Remove specific missing values in r

Question

my data set has missing values marked as 'XXX'

I have tried na.omit(mydata)

df <- data.frame(X=factor(c(0.2, "XXX", 0.4, 0.1)), Y=factor(c(0.8, 1, 0.9, "XXX")))

here X and Y are factors. I found that the missing data is encoded as "XXX" by checking the levels of the factor.

I want to remove row "2" and row "4". can someone help, I have been trying for a while now.

Please update your question with the desired behavior, specific problems, and code to reproduce it. See: [How to create a Minimal, Complete, and Verifiable example](https://stackoverflow.com/help/mcve) — abestrad, Jun 11 '20 at 18:10
Does this answer your question? [how to filter data frame with conditions of two columns?](https://stackoverflow.com/questions/20084462/how-to-filter-data-frame-with-conditions-of-two-columns) — Matt, Jun 11 '20 at 18:17
If any answers have solved your question, please mark the preferable one as "accepted" by clicking the check mark next to it. Thank you! — Darren Tsai, Jun 12 '20 at 09:02

score 2 · Answer 1 · answered Jun 11 '20 at 18:22

2

You can also filter for complete cases like this:

library(dplyr)
library(magrittr)
df %>% replace(.=="XXX", NA_character_) %>% filter(complete.cases(.))

The output is:

> df %>% replace(.=="XXX", NA_character_) %>% filter(complete.cases(.))
    X   Y
1 0.2 0.8  
2 0.4 0.9

answered Jun 11 '20 at 18:22

mysteRious

`library(magrittr)` is not necessary. – Darren Tsai Jun 11 '20 at 19:11
There are pipes in the example. – mysteRious Jun 11 '20 at 19:14
`dplyr` has already loaded the pipe function. You can try to restart your R session, `library(dplyr)` and run your code again, it works without the need of `magrittr`. – Darren Tsai Jun 11 '20 at 19:53

score 1 · Answer 2 · answered Jun 11 '20 at 18:23

1

Two base R solutions:

df <- subset(df, X != "XXX" & Y != "XXX")

or

df <- df[df$X != "XXX" & df$Y != "XXX",]

dplyr solution:

library(dplyr)

df <- df %>% filter(X != "XXX" & Y != "XXX")

Gives us:

    X   Y
1 0.2 0.8
3 0.4 0.9

answered Jun 11 '20 at 18:23

Matt

score 1 · Answer 3 · answered Jun 11 '20 at 18:25

1

Another option using tidyverse:

df %>%
  mutate(across(everything(), str_replace, "XXX", NA_character_)) %>%
  drop_na()

#     X   Y
# 1 0.2 0.8
# 2 0.4 0.9

answered Jun 11 '20 at 18:25

Martin Gal

Darren Tsai · Answer 4 · 2020-06-11T19:11:03.950

1

You don't need to convert "XXX" to NA. Just filter "XXX" directly.

library(dplyr)

df %>% filter(across(everything(), ~ . != "XXX"))

#     X   Y
# 1 0.2 0.8
# 2 0.4 0.9

The corresponding version using filter_all().

df %>% filter_all(all_vars(. != "XXX"))

A base R solution.

df[rowSums(df == "XXX") == 0, ]

edited Jun 11 '20 at 19:11

answered Jun 11 '20 at 19:04

Darren Tsai

4 Answers4