elsewhere in which Hadley himself said they wanted to move away from replace_na toward a more SQL adjacent command coalesce(). The solution involves both across and coalesce. In my case, I used a different variable from the dataframe to supply the missing values ad hoc. You can also specify a fixed value.

Here's an example of what I just did in my work:

	Varname1	Varname2
1	Yes	Yes
2	NA	No
3	NA	Yes
4	No	No

df %>%
  mutate(across(Varname1, coalesce, Varname2))

	Varname1	Varname2
1	Yes	Yes
2	No	No
3	Yes	Yes
4	No	No

Answer 4

I don't have enough reputation points to add a comment to someone's answer, but above it was said:

replace_na() will not work if the variable is a factor, and the replacement is not >already a level for your factor. If this is the issue, you can add another level to >your factor variable for 0 before running replace_na(), or you can convert the >variable to numeric or character first.

This is true. However, you can work around this with fct_explicit_na() from the forcats package.

Answer 5

You can do using base replace :

df<- df %>% 
  mutate(salesContribution = replace(as.numeric(salesContribution), which(is.na(salesContribution)), 0)

Answer 6

If using data.table package you could do something like: You could try something like: x[is.na(field_name)][, field_name := replacement_value]

There should be similar syntax for data.frame as well.